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Hypothesis Testing

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

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

What is Hypothesis Testing in Statistics?

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

Hypothesis Testing Definition

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

Null Hypothesis

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

Alternative Hypothesis

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

Hypothesis Testing P Value

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

Hypothesis Testing Critical region

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

Hypothesis Testing Formula

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

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

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

Types of Hypothesis Testing

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

Hypothesis Testing Z Test

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

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

Hypothesis Testing t Test

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

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

Hypothesis Testing Chi Square

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

One Tailed Hypothesis Testing

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

Right Tailed Hypothesis Testing

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

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

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

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

Left Tailed Hypothesis Testing

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

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

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

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

Two Tailed Hypothesis Testing

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

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

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

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

Hypothesis Testing Steps

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

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

Hypothesis Testing Example

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

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

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

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

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

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

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

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

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

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

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

Hypothesis Testing and Confidence Intervals

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

Related Articles:

• Probability and Statistics
• Data Handling

Important Notes on Hypothesis Testing

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

Examples on Hypothesis Testing

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

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FAQs on Hypothesis Testing

What is hypothesis testing.

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

What is the z Test in Hypothesis Testing?

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

What is the t Test in Hypothesis Testing?

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

What is the formula for z test in Hypothesis Testing?

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

What is the p Value in Hypothesis Testing?

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

What is One Tail Hypothesis Testing?

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

What is the Alpha Level in Two Tail Hypothesis Testing?

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

A statement that could be true, which might then be tested.

Example: Sam has a hypothesis that "large dogs are better at catching tennis balls than small dogs". We can test that hypothesis by having hundreds of different sized dogs try to catch tennis balls.

Sometimes the hypothesis won't be tested, it is simply a good explanation (which could be wrong). Conjecture is a better word for this.

Example: you notice the temperature drops just as the sun rises. Your hypothesis is that the sun warms the air high above you, which rises up and then cooler air comes from the sides.

Note: when someone says "I have a theory" they should say "I have a hypothesis", because in mathematics a theory is actually well proven.

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Hypothesis Testing

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A hypothesis test is a statistical inference method used to test the significance of a proposed (hypothesized) relation between population statistics (parameters) and their corresponding sample estimators . In other words, hypothesis tests are used to determine if there is enough evidence in a sample to prove a hypothesis true for the entire population.

The test considers two hypotheses: the null hypothesis , which is a statement meant to be tested, usually something like "there is no effect" with the intention of proving this false, and the alternate hypothesis , which is the statement meant to stand after the test is performed. The two hypotheses must be mutually exclusive ; moreover, in most applications, the two are complementary (one being the negation of the other). The test works by comparing the \(p\)-value to the level of significance (a chosen target). If the \(p\)-value is less than or equal to the level of significance, then the null hypothesis is rejected.

When analyzing data, only samples of a certain size might be manageable as efficient computations. In some situations the error terms follow a continuous or infinite distribution, hence the use of samples to suggest accuracy of the chosen test statistics. The method of hypothesis testing gives an advantage over guessing what distribution or which parameters the data follows.

Definitions and Methodology

Hypothesis test and confidence intervals.

In statistical inference, properties (parameters) of a population are analyzed by sampling data sets. Given assumptions on the distribution, i.e. a statistical model of the data, certain hypotheses can be deduced from the known behavior of the model. These hypotheses must be tested against sampled data from the population.

The null hypothesis \((\)denoted \(H_0)\) is a statement that is assumed to be true. If the null hypothesis is rejected, then there is enough evidence (statistical significance) to accept the alternate hypothesis \((\)denoted \(H_1).\) Before doing any test for significance, both hypotheses must be clearly stated and non-conflictive, i.e. mutually exclusive, statements. Rejecting the null hypothesis, given that it is true, is called a type I error and it is denoted \(\alpha\), which is also its probability of occurrence. Failing to reject the null hypothesis, given that it is false, is called a type II error and it is denoted \(\beta\), which is also its probability of occurrence. Also, \(\alpha\) is known as the significance level , and \(1-\beta\) is known as the power of the test. \(H_0\) \(\textbf{is true}\)\(\hspace{15mm}\) \(H_0\) \(\textbf{is false}\) \(\textbf{Reject}\) \(H_0\)\(\hspace{10mm}\) Type I error Correct Decision \(\textbf{Reject}\) \(H_1\) Correct Decision Type II error The test statistic is the standardized value following the sampled data under the assumption that the null hypothesis is true, and a chosen particular test. These tests depend on the statistic to be studied and the assumed distribution it follows, e.g. the population mean following a normal distribution. The \(p\)-value is the probability of observing an extreme test statistic in the direction of the alternate hypothesis, given that the null hypothesis is true. The critical value is the value of the assumed distribution of the test statistic such that the probability of making a type I error is small.

Methodologies: Given an estimator \(\hat \theta\) of a population statistic \(\theta\), following a probability distribution \(P(T)\), computed from a sample \(\mathcal{S},\) and given a significance level \(\alpha\) and test statistic \(t^*,\) define \(H_0\) and \(H_1;\) compute the test statistic \(t^*.\) \(p\)-value Approach (most prevalent): Find the \(p\)-value using \(t^*\) (right-tailed). If the \(p\)-value is at most \(\alpha,\) reject \(H_0\). Otherwise, reject \(H_1\). Critical Value Approach: Find the critical value solving the equation \(P(T\geq t_\alpha)=\alpha\) (right-tailed). If \(t^*>t_\alpha\), reject \(H_0\). Otherwise, reject \(H_1\). Note: Failing to reject \(H_0\) only means inability to accept \(H_1\), and it does not mean to accept \(H_0\).

Assume a normally distributed population has recorded cholesterol levels with various statistics computed. From a sample of 100 subjects in the population, the sample mean was 214.12 mg/dL (milligrams per deciliter), with a sample standard deviation of 45.71 mg/dL. Perform a hypothesis test, with significance level 0.05, to test if there is enough evidence to conclude that the population mean is larger than 200 mg/dL. Hypothesis Test We will perform a hypothesis test using the \(p\)-value approach with significance level \(\alpha=0.05:\) Define \(H_0\): \(\mu=200\). Define \(H_1\): \(\mu>200\). Since our values are normally distributed, the test statistic is \(z^*=\frac{\bar X - \mu_0}{\frac{s}{\sqrt{n}}}=\frac{214.12 - 200}{\frac{45.71}{\sqrt{100}}}\approx 3.09\). Using a standard normal distribution, we find that our \(p\)-value is approximately \(0.001\). Since the \(p\)-value is at most \(\alpha=0.05,\) we reject \(H_0\). Therefore, we can conclude that the test shows sufficient evidence to support the claim that \(\mu\) is larger than \(200\) mg/dL.

If the sample size was smaller, the normal and \(t\)-distributions behave differently. Also, the question itself must be managed by a double-tail test instead.

Assume a population's cholesterol levels are recorded and various statistics are computed. From a sample of 25 subjects, the sample mean was 214.12 mg/dL (milligrams per deciliter), with a sample standard deviation of 45.71 mg/dL. Perform a hypothesis test, with significance level 0.05, to test if there is enough evidence to conclude that the population mean is not equal to 200 mg/dL. Hypothesis Test We will perform a hypothesis test using the \(p\)-value approach with significance level \(\alpha=0.05\) and the \(t\)-distribution with 24 degrees of freedom: Define \(H_0\): \(\mu=200\). Define \(H_1\): \(\mu\neq 200\). Using the \(t\)-distribution, the test statistic is \(t^*=\frac{\bar X - \mu_0}{\frac{s}{\sqrt{n}}}=\frac{214.12 - 200}{\frac{45.71}{\sqrt{25}}}\approx 1.54\). Using a \(t\)-distribution with 24 degrees of freedom, we find that our \(p\)-value is approximately \(2(0.068)=0.136\). We have multiplied by two since this is a two-tailed argument, i.e. the mean can be smaller than or larger than. Since the \(p\)-value is larger than \(\alpha=0.05,\) we fail to reject \(H_0\). Therefore, the test does not show sufficient evidence to support the claim that \(\mu\) is not equal to \(200\) mg/dL.

The complement of the rejection on a two-tailed hypothesis test (with significance level \(\alpha\)) for a population parameter \(\theta\) is equivalent to finding a confidence interval \((\)with confidence level \(1-\alpha)\) for the population parameter \(\theta\). If the assumption on the parameter \(\theta\) falls inside the confidence interval, then the test has failed to reject the null hypothesis \((\)with \(p\)-value greater than \(\alpha).\) Otherwise, if \(\theta\) does not fall in the confidence interval, then the null hypothesis is rejected in favor of the alternate \((\)with \(p\)-value at most \(\alpha).\)

• Statistics (Estimation)
• Normal Distribution
• Correlation
• Confidence Intervals

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A hypothesis is a proposition that is consistent with known data, but has been neither verified nor shown to be false.

In statistics, a hypothesis (sometimes called a statistical hypothesis) refers to a statement on which hypothesis testing will be based. Particularly important statistical hypotheses include the null hypothesis and alternative hypothesis .

In symbolic logic , a hypothesis is the first part of an implication (with the second part being known as the predicate ).

In general mathematical usage, "hypothesis" is roughly synonymous with " conjecture ."

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9.1 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.

H 0 , the — null hypothesis: a statement of no difference between sample means or proportions or no difference between a sample mean or proportion and a population mean or proportion. In other words, the difference equals 0.

H a —, the alternative hypothesis: a claim about the population that is contradictory to H 0 and what we conclude when we reject H 0 .

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.

After you have determined which hypothesis the sample supports, you make a decision. There are two options for a decision. They are reject H 0 if the sample information favors the alternative hypothesis or do not reject H 0 or decline to reject H 0 if the sample information is insufficient to reject the null hypothesis.

Mathematical Symbols Used in H 0 and H a :

H 0 always has a symbol with an equal in it. H a never has a symbol with an equal in it. The choice of symbol depends on the wording of the hypothesis test. However, be aware that many researchers use = in the null hypothesis, even with > or < as the symbol in the alternative hypothesis. This practice is acceptable because we only make the decision to reject or not reject the null hypothesis.

Example 9.1

H 0 : No more than 30 percent of the registered voters in Santa Clara County voted in the primary election. p ≤ 30 H a : More than 30 percent of the registered voters in Santa Clara County voted in the primary election. p > 30

A medical trial is conducted to test whether or not a new medicine reduces cholesterol by 25 percent. State the null and alternative hypotheses.

Example 9.2

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 the following: H 0 : μ = 2.0 H a : μ ≠ 2.0

We want to test whether the mean height of eighth graders is 66 inches. State the null and alternative hypotheses. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.

• H 0 : μ __ 66
• H a : μ __ 66

Example 9.3

We want to test if college students take fewer than five years to graduate from college, on the average. The null and alternative hypotheses are the following: H 0 : μ ≥ 5 H a : μ < 5

We want to test if it takes fewer than 45 minutes to teach a lesson plan. State the null and alternative hypotheses. Fill in the correct symbol ( =, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.

• H 0 : μ __ 45
• H a : μ __ 45

Example 9.4

An article on school standards stated that about half of all students in France, Germany, and Israel take advanced placement exams and a third of the students pass. The same article stated that 6.6 percent of U.S. students take advanced placement exams and 4.4 percent pass. Test if the percentage of U.S. students who take advanced placement exams is more than 6.6 percent. State the null and alternative hypotheses. H 0 : p ≤ 0.066 H a : p > 0.066

On a state driver’s test, about 40 percent pass the test on the first try. We want to test if more than 40 percent pass on the first try. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.

• H 0 : p __ 0.40
• H a : p __ 0.40

Collaborative Exercise

Bring to class a newspaper, some news magazines, and some internet articles. In groups, find articles from which your group can write null and alternative hypotheses. Discuss your hypotheses with the rest of the class.

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Hypothesis test

A significance test, also referred to as a statistical hypothesis test, is a method of statistical inference in which observed data is compared to a claim (referred to as a hypothesis) in order to assess the truth of the claim. For example, one might wonder whether age affects the number of apples a person can eat, and may use a significance test to determine whether there is any evidence to suggest that it does.

Generally, the process of statistical hypothesis testing involves the following steps:

• State the null hypothesis.
• State the alternative hypothesis.
• Select the appropriate test statistic and select a significance level.
• Compute the observed value of the test statistic and its corresponding p-value.
• Reject the null hypothesis in favor of the alternative hypothesis, or do not reject the null hypothesis.

The null hypothesis

The null hypothesis, H 0 , is the claim that is being tested in a statistical hypothesis test. It typically is a statement that there is no difference between the populations being studied, or that there is no evidence to support a claim being made. For example, "age has no effect on the number of apples a person can eat."

A significance test is designed to test the evidence against the null hypothesis. This is because it is easier to prove that a claim is false than to prove that it is true; demonstrating that the claim is false in one case is sufficient, while proving that it is true requires that the claim be true in all cases.

The alternative hypothesis

The alternative hypothesis is the opposite of the null hypothesis in that it is a statement that there is some difference between the populations being studied. For example, "younger people can eat more apples than older people."

The alternative hypothesis is typically the hypothesis that researchers are trying to prove. A significance test is meant to determine whether there is sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis. Note that the results of a significance test should either be to reject the null hypothesis in favor of the alternative hypothesis, or to not reject the null hypothesis. The result should not be to reject the alternative hypothesis or to accept the alternative hypothesis.

Test statistics and significance level

A test statistic is a statistic that is calculated as part of hypothesis testing that compares the distribution of observed data to the expected distribution, based on the null hypothesis. Examples of test statistics include the Z-score, T-statistic, F-statistic, and the Chi-square statistic. The test statistic used is dependent on the significance test used, which is dependent on the type of data collected and the type of relationship to be tested.

In many cases, the chosen significance level is 0.05, though 0.01 is also used. A significance level of 0.05 indicates that there is a 5% chance of rejecting the null hypothesis when the null hypothesis is actually true. Thus, a smaller selected significance level will require more evidence if the null hypothesis is to be rejected in favor of the alternative hypothesis.

After the test statistic is computed, the p-value can be determined based on the result of the test statistic. The p-value indicates the probability of obtaining test results that are at least as extreme as the observed results, under the assumption that the null hypothesis is correct. It tells us how likely it is to obtain a result based solely on chance. The smaller the p-value, the less likely a result can occur purely by chance, while a larger p-value makes it more likely. For example, a p-value of 0.01 means that there is a 1% chance that a result occurred solely by chance, given that the null hypothesis is true; a p-value of 0.90 means that there is a 90% chance.

A p-value is significantly affected by sample size. The larger the sample size, the smaller the p-value, even if the difference between populations may not be meaningful. On the other hand, if a sample size is too small, a meaningful difference may not be detected.

The last step in a significance test is to determine whether the p-value provides evidence that the null hypothesis should be rejected in favor of the alternative hypothesis. This is based on the selected significance level. If the p-value is less than or equal to the selected significance level, the null hypothesis is rejected in favor of the alternative hypothesis, and the result is deemed statistically significant. If the p-value is greater than the selected significance level, the null hypothesis is not rejected, and the result is deemed not statistically significant.

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5.2 - writing hypotheses.

The first step in conducting a hypothesis test is to write the hypothesis statements that are going to be tested. For each test you will have a null hypothesis (\(H_0\)) and an alternative hypothesis (\(H_a\)).

When writing hypotheses there are three things that we need to know: (1) the parameter that we are testing (2) the direction of the test (non-directional, right-tailed or left-tailed), and (3) the value of the hypothesized parameter.

• At this point we can write hypotheses for a single mean (\(\mu\)), paired means(\(\mu_d\)), a single proportion (\(p\)), the difference between two independent means (\(\mu_1-\mu_2\)), the difference between two proportions (\(p_1-p_2\)), a simple linear regression slope (\(\beta\)), and a correlation (\(\rho\)). 
• The research question will give us the information necessary to determine if the test is two-tailed (e.g., "different from," "not equal to"), right-tailed (e.g., "greater than," "more than"), or left-tailed (e.g., "less than," "fewer than").
• The research question will also give us the hypothesized parameter value. This is the number that goes in the hypothesis statements (i.e., \(\mu_0\) and \(p_0\)). For the difference between two groups, regression, and correlation, this value is typically 0.

Hypotheses are always written in terms of population parameters (e.g., \(p\) and \(\mu\)).  The tables below display all of the possible hypotheses for the parameters that we have learned thus far. Note that the null hypothesis always includes the equality (i.e., =).

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• Null and Alternative Hypotheses | Definitions & Examples

Null & Alternative Hypotheses | Definitions, Templates & Examples

Published on May 6, 2022 by Shaun Turney . Revised on June 22, 2023.

The null and alternative hypotheses are two competing claims that researchers weigh evidence for and against using a statistical test :

• Null hypothesis ( H 0 ): There’s no effect in the population .
• Alternative hypothesis ( H a or H 1 ) : There’s an effect in the population.

Table of contents

Answering your research question with hypotheses, what is a null hypothesis, what is an alternative hypothesis, similarities and differences between null and alternative hypotheses, how to write null and alternative hypotheses, other interesting articles, frequently asked questions.

The null and alternative hypotheses offer competing answers to your research question . When the research question asks “Does the independent variable affect the dependent variable?”:

• The null hypothesis ( H 0 ) answers “No, there’s no effect in the population.”
• The alternative hypothesis ( H a ) answers “Yes, there is an effect in the population.”

The null and alternative are always claims about the population. That’s because the goal of hypothesis testing is to make inferences about a population based on a sample . Often, we infer whether there’s an effect in the population by looking at differences between groups or relationships between variables in the sample. It’s critical for your research to write strong hypotheses .

You can use a statistical test to decide whether the evidence favors the null or alternative hypothesis. Each type of statistical test comes with a specific way of phrasing the null and alternative hypothesis. However, the hypotheses can also be phrased in a general way that applies to any test.

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The null hypothesis is the claim that there’s no effect in the population.

If the sample provides enough evidence against the claim that there’s no effect in the population ( p ≤ α), then we can reject the null hypothesis . Otherwise, we fail to reject the null hypothesis.

Although “fail to reject” may sound awkward, it’s the only wording that statisticians accept . Be careful not to say you “prove” or “accept” the null hypothesis.

Null hypotheses often include phrases such as “no effect,” “no difference,” or “no relationship.” When written in mathematical terms, they always include an equality (usually =, but sometimes ≥ or ≤).

You can never know with complete certainty whether there is an effect in the population. Some percentage of the time, your inference about the population will be incorrect. When you incorrectly reject the null hypothesis, it’s called a type I error . When you incorrectly fail to reject it, it’s a type II error.

Examples of null hypotheses

The table below gives examples of research questions and null hypotheses. There’s always more than one way to answer a research question, but these null hypotheses can help you get started.

*Note that some researchers prefer to always write the null hypothesis in terms of “no effect” and “=”. It would be fine to say that daily meditation has no effect on the incidence of depression and p 1 = p 2 .

The alternative hypothesis ( H a ) is the other answer to your research question . It claims that there’s an effect in the population.

Often, your alternative hypothesis is the same as your research hypothesis. In other words, it’s the claim that you expect or hope will be true.

The alternative hypothesis is the complement to the null hypothesis. Null and alternative hypotheses are exhaustive, meaning that together they cover every possible outcome. They are also mutually exclusive, meaning that only one can be true at a time.

Alternative hypotheses often include phrases such as “an effect,” “a difference,” or “a relationship.” When alternative hypotheses are written in mathematical terms, they always include an inequality (usually ≠, but sometimes < or >). As with null hypotheses, there are many acceptable ways to phrase an alternative hypothesis.

Examples of alternative hypotheses

The table below gives examples of research questions and alternative hypotheses to help you get started with formulating your own.

Null and alternative hypotheses are similar in some ways:

• They’re both answers to the research question.
• They both make claims about the population.
• They’re both evaluated by statistical tests.

However, there are important differences between the two types of hypotheses, summarized in the following table.

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To help you write your hypotheses, you can use the template sentences below. If you know which statistical test you’re going to use, you can use the test-specific template sentences. Otherwise, you can use the general template sentences.

General template sentences

The only thing you need to know to use these general template sentences are your dependent and independent variables. To write your research question, null hypothesis, and alternative hypothesis, fill in the following sentences with your variables:

Does independent variable affect dependent variable ?

• Null hypothesis ( H 0 ): Independent variable does not affect dependent variable.
• Alternative hypothesis ( H a ): Independent variable affects dependent variable.

Test-specific template sentences

Once you know the statistical test you’ll be using, you can write your hypotheses in a more precise and mathematical way specific to the test you chose. The table below provides template sentences for common statistical tests.

Note: The template sentences above assume that you’re performing one-tailed tests . One-tailed tests are appropriate for most studies.

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

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

Methodology

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

Research bias

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

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

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

The null hypothesis is often abbreviated as H 0 . When the null hypothesis is written using mathematical symbols, it always includes an equality symbol (usually =, but sometimes ≥ or ≤).

The alternative hypothesis is often abbreviated as H a or H 1 . When the alternative hypothesis is written using mathematical symbols, it always includes an inequality symbol (usually ≠, but sometimes < or >).

A research hypothesis is your proposed answer to your research question. The research hypothesis usually includes an explanation (“ x affects y because …”).

A statistical hypothesis, on the other hand, is a mathematical statement about a population parameter. Statistical hypotheses always come in pairs: the null and alternative hypotheses . In a well-designed study , the statistical hypotheses correspond logically to the research hypothesis.

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15 Hypothesis Examples

A hypothesis is defined as a testable prediction , and is used primarily in scientific experiments as a potential or predicted outcome that scientists attempt to prove or disprove (Atkinson et al., 2021; Tan, 2022).

In my types of hypothesis article, I outlined 13 different hypotheses, including the directional hypothesis (which makes a prediction about an effect of a treatment will be positive or negative) and the associative hypothesis (which makes a prediction about the association between two variables).

This article will dive into some interesting examples of hypotheses and examine potential ways you might test each one.

Hypothesis Examples

1. “inadequate sleep decreases memory retention”.

Field: Psychology

Type: Causal Hypothesis A causal hypothesis explores the effect of one variable on another. This example posits that a lack of adequate sleep causes decreased memory retention. In other words, if you are not getting enough sleep, your ability to remember and recall information may suffer.

How to Test:

To test this hypothesis, you might devise an experiment whereby your participants are divided into two groups: one receives an average of 8 hours of sleep per night for a week, while the other gets less than the recommended sleep amount.

During this time, all participants would daily study and recall new, specific information. You’d then measure memory retention of this information for both groups using standard memory tests and compare the results.

Should the group with less sleep have statistically significant poorer memory scores, the hypothesis would be supported.

Ensuring the integrity of the experiment requires taking into account factors such as individual health differences, stress levels, and daily nutrition.

Relevant Study: Sleep loss, learning capacity and academic performance (Curcio, Ferrara & De Gennaro, 2006)

2. “Increase in Temperature Leads to Increase in Kinetic Energy”

Field: Physics

Type: Deductive Hypothesis The deductive hypothesis applies the logic of deductive reasoning – it moves from a general premise to a more specific conclusion. This specific hypothesis assumes that as temperature increases, the kinetic energy of particles also increases – that is, when you heat something up, its particles move around more rapidly.

This hypothesis could be examined by heating a gas in a controlled environment and capturing the movement of its particles as a function of temperature.

You’d gradually increase the temperature and measure the kinetic energy of the gas particles with each increment. If the kinetic energy consistently rises with the temperature, your hypothesis gets supporting evidence.

Variables such as pressure and volume of the gas would need to be held constant to ensure validity of results.

3. “Children Raised in Bilingual Homes Develop Better Cognitive Skills”

Field: Psychology/Linguistics

Type: Comparative Hypothesis The comparative hypothesis posits a difference between two or more groups based on certain variables. In this context, you might propose that children raised in bilingual homes have superior cognitive skills compared to those raised in monolingual homes.

Testing this hypothesis could involve identifying two groups of children: those raised in bilingual homes, and those raised in monolingual homes.

Cognitive skills in both groups would be evaluated using a standard cognitive ability test at different stages of development. The examination would be repeated over a significant time period for consistency.

If the group raised in bilingual homes persistently scores higher than the other, the hypothesis would thereby be supported.

The challenge for the researcher would be controlling for other variables that could impact cognitive development, such as socio-economic status, education level of parents, and parenting styles.

Relevant Study: The cognitive benefits of being bilingual (Marian & Shook, 2012)

4. “High-Fiber Diet Leads to Lower Incidences of Cardiovascular Diseases”

Field: Medicine/Nutrition

Type: Alternative Hypothesis The alternative hypothesis suggests an alternative to a null hypothesis. In this context, the implied null hypothesis could be that diet has no effect on cardiovascular health, which the alternative hypothesis contradicts by suggesting that a high-fiber diet leads to fewer instances of cardiovascular diseases.

To test this hypothesis, a longitudinal study could be conducted on two groups of participants; one adheres to a high-fiber diet, while the other follows a diet low in fiber.

After a fixed period, the cardiovascular health of participants in both groups could be analyzed and compared. If the group following a high-fiber diet has a lower number of recorded cases of cardiovascular diseases, it would provide evidence supporting the hypothesis.

Control measures should be implemented to exclude the influence of other lifestyle and genetic factors that contribute to cardiovascular health.

Relevant Study: Dietary fiber, inflammation, and cardiovascular disease (King, 2005)

5. “Gravity Influences the Directional Growth of Plants”

Field: Agronomy / Botany

Type: Explanatory Hypothesis An explanatory hypothesis attempts to explain a phenomenon. In this case, the hypothesis proposes that gravity affects how plants direct their growth – both above-ground (toward sunlight) and below-ground (towards water and other resources).

The testing could be conducted by growing plants in a rotating cylinder to create artificial gravity.

Observations on the direction of growth, over a specified period, can provide insights into the influencing factors. If plants consistently direct their growth in a manner that indicates the influence of gravitational pull, the hypothesis is substantiated.

It is crucial to ensure that other growth-influencing factors, such as light and water, are uniformly distributed so that only gravity influences the directional growth.

6. “The Implementation of Gamified Learning Improves Students’ Motivation”

Field: Education

Type: Relational Hypothesis The relational hypothesis describes the relation between two variables. Here, the hypothesis is that the implementation of gamified learning has a positive effect on the motivation of students.

To validate this proposition, two sets of classes could be compared: one that implements a learning approach with game-based elements, and another that follows a traditional learning approach.

The students’ motivation levels could be gauged by monitoring their engagement, performance, and feedback over a considerable timeframe.

If the students engaged in the gamified learning context present higher levels of motivation and achievement, the hypothesis would be supported.

Control measures ought to be put into place to account for individual differences, including prior knowledge and attitudes towards learning.

Relevant Study: Does educational gamification improve students’ motivation? (Chapman & Rich, 2018)

7. “Mathematics Anxiety Negatively Affects Performance”

Field: Educational Psychology

Type: Research Hypothesis The research hypothesis involves making a prediction that will be tested. In this case, the hypothesis proposes that a student’s anxiety about math can negatively influence their performance in math-related tasks.

To assess this hypothesis, researchers must first measure the mathematics anxiety levels of a sample of students using a validated instrument, such as the Mathematics Anxiety Rating Scale.

Then, the students’ performance in mathematics would be evaluated through standard testing. If there’s a negative correlation between the levels of math anxiety and math performance (meaning as anxiety increases, performance decreases), the hypothesis would be supported.

It would be crucial to control for relevant factors such as overall academic performance and previous mathematical achievement.

8. “Disruption of Natural Sleep Cycle Impairs Worker Productivity”

Field: Organizational Psychology

Type: Operational Hypothesis The operational hypothesis involves defining the variables in measurable terms. In this example, the hypothesis posits that disrupting the natural sleep cycle, for instance through shift work or irregular working hours, can lessen productivity among workers.

To test this hypothesis, you could collect data from workers who maintain regular working hours and those with irregular schedules.

Measuring productivity could involve examining the worker’s ability to complete tasks, the quality of their work, and their efficiency.

If workers with interrupted sleep cycles demonstrate lower productivity compared to those with regular sleep patterns, it would lend support to the hypothesis.

Consideration should be given to potential confounding variables such as job type, worker age, and overall health.

9. “Regular Physical Activity Reduces the Risk of Depression”

Field: Health Psychology

Type: Predictive Hypothesis A predictive hypothesis involves making a prediction about the outcome of a study based on the observed relationship between variables. In this case, it is hypothesized that individuals who engage in regular physical activity are less likely to suffer from depression.

Longitudinal studies would suit to test this hypothesis, tracking participants’ levels of physical activity and their mental health status over time.

The level of physical activity could be self-reported or monitored, while mental health status could be assessed using standard diagnostic tools or surveys.

If data analysis shows that participants maintaining regular physical activity have a lower incidence of depression, this would endorse the hypothesis.

However, care should be taken to control other lifestyle and behavioral factors that could intervene with the results.

Relevant Study: Regular physical exercise and its association with depression (Kim, 2022)

10. “Regular Meditation Enhances Emotional Stability”

Type: Empirical Hypothesis In the empirical hypothesis, predictions are based on amassed empirical evidence . This particular hypothesis theorizes that frequent meditation leads to improved emotional stability, resonating with numerous studies linking meditation to a variety of psychological benefits.

Earlier studies reported some correlations, but to test this hypothesis directly, you’d organize an experiment where one group meditates regularly over a set period while a control group doesn’t.

Both groups’ emotional stability levels would be measured at the start and end of the experiment using a validated emotional stability assessment.

If regular meditators display noticeable improvements in emotional stability compared to the control group, the hypothesis gains credit.

You’d have to ensure a similar emotional baseline for all participants at the start to avoid skewed results.

11. “Children Exposed to Reading at an Early Age Show Superior Academic Progress”

Type: Directional Hypothesis The directional hypothesis predicts the direction of an expected relationship between variables. Here, the hypothesis anticipates that early exposure to reading positively affects a child’s academic advancement.

A longitudinal study tracking children’s reading habits from an early age and their consequent academic performance could validate this hypothesis.

Parents could report their children’s exposure to reading at home, while standardized school exam results would provide a measure of academic achievement.

If the children exposed to early reading consistently perform better acadically, it gives weight to the hypothesis.

However, it would be important to control for variables that might impact academic performance, such as socioeconomic background, parental education level, and school quality.

12. “Adopting Energy-efficient Technologies Reduces Carbon Footprint of Industries”

Field: Environmental Science

Type: Descriptive Hypothesis A descriptive hypothesis predicts the existence of an association or pattern related to variables. In this scenario, the hypothesis suggests that industries adopting energy-efficient technologies will resultantly show a reduced carbon footprint.

Global industries making use of energy-efficient technologies could track their carbon emissions over time. At the same time, others not implementing such technologies continue their regular tracking.

After a defined time, the carbon emission data of both groups could be compared. If industries that adopted energy-efficient technologies demonstrate a notable reduction in their carbon footprints, the hypothesis would hold strong.

In the experiment, you would exclude variations brought by factors such as industry type, size, and location.

13. “Reduced Screen Time Improves Sleep Quality”

Type: Simple Hypothesis The simple hypothesis is a prediction about the relationship between two variables, excluding any other variables from consideration. This example posits that by reducing time spent on devices like smartphones and computers, an individual should experience improved sleep quality.

A sample group would need to reduce their daily screen time for a pre-determined period. Sleep quality before and after the reduction could be measured using self-report sleep diaries and objective measures like actigraphy, monitoring movement and wakefulness during sleep.

If the data shows that sleep quality improved post the screen time reduction, the hypothesis would be validated.

Other aspects affecting sleep quality, like caffeine intake, should be controlled during the experiment.

Relevant Study: Screen time use impacts low‐income preschool children’s sleep quality, tiredness, and ability to fall asleep (Waller et al., 2021)

14. Engaging in Brain-Training Games Improves Cognitive Functioning in Elderly

Field: Gerontology

Type: Inductive Hypothesis Inductive hypotheses are based on observations leading to broader generalizations and theories. In this context, the hypothesis deduces from observed instances that engaging in brain-training games can help improve cognitive functioning in the elderly.

A longitudinal study could be conducted where an experimental group of elderly people partakes in regular brain-training games.

Their cognitive functioning could be assessed at the start of the study and at regular intervals using standard neuropsychological tests.

If the group engaging in brain-training games shows better cognitive functioning scores over time compared to a control group not playing these games, the hypothesis would be supported.

15. Farming Practices Influence Soil Erosion Rates

Type: Null Hypothesis A null hypothesis is a negative statement assuming no relationship or difference between variables. The hypothesis in this context asserts there’s no effect of different farming practices on the rates of soil erosion.

Comparing soil erosion rates in areas with different farming practices over a considerable timeframe could help test this hypothesis.

If, statistically, the farming practices do not lead to differences in soil erosion rates, the null hypothesis is accepted.

However, if marked variation appears, the null hypothesis is rejected, meaning farming practices do influence soil erosion rates. It would be crucial to control for external factors like weather, soil type, and natural vegetation.

The variety of hypotheses mentioned above underscores the diversity of research constructs inherent in different fields, each with its unique purpose and way of testing.

While researchers may develop hypotheses primarily as tools to define and narrow the focus of the study, these hypotheses also serve as valuable guiding forces for the data collection and analysis procedures, making the research process more efficient and direction-focused.

Hypotheses serve as a compass for any form of academic research. The diverse examples provided, from Psychology to Educational Studies, Environmental Science to Gerontology, clearly demonstrate how certain hypotheses suit specific fields more aptly than others.

It is important to underline that although these varied hypotheses differ in their structure and methods of testing, each endorses the fundamental value of empiricism in research. Evidence-based decision making remains at the heart of scholarly inquiry, regardless of the research field, thus aligning all hypotheses to the core purpose of scientific investigation.

Testing hypotheses is an essential part of the scientific method . By doing so, researchers can either confirm their predictions, giving further validity to an existing theory, or they might uncover new insights that could potentially shift the field’s understanding of a particular phenomenon. In either case, hypotheses serve as the stepping stones for scientific exploration and discovery.

Atkinson, P., Delamont, S., Cernat, A., Sakshaug, J. W., & Williams, R. A. (2021).  SAGE research methods foundations . SAGE Publications Ltd.

Curcio, G., Ferrara, M., & De Gennaro, L. (2006). Sleep loss, learning capacity and academic performance.  Sleep medicine reviews ,  10 (5), 323-337.

Kim, J. H. (2022). Regular physical exercise and its association with depression: A population-based study short title: Exercise and depression.  Psychiatry Research ,  309 , 114406.

King, D. E. (2005). Dietary fiber, inflammation, and cardiovascular disease.  Molecular nutrition & food research ,  49 (6), 594-600.

Marian, V., & Shook, A. (2012, September). The cognitive benefits of being bilingual. In Cerebrum: the Dana forum on brain science (Vol. 2012). Dana Foundation.

Tan, W. C. K. (2022). Research Methods: A Practical Guide For Students And Researchers (Second Edition) . World Scientific Publishing Company.

Waller, N. A., Zhang, N., Cocci, A. H., D’Agostino, C., Wesolek‐Greenson, S., Wheelock, K., … & Resnicow, K. (2021). Screen time use impacts low‐income preschool children’s sleep quality, tiredness, and ability to fall asleep. Child: care, health and development, 47 (5), 618-626.

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What Are Examples of a Hypothesis?

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A hypothesis is an explanation for a set of observations. Here are examples of a scientific hypothesis.

Although you could state a scientific hypothesis in various ways, most hypotheses are either "If, then" statements or forms of the null hypothesis . The null hypothesis is sometimes called the "no difference" hypothesis. The null hypothesis is good for experimentation because it's simple to disprove. If you disprove a null hypothesis, that is evidence for a relationship between the variables you are examining.

Examples of Null Hypotheses

• Hyperactivity is unrelated to eating sugar.
• All daisies have the same number of petals.
• The number of pets in a household is unrelated to the number of people living in it.
• A person's preference for a shirt is unrelated to its color.

Examples of If, Then Hypotheses

• If you get at least 6 hours of sleep, you will do better on tests than if you get less sleep.
• If you drop a ball, it will fall toward the ground.
• If you drink coffee before going to bed, then it will take longer to fall asleep.
• If you cover a wound with a bandage, then it will heal with less scarring.

Improving a Hypothesis to Make It Testable

You may wish to revise your first hypothesis in order to make it easier to design an experiment to test. For example, let's say you have a bad breakout the morning after eating a lot of greasy food. You may wonder if there is a correlation between eating greasy food and getting pimples. You propose the hypothesis:

Eating greasy food causes pimples.

Next, you need to design an experiment to test this hypothesis. Let's say you decide to eat greasy food every day for a week and record the effect on your face. Then, as a control, you'll avoid greasy food for the next week and see what happens. Now, this is not a good experiment because it does not take into account other factors such as hormone levels, stress, sun exposure, exercise, or any number of other variables that might conceivably affect your skin.

The problem is that you cannot assign cause to your effect . If you eat french fries for a week and suffer a breakout, can you definitely say it was the grease in the food that caused it? Maybe it was the salt. Maybe it was the potato. Maybe it was unrelated to diet. You can't prove your hypothesis. It's much easier to disprove a hypothesis.

So, let's restate the hypothesis to make it easier to evaluate the data:

Getting pimples is unaffected by eating greasy food.

So, if you eat fatty food every day for a week and suffer breakouts and then don't break out the week that you avoid greasy food, you can be pretty sure something is up. Can you disprove the hypothesis? Probably not, since it is so hard to assign cause and effect. However, you can make a strong case that there is some relationship between diet and acne.

If your skin stays clear for the entire test, you may decide to accept your hypothesis . Again, you didn't prove or disprove anything, which is fine

• Null Hypothesis Definition and Examples
• What Is a Hypothesis? (Science)
• What Are the Elements of a Good Hypothesis?
• Understanding Simple vs Controlled Experiments
• What Is a Testable Hypothesis?
• What 'Fail to Reject' Means in a Hypothesis Test
• Null Hypothesis Examples
• How To Design a Science Fair Experiment
• Scientific Method Vocabulary Terms
• Scientific Hypothesis Examples
• Six Steps of the Scientific Method
• An Example of a Hypothesis Test
• Definition of a Hypothesis
• Scientific Method Flow Chart
• Null Hypothesis and Alternative Hypothesis

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9.5: Additional Information and Full Hypothesis Test Examples

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• In a hypothesis test problem, you may see words such as "the level of significance is 1%." The "1%" is the preconceived or preset \(\alpha\).
• The statistician setting up the hypothesis test selects the value of α to use before collecting the sample data.
• If no level of significance is given, a common standard to use is \(\alpha = 0.05\).
• When you calculate the \(p\)-value and draw the picture, the \(p\)-value is the area in the left tail, the right tail, or split evenly between the two tails. For this reason, we call the hypothesis test left, right, or two tailed.
• The alternative hypothesis, \(H_{a}\), tells you if the test is left, right, or two-tailed. It is the key to conducting the appropriate test.
• \(H_{a}\) never has a symbol that contains an equal sign.
• Thinking about the meaning of the \(p\)-value: A data analyst (and anyone else) should have more confidence that he made the correct decision to reject the null hypothesis with a smaller \(p\)-value (for example, 0.001 as opposed to 0.04) even if using the 0.05 level for alpha. Similarly, for a large p -value such as 0.4, as opposed to a \(p\)-value of 0.056 (\(\alpha = 0.05\) is less than either number), a data analyst should have more confidence that she made the correct decision in not rejecting the null hypothesis. This makes the data analyst use judgment rather than mindlessly applying rules.

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 \(\pageindex{4}\).

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. Assume that the swim times for the 25-yard freestyle are normal.

Set up the Hypothesis Test:

Since the problem is about a mean, this is a test of a single population mean .

\(H_{0}: \mu = 16.43, H_{a}: \mu < 16.43\)

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:

Random variable: \(\bar{X} =\) the mean time to swim the 25-yard freestyle.

Distribution for the test: \(\bar{X}\) is normal (population standard deviation is known: \(\sigma = 0.8\))

\(\bar{X} - N \left(\mu, \frac{\sigma_{x}}{\sqrt{n}}\right)\) Therefore, \(\bar{X} - N\left(16.43, \frac{0.8}{\sqrt{15}}\right)\)

\(\mu = 16.43\) comes from \(H_{0}\) and not the data. \(\sigma = 0.8\), and \(n = 15\).

Calculate the \(p-\text{value}\) using the normal distribution for a mean:

\(p\text{-value} = P(\bar{x} < 16) = 0.0187\) where the sample mean in the problem is given as 16.

\(p\text{-value} = 0.0187\) (This is called the actual level of significance .) The \(p-\text{value}\) is the area to the left of the sample mean is given as 16.

\(\mu = 16.43\) comes from \(H_{0}\). Our assumption is \(\mu = 16.43\).

Interpretation of the \(p-\text{value}\): If \(H_{0}\) is true , there is a 0.0187 probability (1.87%) that Jeffrey's mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87% chance is small, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event.

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

\(\alpha = 0.05 p\text{-value} = 0.0187 \alpha > p\text{-value}\)

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

This means that you reject \(\mu = 16.43\). In other words, you do not think Jeffrey swims the 25-yard freestyle in 16.43 seconds but faster with the new goggles.

Conclusion: At the 5% significance level, we conclude that Jeffrey swims faster using the new goggles. The sample data show there is sufficient evidence that Jeffrey's mean time to swim the 25-yard freestyle is less than 16.43 seconds.

The p -value can easily be calculated.

Press STAT and arrow over to TESTS . Press 1:Z-Test . Arrow over to Stats and press ENTER . Arrow down and enter 16.43 for \(\mu_{0}\) (null hypothesis), .8 for σ , 16 for the sample mean, and 15 for n . Arrow down to \(\mu\) : (alternate hypothesis) and arrow over to \(< \mu_{0}\). Press ENTER . Arrow down to Calculate and press ENTER . The calculator not only calculates the p -value (\(p = 0.0187\)) but it also calculates the test statistic ( z -score) for the sample mean. \(\mu < 16.43\) is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with \(z = -2.08\) (test statistic) and \(p = 0.0187\) (\(p-\text{value}\)). Make sure when you use Draw that no other equations are highlighted in \(Y =\) and the plots are turned off.

When the calculator does a \(Z\)-Test, the Z-Test function finds the p -value by doing a normal probability calculation using the central limit theorem:

\(P(\bar{X} < 16)\) 2nd DISTR normcdf (\((−10^{99},16,16.43,\frac{0.8}{\sqrt{15}})\).

The Type I and Type II errors for this problem are as follows:

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.)

The Type II error is that there is not evidence to conclude that Jeffrey swims the 25-yard free-style, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25-yard free-style, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.)

Exercise \(\PageIndex{4}\)

The mean throwing distance of a football for a Marco, a high school freshman quarterback, is 40 yards, with a standard deviation of two yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, Marco’s mean distance was 45 yards. The coach thought the different grip helped Marco throw farther than 40 yards. Conduct a hypothesis test using a preset \(\alpha = 0.05\). Assume the throw distances for footballs are normal.

First, determine what type of test this is, set up the hypothesis test, find the p -value, sketch the graph, and state your conclusion.

Press STAT and arrow over to TESTS. Press 1: \(Z\)-Test. Arrow over to Stats and press ENTER. Arrow down and enter 40 for \(\mu_{0}\) (null hypothesis), 2 for \(\sigma\), 45 for the sample mean, and 20 for \(n\). Arrow down to \(\mu\): (alternative hypothesis) and set it either as \(<\), \(\neq\), or \(>\). Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p -value but it also calculates the test statistic ( z -score) for the sample mean. Select \(<\), \(\neq\), or \(>\) for the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with test statistic and \(p\)-value. Make sure when you use Draw that no other equations are highlighted in \(Y =\) and the plots are turned off.

Since the problem is about a mean, this is a test of a single population mean.

• \(H_{0}: \mu = 40\)
• \(H_{a}: \mu > 40\)
• \(p = 0.0062\)

Because \(p < \alpha\), we reject the null hypothesis. There is sufficient evidence to suggest that the change in grip improved Marco’s throwing distance.

Historical Note

The traditional way to compare the two probabilities, \(\alpha\) and the \(p-\text{value}\), is to compare the critical value (\(z\)-score from \(\alpha\)) to the test statistic (\(z\)-score from data). The calculated test statistic for the \(p\)-value is –2.08. (From the Central Limit Theorem, the test statistic formula is \(z = \frac{\bar{x}-\mu_{x}}{\left(\frac{\sigma_{x}}{\sqrt{n}}\right)}\). For this problem, \(\bar{x} = 16\), \(\mu_{x} = 16.43\) from the null hypotheses is, \(\sigma_{x} = 0.8\), and \(n = 15\).) You can find the critical value for \(\alpha = 0.05\) in the normal table (see 15.Tables in the Table of Contents). The \(z\)-score for an area to the left equal to 0.05 is midway between –1.65 and –1.64 (0.05 is midway between 0.0505 and 0.0495). The \(z\)-score is –1.645. Since –1.645 > –2.08 (which demonstrates that \(\alpha > p-\text{value}\)), reject \(H_{0}\). Traditionally, the decision to reject or not reject was done in this way. Today, comparing the two probabilities \(\alpha\) and the \(p\)-value is very common. For this problem, the \(p-\text{value}\), 0.0187 is considerably smaller than \(\alpha = 0.05\). You can be confident about your decision to reject. The graph shows \(\alpha\), the \(p-\text{value}\), and the test statistics and the critical value.

Example \(\PageIndex{5}\)

A college football coach thought that his players could bench press a mean weight of 275 pounds . It is known that the standard deviation is 55 pounds . Three of his players thought that the mean weight was more than that amount. They asked 30 of their teammates for their estimated maximum lift on the bench press exercise. The data ranged from 205 pounds to 385 pounds. The actual different weights were (frequencies are in parentheses) 205(3); 215(3); 225(1); 241(2); 252(2); 265(2); 275(2); 313(2); 316(5); 338(2); 341(1); 345(2); 368(2); 385(1).

Conduct a hypothesis test using a 2.5% level of significance to determine if the bench press mean is more than 275 pounds.

Since the problem is about a mean weight, this is a test of a single population mean.

• \(H_{0}: \mu = 275\)
• \(H_{a}: \mu > 275\)

This is a right-tailed test.

Calculating the distribution needed:

Random variable: \(\bar{X} =\) the mean weight, in pounds, lifted by the football players.

Distribution for the test: It is normal because \(\sigma\) is known.

• \(\bar{X} - N\left(275, \frac{55}{\sqrt{30}}\right)\)
• \(\bar{x} = 286.2\) pounds (from the data).
• \(\sigma = 55\) pounds (Always use \(\sigma\) if you know it.) We assume \(\mu = 275\) pounds unless our data shows us otherwise.

Calculate the p -value using the normal distribution for a mean and using the sample mean as input (see [link] for using the data as input):

\[p\text{-value} = P(\bar{x} > 286.2) = 0.1323.\nonumber \]

Interpretation of the p -value: If \(H_{0}\) is true, then there is a 0.1331 probability (13.23%) that the football players can lift a mean weight of 286.2 pounds or more. Because a 13.23% chance is large enough, a mean weight lift of 286.2 pounds or more is not a rare event.

\(\alpha = 0.025 p-value = 0.1323\)

Make a decision: Since \(\alpha < p\text{-value}\), do not reject \(H_{0}\).

Conclusion: At the 2.5% level of significance, from the sample data, there is not sufficient evidence to conclude that the true mean weight lifted is more than 275 pounds.

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

Put the data and frequencies into lists. Press STAT and arrow over to TESTS . Press 1:Z-Test . Arrow over to Data and press ENTER . Arrow down and enter 275 for \(\mu_{0}\), 55 for \(\sigma\), the name of the list where you put the data, and the name of the list where you put the frequencies. 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 \(p-\text{value}\) (\(p = 0.1331\)), a little different from the previous calculation - in it we used the sample mean rounded to one decimal place instead of the data) but it also calculates the test statistic ( z -score) for the sample mean, the sample mean, and the sample standard deviation. \(\mu > 275\) is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with \(z = 1.112\) (test statistic) and \(p = 0.1331\) (\(p-\text{value})\). Make sure when you use Draw that no other equations are highlighted in \(Y =\) and the plots are turned off.

Example \(\PageIndex{6}\)

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.

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 \(df = 10 - 1 = 9\).

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

\(p\text{-value} = P(\bar{x} > 67) = 0.0396\) where the sample mean and sample standard deviation are calculated as 67 and 3.1972 from the data.

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.

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

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 \(\PageIndex{6}\)

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 \(\PageIndex{7}\)

Joon believes that 50% of first-time brides in the United States are younger than their grooms. She performs a hypothesis test to determine if the percentage is the same or different from 50% . Joon samples 100 first-time brides and 53 reply that they are younger than their grooms. For the hypothesis test, she uses a 1% level of significance.

The 1% level of significance means that α = 0.01. This is a test of a single population proportion .

\(H_{0}: p = 0.50\)  \(H_{a}: p \neq 0.50\)

The words "is the same or different from" tell you this is a two-tailed test.

Calculate the distribution needed:

Random variable: \(P′ =\) the percent of of first-time brides who are younger than their grooms.

Distribution for the test: The problem contains no mention of a mean. The information is given in terms of percentages. Use the distribution for P′ , the estimated proportion.

\[P' - N\left(p, \sqrt{\frac{p-q}{n}}\right)\nonumber \]

\[P' - N\left(0.5, \sqrt{\frac{0.5-0.5}{100}}\right)\nonumber \]

where \(p = 0.50, q = 1−p = 0.50\), and \(n = 100\)

Calculate the p -value using the normal distribution for proportions:

\[p\text{-value} = P(p′ < 0.47 \space or \space p′ > 0.53) = 0.5485\nonumber \]

where \[x = 53, p' = \frac{x}{n} = \frac{53}{100} = 0.53\nonumber \].

Interpretation of the p-value: If the null hypothesis is true, there is 0.5485 probability (54.85%) that the sample (estimated) proportion \(p'\) is 0.53 or more OR 0.47 or less (see the graph in Figure).

\(\mu = p = 0.50\) comes from \(H_{0}\), the null hypothesis.

\(p′ = 0.53\). Since the curve is symmetrical and the test is two-tailed, the \(p′\) for the left tail is equal to \(0.50 – 0.03 = 0.47\) where \(\mu = p = 0.50\). (0.03 is the difference between 0.53 and 0.50.)

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

Since \(\alpha = 0.01\) and \(p\text{-value} = 0.5485\). \(\alpha < p\text{-value}\).

Make a decision: Since \(\alpha < p\text{-value}\), you cannot reject \(H_{0}\).

Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of first-time brides who are younger than their grooms is different from 50%.

Press STAT and arrow over to TESTS . Press 5:1-PropZTest . Enter .5 for \(p_{0}\), 53 for \(x\) and 100 for \(n\). Arrow down to Prop and arrow to not equals \(p_{0}\). Press ENTER . Arrow down to Calculate and press ENTER . The calculator calculates the \(p\text{-value}\) (\(p = 0.5485\)) and the test statistic (\(z\)-score). Prop not equals .5 is the alternate hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with \(z = 0.6\) (test statistic) and \(p = 0.5485\) (\(p\text{-value}\)). Make sure when you use Draw that no other equations are highlighted in \(Y =\) and the plots are turned off.

The Type I and Type II errors are as follows:

The Type I error is to conclude that the proportion of first-time brides who are younger than their grooms is different from 50% when, in fact, the proportion is actually 50%. (Reject the null hypothesis when the null hypothesis is true).

The Type II error is there is not enough evidence to conclude that the proportion of first time brides who are younger than their grooms differs from 50% when, in fact, the proportion does differ from 50%. (Do not reject the null hypothesis when the null hypothesis is false.)

Exercise \(\PageIndex{7}\)

A teacher believes that 85% of students in the class will want to go on a field trip to the local zoo. She performs a hypothesis test to determine if the percentage is the same or different from 85%. The teacher samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1% level of significance.

First, determine what type of test this is, set up the hypothesis test, find the \(p\text{-value}\), sketch the graph, and state your conclusion.

Since the problem is about percentages, this is a test of single population proportions.

• \(H_{0} : p = 0.85\)
• \(H_{a}: p \neq 0.85\)
• \(p = 0.7554\)

Because \(p > \alpha\), we fail to reject the null hypothesis. There is not sufficient evidence to suggest that the proportion of students that want to go to the zoo is not 85%.

Example \(\PageIndex{8}\)

Suppose a consumer group suspects that the proportion of households that have three cell phones is 30%. A cell phone company has reason to believe that the proportion is not 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three cell phones.

\(H_{0}: p = 0.30, H_{a}: p \neq 0.30\)

The random variable is \(P′ =\) proportion of households that have three cell phones.

The distribution for the hypothesis test is \(P' - N\left(0.30, \sqrt{\frac{(0.30 \cdot 0.70)}{150}}\right)\)

Exercise \(\PageIndex{8}\).2

a. The value that helps determine the \(p\text{-value}\) is \(p′\). Calculate \(p′\).

a. \(p' = \frac{x}{n}\) where \(x\) is the number of successes and \(n\) is the total number in the sample.

\(x = 43, n = 150\)

\(p′ = 43150\)

Exercise \(\PageIndex{8}\).3

b. What is a success for this problem?

b. A success is having three cell phones in a household.

Exercise \(\PageIndex{8}\).4

c. What is the level of significance?

c. The level of significance is the preset \(\alpha\). Since \(\alpha\) is not given, assume that \(\alpha = 0.05\).

Exercise \(\PageIndex{8}\).5

d. Draw the graph for this problem. Draw the horizontal axis. Label and shade appropriately.

Calculate the \(p\text{-value}\).

d. \(p\text{-value} = 0.7216\)

Exercise \(\PageIndex{8}\).6

e. Make a decision. _____________(Reject/Do not reject) \(H_{0}\) because____________.

e. Assuming that \(\alpha = 0.05, \alpha < p\text{-value}\). The decision is do not reject \(H_{0}\) because there is not sufficient evidence to conclude that the proportion of households that have three cell phones is not 30%.

Exercise \(\PageIndex{8}\)

Marketers believe that 92% of adults in the United States own a cell phone. A cell phone manufacturer believes that number is actually lower. 200 American adults are surveyed, of which, 174 report having cell phones. Use a 5% level of significance. State the null and alternative hypothesis, find the p -value, state your conclusion, and identify the Type I and Type II errors.

• \(H_{0}: p = 0.92\)
• \(H_{a}: p < 0.92\)
• \(p\text{-value} = 0.0046\)

Because \(p < 0.05\), we reject the null hypothesis. There is sufficient evidence to conclude that fewer than 92% of American adults own cell phones.

• Type I Error: To conclude that fewer than 92% of American adults own cell phones when, in fact, 92% of American adults do own cell phones (reject the null hypothesis when the null hypothesis is true).
• Type II Error: To conclude that 92% of American adults own cell phones when, in fact, fewer than 92% of American adults own cell phones (do not reject the null hypothesis when the null hypothesis is false).

The next example is a poem written by a statistics student named Nicole Hart. The solution to the problem follows the poem. Notice that the hypothesis test is for a single population proportion. This means that the null and alternate hypotheses use the parameter \(p\). The distribution for the test is normal. The estimated proportion \(p′\) is the proportion of fleas killed to the total fleas found on Fido. This is sample information. The problem gives a preconceived \(\alpha = 0.01\), for comparison, and a 95% confidence interval computation. The poem is clever and humorous, so please enjoy it!

Example \(\PageIndex{9}\)

My dog has so many fleas,

They do not come off with ease. As for shampoo, I have tried many types Even one called Bubble Hype, Which only killed 25% of the fleas, Unfortunately I was not pleased.

I've used all kinds of soap, Until I had given up hope Until one day I saw An ad that put me in awe.

A shampoo used for dogs Called GOOD ENOUGH to Clean a Hog Guaranteed to kill more fleas.

I gave Fido a bath And after doing the math His number of fleas Started dropping by 3's! Before his shampoo I counted 42.

At the end of his bath, I redid the math And the new shampoo had killed 17 fleas. So now I was pleased.

Now it is time for you to have some fun With the level of significance being .01, You must help me figure out

Use the new shampoo or go without?

\(H_{0}: p \leq 0.25\)   \(H_{a}: p > 0.25\)

In words, CLEARLY state what your random variable \(\bar{X}\) or \(P′\) represents.

\(P′ =\) The proportion of fleas that are killed by the new shampoo

State the distribution to use for the test.

\[N\left(0.25, \sqrt{\frac{(0.25){1-0.25}}{42}}\right)\nonumber \]

Test Statistic: \(z = 2.3163\)

Calculate the \(p\text{-value}\) using the normal distribution for proportions:

\[p\text{-value} = 0.0103\nonumber \]

In one to two complete sentences, explain what the p -value means for this problem.

If the null hypothesis is true (the proportion is 0.25), then there is a 0.0103 probability that the sample (estimated) proportion is 0.4048 \(\left(\frac{17}{42}\right)\) or more.

Use the previous information to sketch a picture of this situation. CLEARLY, label and scale the horizontal axis and shade the region(s) corresponding to the \(p\text{-value}\).

Indicate the correct decision (“reject” or “do not reject” the null hypothesis), the reason for it, and write an appropriate conclusion, using complete sentences.

Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of fleas that are killed by the new shampoo is more than 25%.

Construct a 95% confidence interval for the true mean or proportion. Include a sketch of the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval.

Confidence Interval: (0.26,0.55) We are 95% confident that the true population proportion p of fleas that are killed by the new shampoo is between 26% and 55%.

This test result is not very definitive since the \(p\text{-value}\) is very close to alpha. In reality, one would probably do more tests by giving the dog another bath after the fleas have had a chance to return.

Example \(\PageIndex{10}\)

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 : We will input the sample data into the TI-83 as follows.

4. State the Conclusions : Since the \(p\text{-value} (p = 0.036)\) 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.

Example \(\PageIndex{11}\)

In a study of 420,019 cell phone users, 172 of the subjects developed brain cancer. Test the claim that cell phone users developed brain cancer at a greater rate than that for non-cell phone users (the rate of brain cancer for non-cell phone users is 0.0340%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.

We will follow the four-step process.

• \(H_{0}: p \leq 0.00034\)
• \(H_{a}: p > 0.00034\)

If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes cancer-causing environments, we want to minimize the chances of incorrectly identifying causes of cancer.

• We will be testing a sample proportion with \(x = 172\) and \(n = 420,019\). The sample is sufficiently large because we have \(np = 420,019(0.00034) = 142.8\), \(nq = 420,019(0.99966) = 419,876.2\), two independent outcomes, and a fixed probability of success \(p = 0.00034\). Thus we will be able to generalize our results to the population.

Figure \(\PageIndex{11}\).

Figure \(\PageIndex{12}\).

• Since the \(p\text{-value} = 0.0073\) is greater than our alpha value \(= 0.005\), we cannot reject the null. Therefore, we conclude that there is not enough evidence to support the claim of higher brain cancer rates for the cell phone users.

Example \(\PageIndex{12}\)

According to the US Census there are approximately 268,608,618 residents aged 12 and older. Statistics from the Rape, Abuse, and Incest National Network indicate that, on average, 207,754 rapes occur each year (male and female) for persons aged 12 and older. This translates into a percentage of sexual assaults of 0.078%. In Daviess County, KY, there were reported 11 rapes for a population of 37,937. Conduct an appropriate hypothesis test to determine if there is a statistically significant difference between the local sexual assault percentage and the national sexual assault percentage. Use a significance level of 0.01.

We will follow the four-step plan.

• We need to test whether the proportion of sexual assaults in Daviess County, KY is significantly different from the national average.
• \(H_{0}: p = 0.00078\)
• \(H_{a}: p \neq 0.00078\)

Figure \(\PageIndex{13}\).

Figure \(\PageIndex{14}\).

• Since the \(p\text{-value}\), \(p = 0.00063\), is less than the alpha level of 0.01, the sample data indicates that we should reject the null hypothesis. In conclusion, the sample data support the claim that the proportion of sexual assaults in Daviess County, Kentucky is different from the national average proportion.

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 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 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.

Assume \(H_{0}: \mu = 9\) and \(H_{a}: \mu < 9\). Is this a left-tailed, right-tailed, or two-tailed test?

This is a left-tailed test.

Exercise \(\PageIndex{9}\)

Assume \(H_{0}: \mu \leq 6\) and \(H_{a}: \mu > 6\). Is this a left-tailed, right-tailed, or two-tailed test?

Exercise \(\PageIndex{10}\)

Assume \(H_{0}: p = 0.25\) and \(H_{a}: p \neq 0.25\). Is this a left-tailed, right-tailed, or two-tailed test?

This is a two-tailed test.

Exercise \(\PageIndex{11}\)

Draw the general graph of a left-tailed test.

Exercise \(\PageIndex{12}\)

Draw the graph of a two-tailed test.

Exercise \(\PageIndex{13}\)

A bottle of water is labeled as containing 16 fluid ounces of water. You believe it is less than that. What type of test would you use?

Exercise \(\PageIndex{14}\)

Your friend claims that his mean golf score is 63. You want to show that it is higher than that. What type of test would you use?

a right-tailed test

Exercise \(\PageIndex{15}\)

A bathroom scale claims to be able to identify correctly any weight within a pound. You think that it cannot be that accurate. What type of test would you use?

Exercise \(\PageIndex{16}\)

You flip a coin and record whether it shows heads or tails. You know the probability of getting heads is 50%, but you think it is less for this particular coin. What type of test would you use?

a left-tailed test

Exercise \(\PageIndex{17}\)

If the alternative hypothesis has a not equals ( \(\neq\) ) symbol, you know to use which type of test?

Exercise \(\PageIndex{18}\)

Assume the null hypothesis states that the mean is at least 18. Is this a left-tailed, right-tailed, or two-tailed test?

Exercise \(\PageIndex{19}\)

Assume the null hypothesis states that the mean is at most 12. Is this a left-tailed, right-tailed, or two-tailed test?

Exercise \(\PageIndex{20}\)

Assume the null hypothesis states that the mean is equal to 88. The alternative hypothesis states that the mean is not equal to 88. Is this a left-tailed, right-tailed, or two-tailed test?

• Data from Amit Schitai. Director of Instructional Technology and Distance Learning. LBCC.
• Data from Bloomberg Businessweek . Available online at www.businessweek.com/news/2011- 09-15/nyc-smoking-rate-falls-to-record-low-of-14-bloomberg-says.html.
• Data from energy.gov. Available online at http://energy.gov (accessed June 27. 2013).
• Data from Gallup®. Available online at www.gallup.com (accessed June 27, 2013).
• Data from Growing by Degrees by Allen and Seaman.
• Data from La Leche League International. Available online at www.lalecheleague.org/Law/BAFeb01.html.
• Data from the American Automobile Association. Available online at www.aaa.com (accessed June 27, 2013).
• Data from the American Library Association. Available online at www.ala.org (accessed June 27, 2013).
• Data from the Bureau of Labor Statistics. Available online at http://www.bls.gov/oes/current/oes291111.htm .
• Data from the Centers for Disease Control and Prevention. Available online at www.cdc.gov (accessed June 27, 2013)
• Data from the U.S. Census Bureau, available online at quickfacts.census.gov/qfd/states/00000.html (accessed June 27, 2013).
• Data from the United States Census Bureau. Available online at www.census.gov/hhes/socdemo/language/.
• Data from Toastmasters International. Available online at http://toastmasters.org/artisan/deta...eID=429&Page=1 .
• Data from Weather Underground. Available online at www.wunderground.com (accessed June 27, 2013).
• Federal Bureau of Investigations. “Uniform Crime Reports and Index of Crime in Daviess in the State of Kentucky enforced by Daviess County from 1985 to 2005.” Available online at http://www.disastercenter.com/kentucky/crime/3868.htm (accessed June 27, 2013).
• “Foothill-De Anza Community College District.” De Anza College, Winter 2006. Available online at research.fhda.edu/factbook/DA...t_da_2006w.pdf.
• Johansen, C., J. Boice, Jr., J. McLaughlin, J. Olsen. “Cellular Telephones and Cancer—a Nationwide Cohort Study in Denmark.” Institute of Cancer Epidemiology and the Danish Cancer Society, 93(3):203-7. Available online at http://www.ncbi.nlm.nih.gov/pubmed/11158188 (accessed June 27, 2013).
• Rape, Abuse & Incest National Network. “How often does sexual assault occur?” RAINN, 2009. Available online at www.rainn.org/get-information...sexual-assault (accessed June 27, 2013).

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How to Write a Great Hypothesis

Hypothesis Definition, Format, Examples, and Tips

Kendra Cherry, MS, is a psychosocial rehabilitation specialist, psychology educator, and author of the "Everything Psychology Book."

Amy Morin, LCSW, is a psychotherapist and international bestselling author. Her books, including "13 Things Mentally Strong People Don't Do," have been translated into more than 40 languages. Her TEDx talk,  "The Secret of Becoming Mentally Strong," is one of the most viewed talks of all time.

Verywell / Alex Dos Diaz

• The Scientific Method

Hypothesis Format

Falsifiability of a hypothesis.

• Operationalization

Hypothesis Types

Hypotheses examples.

• Collecting Data

Frequently Asked Questions

A  hypothesis  is a tentative statement about the relationship between two or more variables. It is a specific, testable prediction about what you expect to happen in a study. It is a preliminary answer to your question that helps guide the research process.

Consider a study designed to examine the relationship between sleep deprivation and test performance. The hypothesis might be: "This study is designed to assess the hypothesis that sleep-deprived people will perform worse on a test than individuals who are not sleep-deprived."

At a Glance

A hypothesis is crucial to scientific research because it offers a clear direction for what the researchers are looking to find. This allows them to design experiments to test their predictions and add to our scientific knowledge about the world. This article explores how a hypothesis is used in psychology research, how to write a good hypothesis, and the different types of hypotheses you might use.

The Hypothesis in the Scientific Method

In the scientific method , whether it involves research in psychology, biology, or some other area, a hypothesis represents what the researchers think will happen in an experiment. The scientific method involves the following steps:

• Forming a question
• Performing background research
• Creating a hypothesis
• Designing an experiment
• Collecting data
• Analyzing the results
• Drawing conclusions
• Communicating the results

The hypothesis is a prediction, but it involves more than a guess. Most of the time, the hypothesis begins with a question which is then explored through background research. At this point, researchers then begin to develop a testable hypothesis.

Unless you are creating an exploratory study, your hypothesis should always explain what you  expect  to happen.

In a study exploring the effects of a particular drug, the hypothesis might be that researchers expect the drug to have some type of effect on the symptoms of a specific illness. In psychology, the hypothesis might focus on how a certain aspect of the environment might influence a particular behavior.

Remember, a hypothesis does not have to be correct. While the hypothesis predicts what the researchers expect to see, the goal of the research is to determine whether this guess is right or wrong. When conducting an experiment, researchers might explore numerous factors to determine which ones might contribute to the ultimate outcome.

In many cases, researchers may find that the results of an experiment  do not  support the original hypothesis. When writing up these results, the researchers might suggest other options that should be explored in future studies.

In many cases, researchers might draw a hypothesis from a specific theory or build on previous research. For example, prior research has shown that stress can impact the immune system. So a researcher might hypothesize: "People with high-stress levels will be more likely to contract a common cold after being exposed to the virus than people who have low-stress levels."

In other instances, researchers might look at commonly held beliefs or folk wisdom. "Birds of a feather flock together" is one example of folk adage that a psychologist might try to investigate. The researcher might pose a specific hypothesis that "People tend to select romantic partners who are similar to them in interests and educational level."

Elements of a Good Hypothesis

So how do you write a good hypothesis? When trying to come up with a hypothesis for your research or experiments, ask yourself the following questions:

• Is your hypothesis based on your research on a topic?
• Can your hypothesis be tested?
• Does your hypothesis include independent and dependent variables?

Before you come up with a specific hypothesis, spend some time doing background research. Once you have completed a literature review, start thinking about potential questions you still have. Pay attention to the discussion section in the  journal articles you read . Many authors will suggest questions that still need to be explored.

How to Formulate a Good Hypothesis

To form a hypothesis, you should take these steps:

• Collect as many observations about a topic or problem as you can.
• Evaluate these observations and look for possible causes of the problem.
• Create a list of possible explanations that you might want to explore.
• After you have developed some possible hypotheses, think of ways that you could confirm or disprove each hypothesis through experimentation. This is known as falsifiability.

In the scientific method ,  falsifiability is an important part of any valid hypothesis. In order to test a claim scientifically, it must be possible that the claim could be proven false.

Students sometimes confuse the idea of falsifiability with the idea that it means that something is false, which is not the case. What falsifiability means is that  if  something was false, then it is possible to demonstrate that it is false.

One of the hallmarks of pseudoscience is that it makes claims that cannot be refuted or proven false.

The Importance of Operational Definitions

A variable is a factor or element that can be changed and manipulated in ways that are observable and measurable. However, the researcher must also define how the variable will be manipulated and measured in the study.

Operational definitions are specific definitions for all relevant factors in a study. This process helps make vague or ambiguous concepts detailed and measurable.

For example, a researcher might operationally define the variable " test anxiety " as the results of a self-report measure of anxiety experienced during an exam. A "study habits" variable might be defined by the amount of studying that actually occurs as measured by time.

These precise descriptions are important because many things can be measured in various ways. Clearly defining these variables and how they are measured helps ensure that other researchers can replicate your results.

Replicability

One of the basic principles of any type of scientific research is that the results must be replicable.

Replication means repeating an experiment in the same way to produce the same results. By clearly detailing the specifics of how the variables were measured and manipulated, other researchers can better understand the results and repeat the study if needed.

Some variables are more difficult than others to define. For example, how would you operationally define a variable such as aggression ? For obvious ethical reasons, researchers cannot create a situation in which a person behaves aggressively toward others.

To measure this variable, the researcher must devise a measurement that assesses aggressive behavior without harming others. The researcher might utilize a simulated task to measure aggressiveness in this situation.

Hypothesis Checklist

• Does your hypothesis focus on something that you can actually test?
• Does your hypothesis include both an independent and dependent variable?
• Can you manipulate the variables?
• Can your hypothesis be tested without violating ethical standards?

The hypothesis you use will depend on what you are investigating and hoping to find. Some of the main types of hypotheses that you might use include:

• Simple hypothesis : This type of hypothesis suggests there is a relationship between one independent variable and one dependent variable.
• Complex hypothesis : This type suggests a relationship between three or more variables, such as two independent and dependent variables.
• Null hypothesis : This hypothesis suggests no relationship exists between two or more variables.
• Alternative hypothesis : This hypothesis states the opposite of the null hypothesis.
• Statistical hypothesis : This hypothesis uses statistical analysis to evaluate a representative population sample and then generalizes the findings to the larger group.
• Logical hypothesis : This hypothesis assumes a relationship between variables without collecting data or evidence.

A hypothesis often follows a basic format of "If {this happens} then {this will happen}." One way to structure your hypothesis is to describe what will happen to the  dependent variable  if you change the  independent variable .

The basic format might be: "If {these changes are made to a certain independent variable}, then we will observe {a change in a specific dependent variable}."

A few examples of simple hypotheses:

• "Students who eat breakfast will perform better on a math exam than students who do not eat breakfast."
• "Students who experience test anxiety before an English exam will get lower scores than students who do not experience test anxiety."​
• "Motorists who talk on the phone while driving will be more likely to make errors on a driving course than those who do not talk on the phone."
• "Children who receive a new reading intervention will have higher reading scores than students who do not receive the intervention."

Examples of a complex hypothesis include:

• "People with high-sugar diets and sedentary activity levels are more likely to develop depression."
• "Younger people who are regularly exposed to green, outdoor areas have better subjective well-being than older adults who have limited exposure to green spaces."

Examples of a null hypothesis include:

• "There is no difference in anxiety levels between people who take St. John's wort supplements and those who do not."
• "There is no difference in scores on a memory recall task between children and adults."
• "There is no difference in aggression levels between children who play first-person shooter games and those who do not."

Examples of an alternative hypothesis:

• "People who take St. John's wort supplements will have less anxiety than those who do not."
• "Adults will perform better on a memory task than children."
• "Children who play first-person shooter games will show higher levels of aggression than children who do not." 

Collecting Data on Your Hypothesis

Once a researcher has formed a testable hypothesis, the next step is to select a research design and start collecting data. The research method depends largely on exactly what they are studying. There are two basic types of research methods: descriptive research and experimental research.

Descriptive Research Methods

Descriptive research such as  case studies ,  naturalistic observations , and surveys are often used when  conducting an experiment is difficult or impossible. These methods are best used to describe different aspects of a behavior or psychological phenomenon.

Once a researcher has collected data using descriptive methods, a  correlational study  can examine how the variables are related. This research method might be used to investigate a hypothesis that is difficult to test experimentally.

Experimental Research Methods

Experimental methods  are used to demonstrate causal relationships between variables. In an experiment, the researcher systematically manipulates a variable of interest (known as the independent variable) and measures the effect on another variable (known as the dependent variable).

Unlike correlational studies, which can only be used to determine if there is a relationship between two variables, experimental methods can be used to determine the actual nature of the relationship—whether changes in one variable actually  cause  another to change.

The hypothesis is a critical part of any scientific exploration. It represents what researchers expect to find in a study or experiment. In situations where the hypothesis is unsupported by the research, the research still has value. Such research helps us better understand how different aspects of the natural world relate to one another. It also helps us develop new hypotheses that can then be tested in the future.

Some examples of how to write a hypothesis include:

• "Staying up late will lead to worse test performance the next day."
• "People who consume one apple each day will visit the doctor fewer times each year."
• "Breaking study sessions up into three 20-minute sessions will lead to better test results than a single 60-minute study session."

The four parts of a hypothesis are:

(1) The research question

(2) The independent variable (IV)

(3) The dependent variable (DV)

(4) The proposed relationship between the IV and DV

No, a hypothesis and a theory are not the same thing. A hypothesis is a testable prediction about a specific research question. A theory, on the other hand, is an explanation supported by an existing body of scientific research.

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By Kendra Cherry, MSEd Kendra Cherry, MS, is a psychosocial rehabilitation specialist, psychology educator, and author of the "Everything Psychology Book."