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11 Hypothesis Testing with One Sample

Student learning outcomes.

By the end of this chapter, the student should be able to:

• Be able to identify and develop the null and alternative hypothesis
• Identify the consequences of Type I and Type II error.
• Be able to perform an one-tailed and two-tailed hypothesis test using the critical value method
• Be able to perform a hypothesis test using the p-value method
• Be able to write conclusions based on hypothesis tests.

Introduction

Now we are down to the bread and butter work of the statistician: developing and testing hypotheses. It is important to   put this material in a broader context so that the method by which a hypothesis is formed is understood completely. Using textbook examples often clouds the real source of statistical hypotheses.

Statistical testing is part of a much larger process known as the scientific method. This method was developed more than two centuries ago as the accepted way that new knowledge could be created. Until then, and unfortunately even today, among some, “knowledge” could be created simply by some authority saying something was so, ipso dicta . Superstition and conspiracy theories were (are?) accepted uncritically.

The scientific method, briefly, states that only by following a careful and specific process can some assertion be included in the accepted body of knowledge. This process begins with a set of assumptions upon which a theory, sometimes called a model, is built. This theory, if it has any validity, will lead to predictions; what we call hypotheses.

As an example, in Microeconomics the theory of consumer choice begins with certain assumption concerning human behavior. From these assumptions a theory of how consumers make choices using indifference curves and the budget line. This theory gave rise to a very important prediction, namely, that there was an inverse relationship between price and quantity demanded. This relationship was known as the demand curve. The negative slope of the demand curve is really just a prediction, or a hypothesis, that can be tested with statistical tools.

Unless hundreds and hundreds of statistical tests of this hypothesis had not confirmed this relationship, the so-called Law of Demand would have been discarded years ago. This is the role of statistics, to test the hypotheses of various theories to determine if they should be admitted into the accepted body of knowledge; how we understand our world. Once admitted, however, they may be later discarded if new theories come along that make better predictions.

Not long ago two scientists claimed that they could get more energy out of a process than was put in. This caused a tremendous stir for obvious reasons. They were on the cover of Time and were offered extravagant sums to bring their research work to private industry and any number of universities. It was not long until their work was subjected to the rigorous tests of the scientific method and found to be a failure. No other lab could replicate their findings. Consequently they have sunk into obscurity and their theory discarded. It may surface again when someone can pass the tests of the hypotheses required by the scientific method, but until then it is just a curiosity. Many pure frauds have been attempted over time, but most have been found out by applying the process of the scientific method.

This discussion is meant to show just where in this process statistics falls. Statistics and statisticians are not necessarily in the business of developing theories, but in the business of testing others’ theories. Hypotheses come from these theories based upon an explicit set of assumptions and sound logic. The hypothesis comes first, before any data are gathered. Data do not create hypotheses; they are used to test them. If we bear this in mind as we study this section the process of forming and testing hypotheses will make more sense.

One job of a statistician is to make statistical inferences about populations based on samples taken from the population. Confidence intervals are one way to estimate a population parameter. Another way to make a statistical inference is to make a decision about the value of a specific parameter. For instance, a car dealer advertises that its new small truck gets 35 miles per gallon, on average. A tutoring service claims that its method of tutoring helps 90% of its students get an A or a B. A company says that women managers in their company earn an average of $60,000 per year.

A statistician will make a decision about these claims. This process is called ” hypothesis testing .” A hypothesis test involves collecting data from a sample and evaluating the data. Then, the statistician makes a decision as to whether or not there is sufficient evidence, based upon analyses of the data, to reject the null hypothesis.

In this chapter, you will conduct hypothesis tests on single means and single proportions. You will also learn about the errors associated with these tests.

Null and Alternative Hypotheses

The actual test begins by considering two hypotheses . They are called the null hypothesis and the alternative hypothesis . These hypotheses contain opposing viewpoints.

Since the null and alternative hypotheses are contradictory, you must examine evidence to decide if you have enough evidence to reject the null hypothesis or not. The evidence is in the form of sample data.

Table 1 presents the various hypotheses in the relevant pairs. For example, if the null hypothesis is equal to some value, the alternative has to be not equal to that value.

NOTE                                                                             

We want to test whether the mean GPA of students in American colleges is different from 2.0 (out of 4.0). The null and alternative hypotheses are:

We want to test if college students take less than five years to graduate from college, on the average. The null and alternative hypotheses are:

Outcomes and the Type I and Type II Errors

The four possible outcomes in the table are:

Each of the errors occurs with a particular probability. The Greek letters α and β represent the probabilities.

By way of example, the American judicial system begins with the concept that a defendant is “presumed innocent”. This is the status quo and is the null hypothesis. The judge will tell the jury that they can not find the defendant guilty unless the evidence indicates guilt beyond a “reasonable doubt” which is usually defined in criminal cases as 95% certainty of guilt. If the jury cannot accept the null, innocent, then action will be taken, jail time. The burden of proof always lies with the alternative hypothesis. (In civil cases, the jury needs only to be more than 50% certain of wrongdoing to find culpability, called “a preponderance of the evidence”).

The example above was for a test of a mean, but the same logic applies to tests of hypotheses for all statistical parameters one may wish to test.

The following are examples of Type I and Type II errors.

Type I error : Frank thinks that his rock climbing equipment may not be safe when, in fact, it really is safe.

Type II error : Frank thinks that his rock climbing equipment may be safe when, in fact, it is not safe.

Notice that, in this case, the error with the greater consequence is the Type II error. (If Frank thinks his rock climbing equipment is safe, he will go ahead and use it.)

This is a situation described as “accepting a false null”.

Type I error : The emergency crew thinks that the victim is dead when, in fact, the victim is alive. Type II error : The emergency crew does not know if the victim is alive when, in fact, the victim is dead.

The error with the greater consequence is the Type I error. (If the emergency crew thinks the victim is dead, they will not treat him.)

Distribution Needed for Hypothesis Testing

Particular distributions are associated with hypothesis testing.We will perform hypotheses tests of a population mean using a normal distribution or a Student’s t -distribution. (Remember, use a Student’s t -distribution when the population standard deviation is unknown and the sample size is small, where small is considered to be less than 30 observations.) We perform tests of a population proportion using a normal distribution when we can assume that the distribution is normally distributed. We consider this to be true if the sample proportion, p ‘ , times the sample size is greater than 5 and 1- p ‘ times the sample size is also greater then 5. This is the same rule of thumb we used when developing the formula for the confidence interval for a population proportion.

Hypothesis Test for the Mean

Going back to the standardizing formula we can derive the test statistic for testing hypotheses concerning means.

This gives us the decision rule for testing a hypothesis for a two-tailed test:

P-Value Approach

Both decision rules will result in the same decision and it is a matter of preference which one is used.

One and Two-tailed Tests

The claim would be in the alternative hypothesis. The burden of proof in hypothesis testing is carried in the alternative. This is because failing to reject the null, the status quo, must be accomplished with 90 or 95 percent significance that it cannot be maintained. Said another way, we want to have only a 5 or 10 percent probability of making a Type I error, rejecting a good null; overthrowing the status quo.

Figure 5 shows the two possible cases and the form of the null and alternative hypothesis that give rise to them.

Effects of Sample Size on Test Statistic

Table 3 summarizes test statistics for varying sample sizes and population standard deviation known and unknown.

A Systematic Approach for Testing A Hypothesis

A systematic approach to hypothesis testing follows the following steps and in this order. This template will work for all hypotheses that you will ever test.

• Set up the null and alternative hypothesis. This is typically the hardest part of the process. Here the question being asked is reviewed. What parameter is being tested, a mean, a proportion, differences in means, etc. Is this a one-tailed test or two-tailed test? Remember, if someone is making a claim it will always be a one-tailed test.
• Decide the level of significance required for this particular case and determine the critical value. These can be found in the appropriate statistical table. The levels of confidence typical for the social sciences are 90, 95 and 99. However, the level of significance is a policy decision and should be based upon the risk of making a Type I error, rejecting a good null. Consider the consequences of making a Type I error.
• Take a sample(s) and calculate the relevant parameters: sample mean, standard deviation, or proportion. Using the formula for the test statistic from above in step 2, now calculate the test statistic for this particular case using the parameters you have just calculated.
• Compare the calculated test statistic and the critical value. Marking these on the graph will give a good visual picture of the situation. There are now only two situations:

a.     The test statistic is in the tail: Cannot Accept the null, the probability that this sample mean (proportion) came from the hypothesized distribution is too small to believe that it is the real home of these sample data.

b.   The test statistic is not in the tail: Cannot Reject the null, the sample data are compatible with the hypothesized population parameter.

• Reach a conclusion. It is best to articulate the conclusion two different ways. First a formal statistical conclusion such as “With a 95 % level of significance we cannot accept the null hypotheses that the population mean is equal to XX (units of measurement)”. The second statement of the conclusion is less formal and states the action, or lack of action, required. If the formal conclusion was that above, then the informal one might be, “The machine is broken and we need to shut it down and call for repairs”.

All hypotheses tested will go through this same process. The only changes are the relevant formulas and those are determined by the hypothesis required to answer the original question.

Full Hypothesis Test Examples

Tests on means.

Jeffrey, as an eight-year old, established a mean time of 16.43 seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds . His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25-yard freestyle swims . For the 15 swims, Jeffrey’s mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds. Conduct a hypothesis test using a preset α = 0.05.

Solution – Example 6

Set up the Hypothesis Test:

Since the problem is about a mean, this is a test of a single population mean . Set the null and alternative hypothesis:

In this case there is an implied challenge or claim. This is that the goggles will reduce the swimming time. The effect of this is to set the hypothesis as a one-tailed test. The claim will always be in the alternative hypothesis because the burden of proof always lies with the alternative. Remember that the status quo must be defeated with a high degree of confidence, in this case 95 % confidence. The null and alternative hypotheses are thus:

For Jeffrey to swim faster, his time will be less than 16.43 seconds. The “<” tells you this is left-tailed. Determine the distribution needed:

Distribution for the test statistic:

The sample size is less than 30 and we do not know the population standard deviation so this is a t-test and the proper formula is:

Our step 2, setting the level of significance, has already been determined by the problem, .05 for a 95 % significance level. It is worth thinking about the meaning of this choice. The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.) For this case the only concern with a Type I error would seem to be that Jeffery’s dad may fail to bet on his son’s victory because he does not have appropriate confidence in the effect of the goggles.

To find the critical value we need to select the appropriate test statistic. We have concluded that this is a t-test on the basis of the sample size and that we are interested in a population mean. We can now draw the graph of the t-distribution and mark the critical value (Figure 6). For this problem the degrees of freedom are n-1, or 14. Looking up 14 degrees of freedom at the 0.05 column of the t-table we find 1.761. This is the critical value and we can put this on our graph.

Step 3 is the calculation of the test statistic using the formula we have selected.

We find that the calculated test statistic is 2.08, meaning that the sample mean is 2.08 standard deviations away from the hypothesized mean of 16.43.

Step 4 has us compare the test statistic and the critical value and mark these on the graph. We see that the test statistic is in the tail and thus we move to step 4 and reach a conclusion. The probability that an average time of 16 minutes could come from a distribution with a population mean of 16.43 minutes is too unlikely for us to accept the null hypothesis. We cannot accept the null.

Step 5 has us state our conclusions first formally and then less formally. A formal conclusion would be stated as: “With a 95% level of significance we cannot accept the null hypothesis that the swimming time with goggles comes from a distribution with a population mean time of 16.43 minutes.” Less formally, “With 95% significance we believe that the goggles improves swimming speed”

If we wished to use the p-value system of reaching a conclusion we would calculate the statistic and take the additional step to find the probability of being 2.08 standard deviations from the mean on a t-distribution. This value is .0187. Comparing this to the α-level of .05 we see that we cannot accept the null. The p-value has been put on the graph as the shaded area beyond -2.08 and it shows that it is smaller than the hatched area which is the alpha level of 0.05. Both methods reach the same conclusion that we cannot accept the null hypothesis.

Jane has just begun her new job as on the sales force of a very competitive company. In a sample of 16 sales calls it was found that she closed the contract for an average value of $108 with a standard deviation of 12 dollars. Test at 5% significance that the population mean is at least $100 against the alternative that it is less than 100 dollars. Company policy requires that new members of the sales force must exceed an average of $100 per contract during the trial employment period. Can we conclude that Jane has met this requirement at the significance level of 95%?

Solution – Example 7

STEP 1 : Set the Null and Alternative Hypothesis.

STEP 2 : Decide the level of significance and draw the graph (Figure 7) showing the critical value.

STEP 3 : Calculate sample parameters and the test statistic.

STEP 4 : Compare test statistic and the critical values

STEP 5 : Reach a Conclusion

The test statistic is a Student’s t because the sample size is below 30; therefore, we cannot use the normal distribution. Comparing the calculated value of the test statistic and the critical value of t ( t a ) at a 5% significance level, we see that the calculated value is in the tail of the distribution. Thus, we conclude that 108 dollars per contract is significantly larger than the hypothesized value of 100 and thus we cannot accept the null hypothesis. There is evidence that supports Jane’s performance meets company standards.

Again we will follow the steps in our analysis of this problem.

Solution – Example 8

STEP 1 : Set the Null and Alternative Hypothesis. The random variable is the quantity of fluid placed in the bottles. This is a continuous random variable and the parameter we are interested in is the mean. Our hypothesis therefore is about the mean. In this case we are concerned that the machine is not filling properly. From what we are told it does not matter if the machine is over-filling or under-filling, both seem to be an equally bad error. This tells us that this is a two-tailed test: if the machine is malfunctioning it will be shutdown regardless if it is from over-filling or under-filling. The null and alternative hypotheses are thus:

STEP 2 : Decide the level of significance and draw the graph showing the critical value.

This problem has already set the level of significance at 99%. The decision seems an appropriate one and shows the thought process when setting the significance level. Management wants to be very certain, as certain as probability will allow, that they are not shutting down a machine that is not in need of repair. To draw the distribution and the critical value, we need to know which distribution to use. Because this is a continuous random variable and we are interested in the mean, and the sample size is greater than 30, the appropriate distribution is the normal distribution and the relevant critical value is 2.575 from the normal table or the t-table at 0.005 column and infinite degrees of freedom. We draw the graph and mark these points (Figure 8).

STEP 3 : Calculate sample parameters and the test statistic. The sample parameters are provided, the sample mean is 7.91 and the sample variance is .03 and the sample size is 35. We need to note that the sample variance was provided not the sample standard deviation, which is what we need for the formula. Remembering that the standard deviation is simply the square root of the variance, we therefore know the sample standard deviation, s, is 0.173. With this information we calculate the test statistic as -3.07, and mark it on the graph.

STEP 4 : Compare test statistic and the critical values Now we compare the test statistic and the critical value by placing the test statistic on the graph. We see that the test statistic is in the tail, decidedly greater than the critical value of 2.575. We note that even the very small difference between the hypothesized value and the sample value is still a large number of standard deviations. The sample mean is only 0.08 ounces different from the required level of 8 ounces, but it is 3 plus standard deviations away and thus we cannot accept the null hypothesis.

Three standard deviations of a test statistic will guarantee that the test will fail. The probability that anything is within three standard deviations is almost zero. Actually it is 0.0026 on the normal distribution, which is certainly almost zero in a practical sense. Our formal conclusion would be “ At a 99% level of significance we cannot accept the hypothesis that the sample mean came from a distribution with a mean of 8 ounces” Or less formally, and getting to the point, “At a 99% level of significance we conclude that the machine is under filling the bottles and is in need of repair”.

Media Attributions

• Type1Type2Error
• HypTestFig2
• HypTestFig3
• HypTestPValue
• OneTailTestFig5
• HypTestExam7
• HypTestExam8

Quantitative Analysis for Business Copyright © by Margo Bergman is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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Statistics Made Easy

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

A  one sample t-test  is used to test whether or not the mean of a population is equal to some value.

This tutorial explains the following:

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

One Sample t-test: Motivation

Suppose we want to know whether or not the mean weight of a certain species of turtle in Florida is equal to 310 pounds. Since there are thousands of turtles in Florida, it would be extremely time-consuming and costly to go around and weigh each individual turtle.

Instead, we might take a simple random sample of 40 turtles and use the mean weight of the turtles in this sample to estimate the true population mean:

However, it’s virtually guaranteed that the mean weight of turtles in our sample will differ from 310 pounds. The question is whether or not this difference is statistically significant . Fortunately, a one sample t-test allows us to answer this question.

One Sample t-test: Formula

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

• H 0 :  μ = μ 0 (population mean is equal to some hypothesized value μ 0 )

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

• H 1 (two-tailed):  μ ≠ μ 0 (population mean is not equal to some hypothesized value μ 0 )
• H 1 (left-tailed):  μ < μ 0 (population mean is less than some hypothesized value μ 0 )
• H 1 (right-tailed):  μ > μ 0 (population mean is greater than some hypothesized value μ 0 )

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

t = ( x  – μ) / (s/√ n )

• x : sample mean
• μ 0 : hypothesized population mean
• s:  sample standard deviation
• n:  sample size

If the p-value that corresponds to the test statistic t with (n-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.

One Sample t-test: Assumptions

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

• The variable under study should be either an interval or ratio variable .
• The observations in the sample should be independent .
• The variable under study should be approximately normally distributed.  You can check this assumption by creating a histogram and visually checking if the distribution has roughly a “bell shape.”
• The variable under study should have no outliers. You can check this assumption by creating a boxplot and visually checking for outliers.

One Sample t-test : Example

Suppose we want to know whether or not the mean weight of a certain species of turtle is equal to 310 pounds. To test this, will perform a one-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 with the following information:

• Sample size n = 40
• Sample mean weight  x  = 300
• Sample standard deviation s = 18.5

Step 2: Define the hypotheses.

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

• H 0 :  μ = 310 (population mean is equal to 310 pounds)
• H 1 :  μ ≠ 310 (population mean is not equal to 310 pounds)

Step 3: Calculate the test statistic  t .

t = ( x  – μ) / (s/√ n ) = (300-310) / (18.5/√ 40 ) =  -3.4187

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 = -3.4817 and degrees of freedom = n-1 = 40-1 = 39 is  0.00149 .

Step 5: Draw a conclusion.

Since this p-value is less than our significance level α = 0.05, we reject the null hypothesis. We have sufficient evidence to say that the mean weight of this species of turtle is not equal to 310 pounds.

Note:  You can also perform this entire one sample t-test by simply using the One Sample t-test calculator .

Additional Resources

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

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

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JMP | Statistical Discovery.™ From SAS.

Statistics Knowledge Portal

A free online introduction to statistics

The One-Sample t -Test

What is the one-sample t -test.

The one-sample t-test is a statistical hypothesis test used to determine whether an unknown population mean is different from a specific value.

When can I use the test?

You can use the test for continuous data. Your data should be a random sample from a normal population.

What if my data isn’t nearly normally distributed?

If your sample sizes are very small, you might not be able to test for normality. You might need to rely on your understanding of the data. When you cannot safely assume normality, you can perform a nonparametric test that doesn’t assume normality.

Using the one-sample t -test

See how to perform a one-sample t -test using statistical software.

• Download JMP to follow along using the sample data included with the software.
• To see more JMP tutorials, visit the JMP Learning Library .

The sections below discuss what we need for the test, checking our data, performing the test, understanding test results and statistical details.

What do we need?

For the one-sample t -test, we need one variable.

We also have an idea, or hypothesis, that the mean of the population has some value. Here are two examples:

• A hospital has a random sample of cholesterol measurements for men. These patients were seen for issues other than cholesterol. They were not taking any medications for high cholesterol. The hospital wants to know if the unknown mean cholesterol for patients is different from a goal level of 200 mg.
• We measure the grams of protein for a sample of energy bars. The label claims that the bars have 20 grams of protein. We want to know if the labels are correct or not.

One-sample t -test assumptions

For a valid test, we need data values that are:

• Independent (values are not related to one another).
• Continuous.
• Obtained via a simple random sample from the population.

Also, the population is assumed to be normally distributed .

One-sample t -test example

Imagine we have collected a random sample of 31 energy bars from a number of different stores to represent the population of energy bars available to the general consumer. The labels on the bars claim that each bar contains 20 grams of protein.

Table 1: Grams of protein in random sample of energy bars

If you look at the table above, you see that some bars have less than 20 grams of protein. Other bars have more. You might think that the data support the idea that the labels are correct. Others might disagree. The statistical test provides a sound method to make a decision, so that everyone makes the same decision on the same set of data values. 

Checking the data

Let’s start by answering: Is the t -test an appropriate method to test that the energy bars have 20 grams of protein ? The list below checks the requirements for the test.

• The data values are independent. The grams of protein in one energy bar do not depend on the grams in any other energy bar. An example of dependent values would be if you collected energy bars from a single production lot. A sample from a single lot is representative of that lot, not energy bars in general.
• The data values are grams of protein. The measurements are continuous.
• We assume the energy bars are a simple random sample from the population of energy bars available to the general consumer (i.e., a mix of lots of bars).
• We assume the population from which we are collecting our sample is normally distributed, and for large samples, we can check this assumption.

We decide that the t -test is an appropriate method.

Before jumping into analysis, we should take a quick look at the data. The figure below shows a histogram and summary statistics for the energy bars.

From a quick look at the histogram, we see that there are no unusual points, or outliers . The data look roughly bell-shaped, so our assumption of a normal distribution seems reasonable.

From a quick look at the statistics, we see that the average is 21.40, above 20. Does this  average from our sample of 31 bars invalidate the label's claim of 20 grams of protein for the unknown entire population mean? Or not?

How to perform the one-sample t -test

For the t -test calculations we need the mean, standard deviation and sample size. These are shown in the summary statistics section of Figure 1 above.

We round the statistics to two decimal places. Software will show more decimal places, and use them in calculations. (Note that Table 1 shows only two decimal places; the actual data used to calculate the summary statistics has more.)

We start by finding the difference between the sample mean and 20:

$ 21.40-20\ =\ 1.40$

Next, we calculate the standard error for the mean. The calculation is:

Standard Error for the mean = $ \frac{s}{\sqrt{n}}= \frac{2.54}{\sqrt{31}}=0.456 $

This matches the value in Figure 1 above.

We now have the pieces for our test statistic. We calculate our test statistic as:

$ t =  \frac{\text{Difference}}{\text{Standard Error}}= \frac{1.40}{0.456}=3.07 $

To make our decision, we compare the test statistic to a value from the t- distribution. This activity involves four steps.

• We calculate a test statistic. Our test statistic is 3.07.
• We decide on the risk we are willing to take for declaring a difference when there is not a difference. For the energy bar data, we decide that we are willing to take a 5% risk of saying that the unknown population mean is different from 20 when in fact it is not. In statistics-speak, we set α = 0.05. In practice, setting your risk level (α) should be made before collecting the data.

We find the value from the t- distribution based on our decision. For a t -test, we need the degrees of freedom to find this value. The degrees of freedom are based on the sample size. For the energy bar data:

degrees of freedom = $ n - 1 = 31 - 1 = 30 $

The critical value of t with α = 0.05 and 30 degrees of freedom is +/- 2.043. Most statistics books have look-up tables for the distribution. You can also find tables online. The most likely situation is that you will use software and will not use printed tables.

We compare the value of our statistic (3.07) to the t value. Since 3.07 > 2.043, we reject the null hypothesis that the mean grams of protein is equal to 20. We make a practical conclusion that the labels are incorrect, and the population mean grams of protein is greater than 20.

Statistical details

Let’s look at the energy bar data and the 1-sample t -test using statistical terms.

Our null hypothesis is that the underlying population mean is equal to 20. The null hypothesis is written as:

$ H_o:  \mathrm{\mu} = 20 $

The alternative hypothesis is that the underlying population mean is not equal to 20. The labels claiming 20 grams of protein would be incorrect. This is written as:

$ H_a:  \mathrm{\mu} ≠ 20 $

This is a two-sided test. We are testing if the population mean is different from 20 grams in either direction. If we can reject the null hypothesis that the mean is equal to 20 grams, then we make a practical conclusion that the labels for the bars are incorrect. If we cannot reject the null hypothesis, then we make a practical conclusion that the labels for the bars may be correct.

We calculate the average for the sample and then calculate the difference with the population mean, mu:

$  \overline{x} - \mathrm{\mu} $

We calculate the standard error as:

$ \frac{s}{ \sqrt{n}} $

The formula shows the sample standard deviation as s and the sample size as n .  

The test statistic uses the formula shown below:

$  \dfrac{\overline{x} - \mathrm{\mu}} {s / \sqrt{n}} $

We compare the test statistic to a t value with our chosen alpha value and the degrees of freedom for our data. Using the energy bar data as an example, we set α = 0.05. The degrees of freedom ( df ) are based on the sample size and are calculated as:

$ df = n - 1 = 31 - 1 = 30 $

Statisticians write the t value with α = 0.05 and 30 degrees of freedom as:

$ t_{0.05,30} $

The t value for a two-sided test with α = 0.05 and 30 degrees of freedom is +/- 2.042. There are two possible results from our comparison:

• The test statistic is less extreme than the critical  t  values; in other words, the test statistic is not less than -2.042, or is not greater than +2.042. You fail to reject the null hypothesis that the mean is equal to the specified value. In our example, you would be unable to conclude that the label for the protein bars should be changed.
• The test statistic is more extreme than the critical  t  values; in other words, the test statistic is less than -2.042, or is greater than +2.042. You reject the null hypothesis that the mean is equal to the specified value. In our example, you conclude that either the label should be updated or the production process should be improved to produce, on average, bars with 20 grams of protein.

Testing for normality

The normality assumption is more important for small sample sizes than for larger sample sizes.

Normal distributions are symmetric, which means they are “even” on both sides of the center. Normal distributions do not have extreme values, or outliers. You can check these two features of a normal distribution with graphs. Earlier, we decided that the energy bar data was “close enough” to normal to go ahead with the assumption of normality. The figure below shows a normal quantile plot for the data, and supports our decision.

You can also perform a formal test for normality using software. The figure below shows results of testing for normality with JMP software. We cannot reject the hypothesis of a normal distribution. 

We can go ahead with the assumption that the energy bar data is normally distributed.

What if my data are not from a Normal distribution?

If your sample size is very small, it is hard to test for normality. In this situation, you might need to use your understanding of the measurements. For example, for the energy bar data, the company knows that the underlying distribution of grams of protein is normally distributed. Even for a very small sample, the company would likely go ahead with the t -test and assume normality.

What if you know the underlying measurements are not normally distributed? Or what if your sample size is large and the test for normality is rejected? In this situation, you can use a nonparametric test. Nonparametric  analyses do not depend on an assumption that the data values are from a specific distribution. For the one-sample t ­-test, the one possible nonparametric test is the Wilcoxon Signed Rank test. 

Understanding p-values

Using a visual, you can check to see if your test statistic is more extreme than a specified value in the distribution. The figure below shows a t- distribution with 30 degrees of freedom.

Since our test is two-sided and we set α = 0.05, the figure shows that the value of 2.042 “cuts off” 5% of the data in the tails combined.

The next figure shows our results. You can see the test statistic falls above the specified critical value. It is far enough “out in the tail” to reject the hypothesis that the mean is equal to 20.

Putting it all together with Software

You are likely to use software to perform a t -test. The figure below shows results for the 1-sample t -test for the energy bar data from JMP software.  

The software shows the null hypothesis value of 20 and the average and standard deviation from the data. The test statistic is 3.07. This matches the calculations above.

The software shows results for a two-sided test and for one-sided tests. We want the two-sided test. Our null hypothesis is that the mean grams of protein is equal to 20. Our alternative hypothesis is that the mean grams of protein is not equal to 20.  The software shows a p- value of 0.0046 for the two-sided test. This p- value describes the likelihood of seeing a sample average as extreme as 21.4, or more extreme, when the underlying population mean is actually 20; in other words, the probability of observing a sample mean as different, or even more different from 20, than the mean we observed in our sample. A p -value of 0.0046 means there is about 46 chances out of 10,000. We feel confident in rejecting the null hypothesis that the population mean is equal to 20.

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Unit 12: Significance tests (hypothesis testing)

About this unit, the idea of significance tests.

• Simple hypothesis testing (Opens a modal)
• Idea behind hypothesis testing (Opens a modal)
• Examples of null and alternative hypotheses (Opens a modal)
• P-values and significance tests (Opens a modal)
• Comparing P-values to different significance levels (Opens a modal)
• Estimating a P-value from a simulation (Opens a modal)
• Using P-values to make conclusions (Opens a modal)
• Simple hypothesis testing Get 3 of 4 questions to level up!
• Writing null and alternative hypotheses Get 3 of 4 questions to level up!
• Estimating P-values from simulations Get 3 of 4 questions to level up!

Error probabilities and power

• Introduction to Type I and Type II errors (Opens a modal)
• Type 1 errors (Opens a modal)
• Examples identifying Type I and Type II errors (Opens a modal)
• Introduction to power in significance tests (Opens a modal)
• Examples thinking about power in significance tests (Opens a modal)
• Consequences of errors and significance (Opens a modal)
• Type I vs Type II error Get 3 of 4 questions to level up!
• Error probabilities and power Get 3 of 4 questions to level up!

Tests about a population proportion

• Constructing hypotheses for a significance test about a proportion (Opens a modal)
• Conditions for a z test about a proportion (Opens a modal)
• Reference: Conditions for inference on a proportion (Opens a modal)
• Calculating a z statistic in a test about a proportion (Opens a modal)
• Calculating a P-value given a z statistic (Opens a modal)
• Making conclusions in a test about a proportion (Opens a modal)
• Writing hypotheses for a test about a proportion Get 3 of 4 questions to level up!
• Conditions for a z test about a proportion Get 3 of 4 questions to level up!
• Calculating the test statistic in a z test for a proportion Get 3 of 4 questions to level up!
• Calculating the P-value in a z test for a proportion Get 3 of 4 questions to level up!
• Making conclusions in a z test for a proportion Get 3 of 4 questions to level up!

Tests about a population mean

• Writing hypotheses for a significance test about a mean (Opens a modal)
• Conditions for a t test about a mean (Opens a modal)
• Reference: Conditions for inference on a mean (Opens a modal)
• When to use z or t statistics in significance tests (Opens a modal)
• Example calculating t statistic for a test about a mean (Opens a modal)
• Using TI calculator for P-value from t statistic (Opens a modal)
• Using a table to estimate P-value from t statistic (Opens a modal)
• Comparing P-value from t statistic to significance level (Opens a modal)
• Free response example: Significance test for a mean (Opens a modal)
• Writing hypotheses for a test about a mean Get 3 of 4 questions to level up!
• Conditions for a t test about a mean Get 3 of 4 questions to level up!
• Calculating the test statistic in a t test for a mean Get 3 of 4 questions to level up!
• Calculating the P-value in a t test for a mean Get 3 of 4 questions to level up!
• Making conclusions in a t test for a mean Get 3 of 4 questions to level up!

More significance testing videos

• Hypothesis testing and p-values (Opens a modal)
• One-tailed and two-tailed tests (Opens a modal)
• Z-statistics vs. T-statistics (Opens a modal)
• Small sample hypothesis test (Opens a modal)
• Large sample proportion hypothesis testing (Opens a modal)

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• Knowledge Base
• Choosing the Right Statistical Test | Types & Examples

Choosing the Right Statistical Test | Types & Examples

Published on January 28, 2020 by Rebecca Bevans . Revised on June 22, 2023.

Statistical tests are used in hypothesis testing . They can be used to:

• determine whether a predictor variable has a statistically significant relationship with an outcome variable.
• estimate the difference between two or more groups.

Statistical tests assume a null hypothesis of no relationship or no difference between groups. Then they determine whether the observed data fall outside of the range of values predicted by the null hypothesis.

If you already know what types of variables you’re dealing with, you can use the flowchart to choose the right statistical test for your data.

Statistical tests flowchart

Table of contents

What does a statistical test do, when to perform a statistical test, choosing a parametric test: regression, comparison, or correlation, choosing a nonparametric test, flowchart: choosing a statistical test, other interesting articles, frequently asked questions about statistical tests.

Statistical tests work by calculating a test statistic – a number that describes how much the relationship between variables in your test differs from the null hypothesis of no relationship.

It then calculates a p value (probability value). The p -value estimates how likely it is that you would see the difference described by the test statistic if the null hypothesis of no relationship were true.

If the value of the test statistic is more extreme than the statistic calculated from the null hypothesis, then you can infer a statistically significant relationship between the predictor and outcome variables.

If the value of the test statistic is less extreme than the one calculated from the null hypothesis, then you can infer no statistically significant relationship between the predictor and outcome variables.

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You can perform statistical tests on data that have been collected in a statistically valid manner – either through an experiment , or through observations made using probability sampling methods .

For a statistical test to be valid , your sample size needs to be large enough to approximate the true distribution of the population being studied.

To determine which statistical test to use, you need to know:

• whether your data meets certain assumptions.
• the types of variables that you’re dealing with.

Statistical assumptions

Statistical tests make some common assumptions about the data they are testing:

• Independence of observations (a.k.a. no autocorrelation): The observations/variables you include in your test are not related (for example, multiple measurements of a single test subject are not independent, while measurements of multiple different test subjects are independent).
• Homogeneity of variance : the variance within each group being compared is similar among all groups. If one group has much more variation than others, it will limit the test’s effectiveness.
• Normality of data : the data follows a normal distribution (a.k.a. a bell curve). This assumption applies only to quantitative data .

If your data do not meet the assumptions of normality or homogeneity of variance, you may be able to perform a nonparametric statistical test , which allows you to make comparisons without any assumptions about the data distribution.

If your data do not meet the assumption of independence of observations, you may be able to use a test that accounts for structure in your data (repeated-measures tests or tests that include blocking variables).

Types of variables

The types of variables you have usually determine what type of statistical test you can use.

Quantitative variables represent amounts of things (e.g. the number of trees in a forest). Types of quantitative variables include:

• Continuous (aka ratio variables): represent measures and can usually be divided into units smaller than one (e.g. 0.75 grams).
• Discrete (aka integer variables): represent counts and usually can’t be divided into units smaller than one (e.g. 1 tree).

Categorical variables represent groupings of things (e.g. the different tree species in a forest). Types of categorical variables include:

• Ordinal : represent data with an order (e.g. rankings).
• Nominal : represent group names (e.g. brands or species names).
• Binary : represent data with a yes/no or 1/0 outcome (e.g. win or lose).

Choose the test that fits the types of predictor and outcome variables you have collected (if you are doing an experiment , these are the independent and dependent variables ). Consult the tables below to see which test best matches your variables.

Parametric tests usually have stricter requirements than nonparametric tests, and are able to make stronger inferences from the data. They can only be conducted with data that adheres to the common assumptions of statistical tests.

The most common types of parametric test include regression tests, comparison tests, and correlation tests.

Regression tests

Regression tests look for cause-and-effect relationships . They can be used to estimate the effect of one or more continuous variables on another variable.

Comparison tests

Comparison tests look for differences among group means . They can be used to test the effect of a categorical variable on the mean value of some other characteristic.

T-tests are used when comparing the means of precisely two groups (e.g., the average heights of men and women). ANOVA and MANOVA tests are used when comparing the means of more than two groups (e.g., the average heights of children, teenagers, and adults).

Correlation tests

Correlation tests check whether variables are related without hypothesizing a cause-and-effect relationship.

These can be used to test whether two variables you want to use in (for example) a multiple regression test are autocorrelated.

Non-parametric tests don’t make as many assumptions about the data, and are useful when one or more of the common statistical assumptions are violated. However, the inferences they make aren’t as strong as with parametric tests.

This flowchart helps you choose among parametric tests. For nonparametric alternatives, check the table above.

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
• Null hypothesis

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

Statistical tests commonly assume that:

• the data are normally distributed
• the groups that are being compared have similar variance
• the data are independent

If your data does not meet these assumptions you might still be able to use a nonparametric statistical test , which have fewer requirements but also make weaker inferences.

A test statistic is a number calculated by a  statistical test . It describes how far your observed data is from the  null hypothesis  of no relationship between  variables or no difference among sample groups.

The test statistic tells you how different two or more groups are from the overall population mean , or how different a linear slope is from the slope predicted by a null hypothesis . Different test statistics are used in different statistical tests.

Statistical significance is a term used by researchers to state that it is unlikely their observations could have occurred under the null hypothesis of a statistical test . Significance is usually denoted by a p -value , or probability value.

Statistical significance is arbitrary – it depends on the threshold, or alpha value, chosen by the researcher. The most common threshold is p < 0.05, which means that the data is likely to occur less than 5% of the time under the null hypothesis .

When the p -value falls below the chosen alpha value, then we say the result of the test is statistically significant.

Quantitative variables are any variables where the data represent amounts (e.g. height, weight, or age).

Categorical variables are any variables where the data represent groups. This includes rankings (e.g. finishing places in a race), classifications (e.g. brands of cereal), and binary outcomes (e.g. coin flips).

You need to know what type of variables you are working with to choose the right statistical test for your data and interpret your results .

Discrete and continuous variables are two types of quantitative variables :

• Discrete variables represent counts (e.g. the number of objects in a collection).
• Continuous variables represent measurable amounts (e.g. water volume or weight).

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5.3 - hypothesis testing for one-sample mean.

In the previous section, we learned how to perform a hypothesis test for one proportion. The concepts of hypothesis testing remain constant for any hypothesis test. In these next few sections, we will present the hypothesis test for one mean. We start with our knowledge of the sampling distribution of the sample mean.

Hypothesis Test for One-Sample Mean Section  

Recall that under certain conditions, the sampling distribution of the sample mean, \(\bar{x} \), is approximately normal with mean, \(\mu \), standard error \(\dfrac{\sigma}{\sqrt{n}} \), and estimated standard error \(\dfrac{s}{\sqrt{n}} \).

\(H_0\colon \mu=\mu_0\)

Conditions:

• The distribution of the population is Normal
• The sample size is large \( n>30 \).

Test Statistic:

If at least one of conditions are satisfied, then...

\( t=\dfrac{\bar{x}-\mu_0}{\frac{s}{\sqrt{n}}} \)

will follow a t-distribution with \(n-1 \) degrees of freedom.

Notice when working with continuous data we are going to use a t statistic as opposed to the z statistic. This is due to the fact that the sample size impacts the sampling distribution and needs to be taken into account. We do this by recognizing “degrees of freedom”. We will not go into too much detail about degrees of freedom in this course.

Let’s look at an example.

Example 5-1 Section  

This depends on the standard deviation of \(\bar{x} \) . 

\begin{align} t^*&=\dfrac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}\\&=\dfrac{8.3-8.5}{\frac{1.2}{\sqrt{61}}}\\&=-1.3 \end{align} 

Thus, we are asking if \(-1.3\) is very far away from zero, since that corresponds to the case when \(\bar{x}\) is equal to \(\mu_0 \). If it is far away, then it is unlikely that the null hypothesis is true and one rejects it. Otherwise, one cannot reject the null hypothesis. 

What hypothesis test should be used to test One-sample test of means One-sample test of proportions One-sample test of variances Two-sample test of means (independent samples) Two-sample test of means (paired samples) Two-sample test of proportions Two-sample test of variances

To determine the appropriate hypothesis test for different situations, consider the following: 1. One-sample test of means: Use a t-test when comparing the mean of a single sample to a known population mean , and the population standard deviation is unknown. 2. One-sample test of proportions: Use a z-test when comparing the proportion of a single sample to a known population proportion. 3. One-sample test of variances: Use the chi-square test to determine if a single sample's variance differs significantly from a known population variance. 4. Two-sample test of means (independent samples): Use an independent samples t-test when comparing the means of two independent samples to determine if there is a significant difference between them. 5. Two-sample test of means (paired samples): Use a paired samples t-test when comparing the means of two related samples to determine if there is a significant difference between them. 6. Two-sample test of proportions: Use a z-test for comparing proportions when determining if there is a significant difference between the proportions of two independent samples. 7. Two-sample test of variances: Use an F-test to compare the variances of two independent samples to determine if there is a significant difference between them. Each test serves a specific purpose based on the data and the research question. Understanding these tests will help you choose the appropriate hypothesis test for your analysis.

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Related Questions

Convert the augmented matrix [3 2 -5 -2 -1 5 0 -8] to the equivalent linear system. Use x1, x2, and x3 to enter the variables x_1, x_2, and x_3. _____________ = __________ _____________ = __________

The values of x1, x2, and x3 satisfy all three equations. To convert the augmented matrix [3 2 -5 -2 -1 5 0 -8] to the equivalent linear system.

We start by writing the coefficients of the variables x1, x2, and x3 in a matrix form, as follows: | 3  2 -5 |   | x1 |   | -2 | | -2  -1  5 | x | x2 | = | -8 | | 0   1  -3 |   | x3 |   |  0 | Each row of the matrix corresponds to an equation in the linear system, and the last column contains the constants on the right-hand side of the equations. To solve for the variables, we can use row operations to transform the matrix into row echelon form or reduced row echelon form. However, since we only need to express the linear system in terms of x1, x2, and x3, we can directly read off the equations from the matrix: 3x1 + 2x2 - 5x3 = -2 -2x1 - x2 + 5x3 = -8 x2 - 3x3 = 0 Therefore, the equivalent linear system is: 3x1 + 2x2 - 5x3 = -2 -2x1 - x2 + 5x3 = -8 x2 - 3x3 = 0 We can solve this system by substitution or elimination to find the values of x1, x2, and x3 that satisfy all three equations.

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Kareem is trying to decide which college to attend full time next year. Kareem believes there is a 55% chance that he will attend State College and a 33% chance that he will attend Northern University. The probability that Kareem will attend either State or Northern is (state your answer as a decimal and round your answer to two decimal places).

The probability that Kareem will attend either State or Northern is 0.88, which is 88% as a percentage. Rounded to two decimal places, the answer is 0.88.

The probability that Kareem will attend either State or Northern is the sum of the individual probabilities of attending each college.

Probability is a measure of the likelihood or chance of an event occurring. It is usually expressed as a number between 0 and 1, where 0 indicates that an event is impossible and 1 indicates that an event is certain.

The probability of an event A is calculated as the number of outcomes that result in A divided by the total number of possible outcomes. This is known as the classical definition of probability.

P(State or Northern) = P(State) + P(Northern) = 0.55 + 0.33 = 0.88

So the probability that Kareem will attend either State or Northern is 0.88, which is 88% as a percentage. Rounded to two decimal places, the answer is 0.88.

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how many minutes does it take to wash, dry, and fold four loads of laundry using a pipelining approach

It will take 180 minutes to wash, dry, and fold four loads of laundry using a pipelining approach .

Using a pipelining approach for laundry, we can complete the tasks more efficiently by dividing them into stages. Let's break it down step by step: 1. Washing: Let's assume it takes 30 minutes to wash a single load of laundry. 2. Drying: Let's assume it takes 45 minutes to dry a single load of laundry. 3. Folding: Let's assume it takes 15 minutes to fold a single load of laundry. Now, using the pipelining approach: 1. Load 1: Wash (30 minutes) → Dry (45 minutes) → Fold (15 minutes ) = 90 minutes 2. Load 2: Wash (30 minutes) → Dry (45 minutes) → Fold (15 minutes) 3. Load 3: Wash (30 minutes) → Dry (45 minutes) → Fold (15 minutes) 4. Load 4: Wash (30 minutes) → Dry (45 minutes) → Fold (15 minutes) The total time taken for four loads of laundry using a pipelining approach will be: - Time to wash, dry, and fold Load 1: 90 minutes - Time to wash Load 2: 30 minutes (as drying and folding of Load 1 can be done simultaneously) - Time to wash Load 3: 30 minutes (as drying and folding of Load 2 can be done simultaneously) - Time to wash Load 4: 30 minutes (as drying and folding of Load 3 can be done simultaneously) Total time = 90 + 30 + 30 + 30 = 180 minutes It will take 180 minutes to wash, dry, and fold four loads of laundry using a pipelining approach.

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A truck left Town A for Town B at a speed of 80 km h. Two hours later, a car travelling at 120 km/h also left Town A for Town B. The car caught up with the truck 30 km away from Town B. Find the distance between the two towns.

The distance between Town A and Town B is calculated as 368 km .

Distance is described as a numerical or occasionally qualitative measurement of how far apart objects or points are.

we have then equation that:

80 km/h x (t + 2) h = 120 km/h x t h + 30 km

we simplify the above equation :

80t + 160 = 120t + 30

t = 2.6 hours

Therefore, the distance between Town A and Town B will be the  distance traveled by truck

= 80 km/h x (t + 2) h

= 80 km/h x 4.6 h

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Suppose b1, b2, b3, ... is a sequence defined as follows: b1 = 4, b2 = 12 bk = bk–2 + bk–1 for all integers k ≥ 3. Prove that bn is divisible by 4 for all integers n ≥ 1.

We have proven that bn is divisible by 4 for all integers n ≥ 1 .To prove that bn is divisible by 4 for all integers n ≥ 1, we will use mathematical induction. Base case: We know that b1 = 4, which is divisible by 4. We also know that b2 = 12, which is divisible by 4. Therefore, the base case is true. I nductive step: Assume that bn-1 and bn-2 are both divisible by 4 for some integer n ≥ 3. We want to show that bn is also divisible by 4. From the definition of the sequence, we know that bk = bk-2 + bk-1 for all integers k ≥ 3. Therefore, bn = bn-2 + bn-1. Since bn-1 and bn-2 are both divisible by 4 (by the induction hypothesis), we know that they can be written as 4m and 4n, where m and n are integers. Substituting into the equation for bn, we get: bn = bn-2 + bn-1 bn = 4n + 4m bn = 4(m + n) Since m + n is an integer, we have shown that bn can be written as 4 times an integer and therefore is divisible by 4. Therefore, by mathematical induction, we have proven that bn is divisible by 4 for all integers n ≥ 1.

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movie has been downloading for 4 min and has downloaded 25%. how many minutes are needed for the remaining 75%

It will take 12 more minutes for the remaining 75% of the movie to download using algebra .

To determine how many minutes are needed for the remaining 75% of the movie to download, we will follow these steps: 1. Observe that the movie has been downloading for 4 minutes and has downloaded 25%. This means that the time taken to download 25% of the movie is 4 minutes. 2. Now, we need to determine the time taken to download the remaining 75% of the movie. Since we know the time taken for 25%, we can use this information to find the time for 75%. 3. We can set up a proportion: (time taken for 25%)/(time taken for 75%) = 25%/75%. This proportion helps us understand the relationship between the time taken for both percentages. 4. Plug in the known value: 4 minutes/(time taken for 75%) = 25%/75%. Now, we need to solve for the unknown variable (time taken for 75%). 5. To solve for the unknown variable, we can cross-multiply. Multiply 4 minutes by 75% and 25% by the time taken for 75%: (4 minutes x 75%) = (25% x time taken for 75%). 6. Calculate 4 minutes x 75%: 4 minutes x 0.75 = 3 minutes. So, 3 minutes = (25% x time taken for 75%). 7. Now, divide both sides of the equation by 25% (or 0.25) to find the time taken for 75%: 3 minutes ÷ 0.25 = 12 minutes. So, it will take 12 more minutes for the remaining 75% of the movie to download.

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An ice cream store has 5 cartons of ice cream on hand each day. The cdf for the amount of ice cream actually sold is F(x) = 0 if x < 0 F(x) = (x3 + x)/130 if 0 SrS5 F(x) = 1 if x > 5 = Answer the following two questions, rounding your answer to 3 decimals What is the expected number of cartons sold in a day? What is the probability that the ice cream store sells more than half of its inventory in a day?

1 the expected number of cartons sold in a day is approximately 3.339. 2 :The probability that the ice cream store sells more than half of its inventory in a day is 0.533, or 53.3%.(rounded to 3 decimals).

1. To find the expected number of cartons sold in a day, we'll need to calculate the expected value (E(x)) using the pdf, which is the derivative of the cdf, F(x). First, we'll find the pdf, f(x): f(x) = dF(x)/dx = (3x^2 + 1)/130 for 0 ≤ x ≤ 5 Now, we can calculate E(x): E(x) = ∫(x * f(x) dx) from 0 to 5 E(x) = ∫(x * (3x^2 + 1)/130 dx) from 0 to 5 After solving the integral and evaluating the limits, we get: E(x) ≈ 3.339 So, the expected number of cartons sold in a day is approximately 3.339. 2. To find the probability that the ice cream s tore sells more than half of its inventory in a day, we'll use the cdf, F(x): P(X > 2.5) = 1 - F(2.5) Using the given cdf function for 0 ≤ x ≤ 5: F(2.5) = ((2.5)^3 + 2.5) / 130 ≈ 0.467 Now, we can find the probability: P(X > 2.5) = 1 - 0.467 ≈ 0.533 So, the probability that the ice cream store sells more than half of its inventory in a day is approximately 0.533, or 53.3%.

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The volume of a gas is inversely proportional to the pressure. If a pressure of 21 pounds per square inch corresponds to a volume of 20 cubic feet, what pressure is needed to produce a volume of 30 cubic feet

A pressure of 14 pounds per square inch is needed to produce a volume of 30 cubic feet, assuming that the volume of the gas is inversely proportional to the pressure.

If the volume of a gas is inversely proportional to the pressure, we can use the formula:

P1 x V1 = P2 x V2

P1 and V1 are the initial pressure and volume, and P2 and V2 are the new pressure and volume.

P1 = 21 pounds per square inch and V1 = 20 cubic feet.

To find P2 when V2 = 30 cubic feet.

Plugging in the values we have:

21 x 20 = P2 x 30

Simplifying:

Dividing both sides by 30:

P2 = 14 pounds per square inch

We may apply the formula: if the volume of a gas is inversely proportional to the pressure.

P1 x V1 equals P2 x V2

The original pressure and volume are P1 and V1, whereas the new pressure and volume are P2 and V2.

V1 is 20 cubic feet, and P1 is 21 pounds per square inch.

when V2 = 30 cubic feet, to determine P2.

When we enter the values we have:

Condensing: 420 = 30P2

30 divided by both sides:

14 pounds per square inch is P2.

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One diagonal of a rhombus is twice as long as the other diagonal. If the area of the rhombus is 169 square millimeters, what are the lengths of the diagonals

The lengths of the diagonals are 13 and 26 millimeters.

Let the length of the shorter diagonal be x.

Then, the length of the longer diagonal is 2x.

The area of a rhombus is given by (1/2) * d1 * d2, where d1 and d2 are the diagonals.

So we have:

(1/2) * x * 2x = 169

Simplifying this equation, we get:

[tex]x^2[/tex] = 169

Taking the square root of both sides, we get:

Therefore, the length of the shorter diagonal is 13.

And the length of the longer diagonal is 2x = 26.

Hence, the lengths of the diagonals are 13 and 26 millimeters.

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Marcos has 2003 pairs of shoes. In how many different ways can he select a left then a right shoe?

Step-by-step explanation:

The size of a certain insect population is given by ​P(t)​, where t is measured in days. ​(a) How many insects were present​ initially? ​(b) Give a differential equation satisfied by​ P(t). ​(c) At what time will the population​ double? ​(d) At what time will the population equal ​?

(a) Without more information, we cannot determine the initial number of insects. (b) The differential equation satisfied by P(t) is: dP/dt = kP, where k is the growth rate of the insect population . (c) To find the time it takes for the population to double, we can use the formula: 2P(0) = P(0)e^(kt) where P(0) is the initial population size . Solving for t, we get: t = ln(2)/k (d) Without more information, we cannot determine the time at which the population will equal a certain value. Hi! To answer your question, I need the specific function P(t). However, I can provide you with a general framework to answer each part of your question once you have the function. (a) To find the initial number of insects, evaluate P(t) at t=0: P(0) = [Insert the function with t=0] (b) To find the differential equation satisfied by P(t), differentiate P(t) with respect to t: dP(t)/dt = [Insert the derivative of the function] (c) To find the time at which the population doubles, first determine the initial population, P(0), then solve for t when P(t) is twice that value: 2*P(0) = P(t) Solve for t: [Insert the solution for t] (d) To find the time at which the population equals a specific value (let's call it N), set P(t) equal to N and solve for t: N = P(t) Solve for t: [Insert the solution for t] Once you have the specific function P(t), you can follow these steps to find the answers to each part of your question.

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100 point question Trigonometry Find Sin of G

Step-by-step explanation

Linear programming models have three important properties. They are: a. proportionaity, additivity and divisibility b. optimality, additivity and sensitivity c. optimality, linearity and divisibility d. divisibility, linearity and non-negativity e. proportionality, additivity and linearity

The correct answer is e. Linear programming models have three important properties: proportionality, additivity, and linearity .

These properties allow for the creation of efficient optimization models that can be used to solve complex problems in various industries . Proportionality refers to the relationship between the input variables and the output, while additivity refers to the ability to combine multiple variables into a single equation. Linearity refers to the fact that the output is directly proportional to the input, making it easier to analyze and optimize the model.

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Consider the simple linear regression model: yi = β0 + β1xi + εi Show that minimizing the sum of squared residuals lead to the following least squares coefficient estimates: βˆ 0 = ¯y − βˆ 1x, ¯ βˆ 1 = Pn i=1(xi − x¯)(yi − y¯) Pn i=1(xi − x¯) 2 , where y¯ = 1 n Pn i=1 yi and x¯ = 1 n Pn i=1 xi .

The simple linear regression model is given by yi = β0 + β1xi + εi, where β0 is the intercept, β1 is the slope, xi is the predictor variable, yi is the response variable, and εi is the error term. These are the least squares coefficient estimates for the simple linear regression model.

The goal of least squares regression is to find the values of β0 and β1 that minimize the sum of the squared residuals. To find the least squares coefficient estimates, we need to minimize the sum of the squared residuals. The residual is the difference between the observed value of yi and the predicted value of yi. The predicted value of yi is given by β0 + β1xi. Therefore, the residual can be written as yi - (β0 + β1xi). The sum of squared residuals is given by: Σi=1n (yi - β0 - β1xi)² To find the values of β0 and β1 that minimize this sum, we take the partial derivatives with respect to β0 and β1 and set them equal to zero: ∂/∂β0 Σi=1n (yi - β0 - β1xi)² = 0 ∂/∂β1 Σi=1n (yi - β0 - β1xi)² = 0 Solving these equations yields: βˆ 0 = ¯y − βˆ 1x and βˆ 1 = Pn i=1(xi − x¯)(yi − y¯) / Pn i=1(xi − x¯)² where y¯ = 1 n Pn i=1 yi and x¯ = 1 n Pn i=1 xi. These are the least squares coefficient estimates for the simple linear regression model.

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b) You watch a roulette wheel spin 210 consecutive times and the ball lands on a red slot each time. What is the probability that the ball will land on a red slot on the next spin

The probability of the ball landing on a red slot on the next spin is still 18/38 or approximately 0.4737 or 47.37%.

The probability of the ball landing on a red slot on a single spin of a standard roulette wheel is 18/38 or approximately 0.4737 or 47.37%. This is because there are 18 red slots out of a total of 38 slots on the wheel.

The outcome of the previous 210 spins has no effect on the probability of the ball landing on a red slot on the next spin. Each spin is an independent event, and the probability of the ball landing on a red slot remains the same for each spin .

Therefore, even though the ball has landed on a red slot for the past 210 spins, the probability of it landing on a red slot on the next spin is still 18/38 or approximately 0.4737 or 47.37%. To know more about probability , refer here:

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Let f(x) = 1+7x / 3x-5 for x ≠ 5/3(a) Determine f^-1(x), the inverse function of f(x). (b) Find the largest possible domain and the range of f(x). (c) Find the function g such that (gºf)(x) = cos(x²)

The   function g such that (gºf)(x) = cos(x²) is g(y) = cos(y^2), where y = (1 + 7x)/(3x - 5).

(a) To find the inverse function of f(x), we need to solve for x in terms of f(x). Let y = f(x), then we have: y = (1 + 7x)/(3x - 5) Multiplying both sides by (3x - 5), we get: y(3x - 5) = 1 + 7x Expanding and rearranging , we get: (3y - 7)x = y + 5 Dividing both sides by (3y - 7), we get: x = (y + 5)/(3y - 7) Therefore, the inverse function of f(x) is: f^-1(x) = (x + 5)/(3x - 7) (b) The function f(x) is defined for all x except x = 5/3, because the denominator 3x - 5 becomes zero at x = 5/3. Therefore, the largest possible domain of f(x) is (-∞, 5/3) U (5/3, ∞). To find the range of f(x), we can use calculus. Taking the derivative of f(x), we get: f'(x) = (16 - 21x)/(3x - 5)^2 The derivative is zero when 16 - 21x = 0, or x = 16/21. This is a local maximum of f(x), because f''(x) = 126/(3x - 5)^3 is positive when x < 5/3 and negative when x > 5/3. Therefore, the maximum value of f(x) is: f(16/21) = (1 + 7(16/21))/(3(16/21) - 5) = 11/2 Since f(x) approaches positive infinity as x approaches 5/3 from the left and negative infinity as x approaches 5/3 from the right, the range of f(x) is (-∞, 11/2) U (11/2, ∞). (c) Let g(y) = cos(y^2). Then, we have: (gºf)(x) = g(f(x)) = g((1 + 7x)/(3x - 5)) = cos(((1 + 7x)/(3x - 5))^2) Therefore, the function g such that (gºf)(x) = cos(x²) is g(y) = cos(y^2), where y = (1 + 7x)/(3x - 5).

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i need help. a lot of it too.

8.) The surface area of the given shape would be = 532ft²

9.) The surface area of the given circle = 24.62cm²

For question 8.)

To calculate the surface area of the square based pyramid the formula given below is used;

S.A = b² + 2bs

S.A = 14² + 2(14×12)

= 196+ 2(168)

For question 9:

To calculate the area of the circle , the formula that should be used is given as follows:

S.A = 3.14×2.8×2.8

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e. Find the standardized values for students scoring 540, 600, 650, and 700 on the test. Explain what these mean.

This information can be useful for comparing and analyzing test scores, especially when comparing scores from different tests or different populations .

To find the standardized values for students scoring 540, 600, 650, and 700 on the test, we need to use the formula for z-score: z = (x - mean) / standard deviation Assuming that the test scores follow a normal distribution with a mean of 500 and a standard deviation of 100 (which are common values for standardized tests ), we can calculate the z-scores as follows: For a score of 540: z = (540 - 500) / 100 = 0.4 For a score of 600: z = (600 - 500) / 100 = 1 For a score of 650: z = (650 - 500) / 100 = 1.5 For a score of 700: z = (700 - 500) / 100 = 2 These standardized values represent the number of standard deviations that each score is away from the mean. A z-score of 0 means that the score is exactly at the mean, while a z-score of 1 means that the score is one standard deviation above the mean , and so on. Therefore, we can interpret these standardized values as follows: - A score of 540 is 0.4 standard deviations above the mean. - A score of 600 is 1 standard deviation above the mean. - A score of 650 is 1.5 standard deviations above the mean. - A score of 700 is 2 standard deviations above the mean.

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Manuel just bought a new television for $629.00. He made a down payment of $57.00 and will pay monthly payments of $26.00 until it is paid off. How many months will Manuel be paying

Manuel will be paying off his new television for a total of 20 months, with a down payment of $57.00 and monthly payments of $26.00.

Manuel's new television costs $629.00, and he made a down payment of $57.00. This means he still owes $629.00 - $57.00 = $572.00. Manuel will be paying this off through monthly payments of $26.00. To calculate the number of months it will take for Manuel to pay off the television, we can use the following formula: Number of months = (Total amount owed - Down payment) ÷ Monthly payment Plugging in Manuel's numbers, we get: Number of months = ($572.00 - $57.00) ÷ $26.00 Number of months = $515.00 ÷ $26.00 Number of months = 19.81 Since we can't have a fraction of a month, we'll round up to the nearest whole number. Therefore, Manuel will be paying off his new television for 20 months.

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The force exerted by an electric charge at the origin on a charged particle at a point (x,y,z) with position vector r = (x,y,z) is F(r) =Kr/||r||^3. Where k is a constant. Find the work done as the particle moves along a straight line from (2, 0, 0) to (2,3,5)

To find the work done, we need to integrate the force F(r) along the path that the particle moves. Since the path is a straight line, we can parametrize it as r(t) = (2, 0, 0) + t((2,3,5)-(2,0,0)) = (2+2t, 3t, 5t),

where 0 <= t <= 1.

Then, the force F(r(t)) is given by F(t) = K(2+2t, 3t, 5t)/[(2+2t)^2 + (3t)^2 + (5t)^2]^(3/2).

The work done by the force as the particle moves from (2,0,0) to (2,3,5) is given by the line integral:

W = ∫ F(r(t)) · dr(t) from t=0 to t=1

where dr(t) is the differential of r(t) with respect to t.

Now, we need to evaluate the dot product F(r(t)) · dr(t).

Note that dr(t) = (2,3,5) dt, since the path is a straight line.

Therefore: F(r(t)) · dr(t) = K(2+2t, 3t, 5t)/[(2+2t)^2 + (3t)^2 + (5t)^2]^(3/2) · (2,3,5) dt

= K(2+2t)(2) + 3t(3) + 5t(5) / [(2+2t)^2 + (3t)^2 + (5t)^2]^(3/2) dt

= K(4+4t + 9t + 25t) / [(2+2t)^2 + (3t)^2 + (5t)^2]^(3/2) dt

= 38Kt / [(2+2t)^2 + (3t)^2 + (5t)^2]^(3/2) dt

Thus, the work done is:

= ∫0^1 38Kt / [(2+2t)^2 + (3t)^2 + (5t)^2]^(3/2) dt

This integral is difficult to solve exactly, so we can use numerical methods or software to obtain an approximation .

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) What is the probability that a randomly chosen miniature Tootsie Roll will weigh more than 3.50 grams

The probability that a randomly chosen miniature Tootsie Roll will weigh more than 3.50 grams is 0.1587 or 16%.

The probability of a randomly chosen miniature Tootsie Roll weighing more than 3.50 grams can be determined through statistical analysis.

To do this, we need to consider the mean and standard deviation of the weight of Tootsie Rolls. Assuming that the weight of Tootsie Rolls follows a normal distribution , we can use the z-score formula to find the probability of a randomly chosen Tootsie Roll weighing more than 3.50 grams. The formula for calculating the z-score is: z = (x - μ) / σ where x is the observed weight , μ is the mean weight, and σ is the standard deviation. Let's assume that the mean weight of miniature Tootsie Rolls is 3 grams and the standard deviation is 0.5 grams.

To find the z-score for a weight of 3.5 grams, we can plug in the values: z = (3.5 - 3) / 0.5 z = 1

Using a z-score table, we can find that the probability of a z-score of 1 (or a Tootsie Roll weighing more than 3.5 grams) is 0.1587. Therefore, the probability of a randomly chosen miniature Tootsie Roll weighing more than 3.50 grams is 0.1587 or approximately 16%. It is important to note that this is an estimate based on assumptions about the distribution of Tootsie Roll weights. The actual probability may differ depending on factors such as batch variability, production methods , and storage conditions.

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To construct a frequency distribution for categorical data, the: Group of answer choices number of observations that appear in each category must be counted. observations in each category must be multiplied by observations in the corresponding

To construct a frequency distribution for categorical data, you need to follow a systematic approach. Firstly, identify the categories or groups present in the data set.

Then, count the number of observations that appear in each category. This step is crucial as it helps you understand the distribution and frequency of each category within the dataset. Once you have the count of observations for each category, you can proceed to calculate the relative frequencies, which are obtained by dividing the number of observations in a specific category by the total number of observations in the dataset . This step enables you to comprehend the proportion of each category in relation to the entire dataset. Lastly, organize and present the data in a meaningful format, such as a table or a chart, to help visualize the distribution of the categorical data effectively. By following these steps, you can successfully construct a frequency distribution for categorical data and analyze the trends and patterns that emerge.

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1) Find A ∩ B. (Enter your answer in roster notation. Enter EMPTY or ∅ for the empty set.) A = {a, d, j, o, z} and B = {a, d, f, g, o,u} A ∩ B = 2) Let A = {5, 3, 4, 1, 2, 7} B = {6, 3, 1, 9} and U be the universal set of natural numbers less than 11. Find the following. (Enter your answers as a comma-separated list. Enter EMPTY or for the empty set.) (A ∩ B)' = 3) Let A = {5, 3, 4, 1, 2, 7} B = {6, 3, 1, 9} and U be the universal set of natural numbers less than 11. Find the following. (Enter your answers as a comma-separated list. Enter EMPTY or ∅ for the empty set.) (A ∩ B)' =

1) To find A ∩ B, we need to identify the elements that are common to both sets A and B. A = {a, d, j, o, z} and B = {a, d, f, g, o, u}. A ∩ B = {a, d, o} 2) Let A = {5, 3, 4, 1, 2, 7}, B = {6, 3, 1, 9}, and U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. First, find A ∩ B, which is the set of elements common to both A and B. A ∩ B = {3, 1}. To find (A ∩ B)', we need to identify the elements in the universal set U that are not in the intersection A ∩ B. (A ∩ B)' = {2, 4, 5, 6, 7, 8, 9, 10}. 3) This question is identical to question 2, so the answer is the same. (A ∩ B)' = {2, 4, 5, 6, 7, 8, 9, 10}.

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Even numbers are usually more powerful than odd numbers when presenting or designing a product or message. This statement is:

It is important to base design decisions on objective criteria such as user research, usability testing, and data analysis, rather than subjective biases or unfounded beliefs about numbers.

Not supported by any empirical evidence or logical reasoning. There is no inherent power or superiority associated with even numbers over odd numbers in the context of presenting or designing a product or message.

While some people may have personal preferences or cultural associations with even or odd numbers, the power or effectiveness of a product or message is determined by various factors such as its content, design, target audience, and context of use.

Therefore, it is important to base design decisions on objective criteria such as user research, usability testing, and data analysis , rather than subjective biases or unfounded beliefs about numbers.

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you wish to compute the 95% confidence interval. how large a sample size should you draw to ensure that the sample proportion does not deviate from the popluation

The size of the sample has to be 119 since you do not want to  deviate from the population

We have to assume that the estimated proportion is given as 0.5

From the standard normal table , we have to solve for the z critical value

a=0.05, Z(0.025) =1.96

The formula for n can be gotten through  n=(Z/E)^2*p*(1-p)

When we put in the values we will have

=(1.96/0.09)^2*0.5*0.5

This is approximated as n = 119

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Normal probability distribution is applied to: A. a subjective random variable B. a discrete random variable C. any random variable D. a continuous random variable

Normal probability distribution is applied to a continuous random variable . The correct option is D.

The normal probability distribution , also known as the Gaussian distribution, is a probability distribution that is commonly used in statistics and probability theory. It is a continuous probability distribution that is often used to model the behavior of a wide range of variables , such as physical measurements like height, weight, and temperature.

The normal distribution is characterized by two parameters: the mean (μ) and the s tandard deviation (σ). It is a bell-shaped curve that is symmetrical around the mean, with the highest point of the curve being located at the mean. The standard deviation determines the width of the curve, and 68% of the data falls within one standard deviation of the mean, while 95% falls within two standard deviations.

The normal distribution is widely used in statistical inference and hypothesis testing, as many test statistics are approximately normally distributed under certain conditions. It is also used in modeling various phenomena, including financial markets, population growth, and natural phenomena like earthquakes and weather patterns.

Overall, the normal probability distribution is a powerful tool for modeling and analyzing a wide range of continuous random variables in a variety of fields.

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Let X have the Chi-Square pdf with 10 degrees of freedom. What is the probability that X is equal to 3.94

The probability that X is equal to 3.94 is approximately 0.0286.

Since X follows a chi-square distribution with 10 degrees of freedom, its probability density function (pdf) is given by:

[tex]f(x) = (1/2^(10/2) * Gamma(10/2)) * x^(10/2 - 1) * e^(-x/2)[/tex]

where Gamma() is the gamma function.

To find the probability that X is equal to 3.94, we need to evaluate the pdf at that value, i.e., we need to find f(3.94). Plugging in the values, we get:

[tex]f(3.94) = (1/2^(10/2) * Gamma(10/2)) * (3.94)^(10/2 - 1) * e^(-3.94/2)[/tex]≈ 0.0286

So the probability that X is equal to 3.94 is approximately 0.0286. Note that since X is a continuous random variable, the probability of it taking any particular value is zero. However, we can still talk about the probability of X being within a certain range or interval.

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22. If a pilot elects to proceed to the selected alternate, the landing minimums used at that airport should be

If a pilot elects to proceed to the selected alternate, the landing minimums used at that airport should be equal to or higher than the minimums required for the original destination airport.

When a pilot chooses to proceed to the selected alternate airport, they must consider the weather conditions at that airport, particularly the landing minimums. The landing minimums are the lowest weather conditions in which an aircraft can safely land at an airport. They are determined by various factors, such as the visibility range and the height of the cloud ceiling. Therefore, if a pilot decides to go to the alternate airport , they should use the landing minimums for that airport to ensure a safe landing. The landing minimums for the alternate airport should be equal to or higher than the minimums required for the original destination airport. It's important to note that the pilot must have knowledge of the landing minimums for the alternate airport before deciding to proceed there. If they don't have the necessary information, they should consider other options, such as returning to the departure airport or diverting to a different airport with suitable weather conditions. In summary, when a pilot chooses to proceed to an alternate airport, they should use the landing minimums for that airport to ensure a safe landing. The pilot must have knowledge of the minimums before proceeding to the alternate airport.

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Tiya flipped a coin 40 times. The coin landed heads up 16 times and tails up 24 times. Part A: Based on the results, what is the experimental probability of the coin landing heads up

The experimental probability of the coin landing heads up is calculated by dividing the number of times the coin landed heads up (16) by the total number of flips (40). So the experimental probability of the coin landing heads up is: P(heads up) = 16/40 Simplifying the fraction by dividing both the numerator and denominator by 8, we get: P(heads up) = 2/5 or 0.4 Therefore, based on the results, the experimental probability of the coin landing heads up is 0.4 or 2/5. To find the experimental probability of the coin landing heads up, you'll need to use the following formula: Experimental probability = ( Number of successful outcomes ) / ( Total number of trials ) In this case, the successful outcome is the coin landing heads up, which occurred 16 times. The total number of trials is 40 flips. So, the experimental probability would be: Experimental probability (heads up) = (16 successful outcomes) / (40 total flips) Now, divide 16 by 40 to get the probability: Experimental probability (heads up) = 16/40 = 0.4 or 40% So, based on the results, the experimental probability of the coin landing heads up is 40%.

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The radius of a ball was measured and found to be 25 cm with a possible error in measurement of at most 0.01cm. What is the maximum error in using this value of the radius to compute the volume of the ball

here you go . ..........................................

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9: Hypothesis Testing with One Sample

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One job of a statistician is to make statistical inferences about populations based on samples taken from the population. Confidence intervals are one way to estimate a population parameter. Another way to make a statistical inference is to make a decision about a parameter. For instance, a car dealer advertises that its new small truck gets 35 miles per gallon, on average. A tutoring service claims that its method of tutoring helps 90% of its students get an A or a B. A company says that women managers in their company earn an average of $60,000 per year.

• 9.1: Prelude to Hypothesis Testing A statistician will make a decision about claims via a process called "hypothesis testing." A hypothesis test involves collecting data from a sample and evaluating the data. Then, the statistician makes a decision as to whether or not there is sufficient evidence, based upon analysis of the data, to reject the null hypothesis.
• 9.1E: Null and Alternative Hypotheses (Exercises)
• 9.2E: Outcomes and the Type I and Type II Errors (Exercises)
• 9.3E: Distribution Needed for Hypothesis Testing (Exercises)
• 9.4E: Rare Events, the Sample, Decision and Conclusion (Exercises)
• 9.6: Additional Information and Full Hypothesis Test Examples The hypothesis test itself has an established process. This can be summarized as follows: Determine H0 and Ha. Remember, they are contradictory. Determine the random variable. Determine the distribution for the test. Draw a graph, calculate the test statistic, and use the test statistic to calculate the p-value. (A z-score and a t-score are examples of test statistics.) Compare the preconceived α with the p-value, make a decision (reject or do not reject H0), and write a clear conclusion.
• 9.7: Hypothesis Testing of a Single Mean and Single Proportion (Worksheet) A statistics Worksheet:  The student will select the appropriate distributions to use in each case. The student will conduct hypothesis tests and interpret the results.
• 9.E: Hypothesis Testing with One Sample (Exercises) These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.

Contributors

Barbara Illowsky and Susan Dean (De Anza College) with many other contributing authors. Content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] .

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8.6: Hypothesis Test of a Single Population Mean with Examples

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Steps for performing Hypothesis Test of a Single Population Mean

Step 1: State your hypotheses about the population mean. Step 2: Summarize the data. State a significance level. State and check conditions required for the procedure

• Find or identify the sample size, n, the sample mean, \(\bar{x}\) and the sample standard deviation, s .

The sampling distribution for the one-mean test statistic is, approximately, T- distribution if the following conditions are met

• Sample is random with independent observations .
• Sample is large. The population must be Normal or the sample size must be at least 30.

Step 3: Perform the procedure based on the assumption that \(H_{0}\) is true

• Find the Estimated Standard Error: \(SE=\frac{s}{\sqrt{n}}\).
• Compute the observed value of the test statistic: \(T_{obs}=\frac{\bar{x}-\mu_{0}}{SE}\).
• Check the type of the test (right-, left-, or two-tailed)
• Find the p-value in order to measure your level of surprise.

Step 4: Make a decision about \(H_{0}\) and \(H_{a}\)

• Do you reject or not reject your null hypothesis?

Step 5: Make a conclusion

• What does this mean in the context of the data?

The following examples illustrate a left-, right-, and two-tailed test.

Example \(\pageindex{1}\).

\(H_{0}: \mu = 5, H_{a}: \mu < 5\)

Test of a single population mean. \(H_{a}\) tells you the test is left-tailed. The picture of the \(p\)-value is as follows:

Exercise \(\PageIndex{1}\)

\(H_{0}: \mu = 10, H_{a}: \mu < 10\)

Assume the \(p\)-value is 0.0935. What type of test is this? Draw the picture of the \(p\)-value.

left-tailed test

Example \(\PageIndex{2}\)

\(H_{0}: \mu \leq 0.2, H_{a}: \mu > 0.2\)

This is a test of a single population proportion. \(H_{a}\) tells you the test is right-tailed . The picture of the p -value is as follows:

Exercise \(\PageIndex{2}\)

\(H_{0}: \mu \leq 1, H_{a}: \mu > 1\)

Assume the \(p\)-value is 0.1243. What type of test is this? Draw the picture of the \(p\)-value.

right-tailed test

Example \(\PageIndex{3}\)

\(H_{0}: \mu = 50, H_{a}: \mu \neq 50\)

This is a test of a single population mean. \(H_{a}\) tells you the test is two-tailed . The picture of the \(p\)-value is as follows.

Exercise \(\PageIndex{3}\)

\(H_{0}: \mu = 0.5, H_{a}: \mu \neq 0.5\)

Assume the p -value is 0.2564. What type of test is this? Draw the picture of the \(p\)-value.

two-tailed test

Full Hypothesis Test Examples

Example \(\pageindex{4}\).

Statistics students believe that the mean score on the first statistics test is 65. A statistics instructor thinks the mean score is higher than 65. He samples ten statistics students and obtains the scores 65 65 70 67 66 63 63 68 72 71. He performs a hypothesis test using a 5% level of significance. The data are assumed to be from a normal distribution.

Set up the hypothesis test:

A 5% level of significance means that \(\alpha = 0.05\). This is a test of a single population mean .

\(H_{0}: \mu = 65  H_{a}: \mu > 65\)

Since the instructor thinks the average score is higher, use a "\(>\)". The "\(>\)" means the test is right-tailed.

Determine the distribution needed:

Random variable: \(\bar{X} =\) average score on the first statistics test.

Distribution for the test: If you read the problem carefully, you will notice that there is no population standard deviation given . You are only given \(n = 10\) sample data values. Notice also that the data come from a normal distribution. This means that the distribution for the test is a student's \(t\).

Use \(t_{df}\). Therefore, the distribution for the test is \(t_{9}\) where \(n = 10\) and \(df = 10 - 1 = 9\).

The sample mean and sample standard deviation are calculated as 67 and 3.1972 from the data.

Calculate the \(p\)-value using the Student's \(t\)-distribution:

\[t_{obs} = \dfrac{\bar{x}-\mu_{\bar{x}}}{\left(\dfrac{s}{\sqrt{n}}\right)}=\dfrac{67-65}{\left(\dfrac{3.1972}{\sqrt{10}}\right)}\]

Use the T-table or Excel's t_dist() function to find p-value:

\(p\text{-value} = P(\bar{x} > 67) =P(T >1.9782 )= 1-0.9604=0.0396\)

Interpretation of the p -value: If the null hypothesis is true, then there is a 0.0396 probability (3.96%) that the sample mean is 65 or more.

Compare \(\alpha\) and the \(p-\text{value}\):

Since \(α = 0.05\) and \(p\text{-value} = 0.0396\). \(\alpha > p\text{-value}\).

Make a decision: Since \(\alpha > p\text{-value}\), reject \(H_{0}\).

This means you reject \(\mu = 65\). In other words, you believe the average test score is more than 65.

Conclusion: At a 5% level of significance, the sample data show sufficient evidence that the mean (average) test score is more than 65, just as the math instructor thinks.

The \(p\text{-value}\) can easily be calculated.

Put the data into a list. Press STAT and arrow over to TESTS . Press 2:T-Test . Arrow over to Data and press ENTER . Arrow down and enter 65 for \(\mu_{0}\), the name of the list where you put the data, and 1 for Freq: . Arrow down to \(\mu\): and arrow over to \(> \mu_{0}\). Press ENTER . Arrow down to Calculate and press ENTER . The calculator not only calculates the \(p\text{-value}\) (p = 0.0396) but it also calculates the test statistic ( t -score) for the sample mean, the sample mean, and the sample standard deviation. \(\mu > 65\) is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with \(t = 1.9781\) (test statistic) and \(p = 0.0396\) (\(p\text{-value}\)). Make sure when you use Draw that no other equations are highlighted in \(Y =\) and the plots are turned off.

Exercise \(\PageIndex{4}\)

It is believed that a stock price for a particular company will grow at a rate of $5 per week with a standard deviation of $1. An investor believes the stock won’t grow as quickly. The changes in stock price is recorded for ten weeks and are as follows: $4, $3, $2, $3, $1, $7, $2, $1, $1, $2. Perform a hypothesis test using a 5% level of significance. State the null and alternative hypotheses, find the p -value, state your conclusion, and identify the Type I and Type II errors.

• \(H_{0}: \mu = 5\)
• \(H_{a}: \mu < 5\)
• \(p = 0.0082\)

Because \(p < \alpha\), we reject the null hypothesis. There is sufficient evidence to suggest that the stock price of the company grows at a rate less than $5 a week.

• Type I Error: To conclude that the stock price is growing slower than $5 a week when, in fact, the stock price is growing at $5 a week (reject the null hypothesis when the null hypothesis is true).
• Type II Error: To conclude that the stock price is growing at a rate of $5 a week when, in fact, the stock price is growing slower than $5 a week (do not reject the null hypothesis when the null hypothesis is false).

Example \(\PageIndex{5}\)

The National Institute of Standards and Technology provides exact data on conductivity properties of materials. Following are conductivity measurements for 11 randomly selected pieces of a particular type of glass.

1.11; 1.07; 1.11; 1.07; 1.12; 1.08; .98; .98 1.02; .95; .95

Is there convincing evidence that the average conductivity of this type of glass is greater than one? Use a significance level of 0.05. Assume the population is normal.

Let’s follow a four-step process to answer this statistical question.

• \(H_{0}: \mu \leq 1\)
• \(H_{a}: \mu > 1\)
• Plan : We are testing a sample mean without a known population standard deviation. Therefore, we need to use a Student's-t distribution. Assume the underlying population is normal.
• Do the calculations : \(p\text{-value} ( = 0.036)\)

4. State the Conclusions : Since the \(p\text{-value} (= 0.036)\) is less than our alpha value, we will reject the null hypothesis. It is reasonable to state that the data supports the claim that the average conductivity level is greater than one.

The hypothesis test itself has an established process. This can be summarized as follows:

• Determine \(H_{0}\) and \(H_{a}\). Remember, they are contradictory.
• Determine the random variable.
• Determine the distribution for the test.
• Draw a graph, calculate the test statistic, and use the test statistic to calculate the \(p\text{-value}\). (A t -score is an example of test statistics.)
• Compare the preconceived α with the p -value, make a decision (reject or do not reject H 0 ), and write a clear conclusion using English sentences.

Notice that in performing the hypothesis test, you use \(\alpha\) and not \(\beta\). \(\beta\) is needed to help determine the sample size of the data that is used in calculating the \(p\text{-value}\). Remember that the quantity \(1 – \beta\) is called the Power of the Test . A high power is desirable. If the power is too low, statisticians typically increase the sample size while keeping α the same.If the power is low, the null hypothesis might not be rejected when it should be.

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