hypothesis testing formula for two means

• school Campus Bookshelves
• menu_book Bookshelves
• perm_media Learning Objects
• login Login
• how_to_reg Request Instructor Account
• hub Instructor Commons
• Download Page (PDF)
• Download Full Book (PDF)
• Periodic Table
• Physics Constants
• Scientific Calculator
• Reference & Cite
• Tools expand_more
• Readability

selected template will load here

This action is not available.

10.29: Hypothesis Test for a Difference in Two Population Means (1 of 2)

• Last updated
• Save as PDF
• Page ID 14167

Learning Objectives

• Under appropriate conditions, conduct a hypothesis test about a difference between two population means. State a conclusion in context.

Using the Hypothesis Test for a Difference in Two Population Means

The general steps of this hypothesis test are the same as always. As expected, the details of the conditions for use of the test and the test statistic are unique to this test (but similar in many ways to what we have seen before.)

Step 1: Determine the hypotheses.

The hypotheses for a difference in two population means are similar to those for a difference in two population proportions. The null hypothesis, H 0 , is again a statement of “no effect” or “no difference.”

• H 0 : μ 1 – μ 2 = 0, which is the same as H 0 : μ 1 = μ 2

The alternative hypothesis, H a , can be any one of the following.

• H a : μ 1 – μ 2 < 0, which is the same as H a : μ 1 < μ 2
• H a : μ 1 – μ 2 > 0, which is the same as H a : μ 1 > μ 2
• H a : μ 1 – μ 2 ≠ 0, which is the same as H a : μ 1 ≠ μ 2

Step 2: Collect the data.

As usual, how we collect the data determines whether we can use it in the inference procedure. We have our usual two requirements for data collection.

• Samples must be random to remove or minimize bias.
• Samples must be representative of the populations in question.

We use this hypothesis test when the data meets the following conditions.

• The two random samples are independent .
• The variable is normally distributed in both populations . If this variable is not known, samples of more than 30 will have a difference in sample means that can be modeled adequately by the t-distribution. As we discussed in “Hypothesis Test for a Population Mean,” t-procedures are robust even when the variable is not normally distributed in the population. If checking normality in the populations is impossible, then we look at the distribution in the samples. If a histogram or dotplot of the data does not show extreme skew or outliers, we take it as a sign that the variable is not heavily skewed in the populations, and we use the inference procedure. (Note: This is the same condition we used for the one-sample t-test in “Hypothesis Test for a Population Mean.”)

Step 3: Assess the evidence.

If the conditions are met, then we calculate the t-test statistic. The t-test statistic has a familiar form.

Since the null hypothesis assumes there is no difference in the population means, the expression (μ 1 – μ 2 ) is always zero.

As we learned in “Estimating a Population Mean,” the t-distribution depends on the degrees of freedom (df) . In the one-sample and matched-pair cases df = n – 1. For the two-sample t-test, determining the correct df is based on a complicated formula that we do not cover in this course. We will either give the df or use technology to find the df . With the t-test statistic and the degrees of freedom, we can use the appropriate t-model to find the P-value, just as we did in “Hypothesis Test for a Population Mean.” We can even use the same simulation.

Step 4: State a conclusion.

To state a conclusion, we follow what we have done with other hypothesis tests. We compare our P-value to a stated level of significance.

• If the P-value ≤ α, we reject the null hypothesis in favor of the alternative hypothesis.
• If the P-value > α, we fail to reject the null hypothesis. We do not have enough evidence to support the alternative hypothesis.

As always, we state our conclusion in context, usually by referring to the alternative hypothesis.

“Context and Calories”

Does the company you keep impact what you eat? This example comes from an article titled “Impact of Group Settings and Gender on Meals Purchased by College Students” (Allen-O’Donnell, M., T. C. Nowak, K. A. Snyder, and M. D. Cottingham, Journal of Applied Social Psychology 49(9), 2011, onlinelibrary.wiley.com/doi/10.1111/j.1559-1816.2011.00804.x/full) . In this study, researchers examined this issue in the context of gender-related theories in their field. For our purposes, we look at this research more narrowly.

Step 1: Stating the hypotheses.

In the article, the authors make the following hypothesis. “The attempt to appear feminine will be empirically demonstrated by the purchase of fewer calories by women in mixed-gender groups than by women in same-gender groups.” We translate this into a simpler and narrower research question: Do women purchase fewer calories when they eat with men compared to when they eat with women?

Here the two populations are “women eating with women” (population 1) and “women eating with men” (population 2). The variable is the calories in the meal. We test the following hypotheses at the 5% level of significance.

The null hypothesis is always H 0 : μ 1 – μ 2 = 0, which is the same as H 0 : μ 1 = μ 2 .

The alternative hypothesis H a : μ 1 – μ 2 > 0, which is the same as H a : μ 1 > μ 2 .

Here μ 1 represents the mean number of calories ordered by women when they were eating with other women, and μ 2 represents the mean number of calories ordered by women when they were eating with men.

Note: It does not matter which population we label as 1 or 2, but once we decide, we have to stay consistent throughout the hypothesis test. Since we expect the number of calories to be greater for the women eating with other women, the difference is positive if “women eating with women” is population 1. If you prefer to work with positive numbers, choose the group with the larger expected mean as population 1. This is a good general tip.

Step 2: Collect Data.

As usual, there are two major things to keep in mind when considering the collection of data.

• Samples need to be representative of the population in question.
• Samples need to be random in order to remove or minimize bias.

Representative Samples?

The researchers state their hypothesis in terms of “women.” We did the same. But the researchers gathered data by watching people eat at the HUB Rock Café II on the campus of Indiana University of Pennsylvania during the Spring semester of 2006. Almost all of the women in the data set were white undergraduates between the ages of 18 and 24, so there are some definite limitations on the scope of this study. These limitations will affect our conclusion (and the specific definition of the population means in our hypotheses.)

Random Samples?

The observations were collected on February 13, 2006, through February 22, 2006, between 11 a.m. and 7 p.m. We can see that the researchers included both lunch and dinner. They also made observations on all days of the week to ensure that weekly customer patterns did not confound their findings. The authors state that “since the time period for observations and the place where [they] observed students were limited, the sample was a convenience sample.” Despite these limitations, the researchers conducted inference procedures with the data, and the results were published in a reputable journal. We will also conduct inference with this data, but we also include a discussion of the limitations of the study with our conclusion. The authors did this, also.

Do the data met the conditions for use of a t-test?

The researchers reported the following sample statistics.

• In a sample of 45 women dining with other women, the average number of calories ordered was 850, and the standard deviation was 252.
• In a sample of 27 women dining with men, the average number of calories ordered was 719, and the standard deviation was 322.

One of the samples has fewer than 30 women. We need to make sure the distribution of calories in this sample is not heavily skewed and has no outliers, but we do not have access to a spreadsheet of the actual data. Since the researchers conducted a t-test with this data, we will assume that the conditions are met. This includes the assumption that the samples are independent.

As noted previously, the researchers reported the following sample statistics.

To compute the t-test statistic, make sure sample 1 corresponds to population 1. Here our population 1 is “women eating with other women.” So x 1 = 850, s 1 = 252, n 1 =45, and so on.

Using technology, we determined that the degrees of freedom are about 45 for this data. To find the P-value, we use our familiar simulation of the t-distribution. Since the alternative hypothesis is a “greater than” statement, we look for the area to the right of T = 1.81. The P-value is 0.0385.

Generic Conclusion

The hypotheses for this test are H 0 : μ 1 – μ 2 = 0 and H a : μ 1 – μ 2 > 0. Since the P-value is less than the significance level (0.0385 < 0.05), we reject H 0 and accept H a .

Conclusion in context

At Indiana University of Pennsylvania, the mean number of calories ordered by undergraduate women eating with other women is greater than the mean number of calories ordered by undergraduate women eating with men (P-value = 0.0385).

Comment about Conclusions

In the conclusion above, we did not generalize the findings to all women. Since the samples included only undergraduate women at one university, we included this information in our conclusion. But our conclusion is a cautious statement of the findings. The authors see the results more broadly in the context of theories in the field of social psychology. In the context of these theories, they write, “Our findings support the assertion that meal size is a tool for influencing the impressions of others. For traditional-age, predominantly White college women, diminished meal size appears to be an attempt to assert femininity in groups that include men.” This viewpoint is echoed in the following summary of the study for the general public on National Public Radio (npr.org).

• Both men and women appear to choose larger portions when they eat with women, and both men and women choose smaller portions when they eat in the company of men, according to new research published in the Journal of Applied Social Psychology . The study, conducted among a sample of 127 college students, suggests that both men and women are influenced by unconscious scripts about how to behave in each other’s company. And these scripts change the way men and women eat when they eat together and when they eat apart.

Should we be concerned that the findings of this study are generalized in this way? Perhaps. But the authors of the article address this concern by including the following disclaimer with their findings: “While the results of our research are suggestive, they should be replicated with larger, representative samples. Studies should be done not only with primarily White, middle-class college students, but also with students who differ in terms of race/ethnicity, social class, age, sexual orientation, and so forth.” This is an example of good statistical practice. It is often very difficult to select truly random samples from the populations of interest. Researchers therefore discuss the limitations of their sampling design when they discuss their conclusions.

In the following activities, you will have the opportunity to practice parts of the hypothesis test for a difference in two population means. On the next page, the activities focus on the entire process and also incorporate technology.

National Health and Nutrition Survey

https://assessments.lumenlearning.co...sessments/3705

https://assessments.lumenlearning.co...sessments/3782

https://assessments.lumenlearning.co...sessments/3706

Contributors and Attributions

• Concepts in Statistics. Provided by : Open Learning Initiative. Located at : http://oli.cmu.edu . License : CC BY: Attribution

Statistics Made Easy

Two Sample t-test: Definition, Formula, and Example

A two sample t-test is used to determine whether or not two population means are equal.

This tutorial explains the following:

• The motivation for performing a two sample t-test.
• The formula to perform a two sample t-test.
• The assumptions that should be met to perform a two sample t-test.
• An example of how to perform a two sample t-test.

Two Sample t-test: Motivation

Suppose we want to know whether or not the mean weight between two different species of turtles is equal. Since there are thousands of turtles in each population, it would be too time-consuming and costly to go around and weigh each individual turtle.

Instead, we might take a simple random sample of 15 turtles from each population and use the mean weight in each sample to determine if the mean weight is equal between the two populations:

However, it’s virtually guaranteed that the mean weight between the two samples will be at least a little different. The question is whether or not this difference is statistically significant . Fortunately, a two sample t-test allows us to answer this question.

Two Sample t-test: Formula

A two-sample t-test always uses the following null hypothesis:

• H 0 : μ 1  = μ 2 (the two population means are equal)

The alternative hypothesis can be either two-tailed, left-tailed, or right-tailed:

• H 1 (two-tailed): μ 1  ≠ μ 2 (the two population means are not equal)
• H 1 (left-tailed): μ 1  < μ 2  (population 1 mean is less than population 2 mean)
• H 1 (right-tailed):  μ 1 > μ 2  (population 1 mean is greater than population 2 mean)

We use the following formula to calculate the test statistic t:

Test statistic:  ( x 1  –  x 2 )  /  s p (√ 1/n 1  + 1/n 2 )

where  x 1  and  x 2 are the sample means, n 1 and n 2  are the sample sizes, and where s p is calculated as:

s p = √  (n 1 -1)s 1 2  +  (n 2 -1)s 2 2  /  (n 1 +n 2 -2)

where s 1 2  and s 2 2  are the sample variances.

If the p-value that corresponds to the test statistic t with (n 1 +n 2 -1) degrees of freedom is less than your chosen significance level (common choices are 0.10, 0.05, and 0.01) then you can reject the null hypothesis.

Two Sample t-test: Assumptions

For the results of a two sample t-test to be valid, the following assumptions should be met:

• The observations in one sample should be independent of the observations in the other sample.
• The data should be approximately normally distributed.
• The two samples should have approximately the same variance. If this assumption is not met, you should instead perform Welch’s t-test .
• The data in both samples was obtained using a random sampling method .

Two Sample t-test : Example

Suppose we want to know whether or not the mean weight between two different species of turtles is equal. To test this, will perform a two sample t-test at significance level α = 0.05 using the following steps:

Step 1: Gather the sample data.

Suppose we collect a random sample of turtles from each population with the following information:

• Sample size n 1 = 40
• Sample mean weight  x 1  = 300
• Sample standard deviation s 1 = 18.5
• Sample size n 2 = 38
• Sample mean weight  x 2  = 305
• Sample standard deviation s 2 = 16.7

Step 2: Define the hypotheses.

We will perform the two sample t-test with the following hypotheses:

• H 0 :  μ 1  = μ 2 (the two population means are equal)
• H 1 :  μ 1  ≠ μ 2 (the two population means are not equal)

Step 3: Calculate the test statistic  t .

First, we will calculate the pooled standard deviation s p :

s p = √  (n 1 -1)s 1 2  +  (n 2 -1)s 2 2  /  (n 1 +n 2 -2)  = √  (40-1)18.5 2  +  (38-1)16.7 2  /  (40+38-2)  = 17.647

Next, we will calculate the test statistic  t :

t = ( x 1  –  x 2 )  /  s p (√ 1/n 1  + 1/n 2 ) =  (300-305) / 17.647(√ 1/40 + 1/38 ) =  -1.2508

Step 4: Calculate the p-value of the test statistic  t .

According to the T Score to P Value Calculator , the p-value associated with t = -1.2508 and degrees of freedom = n 1 +n 2 -2 = 40+38-2 = 76 is  0.21484 .

Step 5: Draw a conclusion.

Since this p-value is not less than our significance level α = 0.05, we fail to reject the null hypothesis. We do not have sufficient evidence to say that the mean weight of turtles between these two populations is different.

Note:  You can also perform this entire two sample t-test by simply using the Two Sample t-test Calculator .

Additional Resources

The following tutorials explain how to perform a two-sample t-test using different statistical programs:

How to Perform a Two Sample t-test in Excel How to Perform a Two Sample t-test in SPSS How to Perform a Two Sample t-test in Stata How to Perform a Two Sample t-test in R How to Perform a Two Sample t-test in Python How to Perform a Two Sample t-test on a TI-84 Calculator

Published by Zach

Leave a reply cancel reply.

Your email address will not be published. Required fields are marked *

Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices.

Inference for Comparing 2 Population Means (HT for 2 Means, independent samples)

More of the good stuff! We will need to know how to label the null and alternative hypothesis, calculate the test statistic, and then reach our conclusion using the critical value method or the p-value method.

The Test Statistic for a Test of 2 Means from Independent Samples:

[latex]t = \displaystyle \frac{(\bar{x_1} - \bar{x_2}) - (\mu_1 - \mu_2)}{\sqrt{\displaystyle \frac{s_1^2}{n_1} + \displaystyle \frac{s_2^2}{n_2}}}[/latex]

What the different symbols mean:

[latex]n_1[/latex] is the sample size for the first group

[latex]n_2[/latex] is the sample size for the second group

[latex]df[/latex], the degrees of freedom, is the smaller of [latex]n_1 - 1[/latex] and [latex]n_2 - 1[/latex]

[latex]\mu_1[/latex] is the population mean from the first group

[latex]\mu_2[/latex] is the population mean from the second group

[latex]\bar{x_1}[/latex] is the sample mean for the first group

[latex]\bar{x_2}[/latex] is the sample mean for the second group

[latex]s_1[/latex] is the sample standard deviation for the first group

[latex]s_2[/latex] is the sample standard deviation for the second group

[latex]\alpha[/latex] is the significance level , usually given within the problem, or if not given, we assume it to be 5% or 0.05

Assumptions when conducting a Test for 2 Means from Independent Samples:

• We do not know the population standard deviations, and we do not assume they are equal
• The two samples or groups are independent
• Both samples are simple random samples
• Both populations are Normally distributed OR both samples are large ([latex]n_1 > 30[/latex] and [latex]n_2 > 30[/latex])

Steps to conduct the Test for 2 Means from Independent Samples:

• Identify all the symbols listed above (all the stuff that will go into the formulas). This includes [latex]n_1[/latex] and [latex]n_2[/latex], [latex]df[/latex], [latex]\mu_1[/latex] and [latex]\mu_2[/latex], [latex]\bar{x_1}[/latex] and [latex]\bar{x_2}[/latex], [latex]s_1[/latex] and [latex]s_2[/latex], and [latex]\alpha[/latex]
• Identify the null and alternative hypotheses
• Calculate the test statistic, [latex]t = \displaystyle \frac{(\bar{x_1} - \bar{x_2}) - (\mu_1 - \mu_2)}{\sqrt{\displaystyle \frac{s_1^2}{n_1} + \displaystyle \frac{s_2^2}{n_2}}}[/latex]
• Find the critical value(s) OR the p-value OR both
• Apply the Decision Rule
• Write up a conclusion for the test

Example 1: Study on the effectiveness of stents for stroke patients [1]

In this study , researchers randomly assigned stroke patients to two groups: one received the current standard care (control) and the other received a stent surgery in addition to the standard care (stent treatment). If the stents work, the treatment group should have a lower average disability score . Do the results give convincing statistical evidence that the stent treatment reduces the average disability from stroke?

Since we are being asked for convincing statistical evidence, a hypothesis test should be conducted. In this case, we are dealing with averages from two samples or groups (the patients with stent treatment and patients receiving the standard care), so we will conduct a Test of 2 Means.

• [latex]n_1 = 98[/latex] is the sample size for the first group
• [latex]n_2 = 93[/latex] is the sample size for the second group
• [latex]df[/latex], the degrees of freedom, is the smaller of [latex]98 - 1 = 97[/latex] and [latex]93 - 1 = 92[/latex], so [latex]df = 92[/latex]
• [latex]\bar{x_1} = 2.26[/latex] is the sample mean for the first group
• [latex]\bar{x_2} = 3.23[/latex] is the sample mean for the second group
• [latex]s_1 = 1.78[/latex] is the sample standard deviation for the first group
• [latex]s_2 = 1.78[/latex] is the sample standard deviation for the second group
• [latex]\alpha = 0.05[/latex] (we were not told a specific value in the problem, so we are assuming it is 5%)
• One additional assumption we extend from the null hypothesis is that [latex]\mu_1 - \mu_2 = 0[/latex]; this means that in our formula, those variables cancel out
• [latex]H_{0}: \mu_1 = \mu_2[/latex]
• [latex]H_{A}: \mu_1 < \mu_2[/latex]
• [latex]t = \displaystyle \frac{(\bar{x_1} - \bar{x_2}) - (\mu_1 - \mu_2)}{\sqrt{\displaystyle \frac{s_1^2}{n_1} + \displaystyle \frac{s_2^2}{n_2}}} = \displaystyle \frac{(2.26 - 3.23) - 0)}{\sqrt{\displaystyle \frac{1.78^2}{98} + \displaystyle \frac{1.78^2}{93}}} = -3.76[/latex]
• StatDisk : We can conduct this test using StatDisk. The nice thing about StatDisk is that it will also compute the test statistic. From the main menu above we click on Analysis, Hypothesis Testing, and then Mean Two Independent Samples. From there enter the 0.05 significance, along with the specific values as outlined in the picture below in Step 2. Notice the alternative hypothesis is the [latex]<[/latex] option. Enter the sample size, mean, and standard deviation for each group, and make sure that unequal variances is selected. Now we click on Evaluate. If you check the values, the test statistic is reported in the Step 3 display, as well as the P-Value of 0.00011.
• Applying the Decision Rule: We now compare this to our significance level, which is 0.05. If the p-value is smaller or equal to the alpha level, we have enough evidence for our claim, otherwise we do not. Here, [latex]p-value = 0.00011[/latex], which is definitely smaller than [latex]\alpha = 0.05[/latex], so we have enough evidence for the alternative hypothesis…but what does this mean?
• Conclusion: Because our p-value  of [latex]0.00011[/latex] is less than our [latex]\alpha[/latex] level of [latex]0.05[/latex], we reject [latex]H_{0}[/latex]. We have convincing statistical evidence that the stent treatment reduces the average disability from stroke.

Example 2: Home Run Distances

In 1998, Sammy Sosa and Mark McGwire (2 players in Major League Baseball) were on pace to set a new home run record. At the end of the season McGwire ended up with 70 home runs, and Sosa ended up with 66. The home run distances were recorded and compared (sometimes a player’s home run distance is used to measure their “power”). Do the results give convincing statistical evidence that the home run distances are different from each other? Who would you say “hit the ball farther” in this comparison?

Since we are being asked for convincing statistical evidence, a hypothesis test should be conducted. In this case, we are dealing with averages from two samples or groups (the home run distances), so we will conduct a Test of 2 Means.

• [latex]n_1 = 70[/latex] is the sample size for the first group
• [latex]n_2 = 66[/latex] is the sample size for the second group
• [latex]df[/latex], the degrees of freedom, is the smaller of [latex]70 - 1 = 69[/latex] and [latex]66 - 1 = 65[/latex], so [latex]df = 65[/latex]
• [latex]\bar{x_1} = 418.5[/latex] is the sample mean for the first group
• [latex]\bar{x_2} = 404.8[/latex] is the sample mean for the second group
• [latex]s_1 = 45.5[/latex] is the sample standard deviation for the first group
• [latex]s_2 = 35.7[/latex] is the sample standard deviation for the second group
• [latex]H_{A}: \mu_1 \neq \mu_2[/latex]
• [latex]t = \displaystyle \frac{(\bar{x_1} - \bar{x_2}) - (\mu_1 - \mu_2)}{\sqrt{\displaystyle \frac{s_1^2}{n_1} + \displaystyle \frac{s_2^2}{n_2}}} = \displaystyle \frac{(418.5 - 404.8) - 0)}{\sqrt{\displaystyle \frac{45.5^2}{70} + \displaystyle \frac{35.7^2}{65}}} = 1.95[/latex]
• StatDisk : We can conduct this test using StatDisk. The nice thing about StatDisk is that it will also compute the test statistic. From the main menu above we click on Analysis, Hypothesis Testing, and then Mean Two Independent Samples. From there enter the 0.05 significance, along with the specific values as outlined in the picture below in Step 2. Notice the alternative hypothesis is the [latex]\neq[/latex] option. Enter the sample size, mean, and standard deviation for each group, and make sure that unequal variances is selected. Now we click on Evaluate. If you check the values, the test statistic is reported in the Step 3 display, as well as the P-Value of 0.05221.
• Applying the Decision Rule: We now compare this to our significance level, which is 0.05. If the p-value is smaller or equal to the alpha level, we have enough evidence for our claim, otherwise we do not. Here, [latex]p-value = 0.05221[/latex], which is larger than [latex]\alpha = 0.05[/latex], so we do not have enough evidence for the alternative hypothesis…but what does this mean?
• Conclusion: Because our p-value  of [latex]0.05221[/latex] is larger than our [latex]\alpha[/latex] level of [latex]0.05[/latex], we fail to reject [latex]H_{0}[/latex]. We do not have convincing statistical evidence that the home run distances are different.
• Follow-up commentary: But what does this mean? There actually was a difference, right? If we take McGwire’s average and subtract Sosa’s average we get a difference of 13.7. What this result indicates is that the difference is not statistically significant; it could be due more to random chance than something meaningful. Other factors, such as sample size, could also be a determining factor (with a larger sample size, the difference may have been more meaningful).
• Adapted from the Skew The Script curriculum ( skewthescript.org ), licensed under CC BY-NC-Sa 4.0 ↵

Basic Statistics Copyright © by Allyn Leon is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

Share This Book

• Study Guides
• Two Sample t test for Comparing Two Means
• Method of Statistical Inference
• Types of Statistics
• Steps in the Process
• Making Predictions
• Comparing Results
• Probability
• Quiz: Introduction to Statistics
• What Are Statistics?
• Quiz: Bar Chart
• Quiz: Pie Chart
• Introduction to Graphic Displays
• Quiz: Dot Plot
• Quiz: Introduction to Graphic Displays
• Frequency Histogram
• Relative Frequency Histogram
• Quiz: Relative Frequency Histogram
• Frequency Polygon
• Quiz: Frequency Polygon
• Frequency Distribution
• Stem-and-Leaf
• Box Plot (Box-and-Whiskers)
• Quiz: Box Plot (Box-and-Whiskers)
• Scatter Plot
• Measures of Central Tendency
• Quiz: Measures of Central Tendency
• Measures of Variability
• Quiz: Measures of Variability
• Measurement Scales
• Quiz: Introduction to Numerical Measures
• Classic Theory
• Relative Frequency Theory
• Probability of Simple Events
• Quiz: Probability of Simple Events
• Independent Events
• Dependent Events
• Introduction to Probability
• Quiz: Introduction to Probability
• Probability of Joint Occurrences
• Quiz: Probability of Joint Occurrences
• Non-Mutually-Exclusive Outcomes
• Quiz: Non-Mutually-Exclusive Outcomes
• Double-Counting
• Conditional Probability
• Quiz: Conditional Probability
• Probability Distributions
• Quiz: Probability Distributions
• The Binomial
• Quiz: The Binomial
• Quiz: Sampling Distributions
• Random and Systematic Error
• Central Limit Theorem
• Quiz: Central Limit Theorem
• Populations, Samples, Parameters, and Statistics
• Properties of the Normal Curve
• Quiz: Populations, Samples, Parameters, and Statistics
• Sampling Distributions
• Quiz: Properties of the Normal Curve
• Normal Approximation to the Binomial
• Quiz: Normal Approximation to the Binomial
• Quiz: Stating Hypotheses
• The Test Statistic
• Quiz: The Test Statistic
• One- and Two-Tailed Tests
• Quiz: One- and Two-Tailed Tests
• Type I and II Errors
• Quiz: Type I and II Errors
• Stating Hypotheses
• Significance
• Quiz: Significance
• Point Estimates and Confidence Intervals
• Quiz: Point Estimates and Confidence Intervals
• Estimating a Difference Score
• Quiz: Estimating a Difference Score
• Univariate Tests: An Overview
• Quiz: Univariate Tests: An Overview
• One-Sample z-test
• Quiz: One-Sample z-test
• One-Sample t-test
• Quiz: One-Sample t-test
• Two-Sample z-test for Comparing Two Means
• Quiz: Introduction to Univariate Inferential Tests
• Quiz: Two-Sample z-test for Comparing Two Means
• Quiz: Two-Sample t-test for Comparing Two Means
• Paired Difference t-test
• Quiz: Paired Difference t-test
• Test for a Single Population Proportion
• Quiz: Test for a Single Population Proportion
• Test for Comparing Two Proportions
• Quiz: Test for Comparing Two Proportions
• Quiz: Simple Linear Regression
• Chi-Square (X2)
• Quiz: Chi-Square (X2)
• Correlation
• Quiz: Correlation
• Simple Linear Regression
• Common Mistakes
• Statistics Tables
• Quiz: Cumulative Review A
• Quiz: Cumulative Review B
• Statistics Quizzes

Requirements : Two normally distributed but independent populations, σ is unknown

Hypothesis test

An experiment is conducted to determine whether intensive tutoring (covering a great deal of material in a fixed amount of time) is more effective than paced tutoring (covering less material in the same amount of time). Two randomly chosen groups are tutored separately and then administered proficiency tests. Use a significance level of α < 0.05.

Let μ 1 represent the population mean for the intensive tutoring group and μ 2 represent the population mean for the paced tutoring group.

null hypothesis : H 0 : μ 1 = μ 2

or H 0 : μ 1 – μ 2 = 0

alternative hypothesis : H a : μ 1 > μ 2

or: H a : μ 1 – μ 2 > 0

The degrees of freedom parameter is the smaller of (12 – 1) and (10 – 1), or 9. Because this is a one‐tailed test, the alpha level (0.05) is not divided by two. The next step is to look up t .05,9 in the t‐ table (Table 3 in "Statistics Tables"), which gives a critical value of 1.833. The computed t of 1.166 does not exceed the tabled value, so the null hypothesis cannot be rejected. This test has not provided statistically significant evidence that intensive tutoring is superior to paced tutoring.

Estimate a 90 percent confidence interval for the difference between the number of raisins per box in two brands of breakfast cereal.

The interval is (–2.81, 19.81).

You can be 90 percent confident that Brand A cereal has between 2.81 fewer and 19.81 more raisins per box than Brand B. The fact that the interval contains 0 means that if you had performed a test of the hypothesis that the two population means are different (using the same significance level), you would not have been able to reject the null hypothesis of no difference.

If the two population distributions can be assumed to have the same variance—and, therefore, the same standard deviation— s 1 and s 2 can be pooled together, each weighted by the number of cases in each sample. Although using pooled variance in a t‐ test is generally more likely to yield significant results than using separate variances, it is often hard to know whether the variances of the two populations are equal. For this reason, the pooled variance method should be used with caution. The formula for the pooled estimator of σ 2 is 

where s 1 and s 2 are the standard deviations of the two samples and n 1 and n 2 are the sizes of the two samples.

The formula for comparing the means of two populations using pooled variance is

df = n 1 + n 2 – 2

Does right‐ or left‐handedness affect how fast people type? Random samples of students from a typing class are given a typing speed test (words per minute), and the results are compared. Significance level for the test: 0.10. Because you are looking for a difference between the groups in either direction (right‐handed faster than left, or vice versa), this is a two‐tailed test.

or: H 0 : μ 1 – μ 2 = 0

alternative hypothesis : H a : μ 1 ≠ μ 2

or: H a : μ 1 – μ 2 ≠ 0

First, calculate the pooled variance: 

Next, calculate the t‐ value:

The degrees‐of ‐ freedom parameter is 16 + 9 – 2, or 23. This test is a two‐tailed one, so you divide the alpha level (0.10) by two. Next, you look up t .05,23 in the t‐ table (Table 3 in "Statistics Tables"), which gives a critical value

of 1.714. This value is larger than the absolute value of the computed t of –1.598, so the null hypothesis of equal population means cannot be rejected. There is no evidence that right‐ or left ‐ handedness has any effect on typing speed.

Previous Quiz: Two-Sample z-test for Comparing Two Means

Next Quiz: Two-Sample t-test for Comparing Two Means

• Online Quizzes for CliffsNotes Statistics QuickReview, 2nd Edition

Two Sample t-test: Definition, Formula, and Example

A two sample t-test is used to determine whether or not two population means are equal.

This tutorial explains the following:

• The motivation for performing a two sample t-test.
• The formula to perform a two sample t-test.
• The assumptions that should be met to perform a two sample t-test.
• An example of how to perform a two sample t-test.

Two Sample t-test: Motivation

Suppose we want to know whether or not the mean weight between two different species of turtles is equal. Since there are thousands of turtles in each population, it would be too time-consuming and costly to go around and weigh each individual turtle.

Instead, we might take a simple random sample of 15 turtles from each population and use the mean weight in each sample to determine if the mean weight is equal between the two populations:

However, it’s virtually guaranteed that the mean weight between the two samples will be at least a little different. The question is whether or not this difference is statistically significant . Fortunately, a two sample t-test allows us to answer this question.

Two Sample t-test: Formula

A two-sample t-test always uses the following null hypothesis:

• H 0 : μ 1  = μ 2 (the two population means are equal)

The alternative hypothesis can be either two-tailed, left-tailed, or right-tailed:

• H 1 (two-tailed): μ 1  ≠ μ 2 (the two population means are not equal)
• H 1 (left-tailed): μ 1  2 (population 1 mean is less than population 2 mean)
• H 1 (right-tailed):  μ 1 > μ 2  (population 1 mean is greater than population 2 mean)

We use the following formula to calculate the test statistic t:

Test statistic:  ( x 1  –  x 2 )  /  s p (√ 1/n 1  + 1/n 2 )

where  x 1  and  x 2 are the sample means, n 1 and n 2  are the sample sizes, and where s p is calculated as:

s p = √  (n 1 -1)s 1 2  +  (n 2 -1)s 2 2  /  (n 1 +n 2 -2)

where s 1 2  and s 2 2  are the sample variances.

If the p-value that corresponds to the test statistic t with (n 1 +n 2 -1) degrees of freedom is less than your chosen significance level (common choices are 0.10, 0.05, and 0.01) then you can reject the null hypothesis.

Two Sample t-test: Assumptions

For the results of a two sample t-test to be valid, the following assumptions should be met:

• The observations in one sample should be independent of the observations in the other sample.
• The data should be approximately normally distributed.
• The two samples should have approximately the same variance. If this assumption is not met, you should instead perform Welch’s t-test .
• The data in both samples was obtained using a random sampling method .

Two Sample t-test : Example

Suppose we want to know whether or not the mean weight between two different species of turtles is equal. To test this, will perform a two sample t-test at significance level α = 0.05 using the following steps:

Step 1: Gather the sample data.

Suppose we collect a random sample of turtles from each population with the following information:

• Sample size n 1 = 40
• Sample mean weight  x 1  = 300
• Sample standard deviation s 1 = 18.5
• Sample size n 2 = 38
• Sample mean weight  x 2  = 305
• Sample standard deviation s 2 = 16.7

Step 2: Define the hypotheses.

We will perform the two sample t-test with the following hypotheses:

• H 0 :  μ 1  = μ 2 (the two population means are equal)
• H 1 :  μ 1  ≠ μ 2 (the two population means are not equal)

Step 3: Calculate the test statistic  t .

First, we will calculate the pooled standard deviation s p :

s p = √  (n 1 -1)s 1 2  +  (n 2 -1)s 2 2  /  (n 1 +n 2 -2)  = √  (40-1)18.5 2  +  (38-1)16.7 2  /  (40+38-2)  = 17.647

Next, we will calculate the test statistic  t :

t = ( x 1  –  x 2 )  /  s p (√ 1/n 1  + 1/n 2 ) =  (300-305) / 17.647(√ 1/40 + 1/38 ) =  -1.2508

Step 4: Calculate the p-value of the test statistic  t .

According to the T Score to P Value Calculator , the p-value associated with t = -1.2508 and degrees of freedom = n 1 +n 2 -2 = 40+38-2 = 76 is  0.21484 .

Step 5: Draw a conclusion.

Since this p-value is not less than our significance level α = 0.05, we fail to reject the null hypothesis. We do not have sufficient evidence to say that the mean weight of turtles between these two populations is different.

Note:  You can also perform this entire two sample t-test by simply using the Two Sample t-test Calculator .

Additional Resources

The following tutorials explain how to perform a two-sample t-test using different statistical programs:

How to Perform a Two Sample t-test in Excel How to Perform a Two Sample t-test in SPSS How to Perform a Two Sample t-test in Stata How to Perform a Two Sample t-test in R How to Perform a Two Sample t-test in Python How to Perform a Two Sample t-test on a TI-84 Calculator

An Introduction to the Binomial Distribution

4 examples of using linear regression in real life, related posts, three-way anova: definition & example, two sample z-test: definition, formula, and example, one sample z-test: definition, formula, and example, how to find a confidence interval for a..., an introduction to the exponential distribution, an introduction to the uniform distribution, the breusch-pagan test: definition & example, population vs. sample: what’s the difference, introduction to multiple linear regression, dunn’s test for multiple comparisons.

If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

Statistics and probability

Course: statistics and probability   >   unit 13.

• Statistical significance of experiment
• Statistical significance on bus speeds
• Hypothesis testing in experiments
• Difference of sample means distribution
• Confidence interval of difference of means
• Clarification of confidence interval of difference of means

Hypothesis test for difference of means

Want to join the conversation.

• Upvote Button navigates to signup page
• Downvote Button navigates to signup page
• Flag Button navigates to signup page

Video transcript

Hypothesis Testing for Means & Proportions

Lisa Sullivan, PhD

Professor of Biostatistics

Boston University School of Public Health

Introduction

This is the first of three modules that will addresses the second area of statistical inference, which is hypothesis testing, in which a specific statement or hypothesis is generated about a population parameter, and sample statistics are used to assess the likelihood that the hypothesis is true. The hypothesis is based on available information and the investigator's belief about the population parameters. The process of hypothesis testing involves setting up two competing hypotheses, the null hypothesis and the alternate hypothesis. One selects a random sample (or multiple samples when there are more comparison groups), computes summary statistics and then assesses the likelihood that the sample data support the research or alternative hypothesis. Similar to estimation, the process of hypothesis testing is based on probability theory and the Central Limit Theorem.  

This module will focus on hypothesis testing for means and proportions. The next two modules in this series will address analysis of variance and chi-squared tests. 

Learning Objectives

After completing this module, the student will be able to:

• Define null and research hypothesis, test statistic, level of significance and decision rule
• Distinguish between Type I and Type II errors and discuss the implications of each
• Explain the difference between one and two sided tests of hypothesis
• Estimate and interpret p-values
• Explain the relationship between confidence interval estimates and p-values in drawing inferences
• Differentiate hypothesis testing procedures based on type of outcome variable and number of sample

Introduction to Hypothesis Testing

Techniques for hypothesis testing  .

The techniques for hypothesis testing depend on

• the type of outcome variable being analyzed (continuous, dichotomous, discrete)
• the number of comparison groups in the investigation
• whether the comparison groups are independent (i.e., physically separate such as men versus women) or dependent (i.e., matched or paired such as pre- and post-assessments on the same participants).

In estimation we focused explicitly on techniques for one and two samples and discussed estimation for a specific parameter (e.g., the mean or proportion of a population), for differences (e.g., difference in means, the risk difference) and ratios (e.g., the relative risk and odds ratio). Here we will focus on procedures for one and two samples when the outcome is either continuous (and we focus on means) or dichotomous (and we focus on proportions).

General Approach: A Simple Example

The Centers for Disease Control (CDC) reported on trends in weight, height and body mass index from the 1960's through 2002. 1 The general trend was that Americans were much heavier and slightly taller in 2002 as compared to 1960; both men and women gained approximately 24 pounds, on average, between 1960 and 2002.   In 2002, the mean weight for men was reported at 191 pounds. Suppose that an investigator hypothesizes that weights are even higher in 2006 (i.e., that the trend continued over the subsequent 4 years). The research hypothesis is that the mean weight in men in 2006 is more than 191 pounds. The null hypothesis is that there is no change in weight, and therefore the mean weight is still 191 pounds in 2006.  

In order to test the hypotheses, we select a random sample of American males in 2006 and measure their weights. Suppose we have resources available to recruit n=100 men into our sample. We weigh each participant and compute summary statistics on the sample data. Suppose in the sample we determine the following:

Do the sample data support the null or research hypothesis? The sample mean of 197.1 is numerically higher than 191. However, is this difference more than would be expected by chance? In hypothesis testing, we assume that the null hypothesis holds until proven otherwise. We therefore need to determine the likelihood of observing a sample mean of 197.1 or higher when the true population mean is 191 (i.e., if the null hypothesis is true or under the null hypothesis). We can compute this probability using the Central Limit Theorem. Specifically,

(Notice that we use the sample standard deviation in computing the Z score. This is generally an appropriate substitution as long as the sample size is large, n > 30. Thus, there is less than a 1% probability of observing a sample mean as large as 197.1 when the true population mean is 191. Do you think that the null hypothesis is likely true? Based on how unlikely it is to observe a sample mean of 197.1 under the null hypothesis (i.e., <1% probability), we might infer, from our data, that the null hypothesis is probably not true.

Suppose that the sample data had turned out differently. Suppose that we instead observed the following in 2006:

How likely it is to observe a sample mean of 192.1 or higher when the true population mean is 191 (i.e., if the null hypothesis is true)? We can again compute this probability using the Central Limit Theorem. Specifically,

There is a 33.4% probability of observing a sample mean as large as 192.1 when the true population mean is 191. Do you think that the null hypothesis is likely true?  

Neither of the sample means that we obtained allows us to know with certainty whether the null hypothesis is true or not. However, our computations suggest that, if the null hypothesis were true, the probability of observing a sample mean >197.1 is less than 1%. In contrast, if the null hypothesis were true, the probability of observing a sample mean >192.1 is about 33%. We can't know whether the null hypothesis is true, but the sample that provided a mean value of 197.1 provides much stronger evidence in favor of rejecting the null hypothesis, than the sample that provided a mean value of 192.1. Note that this does not mean that a sample mean of 192.1 indicates that the null hypothesis is true; it just doesn't provide compelling evidence to reject it.

In essence, hypothesis testing is a procedure to compute a probability that reflects the strength of the evidence (based on a given sample) for rejecting the null hypothesis. In hypothesis testing, we determine a threshold or cut-off point (called the critical value) to decide when to believe the null hypothesis and when to believe the research hypothesis. It is important to note that it is possible to observe any sample mean when the true population mean is true (in this example equal to 191), but some sample means are very unlikely. Based on the two samples above it would seem reasonable to believe the research hypothesis when x̄ = 197.1, but to believe the null hypothesis when x̄ =192.1. What we need is a threshold value such that if x̄ is above that threshold then we believe that H 1 is true and if x̄ is below that threshold then we believe that H 0 is true. The difficulty in determining a threshold for x̄ is that it depends on the scale of measurement. In this example, the threshold, sometimes called the critical value, might be 195 (i.e., if the sample mean is 195 or more then we believe that H 1 is true and if the sample mean is less than 195 then we believe that H 0 is true). Suppose we are interested in assessing an increase in blood pressure over time, the critical value will be different because blood pressures are measured in millimeters of mercury (mmHg) as opposed to in pounds. In the following we will explain how the critical value is determined and how we handle the issue of scale.

First, to address the issue of scale in determining the critical value, we convert our sample data (in particular the sample mean) into a Z score. We know from the module on probability that the center of the Z distribution is zero and extreme values are those that exceed 2 or fall below -2. Z scores above 2 and below -2 represent approximately 5% of all Z values. If the observed sample mean is close to the mean specified in H 0 (here m =191), then Z will be close to zero. If the observed sample mean is much larger than the mean specified in H 0 , then Z will be large.  

In hypothesis testing, we select a critical value from the Z distribution. This is done by first determining what is called the level of significance, denoted α ("alpha"). What we are doing here is drawing a line at extreme values. The level of significance is the probability that we reject the null hypothesis (in favor of the alternative) when it is actually true and is also called the Type I error rate.

α = Level of significance = P(Type I error) = P(Reject H 0 | H 0 is true).

Because α is a probability, it ranges between 0 and 1. The most commonly used value in the medical literature for α is 0.05, or 5%. Thus, if an investigator selects α=0.05, then they are allowing a 5% probability of incorrectly rejecting the null hypothesis in favor of the alternative when the null is in fact true. Depending on the circumstances, one might choose to use a level of significance of 1% or 10%. For example, if an investigator wanted to reject the null only if there were even stronger evidence than that ensured with α=0.05, they could choose a =0.01as their level of significance. The typical values for α are 0.01, 0.05 and 0.10, with α=0.05 the most commonly used value.  

Suppose in our weight study we select α=0.05. We need to determine the value of Z that holds 5% of the values above it (see below).

The critical value of Z for α =0.05 is Z = 1.645 (i.e., 5% of the distribution is above Z=1.645). With this value we can set up what is called our decision rule for the test. The rule is to reject H 0 if the Z score is 1.645 or more.  

With the first sample we have

Because 2.38 > 1.645, we reject the null hypothesis. (The same conclusion can be drawn by comparing the 0.0087 probability of observing a sample mean as extreme as 197.1 to the level of significance of 0.05. If the observed probability is smaller than the level of significance we reject H 0 ). Because the Z score exceeds the critical value, we conclude that the mean weight for men in 2006 is more than 191 pounds, the value reported in 2002. If we observed the second sample (i.e., sample mean =192.1), we would not be able to reject the null hypothesis because the Z score is 0.43 which is not in the rejection region (i.e., the region in the tail end of the curve above 1.645). With the second sample we do not have sufficient evidence (because we set our level of significance at 5%) to conclude that weights have increased. Again, the same conclusion can be reached by comparing probabilities. The probability of observing a sample mean as extreme as 192.1 is 33.4% which is not below our 5% level of significance.

Hypothesis Testing: Upper-, Lower, and Two Tailed Tests

The procedure for hypothesis testing is based on the ideas described above. Specifically, we set up competing hypotheses, select a random sample from the population of interest and compute summary statistics. We then determine whether the sample data supports the null or alternative hypotheses. The procedure can be broken down into the following five steps.  

• Step 1. Set up hypotheses and select the level of significance α.

H 0 : Null hypothesis (no change, no difference);  

H 1 : Research hypothesis (investigator's belief); α =0.05

• Step 2. Select the appropriate test statistic.  

The test statistic is a single number that summarizes the sample information.   An example of a test statistic is the Z statistic computed as follows:

When the sample size is small, we will use t statistics (just as we did when constructing confidence intervals for small samples). As we present each scenario, alternative test statistics are provided along with conditions for their appropriate use.

• Step 3.  Set up decision rule.  

The decision rule is a statement that tells under what circumstances to reject the null hypothesis. The decision rule is based on specific values of the test statistic (e.g., reject H 0 if Z > 1.645). The decision rule for a specific test depends on 3 factors: the research or alternative hypothesis, the test statistic and the level of significance. Each is discussed below.

• The decision rule depends on whether an upper-tailed, lower-tailed, or two-tailed test is proposed. In an upper-tailed test the decision rule has investigators reject H 0 if the test statistic is larger than the critical value. In a lower-tailed test the decision rule has investigators reject H 0 if the test statistic is smaller than the critical value.  In a two-tailed test the decision rule has investigators reject H 0 if the test statistic is extreme, either larger than an upper critical value or smaller than a lower critical value.
• The exact form of the test statistic is also important in determining the decision rule. If the test statistic follows the standard normal distribution (Z), then the decision rule will be based on the standard normal distribution. If the test statistic follows the t distribution, then the decision rule will be based on the t distribution. The appropriate critical value will be selected from the t distribution again depending on the specific alternative hypothesis and the level of significance.  
• The third factor is the level of significance. The level of significance which is selected in Step 1 (e.g., α =0.05) dictates the critical value.   For example, in an upper tailed Z test, if α =0.05 then the critical value is Z=1.645.  

The following figures illustrate the rejection regions defined by the decision rule for upper-, lower- and two-tailed Z tests with α=0.05. Notice that the rejection regions are in the upper, lower and both tails of the curves, respectively. The decision rules are written below each figure.

Rejection Region for Lower-Tailed Z Test (H 1 : μ < μ 0 ) with α =0.05

The decision rule is: Reject H 0 if Z < 1.645.

Rejection Region for Two-Tailed Z Test (H 1 : μ ≠ μ 0 ) with α =0.05

The decision rule is: Reject H 0 if Z < -1.960 or if Z > 1.960.

The complete table of critical values of Z for upper, lower and two-tailed tests can be found in the table of Z values to the right in "Other Resources."

Critical values of t for upper, lower and two-tailed tests can be found in the table of t values in "Other Resources."

• Step 4. Compute the test statistic.  

Here we compute the test statistic by substituting the observed sample data into the test statistic identified in Step 2.

• Step 5. Conclusion.  

The final conclusion is made by comparing the test statistic (which is a summary of the information observed in the sample) to the decision rule. The final conclusion will be either to reject the null hypothesis (because the sample data are very unlikely if the null hypothesis is true) or not to reject the null hypothesis (because the sample data are not very unlikely).  

If the null hypothesis is rejected, then an exact significance level is computed to describe the likelihood of observing the sample data assuming that the null hypothesis is true. The exact level of significance is called the p-value and it will be less than the chosen level of significance if we reject H 0 .

Statistical computing packages provide exact p-values as part of their standard output for hypothesis tests. In fact, when using a statistical computing package, the steps outlined about can be abbreviated. The hypotheses (step 1) should always be set up in advance of any analysis and the significance criterion should also be determined (e.g., α =0.05). Statistical computing packages will produce the test statistic (usually reporting the test statistic as t) and a p-value. The investigator can then determine statistical significance using the following: If p < α then reject H 0 .  

• Step 1. Set up hypotheses and determine level of significance

H 0 : μ = 191 H 1 : μ > 191                 α =0.05

The research hypothesis is that weights have increased, and therefore an upper tailed test is used.

• Step 2. Select the appropriate test statistic.

Because the sample size is large (n > 30) the appropriate test statistic is

• Step 3. Set up decision rule.  

In this example, we are performing an upper tailed test (H 1 : μ> 191), with a Z test statistic and selected α =0.05.   Reject H 0 if Z > 1.645.

We now substitute the sample data into the formula for the test statistic identified in Step 2.  

We reject H 0 because 2.38 > 1.645. We have statistically significant evidence at a =0.05, to show that the mean weight in men in 2006 is more than 191 pounds. Because we rejected the null hypothesis, we now approximate the p-value which is the likelihood of observing the sample data if the null hypothesis is true. An alternative definition of the p-value is the smallest level of significance where we can still reject H 0 . In this example, we observed Z=2.38 and for α=0.05, the critical value was 1.645. Because 2.38 exceeded 1.645 we rejected H 0 . In our conclusion we reported a statistically significant increase in mean weight at a 5% level of significance. Using the table of critical values for upper tailed tests, we can approximate the p-value. If we select α=0.025, the critical value is 1.96, and we still reject H 0 because 2.38 > 1.960. If we select α=0.010 the critical value is 2.326, and we still reject H 0 because 2.38 > 2.326. However, if we select α=0.005, the critical value is 2.576, and we cannot reject H 0 because 2.38 < 2.576. Therefore, the smallest α where we still reject H 0 is 0.010. This is the p-value. A statistical computing package would produce a more precise p-value which would be in between 0.005 and 0.010. Here we are approximating the p-value and would report p < 0.010.                  

Type I and Type II Errors

In all tests of hypothesis, there are two types of errors that can be committed. The first is called a Type I error and refers to the situation where we incorrectly reject H 0 when in fact it is true. This is also called a false positive result (as we incorrectly conclude that the research hypothesis is true when in fact it is not). When we run a test of hypothesis and decide to reject H 0 (e.g., because the test statistic exceeds the critical value in an upper tailed test) then either we make a correct decision because the research hypothesis is true or we commit a Type I error. The different conclusions are summarized in the table below. Note that we will never know whether the null hypothesis is really true or false (i.e., we will never know which row of the following table reflects reality).

Table - Conclusions in Test of Hypothesis

In the first step of the hypothesis test, we select a level of significance, α, and α= P(Type I error). Because we purposely select a small value for α, we control the probability of committing a Type I error. For example, if we select α=0.05, and our test tells us to reject H 0 , then there is a 5% probability that we commit a Type I error. Most investigators are very comfortable with this and are confident when rejecting H 0 that the research hypothesis is true (as it is the more likely scenario when we reject H 0 ).

When we run a test of hypothesis and decide not to reject H 0 (e.g., because the test statistic is below the critical value in an upper tailed test) then either we make a correct decision because the null hypothesis is true or we commit a Type II error. Beta (β) represents the probability of a Type II error and is defined as follows: β=P(Type II error) = P(Do not Reject H 0 | H 0 is false). Unfortunately, we cannot choose β to be small (e.g., 0.05) to control the probability of committing a Type II error because β depends on several factors including the sample size, α, and the research hypothesis. When we do not reject H 0 , it may be very likely that we are committing a Type II error (i.e., failing to reject H 0 when in fact it is false). Therefore, when tests are run and the null hypothesis is not rejected we often make a weak concluding statement allowing for the possibility that we might be committing a Type II error. If we do not reject H 0 , we conclude that we do not have significant evidence to show that H 1 is true. We do not conclude that H 0 is true.

 The most common reason for a Type II error is a small sample size.

Tests with One Sample, Continuous Outcome

Hypothesis testing applications with a continuous outcome variable in a single population are performed according to the five-step procedure outlined above. A key component is setting up the null and research hypotheses. The objective is to compare the mean in a single population to known mean (μ 0 ). The known value is generally derived from another study or report, for example a study in a similar, but not identical, population or a study performed some years ago. The latter is called a historical control. It is important in setting up the hypotheses in a one sample test that the mean specified in the null hypothesis is a fair and reasonable comparator. This will be discussed in the examples that follow.

Test Statistics for Testing H 0 : μ= μ 0

• if n > 30
• if n < 30

Note that statistical computing packages will use the t statistic exclusively and make the necessary adjustments for comparing the test statistic to appropriate values from probability tables to produce a p-value. 

The National Center for Health Statistics (NCHS) published a report in 2005 entitled Health, United States, containing extensive information on major trends in the health of Americans. Data are provided for the US population as a whole and for specific ages, sexes and races.  The NCHS report indicated that in 2002 Americans paid an average of $3,302 per year on health care and prescription drugs. An investigator hypothesizes that in 2005 expenditures have decreased primarily due to the availability of generic drugs. To test the hypothesis, a sample of 100 Americans are selected and their expenditures on health care and prescription drugs in 2005 are measured.   The sample data are summarized as follows: n=100, x̄

=$3,190 and s=$890. Is there statistical evidence of a reduction in expenditures on health care and prescription drugs in 2005? Is the sample mean of $3,190 evidence of a true reduction in the mean or is it within chance fluctuation? We will run the test using the five-step approach. 

• Step 1.  Set up hypotheses and determine level of significance

H 0 : μ = 3,302 H 1 : μ < 3,302           α =0.05

The research hypothesis is that expenditures have decreased, and therefore a lower-tailed test is used.

This is a lower tailed test, using a Z statistic and a 5% level of significance.   Reject H 0 if Z < -1.645.

•   Step 4. Compute the test statistic.  

We do not reject H 0 because -1.26 > -1.645. We do not have statistically significant evidence at α=0.05 to show that the mean expenditures on health care and prescription drugs are lower in 2005 than the mean of $3,302 reported in 2002.  

Recall that when we fail to reject H 0 in a test of hypothesis that either the null hypothesis is true (here the mean expenditures in 2005 are the same as those in 2002 and equal to $3,302) or we committed a Type II error (i.e., we failed to reject H 0 when in fact it is false). In summarizing this test, we conclude that we do not have sufficient evidence to reject H 0 . We do not conclude that H 0 is true, because there may be a moderate to high probability that we committed a Type II error. It is possible that the sample size is not large enough to detect a difference in mean expenditures.      

The NCHS reported that the mean total cholesterol level in 2002 for all adults was 203. Total cholesterol levels in participants who attended the seventh examination of the Offspring in the Framingham Heart Study are summarized as follows: n=3,310, x̄ =200.3, and s=36.8. Is there statistical evidence of a difference in mean cholesterol levels in the Framingham Offspring?

Here we want to assess whether the sample mean of 200.3 in the Framingham sample is statistically significantly different from 203 (i.e., beyond what we would expect by chance). We will run the test using the five-step approach.

H 0 : μ= 203 H 1 : μ≠ 203                       α=0.05

The research hypothesis is that cholesterol levels are different in the Framingham Offspring, and therefore a two-tailed test is used.

•   Step 3. Set up decision rule.  

This is a two-tailed test, using a Z statistic and a 5% level of significance. Reject H 0 if Z < -1.960 or is Z > 1.960.

We reject H 0 because -4.22 ≤ -1. .960. We have statistically significant evidence at α=0.05 to show that the mean total cholesterol level in the Framingham Offspring is different from the national average of 203 reported in 2002.   Because we reject H 0 , we also approximate a p-value. Using the two-sided significance levels, p < 0.0001.  

Statistical Significance versus Clinical (Practical) Significance

This example raises an important concept of statistical versus clinical or practical significance. From a statistical standpoint, the total cholesterol levels in the Framingham sample are highly statistically significantly different from the national average with p < 0.0001 (i.e., there is less than a 0.01% chance that we are incorrectly rejecting the null hypothesis). However, the sample mean in the Framingham Offspring study is 200.3, less than 3 units different from the national mean of 203. The reason that the data are so highly statistically significant is due to the very large sample size. It is always important to assess both statistical and clinical significance of data. This is particularly relevant when the sample size is large. Is a 3 unit difference in total cholesterol a meaningful difference?  

Consider again the NCHS-reported mean total cholesterol level in 2002 for all adults of 203. Suppose a new drug is proposed to lower total cholesterol. A study is designed to evaluate the efficacy of the drug in lowering cholesterol.   Fifteen patients are enrolled in the study and asked to take the new drug for 6 weeks. At the end of 6 weeks, each patient's total cholesterol level is measured and the sample statistics are as follows:   n=15, x̄ =195.9 and s=28.7. Is there statistical evidence of a reduction in mean total cholesterol in patients after using the new drug for 6 weeks? We will run the test using the five-step approach. 

H 0 : μ= 203 H 1 : μ< 203                   α=0.05

•  Step 2. Select the appropriate test statistic.  

Because the sample size is small (n<30) the appropriate test statistic is

This is a lower tailed test, using a t statistic and a 5% level of significance. In order to determine the critical value of t, we need degrees of freedom, df, defined as df=n-1. In this example df=15-1=14. The critical value for a lower tailed test with df=14 and a =0.05 is -2.145 and the decision rule is as follows:   Reject H 0 if t < -2.145.

We do not reject H 0 because -0.96 > -2.145. We do not have statistically significant evidence at α=0.05 to show that the mean total cholesterol level is lower than the national mean in patients taking the new drug for 6 weeks. Again, because we failed to reject the null hypothesis we make a weaker concluding statement allowing for the possibility that we may have committed a Type II error (i.e., failed to reject H 0 when in fact the drug is efficacious).

This example raises an important issue in terms of study design. In this example we assume in the null hypothesis that the mean cholesterol level is 203. This is taken to be the mean cholesterol level in patients without treatment. Is this an appropriate comparator? Alternative and potentially more efficient study designs to evaluate the effect of the new drug could involve two treatment groups, where one group receives the new drug and the other does not, or we could measure each patient's baseline or pre-treatment cholesterol level and then assess changes from baseline to 6 weeks post-treatment. These designs are also discussed here.

Video - Comparing a Sample Mean to Known Population Mean (8:20)

Link to transcript of the video

Tests with One Sample, Dichotomous Outcome

Hypothesis testing applications with a dichotomous outcome variable in a single population are also performed according to the five-step procedure. Similar to tests for means, a key component is setting up the null and research hypotheses. The objective is to compare the proportion of successes in a single population to a known proportion (p 0 ). That known proportion is generally derived from another study or report and is sometimes called a historical control. It is important in setting up the hypotheses in a one sample test that the proportion specified in the null hypothesis is a fair and reasonable comparator.    

In one sample tests for a dichotomous outcome, we set up our hypotheses against an appropriate comparator. We select a sample and compute descriptive statistics on the sample data. Specifically, we compute the sample size (n) and the sample proportion which is computed by taking the ratio of the number of successes to the sample size,

We then determine the appropriate test statistic (Step 2) for the hypothesis test. The formula for the test statistic is given below.

Test Statistic for Testing H 0 : p = p 0

if min(np 0 , n(1-p 0 )) > 5

The formula above is appropriate for large samples, defined when the smaller of np 0 and n(1-p 0 ) is at least 5. This is similar, but not identical, to the condition required for appropriate use of the confidence interval formula for a population proportion, i.e.,

Here we use the proportion specified in the null hypothesis as the true proportion of successes rather than the sample proportion. If we fail to satisfy the condition, then alternative procedures, called exact methods must be used to test the hypothesis about the population proportion.

Example:  

The NCHS report indicated that in 2002 the prevalence of cigarette smoking among American adults was 21.1%.  Data on prevalent smoking in n=3,536 participants who attended the seventh examination of the Offspring in the Framingham Heart Study indicated that 482/3,536 = 13.6% of the respondents were currently smoking at the time of the exam. Suppose we want to assess whether the prevalence of smoking is lower in the Framingham Offspring sample given the focus on cardiovascular health in that community. Is there evidence of a statistically lower prevalence of smoking in the Framingham Offspring study as compared to the prevalence among all Americans?

H 0 : p = 0.211 H 1 : p < 0.211                     α=0.05

We must first check that the sample size is adequate.   Specifically, we need to check min(np 0 , n(1-p 0 )) = min( 3,536(0.211), 3,536(1-0.211))=min(746, 2790)=746. The sample size is more than adequate so the following formula can be used:

This is a lower tailed test, using a Z statistic and a 5% level of significance. Reject H 0 if Z < -1.645.

We reject H 0 because -10.93 < -1.645. We have statistically significant evidence at α=0.05 to show that the prevalence of smoking in the Framingham Offspring is lower than the prevalence nationally (21.1%). Here, p < 0.0001.  

The NCHS report indicated that in 2002, 75% of children aged 2 to 17 saw a dentist in the past year. An investigator wants to assess whether use of dental services is similar in children living in the city of Boston. A sample of 125 children aged 2 to 17 living in Boston are surveyed and 64 reported seeing a dentist over the past 12 months. Is there a significant difference in use of dental services between children living in Boston and the national data?

Calculate this on your own before checking the answer.

Video - Hypothesis Test for One Sample and a Dichotomous Outcome (3:55)

Tests with Two Independent Samples, Continuous Outcome

There are many applications where it is of interest to compare two independent groups with respect to their mean scores on a continuous outcome. Here we compare means between groups, but rather than generating an estimate of the difference, we will test whether the observed difference (increase, decrease or difference) is statistically significant or not. Remember, that hypothesis testing gives an assessment of statistical significance, whereas estimation gives an estimate of effect and both are important.

Here we discuss the comparison of means when the two comparison groups are independent or physically separate. The two groups might be determined by a particular attribute (e.g., sex, diagnosis of cardiovascular disease) or might be set up by the investigator (e.g., participants assigned to receive an experimental treatment or placebo). The first step in the analysis involves computing descriptive statistics on each of the two samples. Specifically, we compute the sample size, mean and standard deviation in each sample and we denote these summary statistics as follows:

for sample 1:

for sample 2:

The designation of sample 1 and sample 2 is arbitrary. In a clinical trial setting the convention is to call the treatment group 1 and the control group 2. However, when comparing men and women, for example, either group can be 1 or 2.  

In the two independent samples application with a continuous outcome, the parameter of interest in the test of hypothesis is the difference in population means, μ 1 -μ 2 . The null hypothesis is always that there is no difference between groups with respect to means, i.e.,

The null hypothesis can also be written as follows: H 0 : μ 1 = μ 2 . In the research hypothesis, an investigator can hypothesize that the first mean is larger than the second (H 1 : μ 1 > μ 2 ), that the first mean is smaller than the second (H 1 : μ 1 < μ 2 ), or that the means are different (H 1 : μ 1 ≠ μ 2 ). The three different alternatives represent upper-, lower-, and two-tailed tests, respectively. The following test statistics are used to test these hypotheses.

Test Statistics for Testing H 0 : μ 1 = μ 2

• if n 1 > 30 and n 2 > 30
• if n 1 < 30 or n 2 < 30

NOTE: The formulas above assume equal variability in the two populations (i.e., the population variances are equal, or s 1 2 = s 2 2 ). This means that the outcome is equally variable in each of the comparison populations. For analysis, we have samples from each of the comparison populations. If the sample variances are similar, then the assumption about variability in the populations is probably reasonable. As a guideline, if the ratio of the sample variances, s 1 2 /s 2 2 is between 0.5 and 2 (i.e., if one variance is no more than double the other), then the formulas above are appropriate. If the ratio of the sample variances is greater than 2 or less than 0.5 then alternative formulas must be used to account for the heterogeneity in variances.    

The test statistics include Sp, which is the pooled estimate of the common standard deviation (again assuming that the variances in the populations are similar) computed as the weighted average of the standard deviations in the samples as follows:

Because we are assuming equal variances between groups, we pool the information on variability (sample variances) to generate an estimate of the variability in the population. Note: Because Sp is a weighted average of the standard deviations in the sample, Sp will always be in between s 1 and s 2 .)

Data measured on n=3,539 participants who attended the seventh examination of the Offspring in the Framingham Heart Study are shown below.  

Suppose we now wish to assess whether there is a statistically significant difference in mean systolic blood pressures between men and women using a 5% level of significance.  

H 0 : μ 1 = μ 2

H 1 : μ 1 ≠ μ 2                       α=0.05

Because both samples are large ( > 30), we can use the Z test statistic as opposed to t. Note that statistical computing packages use t throughout. Before implementing the formula, we first check whether the assumption of equality of population variances is reasonable. The guideline suggests investigating the ratio of the sample variances, s 1 2 /s 2 2 . Suppose we call the men group 1 and the women group 2. Again, this is arbitrary; it only needs to be noted when interpreting the results. The ratio of the sample variances is 17.5 2 /20.1 2 = 0.76, which falls between 0.5 and 2 suggesting that the assumption of equality of population variances is reasonable. The appropriate test statistic is

We now substitute the sample data into the formula for the test statistic identified in Step 2. Before substituting, we will first compute Sp, the pooled estimate of the common standard deviation.

Notice that the pooled estimate of the common standard deviation, Sp, falls in between the standard deviations in the comparison groups (i.e., 17.5 and 20.1). Sp is slightly closer in value to the standard deviation in the women (20.1) as there were slightly more women in the sample.   Recall, Sp is a weight average of the standard deviations in the comparison groups, weighted by the respective sample sizes.  

Now the test statistic:

We reject H 0 because 2.66 > 1.960. We have statistically significant evidence at α=0.05 to show that there is a difference in mean systolic blood pressures between men and women. The p-value is p < 0.010.  

Here again we find that there is a statistically significant difference in mean systolic blood pressures between men and women at p < 0.010. Notice that there is a very small difference in the sample means (128.2-126.5 = 1.7 units), but this difference is beyond what would be expected by chance. Is this a clinically meaningful difference? The large sample size in this example is driving the statistical significance. A 95% confidence interval for the difference in mean systolic blood pressures is: 1.7 + 1.26 or (0.44, 2.96). The confidence interval provides an assessment of the magnitude of the difference between means whereas the test of hypothesis and p-value provide an assessment of the statistical significance of the difference.  

Above we performed a study to evaluate a new drug designed to lower total cholesterol. The study involved one sample of patients, each patient took the new drug for 6 weeks and had their cholesterol measured. As a means of evaluating the efficacy of the new drug, the mean total cholesterol following 6 weeks of treatment was compared to the NCHS-reported mean total cholesterol level in 2002 for all adults of 203. At the end of the example, we discussed the appropriateness of the fixed comparator as well as an alternative study design to evaluate the effect of the new drug involving two treatment groups, where one group receives the new drug and the other does not. Here, we revisit the example with a concurrent or parallel control group, which is very typical in randomized controlled trials or clinical trials (refer to the EP713 module on Clinical Trials).  

A new drug is proposed to lower total cholesterol. A randomized controlled trial is designed to evaluate the efficacy of the medication in lowering cholesterol. Thirty participants are enrolled in the trial and are randomly assigned to receive either the new drug or a placebo. The participants do not know which treatment they are assigned. Each participant is asked to take the assigned treatment for 6 weeks. At the end of 6 weeks, each patient's total cholesterol level is measured and the sample statistics are as follows.

Is there statistical evidence of a reduction in mean total cholesterol in patients taking the new drug for 6 weeks as compared to participants taking placebo? We will run the test using the five-step approach.

H 0 : μ 1 = μ 2 H 1 : μ 1 < μ 2                         α=0.05

Because both samples are small (< 30), we use the t test statistic. Before implementing the formula, we first check whether the assumption of equality of population variances is reasonable. The ratio of the sample variances, s 1 2 /s 2 2 =28.7 2 /30.3 2 = 0.90, which falls between 0.5 and 2, suggesting that the assumption of equality of population variances is reasonable. The appropriate test statistic is:

This is a lower-tailed test, using a t statistic and a 5% level of significance. The appropriate critical value can be found in the t Table (in More Resources to the right). In order to determine the critical value of t we need degrees of freedom, df, defined as df=n 1 +n 2 -2 = 15+15-2=28. The critical value for a lower tailed test with df=28 and α=0.05 is -1.701 and the decision rule is: Reject H 0 if t < -1.701.

Now the test statistic,

We reject H 0 because -2.92 < -1.701. We have statistically significant evidence at α=0.05 to show that the mean total cholesterol level is lower in patients taking the new drug for 6 weeks as compared to patients taking placebo, p < 0.005.

The clinical trial in this example finds a statistically significant reduction in total cholesterol, whereas in the previous example where we had a historical control (as opposed to a parallel control group) we did not demonstrate efficacy of the new drug. Notice that the mean total cholesterol level in patients taking placebo is 217.4 which is very different from the mean cholesterol reported among all Americans in 2002 of 203 and used as the comparator in the prior example. The historical control value may not have been the most appropriate comparator as cholesterol levels have been increasing over time. In the next section, we present another design that can be used to assess the efficacy of the new drug.

Video - Comparison of Two Independent Samples With a Continuous Outcome (8:02)

Tests with Matched Samples, Continuous Outcome

In the previous section we compared two groups with respect to their mean scores on a continuous outcome. An alternative study design is to compare matched or paired samples. The two comparison groups are said to be dependent, and the data can arise from a single sample of participants where each participant is measured twice (possibly before and after an intervention) or from two samples that are matched on specific characteristics (e.g., siblings). When the samples are dependent, we focus on difference scores in each participant or between members of a pair and the test of hypothesis is based on the mean difference, μ d . The null hypothesis again reflects "no difference" and is stated as H 0 : μ d =0 . Note that there are some instances where it is of interest to test whether there is a difference of a particular magnitude (e.g., μ d =5) but in most instances the null hypothesis reflects no difference (i.e., μ d =0).  

The appropriate formula for the test of hypothesis depends on the sample size. The formulas are shown below and are identical to those we presented for estimating the mean of a single sample presented (e.g., when comparing against an external or historical control), except here we focus on difference scores.

Test Statistics for Testing H 0 : μ d =0

A new drug is proposed to lower total cholesterol and a study is designed to evaluate the efficacy of the drug in lowering cholesterol. Fifteen patients agree to participate in the study and each is asked to take the new drug for 6 weeks. However, before starting the treatment, each patient's total cholesterol level is measured. The initial measurement is a pre-treatment or baseline value. After taking the drug for 6 weeks, each patient's total cholesterol level is measured again and the data are shown below. The rightmost column contains difference scores for each patient, computed by subtracting the 6 week cholesterol level from the baseline level. The differences represent the reduction in total cholesterol over 4 weeks. (The differences could have been computed by subtracting the baseline total cholesterol level from the level measured at 6 weeks. The way in which the differences are computed does not affect the outcome of the analysis only the interpretation.)

Because the differences are computed by subtracting the cholesterols measured at 6 weeks from the baseline values, positive differences indicate reductions and negative differences indicate increases (e.g., participant 12 increases by 2 units over 6 weeks). The goal here is to test whether there is a statistically significant reduction in cholesterol. Because of the way in which we computed the differences, we want to look for an increase in the mean difference (i.e., a positive reduction). In order to conduct the test, we need to summarize the differences. In this sample, we have

The calculations are shown below.  

Is there statistical evidence of a reduction in mean total cholesterol in patients after using the new medication for 6 weeks? We will run the test using the five-step approach.

H 0 : μ d = 0 H 1 : μ d > 0                 α=0.05

NOTE: If we had computed differences by subtracting the baseline level from the level measured at 6 weeks then negative differences would have reflected reductions and the research hypothesis would have been H 1 : μ d < 0. 

• Step 2 . Select the appropriate test statistic.

This is an upper-tailed test, using a t statistic and a 5% level of significance. The appropriate critical value can be found in the t Table at the right, with df=15-1=14. The critical value for an upper-tailed test with df=14 and α=0.05 is 2.145 and the decision rule is Reject H 0 if t > 2.145.

We now substitute the sample data into the formula for the test statistic identified in Step 2.

We reject H 0 because 4.61 > 2.145. We have statistically significant evidence at α=0.05 to show that there is a reduction in cholesterol levels over 6 weeks.  

Here we illustrate the use of a matched design to test the efficacy of a new drug to lower total cholesterol. We also considered a parallel design (randomized clinical trial) and a study using a historical comparator. It is extremely important to design studies that are best suited to detect a meaningful difference when one exists. There are often several alternatives and investigators work with biostatisticians to determine the best design for each application. It is worth noting that the matched design used here can be problematic in that observed differences may only reflect a "placebo" effect. All participants took the assigned medication, but is the observed reduction attributable to the medication or a result of these participation in a study.

Video - Hypothesis Testing With a Matched Sample and a Continuous Outcome (3:11)

Tests with Two Independent Samples, Dichotomous Outcome

There are several approaches that can be used to test hypotheses concerning two independent proportions. Here we present one approach - the chi-square test of independence is an alternative, equivalent, and perhaps more popular approach to the same analysis. Hypothesis testing with the chi-square test is addressed in the third module in this series: BS704_HypothesisTesting-ChiSquare.

In tests of hypothesis comparing proportions between two independent groups, one test is performed and results can be interpreted to apply to a risk difference, relative risk or odds ratio. As a reminder, the risk difference is computed by taking the difference in proportions between comparison groups, the risk ratio is computed by taking the ratio of proportions, and the odds ratio is computed by taking the ratio of the odds of success in the comparison groups. Because the null values for the risk difference, the risk ratio and the odds ratio are different, the hypotheses in tests of hypothesis look slightly different depending on which measure is used. When performing tests of hypothesis for the risk difference, relative risk or odds ratio, the convention is to label the exposed or treated group 1 and the unexposed or control group 2.      

For example, suppose a study is designed to assess whether there is a significant difference in proportions in two independent comparison groups. The test of interest is as follows:

H 0 : p 1 = p 2 versus H 1 : p 1 ≠ p 2 .  

The following are the hypothesis for testing for a difference in proportions using the risk difference, the risk ratio and the odds ratio. First, the hypotheses above are equivalent to the following:

• For the risk difference, H 0 : p 1 - p 2 = 0 versus H 1 : p 1 - p 2 ≠ 0 which are, by definition, equal to H 0 : RD = 0 versus H 1 : RD ≠ 0.
• If an investigator wants to focus on the risk ratio, the equivalent hypotheses are H 0 : RR = 1 versus H 1 : RR ≠ 1.
• If the investigator wants to focus on the odds ratio, the equivalent hypotheses are H 0 : OR = 1 versus H 1 : OR ≠ 1.  

Suppose a test is performed to test H 0 : RD = 0 versus H 1 : RD ≠ 0 and the test rejects H 0 at α=0.05. Based on this test we can conclude that there is significant evidence, α=0.05, of a difference in proportions, significant evidence that the risk difference is not zero, significant evidence that the risk ratio and odds ratio are not one. The risk difference is analogous to the difference in means when the outcome is continuous. Here the parameter of interest is the difference in proportions in the population, RD = p 1 -p 2 and the null value for the risk difference is zero. In a test of hypothesis for the risk difference, the null hypothesis is always H 0 : RD = 0. This is equivalent to H 0 : RR = 1 and H 0 : OR = 1. In the research hypothesis, an investigator can hypothesize that the first proportion is larger than the second (H 1 : p 1 > p 2 , which is equivalent to H 1 : RD > 0, H 1 : RR > 1 and H 1 : OR > 1), that the first proportion is smaller than the second (H 1 : p 1 < p 2 , which is equivalent to H 1 : RD < 0, H 1 : RR < 1 and H 1 : OR < 1), or that the proportions are different (H 1 : p 1 ≠ p 2 , which is equivalent to H 1 : RD ≠ 0, H 1 : RR ≠ 1 and H 1 : OR ≠

1). The three different alternatives represent upper-, lower- and two-tailed tests, respectively.  

The formula for the test of hypothesis for the difference in proportions is given below.

Test Statistics for Testing H 0 : p 1 = p

The formula above is appropriate for large samples, defined as at least 5 successes (np > 5) and at least 5 failures (n(1-p > 5)) in each of the two samples. If there are fewer than 5 successes or failures in either comparison group, then alternative procedures, called exact methods must be used to estimate the difference in population proportions.

The following table summarizes data from n=3,799 participants who attended the fifth examination of the Offspring in the Framingham Heart Study. The outcome of interest is prevalent CVD and we want to test whether the prevalence of CVD is significantly higher in smokers as compared to non-smokers.

The prevalence of CVD (or proportion of participants with prevalent CVD) among non-smokers is 298/3,055 = 0.0975 and the prevalence of CVD among current smokers is 81/744 = 0.1089. Here smoking status defines the comparison groups and we will call the current smokers group 1 (exposed) and the non-smokers (unexposed) group 2. The test of hypothesis is conducted below using the five step approach.

H 0 : p 1 = p 2     H 1 : p 1 ≠ p 2                 α=0.05

• Step 2.  Select the appropriate test statistic.  

We must first check that the sample size is adequate. Specifically, we need to ensure that we have at least 5 successes and 5 failures in each comparison group. In this example, we have more than enough successes (cases of prevalent CVD) and failures (persons free of CVD) in each comparison group. The sample size is more than adequate so the following formula can be used:

Reject H 0 if Z < -1.960 or if Z > 1.960.

We now substitute the sample data into the formula for the test statistic identified in Step 2. We first compute the overall proportion of successes:

We now substitute to compute the test statistic.

• Step 5. Conclusion.

We do not reject H 0 because -1.960 < 0.927 < 1.960. We do not have statistically significant evidence at α=0.05 to show that there is a difference in prevalent CVD between smokers and non-smokers.  

A 95% confidence interval for the difference in prevalent CVD (or risk difference) between smokers and non-smokers as 0.0114 + 0.0247, or between -0.0133 and 0.0361. Because the 95% confidence interval for the risk difference includes zero we again conclude that there is no statistically significant difference in prevalent CVD between smokers and non-smokers.    

Smoking has been shown over and over to be a risk factor for cardiovascular disease. What might explain the fact that we did not observe a statistically significant difference using data from the Framingham Heart Study? HINT: Here we consider prevalent CVD, would the results have been different if we considered incident CVD?

A randomized trial is designed to evaluate the effectiveness of a newly developed pain reliever designed to reduce pain in patients following joint replacement surgery. The trial compares the new pain reliever to the pain reliever currently in use (called the standard of care). A total of 100 patients undergoing joint replacement surgery agreed to participate in the trial. Patients were randomly assigned to receive either the new pain reliever or the standard pain reliever following surgery and were blind to the treatment assignment. Before receiving the assigned treatment, patients were asked to rate their pain on a scale of 0-10 with higher scores indicative of more pain. Each patient was then given the assigned treatment and after 30 minutes was again asked to rate their pain on the same scale. The primary outcome was a reduction in pain of 3 or more scale points (defined by clinicians as a clinically meaningful reduction). The following data were observed in the trial.

We now test whether there is a statistically significant difference in the proportions of patients reporting a meaningful reduction (i.e., a reduction of 3 or more scale points) using the five step approach.  

H 0 : p 1 = p 2     H 1 : p 1 ≠ p 2              α=0.05

Here the new or experimental pain reliever is group 1 and the standard pain reliever is group 2.

We must first check that the sample size is adequate. Specifically, we need to ensure that we have at least 5 successes and 5 failures in each comparison group, i.e.,

In this example, we have min(50(0.46), 50(1-0.46), 50(0.22), 50(1-0.22)) = min(23, 27, 11, 39) = 11. The sample size is adequate so the following formula can be used

We reject H 0 because 2.526 > 1960. We have statistically significant evidence at a =0.05 to show that there is a difference in the proportions of patients on the new pain reliever reporting a meaningful reduction (i.e., a reduction of 3 or more scale points) as compared to patients on the standard pain reliever.

A 95% confidence interval for the difference in proportions of patients on the new pain reliever reporting a meaningful reduction (i.e., a reduction of 3 or more scale points) as compared to patients on the standard pain reliever is 0.24 + 0.18 or between 0.06 and 0.42. Because the 95% confidence interval does not include zero we concluded that there was a statistically significant difference in proportions which is consistent with the test of hypothesis result. 

Again, the procedures discussed here apply to applications where there are two independent comparison groups and a dichotomous outcome. There are other applications in which it is of interest to compare a dichotomous outcome in matched or paired samples. For example, in a clinical trial we might wish to test the effectiveness of a new antibiotic eye drop for the treatment of bacterial conjunctivitis. Participants use the new antibiotic eye drop in one eye and a comparator (placebo or active control treatment) in the other. The success of the treatment (yes/no) is recorded for each participant for each eye. Because the two assessments (success or failure) are paired, we cannot use the procedures discussed here. The appropriate test is called McNemar's test (sometimes called McNemar's test for dependent proportions).  

Vide0 - Hypothesis Testing With Two Independent Samples and a Dichotomous Outcome (2:55)

Here we presented hypothesis testing techniques for means and proportions in one and two sample situations. Tests of hypothesis involve several steps, including specifying the null and alternative or research hypothesis, selecting and computing an appropriate test statistic, setting up a decision rule and drawing a conclusion. There are many details to consider in hypothesis testing. The first is to determine the appropriate test. We discussed Z and t tests here for different applications. The appropriate test depends on the distribution of the outcome variable (continuous or dichotomous), the number of comparison groups (one, two) and whether the comparison groups are independent or dependent. The following table summarizes the different tests of hypothesis discussed here.

• Continuous Outcome, One Sample: H0: μ = μ0
• Continuous Outcome, Two Independent Samples: H0: μ1 = μ2
• Continuous Outcome, Two Matched Samples: H0: μd = 0
• Dichotomous Outcome, One Sample: H0: p = p 0
• Dichotomous Outcome, Two Independent Samples: H0: p1 = p2, RD=0, RR=1, OR=1

Once the type of test is determined, the details of the test must be specified. Specifically, the null and alternative hypotheses must be clearly stated. The null hypothesis always reflects the "no change" or "no difference" situation. The alternative or research hypothesis reflects the investigator's belief. The investigator might hypothesize that a parameter (e.g., a mean, proportion, difference in means or proportions) will increase, will decrease or will be different under specific conditions (sometimes the conditions are different experimental conditions and other times the conditions are simply different groups of participants). Once the hypotheses are specified, data are collected and summarized. The appropriate test is then conducted according to the five step approach. If the test leads to rejection of the null hypothesis, an approximate p-value is computed to summarize the significance of the findings. When tests of hypothesis are conducted using statistical computing packages, exact p-values are computed. Because the statistical tables in this textbook are limited, we can only approximate p-values. If the test fails to reject the null hypothesis, then a weaker concluding statement is made for the following reason.

In hypothesis testing, there are two types of errors that can be committed. A Type I error occurs when a test incorrectly rejects the null hypothesis. This is referred to as a false positive result, and the probability that this occurs is equal to the level of significance, α. The investigator chooses the level of significance in Step 1, and purposely chooses a small value such as α=0.05 to control the probability of committing a Type I error. A Type II error occurs when a test fails to reject the null hypothesis when in fact it is false. The probability that this occurs is equal to β. Unfortunately, the investigator cannot specify β at the outset because it depends on several factors including the sample size (smaller samples have higher b), the level of significance (β decreases as a increases), and the difference in the parameter under the null and alternative hypothesis.    

We noted in several examples in this chapter, the relationship between confidence intervals and tests of hypothesis. The approaches are different, yet related. It is possible to draw a conclusion about statistical significance by examining a confidence interval. For example, if a 95% confidence interval does not contain the null value (e.g., zero when analyzing a mean difference or risk difference, one when analyzing relative risks or odds ratios), then one can conclude that a two-sided test of hypothesis would reject the null at α=0.05. It is important to note that the correspondence between a confidence interval and test of hypothesis relates to a two-sided test and that the confidence level corresponds to a specific level of significance (e.g., 95% to α=0.05, 90% to α=0.10 and so on). The exact significance of the test, the p-value, can only be determined using the hypothesis testing approach and the p-value provides an assessment of the strength of the evidence and not an estimate of the effect.

Answers to Selected Problems

Dental services problem - bottom of page 5.

• Step 1: Set up hypotheses and determine the level of significance.

α=0.05

• Step 2: Select the appropriate test statistic.

First, determine whether the sample size is adequate.

Therefore the sample size is adequate, and we can use the following formula:

• Step 3: Set up the decision rule.

Reject H0 if Z is less than or equal to -1.96 or if Z is greater than or equal to 1.96.

• Step 4: Compute the test statistic
• Step 5: Conclusion.

We reject the null hypothesis because -6.15<-1.96. Therefore there is a statistically significant difference in the proportion of children in Boston using dental services compated to the national proportion.

Hypothesis Testing

Hypothesis testing is a tool for making statistical inferences about the population data. It is an analysis tool that tests assumptions and determines how likely something is within a given standard of accuracy. Hypothesis testing provides a way to verify whether the results of an experiment are valid.

A null hypothesis and an alternative hypothesis are set up before performing the hypothesis testing. This helps to arrive at a conclusion regarding the sample obtained from the population. In this article, we will learn more about hypothesis testing, its types, steps to perform the testing, and associated examples.

What is Hypothesis Testing in Statistics?

Hypothesis testing uses sample data from the population to draw useful conclusions regarding the population probability distribution . It tests an assumption made about the data using different types of hypothesis testing methodologies. The hypothesis testing results in either rejecting or not rejecting the null hypothesis.

Hypothesis Testing Definition

Hypothesis testing can be defined as a statistical tool that is used to identify if the results of an experiment are meaningful or not. It involves setting up a null hypothesis and an alternative hypothesis. These two hypotheses will always be mutually exclusive. This means that if the null hypothesis is true then the alternative hypothesis is false and vice versa. An example of hypothesis testing is setting up a test to check if a new medicine works on a disease in a more efficient manner.

Null Hypothesis

The null hypothesis is a concise mathematical statement that is used to indicate that there is no difference between two possibilities. In other words, there is no difference between certain characteristics of data. This hypothesis assumes that the outcomes of an experiment are based on chance alone. It is denoted as \(H_{0}\). Hypothesis testing is used to conclude if the null hypothesis can be rejected or not. Suppose an experiment is conducted to check if girls are shorter than boys at the age of 5. The null hypothesis will say that they are the same height.

Alternative Hypothesis

The alternative hypothesis is an alternative to the null hypothesis. It is used to show that the observations of an experiment are due to some real effect. It indicates that there is a statistical significance between two possible outcomes and can be denoted as \(H_{1}\) or \(H_{a}\). For the above-mentioned example, the alternative hypothesis would be that girls are shorter than boys at the age of 5.

Hypothesis Testing P Value

In hypothesis testing, the p value is used to indicate whether the results obtained after conducting a test are statistically significant or not. It also indicates the probability of making an error in rejecting or not rejecting the null hypothesis.This value is always a number between 0 and 1. The p value is compared to an alpha level, \(\alpha\) or significance level. The alpha level can be defined as the acceptable risk of incorrectly rejecting the null hypothesis. The alpha level is usually chosen between 1% to 5%.

Hypothesis Testing Critical region

All sets of values that lead to rejecting the null hypothesis lie in the critical region. Furthermore, the value that separates the critical region from the non-critical region is known as the critical value.

Hypothesis Testing Formula

Depending upon the type of data available and the size, different types of hypothesis testing are used to determine whether the null hypothesis can be rejected or not. The hypothesis testing formula for some important test statistics are given below:

• z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\). \(\overline{x}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation and n is the size of the sample.
• t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\). s is the sample standard deviation.
• \(\chi ^{2} = \sum \frac{(O_{i}-E_{i})^{2}}{E_{i}}\). \(O_{i}\) is the observed value and \(E_{i}\) is the expected value.

We will learn more about these test statistics in the upcoming section.

Types of Hypothesis Testing

Selecting the correct test for performing hypothesis testing can be confusing. These tests are used to determine a test statistic on the basis of which the null hypothesis can either be rejected or not rejected. Some of the important tests used for hypothesis testing are given below.

Hypothesis Testing Z Test

A z test is a way of hypothesis testing that is used for a large sample size (n ≥ 30). It is used to determine whether there is a difference between the population mean and the sample mean when the population standard deviation is known. It can also be used to compare the mean of two samples. It is used to compute the z test statistic. The formulas are given as follows:

• One sample: z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\).
• Two samples: z = \(\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}\).

Hypothesis Testing t Test

The t test is another method of hypothesis testing that is used for a small sample size (n < 30). It is also used to compare the sample mean and population mean. However, the population standard deviation is not known. Instead, the sample standard deviation is known. The mean of two samples can also be compared using the t test.

• One sample: t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\).
• Two samples: t = \(\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}\).

Hypothesis Testing Chi Square

The Chi square test is a hypothesis testing method that is used to check whether the variables in a population are independent or not. It is used when the test statistic is chi-squared distributed.

One Tailed Hypothesis Testing

One tailed hypothesis testing is done when the rejection region is only in one direction. It can also be known as directional hypothesis testing because the effects can be tested in one direction only. This type of testing is further classified into the right tailed test and left tailed test.

Right Tailed Hypothesis Testing

The right tail test is also known as the upper tail test. This test is used to check whether the population parameter is greater than some value. The null and alternative hypotheses for this test are given as follows:

\(H_{0}\): The population parameter is ≤ some value

\(H_{1}\): The population parameter is > some value.

If the test statistic has a greater value than the critical value then the null hypothesis is rejected

Left Tailed Hypothesis Testing

The left tail test is also known as the lower tail test. It is used to check whether the population parameter is less than some value. The hypotheses for this hypothesis testing can be written as follows:

\(H_{0}\): The population parameter is ≥ some value

\(H_{1}\): The population parameter is < some value.

The null hypothesis is rejected if the test statistic has a value lesser than the critical value.

Two Tailed Hypothesis Testing

In this hypothesis testing method, the critical region lies on both sides of the sampling distribution. It is also known as a non - directional hypothesis testing method. The two-tailed test is used when it needs to be determined if the population parameter is assumed to be different than some value. The hypotheses can be set up as follows:

\(H_{0}\): the population parameter = some value

\(H_{1}\): the population parameter ≠ some value

The null hypothesis is rejected if the test statistic has a value that is not equal to the critical value.

Hypothesis Testing Steps

Hypothesis testing can be easily performed in five simple steps. The most important step is to correctly set up the hypotheses and identify the right method for hypothesis testing. The basic steps to perform hypothesis testing are as follows:

• Step 1: Set up the null hypothesis by correctly identifying whether it is the left-tailed, right-tailed, or two-tailed hypothesis testing.
• Step 2: Set up the alternative hypothesis.
• Step 3: Choose the correct significance level, \(\alpha\), and find the critical value.
• Step 4: Calculate the correct test statistic (z, t or \(\chi\)) and p-value.
• Step 5: Compare the test statistic with the critical value or compare the p-value with \(\alpha\) to arrive at a conclusion. In other words, decide if the null hypothesis is to be rejected or not.

Hypothesis Testing Example

The best way to solve a problem on hypothesis testing is by applying the 5 steps mentioned in the previous section. Suppose a researcher claims that the mean average weight of men is greater than 100kgs with a standard deviation of 15kgs. 30 men are chosen with an average weight of 112.5 Kgs. Using hypothesis testing, check if there is enough evidence to support the researcher's claim. The confidence interval is given as 95%.

Step 1: This is an example of a right-tailed test. Set up the null hypothesis as \(H_{0}\): \(\mu\) = 100.

Step 2: The alternative hypothesis is given by \(H_{1}\): \(\mu\) > 100.

Step 3: As this is a one-tailed test, \(\alpha\) = 100% - 95% = 5%. This can be used to determine the critical value.

1 - \(\alpha\) = 1 - 0.05 = 0.95

0.95 gives the required area under the curve. Now using a normal distribution table, the area 0.95 is at z = 1.645. A similar process can be followed for a t-test. The only additional requirement is to calculate the degrees of freedom given by n - 1.

Step 4: Calculate the z test statistic. This is because the sample size is 30. Furthermore, the sample and population means are known along with the standard deviation.

z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\).

\(\mu\) = 100, \(\overline{x}\) = 112.5, n = 30, \(\sigma\) = 15

z = \(\frac{112.5-100}{\frac{15}{\sqrt{30}}}\) = 4.56

Step 5: Conclusion. As 4.56 > 1.645 thus, the null hypothesis can be rejected.

Hypothesis Testing and Confidence Intervals

Confidence intervals form an important part of hypothesis testing. This is because the alpha level can be determined from a given confidence interval. Suppose a confidence interval is given as 95%. Subtract the confidence interval from 100%. This gives 100 - 95 = 5% or 0.05. This is the alpha value of a one-tailed hypothesis testing. To obtain the alpha value for a two-tailed hypothesis testing, divide this value by 2. This gives 0.05 / 2 = 0.025.

Related Articles:

• Probability and Statistics
• Data Handling

Important Notes on Hypothesis Testing

• Hypothesis testing is a technique that is used to verify whether the results of an experiment are statistically significant.
• It involves the setting up of a null hypothesis and an alternate hypothesis.
• There are three types of tests that can be conducted under hypothesis testing - z test, t test, and chi square test.
• Hypothesis testing can be classified as right tail, left tail, and two tail tests.

Examples on Hypothesis Testing

• Example 1: The average weight of a dumbbell in a gym is 90lbs. However, a physical trainer believes that the average weight might be higher. A random sample of 5 dumbbells with an average weight of 110lbs and a standard deviation of 18lbs. Using hypothesis testing check if the physical trainer's claim can be supported for a 95% confidence level. Solution: As the sample size is lesser than 30, the t-test is used. \(H_{0}\): \(\mu\) = 90, \(H_{1}\): \(\mu\) > 90 \(\overline{x}\) = 110, \(\mu\) = 90, n = 5, s = 18. \(\alpha\) = 0.05 Using the t-distribution table, the critical value is 2.132 t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\) t = 2.484 As 2.484 > 2.132, the null hypothesis is rejected. Answer: The average weight of the dumbbells may be greater than 90lbs
• Example 2: The average score on a test is 80 with a standard deviation of 10. With a new teaching curriculum introduced it is believed that this score will change. On random testing, the score of 38 students, the mean was found to be 88. With a 0.05 significance level, is there any evidence to support this claim? Solution: This is an example of two-tail hypothesis testing. The z test will be used. \(H_{0}\): \(\mu\) = 80, \(H_{1}\): \(\mu\) ≠ 80 \(\overline{x}\) = 88, \(\mu\) = 80, n = 36, \(\sigma\) = 10. \(\alpha\) = 0.05 / 2 = 0.025 The critical value using the normal distribution table is 1.96 z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\) z = \(\frac{88-80}{\frac{10}{\sqrt{36}}}\) = 4.8 As 4.8 > 1.96, the null hypothesis is rejected. Answer: There is a difference in the scores after the new curriculum was introduced.
• Example 3: The average score of a class is 90. However, a teacher believes that the average score might be lower. The scores of 6 students were randomly measured. The mean was 82 with a standard deviation of 18. With a 0.05 significance level use hypothesis testing to check if this claim is true. Solution: The t test will be used. \(H_{0}\): \(\mu\) = 90, \(H_{1}\): \(\mu\) < 90 \(\overline{x}\) = 110, \(\mu\) = 90, n = 6, s = 18 The critical value from the t table is -2.015 t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\) t = \(\frac{82-90}{\frac{18}{\sqrt{6}}}\) t = -1.088 As -1.088 > -2.015, we fail to reject the null hypothesis. Answer: There is not enough evidence to support the claim.

go to slide go to slide go to slide

Book a Free Trial Class

FAQs on Hypothesis Testing

What is hypothesis testing.

Hypothesis testing in statistics is a tool that is used to make inferences about the population data. It is also used to check if the results of an experiment are valid.

What is the z Test in Hypothesis Testing?

The z test in hypothesis testing is used to find the z test statistic for normally distributed data . The z test is used when the standard deviation of the population is known and the sample size is greater than or equal to 30.

What is the t Test in Hypothesis Testing?

The t test in hypothesis testing is used when the data follows a student t distribution . It is used when the sample size is less than 30 and standard deviation of the population is not known.

What is the formula for z test in Hypothesis Testing?

The formula for a one sample z test in hypothesis testing is z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\) and for two samples is z = \(\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}\).

What is the p Value in Hypothesis Testing?

The p value helps to determine if the test results are statistically significant or not. In hypothesis testing, the null hypothesis can either be rejected or not rejected based on the comparison between the p value and the alpha level.

What is One Tail Hypothesis Testing?

When the rejection region is only on one side of the distribution curve then it is known as one tail hypothesis testing. The right tail test and the left tail test are two types of directional hypothesis testing.

What is the Alpha Level in Two Tail Hypothesis Testing?

To get the alpha level in a two tail hypothesis testing divide \(\alpha\) by 2. This is done as there are two rejection regions in the curve.

10.5 Hypothesis Testing for Two Means and Two Proportions

Hypothesis testing for two means and two proportions.

Class Time:

Student Learning Outcomes

• The student will select the appropriate distributions to use in each case.
• The student will conduct hypothesis tests and interpret the results.
• the business section from two consecutive days’ newspapers
• three small packages of M&Ms®
• five small packages of Reese's Pieces®

Increasing Stocks Survey Look at yesterday’s newspaper business section. Conduct a hypothesis test to determine if the proportion of New York Stock Exchange (NYSE) stocks that increased is greater than the proportion of NASDAQ stocks that increased. As randomly as possible, choose 40 NYSE stocks, and 32 NASDAQ stocks and complete the following statements.

• H 0 : _________
• H a : _________
• In words, define the random variable.
• The distribution to use for the test is _____________.
• Calculate the test statistic using your data.
• Calculate the p -value.
• Do you reject or not reject the null hypothesis? Why?
• Write a clear conclusion using a complete sentence.

Decreasing Stocks Survey Randomly pick eight stocks from the newspaper. Using two consecutive days’ business sections, test whether the stocks went down, on average, for the second day.

• H 0 : ________
• H a : ________
• Calculate the p -value:

Candy Survey Buy three small packages of M&Ms and five small packages of Reese's Pieces (same net weight as the M&Ms). Test whether or not the mean number of candy pieces per package is the same for the two brands.

• What distribution should be used for this test?

Shoe Survey Test whether women have, on average, more pairs of shoes than men. Include all forms of sneakers, shoes, sandals, and boots. Use your class as the sample.

• The distribution to use for the test is ________________.

As an Amazon Associate we earn from qualifying purchases.

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/introductory-statistics/pages/1-introduction

• Authors: Barbara Illowsky, Susan Dean
• Publisher/website: OpenStax
• Book title: Introductory Statistics
• Publication date: Sep 19, 2013
• Location: Houston, Texas
• Book URL: https://openstax.org/books/introductory-statistics/pages/1-introduction
• Section URL: https://openstax.org/books/introductory-statistics/pages/10-5-hypothesis-testing-for-two-means-and-two-proportions

© Jun 23, 2022 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

Module 10: Hypothesis Testing With Two Samples

Introduction to hypothesis testing with two samples.

There are other situations that deal with the comparison of two groups. For example, studies compare various diet and exercise programs. Politicians compare the proportion of individuals from different income brackets who might vote for them. Students are interested in whether SAT or GRE preparatory courses really help raise their scores.

You have learned to conduct hypothesis tests on single means and single proportions. You will expand upon that in this chapter. You will compare two means or two proportions to each other. The general procedure is still the same, just expanded.

To compare two means or two proportions, you work with two groups. The groups are classified either as independent or matched pairs . Independent groups consist of two samples that are independent, that is, sample values selected from one population are not related in any way to sample values selected from the other population. Matched pairs consist of two samples that are dependent. The parameter tested using matched pairs is the population mean. The parameters tested using independent groups are either population means or population proportions.

This chapter deals with the following hypothesis tests:

• Test of two population means.
• Test of two population proportions.
• Test of the two population proportions by testing one population mean of differences.
• Introductory Statistics . Authored by : Barbara Illowski, Susan Dean. Provided by : Open Stax. Located at : http://cnx.org/contents/[email protected] . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]

Privacy Policy

Have a language expert improve your writing

Run a free plagiarism check in 10 minutes, generate accurate citations for free.

• Knowledge Base

Hypothesis Testing | A Step-by-Step Guide with Easy Examples

Published on November 8, 2019 by Rebecca Bevans . Revised on June 22, 2023.

Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics . It is most often used by scientists to test specific predictions, called hypotheses, that arise from theories.

There are 5 main steps in hypothesis testing:

• State your research hypothesis as a null hypothesis and alternate hypothesis (H o ) and (H a  or H 1 ).
• Collect data in a way designed to test the hypothesis.
• Perform an appropriate statistical test .
• Decide whether to reject or fail to reject your null hypothesis.
• Present the findings in your results and discussion section.

Though the specific details might vary, the procedure you will use when testing a hypothesis will always follow some version of these steps.

Table of contents

Step 1: state your null and alternate hypothesis, step 2: collect data, step 3: perform a statistical test, step 4: decide whether to reject or fail to reject your null hypothesis, step 5: present your findings, other interesting articles, frequently asked questions about hypothesis testing.

After developing your initial research hypothesis (the prediction that you want to investigate), it is important to restate it as a null (H o ) and alternate (H a ) hypothesis so that you can test it mathematically.

The alternate hypothesis is usually your initial hypothesis that predicts a relationship between variables. The null hypothesis is a prediction of no relationship between the variables you are interested in.

• H 0 : Men are, on average, not taller than women. H a : Men are, on average, taller than women.

Receive feedback on language, structure, and formatting

Professional editors proofread and edit your paper by focusing on:

• Academic style
• Vague sentences
• Style consistency

See an example

For a statistical test to be valid , it is important to perform sampling and collect data in a way that is designed to test your hypothesis. If your data are not representative, then you cannot make statistical inferences about the population you are interested in.

There are a variety of statistical tests available, but they are all based on the comparison of within-group variance (how spread out the data is within a category) versus between-group variance (how different the categories are from one another).

If the between-group variance is large enough that there is little or no overlap between groups, then your statistical test will reflect that by showing a low p -value . This means it is unlikely that the differences between these groups came about by chance.

Alternatively, if there is high within-group variance and low between-group variance, then your statistical test will reflect that with a high p -value. This means it is likely that any difference you measure between groups is due to chance.

Your choice of statistical test will be based on the type of variables and the level of measurement of your collected data .

• an estimate of the difference in average height between the two groups.
• a p -value showing how likely you are to see this difference if the null hypothesis of no difference is true.

Based on the outcome of your statistical test, you will have to decide whether to reject or fail to reject your null hypothesis.

In most cases you will use the p -value generated by your statistical test to guide your decision. And in most cases, your predetermined level of significance for rejecting the null hypothesis will be 0.05 – that is, when there is a less than 5% chance that you would see these results if the null hypothesis were true.

In some cases, researchers choose a more conservative level of significance, such as 0.01 (1%). This minimizes the risk of incorrectly rejecting the null hypothesis ( Type I error ).

Prevent plagiarism. Run a free check.

The results of hypothesis testing will be presented in the results and discussion sections of your research paper , dissertation or thesis .

In the results section you should give a brief summary of the data and a summary of the results of your statistical test (for example, the estimated difference between group means and associated p -value). In the discussion , you can discuss whether your initial hypothesis was supported by your results or not.

In the formal language of hypothesis testing, we talk about rejecting or failing to reject the null hypothesis. You will probably be asked to do this in your statistics assignments.

However, when presenting research results in academic papers we rarely talk this way. Instead, we go back to our alternate hypothesis (in this case, the hypothesis that men are on average taller than women) and state whether the result of our test did or did not support the alternate hypothesis.

If your null hypothesis was rejected, this result is interpreted as “supported the alternate hypothesis.”

These are superficial differences; you can see that they mean the same thing.

You might notice that we don’t say that we reject or fail to reject the alternate hypothesis . This is because hypothesis testing is not designed to prove or disprove anything. It is only designed to test whether a pattern we measure could have arisen spuriously, or by chance.

If we reject the null hypothesis based on our research (i.e., we find that it is unlikely that the pattern arose by chance), then we can say our test lends support to our hypothesis . But if the pattern does not pass our decision rule, meaning that it could have arisen by chance, then we say the test is inconsistent with our hypothesis .

If you want to know more about statistics , methodology , or research bias , make sure to check out some of our other articles with explanations and examples.

• Normal distribution
• Descriptive statistics
• Measures of central tendency
• Correlation coefficient

Methodology

• Cluster sampling
• Stratified sampling
• Types of interviews
• Cohort study
• Thematic analysis

Research bias

• Implicit bias
• Cognitive bias
• Survivorship bias
• Availability heuristic
• Nonresponse bias
• Regression to the mean

Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics. It is used by scientists to test specific predictions, called hypotheses , by calculating how likely it is that a pattern or relationship between variables could have arisen by chance.

A hypothesis states your predictions about what your research will find. It is a tentative answer to your research question that has not yet been tested. For some research projects, you might have to write several hypotheses that address different aspects of your research question.

A hypothesis is not just a guess — it should be based on existing theories and knowledge. It also has to be testable, which means you can support or refute it through scientific research methods (such as experiments, observations and statistical analysis of data).

Null and alternative hypotheses are used in statistical hypothesis testing . The null hypothesis of a test always predicts no effect or no relationship between variables, while the alternative hypothesis states your research prediction of an effect or relationship.

Cite this Scribbr article

If you want to cite this source, you can copy and paste the citation or click the “Cite this Scribbr article” button to automatically add the citation to our free Citation Generator.

Bevans, R. (2023, June 22). Hypothesis Testing | A Step-by-Step Guide with Easy Examples. Scribbr. Retrieved April 3, 2024, from https://www.scribbr.com/statistics/hypothesis-testing/

Is this article helpful?

Rebecca Bevans

Other students also liked, choosing the right statistical test | types & examples, understanding p values | definition and examples, what is your plagiarism score.

Teach yourself statistics

Hypothesis Test for a Mean

This lesson explains how to conduct a hypothesis test of a mean, when the following conditions are met:

• The sampling method is simple random sampling .
• The sampling distribution is normal or nearly normal.

Generally, the sampling distribution will be approximately normally distributed if any of the following conditions apply.

• The population distribution is normal.
• The population distribution is symmetric , unimodal , without outliers , and the sample size is 15 or less.
• The population distribution is moderately skewed , unimodal, without outliers, and the sample size is between 16 and 40.
• The sample size is greater than 40, without outliers.

This approach consists of four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results.

State the Hypotheses

Every hypothesis test requires the analyst to state a null hypothesis and an alternative hypothesis . The hypotheses are stated in such a way that they are mutually exclusive. That is, if one is true, the other must be false; and vice versa.

The table below shows three sets of hypotheses. Each makes a statement about how the population mean μ is related to a specified value M . (In the table, the symbol ≠ means " not equal to ".)

The first set of hypotheses (Set 1) is an example of a two-tailed test , since an extreme value on either side of the sampling distribution would cause a researcher to reject the null hypothesis. The other two sets of hypotheses (Sets 2 and 3) are one-tailed tests , since an extreme value on only one side of the sampling distribution would cause a researcher to reject the null hypothesis.

Formulate an Analysis Plan

The analysis plan describes how to use sample data to accept or reject the null hypothesis. It should specify the following elements.

• Significance level. Often, researchers choose significance levels equal to 0.01, 0.05, or 0.10; but any value between 0 and 1 can be used.
• Test method. Use the one-sample t-test to determine whether the hypothesized mean differs significantly from the observed sample mean.

Analyze Sample Data

Using sample data, conduct a one-sample t-test. This involves finding the standard error, degrees of freedom, test statistic, and the P-value associated with the test statistic.

SE = s * sqrt{ ( 1/n ) * [ ( N - n ) / ( N - 1 ) ] }

SE = s / sqrt( n )

• Degrees of freedom. The degrees of freedom (DF) is equal to the sample size (n) minus one. Thus, DF = n - 1.

t = ( x - μ) / SE

• P-value. The P-value is the probability of observing a sample statistic as extreme as the test statistic. Since the test statistic is a t statistic, use the t Distribution Calculator to assess the probability associated with the t statistic, given the degrees of freedom computed above. (See sample problems at the end of this lesson for examples of how this is done.)

Sample Size Calculator

As you probably noticed, the process of hypothesis testing can be complex. When you need to test a hypothesis about a mean score, consider using the Sample Size Calculator. The calculator is fairly easy to use, and it is free. You can find the Sample Size Calculator in Stat Trek's main menu under the Stat Tools tab. Or you can tap the button below.

Interpret Results

If the sample findings are unlikely, given the null hypothesis, the researcher rejects the null hypothesis. Typically, this involves comparing the P-value to the significance level , and rejecting the null hypothesis when the P-value is less than the significance level.

Test Your Understanding

In this section, two sample problems illustrate how to conduct a hypothesis test of a mean score. The first problem involves a two-tailed test; the second problem, a one-tailed test.

Problem 1: Two-Tailed Test

An inventor has developed a new, energy-efficient lawn mower engine. He claims that the engine will run continuously for 5 hours (300 minutes) on a single gallon of regular gasoline. From his stock of 2000 engines, the inventor selects a simple random sample of 50 engines for testing. The engines run for an average of 295 minutes, with a standard deviation of 20 minutes. Test the null hypothesis that the mean run time is 300 minutes against the alternative hypothesis that the mean run time is not 300 minutes. Use a 0.05 level of significance. (Assume that run times for the population of engines are normally distributed.)

Solution: The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

Null hypothesis: μ = 300

Alternative hypothesis: μ ≠ 300

• Formulate an analysis plan . For this analysis, the significance level is 0.05. The test method is a one-sample t-test .

SE = s / sqrt(n) = 20 / sqrt(50) = 20/7.07 = 2.83

DF = n - 1 = 50 - 1 = 49

t = ( x - μ) / SE = (295 - 300)/2.83 = -1.77

where s is the standard deviation of the sample, x is the sample mean, μ is the hypothesized population mean, and n is the sample size.

Since we have a two-tailed test , the P-value is the probability that the t statistic having 49 degrees of freedom is less than -1.77 or greater than 1.77. We use the t Distribution Calculator to find P(t < -1.77) is about 0.04.

• If you enter 1.77 as the sample mean in the t Distribution Calculator, you will find the that the P(t < 1.77) is about 0.04. Therefore, P(t >  1.77) is 1 minus 0.96 or 0.04. Thus, the P-value = 0.04 + 0.04 = 0.08.
• Interpret results . Since the P-value (0.08) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Note: If you use this approach on an exam, you may also want to mention why this approach is appropriate. Specifically, the approach is appropriate because the sampling method was simple random sampling, the population was normally distributed, and the sample size was small relative to the population size (less than 5%).

Problem 2: One-Tailed Test

Bon Air Elementary School has 1000 students. The principal of the school thinks that the average IQ of students at Bon Air is at least 110. To prove her point, she administers an IQ test to 20 randomly selected students. Among the sampled students, the average IQ is 108 with a standard deviation of 10. Based on these results, should the principal accept or reject her original hypothesis? Assume a significance level of 0.01. (Assume that test scores in the population of engines are normally distributed.)

Null hypothesis: μ >= 110

Alternative hypothesis: μ < 110

• Formulate an analysis plan . For this analysis, the significance level is 0.01. The test method is a one-sample t-test .

SE = s / sqrt(n) = 10 / sqrt(20) = 10/4.472 = 2.236

DF = n - 1 = 20 - 1 = 19

t = ( x - μ) / SE = (108 - 110)/2.236 = -0.894

Here is the logic of the analysis: Given the alternative hypothesis (μ < 110), we want to know whether the observed sample mean is small enough to cause us to reject the null hypothesis.

The observed sample mean produced a t statistic test statistic of -0.894. We use the t Distribution Calculator to find P(t < -0.894) is about 0.19.

• This means we would expect to find a sample mean of 108 or smaller in 19 percent of our samples, if the true population IQ were 110. Thus the P-value in this analysis is 0.19.
• Interpret results . Since the P-value (0.19) is greater than the significance level (0.01), we cannot reject the null hypothesis.