good null hypothesis test

What is The Null Hypothesis & When Do You Reject The Null Hypothesis

Julia Simkus

Editor at Simply Psychology

BA (Hons) Psychology, Princeton University

Julia Simkus is a graduate of Princeton University with a Bachelor of Arts in Psychology. She is currently studying for a Master's Degree in Counseling for Mental Health and Wellness in September 2023. Julia's research has been published in peer reviewed journals.

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BSc (Hons) Psychology, MRes, PhD, University of Manchester

Saul Mcleod, PhD., is a qualified psychology teacher with over 18 years of experience in further and higher education. He has been published in peer-reviewed journals, including the Journal of Clinical Psychology.

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On This Page:

A null hypothesis is a statistical concept suggesting no significant difference or relationship between measured variables. It’s the default assumption unless empirical evidence proves otherwise.

The null hypothesis states no relationship exists between the two variables being studied (i.e., one variable does not affect the other).

The null hypothesis is the statement that a researcher or an investigator wants to disprove.

Testing the null hypothesis can tell you whether your results are due to the effects of manipulating ​ the dependent variable or due to random chance. 

How to Write a Null Hypothesis

Null hypotheses (H0) start as research questions that the investigator rephrases as statements indicating no effect or relationship between the independent and dependent variables.

It is a default position that your research aims to challenge or confirm.

For example, if studying the impact of exercise on weight loss, your null hypothesis might be:

There is no significant difference in weight loss between individuals who exercise daily and those who do not.

Examples of Null Hypotheses

When do we reject the null hypothesis .

We reject the null hypothesis when the data provide strong enough evidence to conclude that it is likely incorrect. This often occurs when the p-value (probability of observing the data given the null hypothesis is true) is below a predetermined significance level.

If the collected data does not meet the expectation of the null hypothesis, a researcher can conclude that the data lacks sufficient evidence to back up the null hypothesis, and thus the null hypothesis is rejected. 

Rejecting the null hypothesis means that a relationship does exist between a set of variables and the effect is statistically significant ( p > 0.05).

If the data collected from the random sample is not statistically significance , then the null hypothesis will be accepted, and the researchers can conclude that there is no relationship between the variables. 

You need to perform a statistical test on your data in order to evaluate how consistent it is with the null hypothesis. A p-value is one statistical measurement used to validate a hypothesis against observed data.

Calculating the p-value is a critical part of null-hypothesis significance testing because it quantifies how strongly the sample data contradicts the null hypothesis.

The level of statistical significance is often expressed as a  p  -value between 0 and 1. The smaller the p-value, the stronger the evidence that you should reject the null hypothesis.

Usually, a researcher uses a confidence level of 95% or 99% (p-value of 0.05 or 0.01) as general guidelines to decide if you should reject or keep the null.

When your p-value is less than or equal to your significance level, you reject the null hypothesis.

In other words, smaller p-values are taken as stronger evidence against the null hypothesis. Conversely, when the p-value is greater than your significance level, you fail to reject the null hypothesis.

In this case, the sample data provides insufficient data to conclude that the effect exists in the population.

Because you can never know with complete certainty whether there is an effect in the population, your inferences about a population will sometimes 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 called a type II error.

Why Do We Never Accept The Null Hypothesis?

The reason we do not say “accept the null” is because we are always assuming the null hypothesis is true and then conducting a study to see if there is evidence against it. And, even if we don’t find evidence against it, a null hypothesis is not accepted.

A lack of evidence only means that you haven’t proven that something exists. It does not prove that something doesn’t exist. 

It is risky to conclude that the null hypothesis is true merely because we did not find evidence to reject it. It is always possible that researchers elsewhere have disproved the null hypothesis, so we cannot accept it as true, but instead, we state that we failed to reject the null. 

One can either reject the null hypothesis, or fail to reject it, but can never accept it.

Why Do We Use The Null Hypothesis?

We can never prove with 100% certainty that a hypothesis is true; We can only collect evidence that supports a theory. However, testing a hypothesis can set the stage for rejecting or accepting this hypothesis within a certain confidence level.

The null hypothesis is useful because it can tell us whether the results of our study are due to random chance or the manipulation of a variable (with a certain level of confidence).

A null hypothesis is rejected if the measured data is significantly unlikely to have occurred and a null hypothesis is accepted if the observed outcome is consistent with the position held by the null hypothesis.

Rejecting the null hypothesis sets the stage for further experimentation to see if a relationship between two variables exists. 

Hypothesis testing is a critical part of the scientific method as it helps decide whether the results of a research study support a particular theory about a given population. Hypothesis testing is a systematic way of backing up researchers’ predictions with statistical analysis.

It helps provide sufficient statistical evidence that either favors or rejects a certain hypothesis about the population parameter. 

Purpose of a Null Hypothesis 

• The primary purpose of the null hypothesis is to disprove an assumption. 
• Whether rejected or accepted, the null hypothesis can help further progress a theory in many scientific cases.
• A null hypothesis can be used to ascertain how consistent the outcomes of multiple studies are.

Do you always need both a Null Hypothesis and an Alternative Hypothesis?

The null (H0) and alternative (Ha or H1) hypotheses are two competing claims that describe the effect of the independent variable on the dependent variable. They are mutually exclusive, which means that only one of the two hypotheses can be true. 

While the null hypothesis states that there is no effect in the population, an alternative hypothesis states that there is statistical significance between two variables. 

The goal of hypothesis testing is to make inferences about a population based on a sample. In order to undertake hypothesis testing, you must express your research hypothesis as a null and alternative hypothesis. Both hypotheses are required to cover every possible outcome of the study. 

What is the difference between a null hypothesis and an alternative hypothesis?

The alternative hypothesis is the complement to the null hypothesis. The null hypothesis states that there is no effect or no relationship between variables, while the alternative hypothesis claims that there is an effect or relationship in the population.

It is the claim that you expect or hope will be true. The null hypothesis and the alternative hypothesis are always mutually exclusive, meaning that only one can be true at a time.

What are some problems with the null hypothesis?

One major problem with the null hypothesis is that researchers typically will assume that accepting the null is a failure of the experiment. However, accepting or rejecting any hypothesis is a positive result. Even if the null is not refuted, the researchers will still learn something new.

Why can a null hypothesis not be accepted?

We can either reject or fail to reject a null hypothesis, but never accept it. If your test fails to detect an effect, this is not proof that the effect doesn’t exist. It just means that your sample did not have enough evidence to conclude that it exists.

We can’t accept a null hypothesis because a lack of evidence does not prove something that does not exist. Instead, we fail to reject it.

Failing to reject the null indicates that the sample did not provide sufficient enough evidence to conclude that an effect exists.

If the p-value is greater than the significance level, then you fail to reject the null hypothesis.

Is a null hypothesis directional or non-directional?

A hypothesis test can either contain an alternative directional hypothesis or a non-directional alternative hypothesis. A directional hypothesis is one that contains the less than (“<“) or greater than (“>”) sign.

A nondirectional hypothesis contains the not equal sign (“≠”).  However, a null hypothesis is neither directional nor non-directional.

A null hypothesis is a prediction that there will be no change, relationship, or difference between two variables.

The directional hypothesis or nondirectional hypothesis would then be considered alternative hypotheses to the null hypothesis.

Gill, J. (1999). The insignificance of null hypothesis significance testing.  Political research quarterly ,  52 (3), 647-674.

Krueger, J. (2001). Null hypothesis significance testing: On the survival of a flawed method.  American Psychologist ,  56 (1), 16.

Masson, M. E. (2011). A tutorial on a practical Bayesian alternative to null-hypothesis significance testing.  Behavior research methods ,  43 , 679-690.

Nickerson, R. S. (2000). Null hypothesis significance testing: a review of an old and continuing controversy.  Psychological methods ,  5 (2), 241.

Rozeboom, W. W. (1960). The fallacy of the null-hypothesis significance test.  Psychological bulletin ,  57 (5), 416.

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Chapter 13: Inferential Statistics

Understanding Null Hypothesis Testing

Learning Objectives

• Explain the purpose of null hypothesis testing, including the role of sampling error.
• Describe the basic logic of null hypothesis testing.
• Describe the role of relationship strength and sample size in determining statistical significance and make reasonable judgments about statistical significance based on these two factors.

The Purpose of Null Hypothesis Testing

As we have seen, psychological research typically involves measuring one or more variables for a sample and computing descriptive statistics for that sample. In general, however, the researcher’s goal is not to draw conclusions about that sample but to draw conclusions about the population that the sample was selected from. Thus researchers must use sample statistics to draw conclusions about the corresponding values in the population. These corresponding values in the population are called  parameters . Imagine, for example, that a researcher measures the number of depressive symptoms exhibited by each of 50 clinically depressed adults and computes the mean number of symptoms. The researcher probably wants to use this sample statistic (the mean number of symptoms for the sample) to draw conclusions about the corresponding population parameter (the mean number of symptoms for clinically depressed adults).

Unfortunately, sample statistics are not perfect estimates of their corresponding population parameters. This is because there is a certain amount of random variability in any statistic from sample to sample. The mean number of depressive symptoms might be 8.73 in one sample of clinically depressed adults, 6.45 in a second sample, and 9.44 in a third—even though these samples are selected randomly from the same population. Similarly, the correlation (Pearson’s  r ) between two variables might be +.24 in one sample, −.04 in a second sample, and +.15 in a third—again, even though these samples are selected randomly from the same population. This random variability in a statistic from sample to sample is called  sampling error . (Note that the term error  here refers to random variability and does not imply that anyone has made a mistake. No one “commits a sampling error.”)

One implication of this is that when there is a statistical relationship in a sample, it is not always clear that there is a statistical relationship in the population. A small difference between two group means in a sample might indicate that there is a small difference between the two group means in the population. But it could also be that there is no difference between the means in the population and that the difference in the sample is just a matter of sampling error. Similarly, a Pearson’s  r  value of −.29 in a sample might mean that there is a negative relationship in the population. But it could also be that there is no relationship in the population and that the relationship in the sample is just a matter of sampling error.

In fact, any statistical relationship in a sample can be interpreted in two ways:

• There is a relationship in the population, and the relationship in the sample reflects this.
• There is no relationship in the population, and the relationship in the sample reflects only sampling error.

The purpose of null hypothesis testing is simply to help researchers decide between these two interpretations.

The Logic of Null Hypothesis Testing

Null hypothesis testing  is a formal approach to deciding between two interpretations of a statistical relationship in a sample. One interpretation is called the   null hypothesis  (often symbolized  H 0  and read as “H-naught”). This is the idea that there is no relationship in the population and that the relationship in the sample reflects only sampling error. Informally, the null hypothesis is that the sample relationship “occurred by chance.” The other interpretation is called the  alternative hypothesis  (often symbolized as  H 1 ). This is the idea that there is a relationship in the population and that the relationship in the sample reflects this relationship in the population.

Again, every statistical relationship in a sample can be interpreted in either of these two ways: It might have occurred by chance, or it might reflect a relationship in the population. So researchers need a way to decide between them. Although there are many specific null hypothesis testing techniques, they are all based on the same general logic. The steps are as follows:

• Assume for the moment that the null hypothesis is true. There is no relationship between the variables in the population.
• Determine how likely the sample relationship would be if the null hypothesis were true.
• If the sample relationship would be extremely unlikely, then reject the null hypothesis  in favour of the alternative hypothesis. If it would not be extremely unlikely, then  retain the null hypothesis .

Following this logic, we can begin to understand why Mehl and his colleagues concluded that there is no difference in talkativeness between women and men in the population. In essence, they asked the following question: “If there were no difference in the population, how likely is it that we would find a small difference of  d  = 0.06 in our sample?” Their answer to this question was that this sample relationship would be fairly likely if the null hypothesis were true. Therefore, they retained the null hypothesis—concluding that there is no evidence of a sex difference in the population. We can also see why Kanner and his colleagues concluded that there is a correlation between hassles and symptoms in the population. They asked, “If the null hypothesis were true, how likely is it that we would find a strong correlation of +.60 in our sample?” Their answer to this question was that this sample relationship would be fairly unlikely if the null hypothesis were true. Therefore, they rejected the null hypothesis in favour of the alternative hypothesis—concluding that there is a positive correlation between these variables in the population.

A crucial step in null hypothesis testing is finding the likelihood of the sample result if the null hypothesis were true. This probability is called the  p value . A low  p  value means that the sample result would be unlikely if the null hypothesis were true and leads to the rejection of the null hypothesis. A high  p  value means that the sample result would be likely if the null hypothesis were true and leads to the retention of the null hypothesis. But how low must the  p  value be before the sample result is considered unlikely enough to reject the null hypothesis? In null hypothesis testing, this criterion is called  α (alpha)  and is almost always set to .05. If there is less than a 5% chance of a result as extreme as the sample result if the null hypothesis were true, then the null hypothesis is rejected. When this happens, the result is said to be  statistically significant . If there is greater than a 5% chance of a result as extreme as the sample result when the null hypothesis is true, then the null hypothesis is retained. This does not necessarily mean that the researcher accepts the null hypothesis as true—only that there is not currently enough evidence to conclude that it is true. Researchers often use the expression “fail to reject the null hypothesis” rather than “retain the null hypothesis,” but they never use the expression “accept the null hypothesis.”

The Misunderstood  p  Value

The  p  value is one of the most misunderstood quantities in psychological research (Cohen, 1994) [1] . Even professional researchers misinterpret it, and it is not unusual for such misinterpretations to appear in statistics textbooks!

The most common misinterpretation is that the  p  value is the probability that the null hypothesis is true—that the sample result occurred by chance. For example, a misguided researcher might say that because the  p  value is .02, there is only a 2% chance that the result is due to chance and a 98% chance that it reflects a real relationship in the population. But this is incorrect . The  p  value is really the probability of a result at least as extreme as the sample result  if  the null hypothesis  were  true. So a  p  value of .02 means that if the null hypothesis were true, a sample result this extreme would occur only 2% of the time.

You can avoid this misunderstanding by remembering that the  p  value is not the probability that any particular  hypothesis  is true or false. Instead, it is the probability of obtaining the  sample result  if the null hypothesis were true.

Role of Sample Size and Relationship Strength

Recall that null hypothesis testing involves answering the question, “If the null hypothesis were true, what is the probability of a sample result as extreme as this one?” In other words, “What is the  p  value?” It can be helpful to see that the answer to this question depends on just two considerations: the strength of the relationship and the size of the sample. Specifically, the stronger the sample relationship and the larger the sample, the less likely the result would be if the null hypothesis were true. That is, the lower the  p  value. This should make sense. Imagine a study in which a sample of 500 women is compared with a sample of 500 men in terms of some psychological characteristic, and Cohen’s  d  is a strong 0.50. If there were really no sex difference in the population, then a result this strong based on such a large sample should seem highly unlikely. Now imagine a similar study in which a sample of three women is compared with a sample of three men, and Cohen’s  d  is a weak 0.10. If there were no sex difference in the population, then a relationship this weak based on such a small sample should seem likely. And this is precisely why the null hypothesis would be rejected in the first example and retained in the second.

Of course, sometimes the result can be weak and the sample large, or the result can be strong and the sample small. In these cases, the two considerations trade off against each other so that a weak result can be statistically significant if the sample is large enough and a strong relationship can be statistically significant even if the sample is small. Table 13.1 shows roughly how relationship strength and sample size combine to determine whether a sample result is statistically significant. The columns of the table represent the three levels of relationship strength: weak, medium, and strong. The rows represent four sample sizes that can be considered small, medium, large, and extra large in the context of psychological research. Thus each cell in the table represents a combination of relationship strength and sample size. If a cell contains the word  Yes , then this combination would be statistically significant for both Cohen’s  d  and Pearson’s  r . If it contains the word  No , then it would not be statistically significant for either. There is one cell where the decision for  d  and  r  would be different and another where it might be different depending on some additional considerations, which are discussed in Section 13.2 “Some Basic Null Hypothesis Tests”

Although Table 13.1 provides only a rough guideline, it shows very clearly that weak relationships based on medium or small samples are never statistically significant and that strong relationships based on medium or larger samples are always statistically significant. If you keep this lesson in mind, you will often know whether a result is statistically significant based on the descriptive statistics alone. It is extremely useful to be able to develop this kind of intuitive judgment. One reason is that it allows you to develop expectations about how your formal null hypothesis tests are going to come out, which in turn allows you to detect problems in your analyses. For example, if your sample relationship is strong and your sample is medium, then you would expect to reject the null hypothesis. If for some reason your formal null hypothesis test indicates otherwise, then you need to double-check your computations and interpretations. A second reason is that the ability to make this kind of intuitive judgment is an indication that you understand the basic logic of this approach in addition to being able to do the computations.

Statistical Significance Versus Practical Significance

Table 13.1 illustrates another extremely important point. A statistically significant result is not necessarily a strong one. Even a very weak result can be statistically significant if it is based on a large enough sample. This is closely related to Janet Shibley Hyde’s argument about sex differences (Hyde, 2007) [2] . The differences between women and men in mathematical problem solving and leadership ability are statistically significant. But the word  significant  can cause people to interpret these differences as strong and important—perhaps even important enough to influence the college courses they take or even who they vote for. As we have seen, however, these statistically significant differences are actually quite weak—perhaps even “trivial.”

This is why it is important to distinguish between the  statistical  significance of a result and the  practical  significance of that result.  Practical significance refers to the importance or usefulness of the result in some real-world context. Many sex differences are statistically significant—and may even be interesting for purely scientific reasons—but they are not practically significant. In clinical practice, this same concept is often referred to as “clinical significance.” For example, a study on a new treatment for social phobia might show that it produces a statistically significant positive effect. Yet this effect still might not be strong enough to justify the time, effort, and other costs of putting it into practice—especially if easier and cheaper treatments that work almost as well already exist. Although statistically significant, this result would be said to lack practical or clinical significance.

Key Takeaways

• Null hypothesis testing is a formal approach to deciding whether a statistical relationship in a sample reflects a real relationship in the population or is just due to chance.
• The logic of null hypothesis testing involves assuming that the null hypothesis is true, finding how likely the sample result would be if this assumption were correct, and then making a decision. If the sample result would be unlikely if the null hypothesis were true, then it is rejected in favour of the alternative hypothesis. If it would not be unlikely, then the null hypothesis is retained.
• The probability of obtaining the sample result if the null hypothesis were true (the  p  value) is based on two considerations: relationship strength and sample size. Reasonable judgments about whether a sample relationship is statistically significant can often be made by quickly considering these two factors.
• Statistical significance is not the same as relationship strength or importance. Even weak relationships can be statistically significant if the sample size is large enough. It is important to consider relationship strength and the practical significance of a result in addition to its statistical significance.
• Discussion: Imagine a study showing that people who eat more broccoli tend to be happier. Explain for someone who knows nothing about statistics why the researchers would conduct a null hypothesis test.
• The correlation between two variables is  r  = −.78 based on a sample size of 137.
• The mean score on a psychological characteristic for women is 25 ( SD  = 5) and the mean score for men is 24 ( SD  = 5). There were 12 women and 10 men in this study.
• In a memory experiment, the mean number of items recalled by the 40 participants in Condition A was 0.50 standard deviations greater than the mean number recalled by the 40 participants in Condition B.
• In another memory experiment, the mean scores for participants in Condition A and Condition B came out exactly the same!
• A student finds a correlation of  r  = .04 between the number of units the students in his research methods class are taking and the students’ level of stress.

Long Descriptions

“Null Hypothesis” long description: A comic depicting a man and a woman talking in the foreground. In the background is a child working at a desk. The man says to the woman, “I can’t believe schools are still teaching kids about the null hypothesis. I remember reading a big study that conclusively disproved it years ago.” [Return to “Null Hypothesis”]

“Conditional Risk” long description: A comic depicting two hikers beside a tree during a thunderstorm. A bolt of lightning goes “crack” in the dark sky as thunder booms. One of the hikers says, “Whoa! We should get inside!” The other hiker says, “It’s okay! Lightning only kills about 45 Americans a year, so the chances of dying are only one in 7,000,000. Let’s go on!” The comic’s caption says, “The annual death rate among people who know that statistic is one in six.” [Return to “Conditional Risk”]

Media Attributions

• Null Hypothesis by XKCD  CC BY-NC (Attribution NonCommercial)
• Conditional Risk by XKCD  CC BY-NC (Attribution NonCommercial)
• Cohen, J. (1994). The world is round: p < .05. American Psychologist, 49 , 997–1003. ↵
• Hyde, J. S. (2007). New directions in the study of gender similarities and differences. Current Directions in Psychological Science, 16 , 259–263. ↵

Values in a population that correspond to variables measured in a study.

The random variability in a statistic from sample to sample.

A formal approach to deciding between two interpretations of a statistical relationship in a sample.

The idea that there is no relationship in the population and that the relationship in the sample reflects only sampling error.

The idea that there is a relationship in the population and that the relationship in the sample reflects this relationship in the population.

When the relationship found in the sample would be extremely unlikely, the idea that the relationship occurred “by chance” is rejected.

When the relationship found in the sample is likely to have occurred by chance, the null hypothesis is not rejected.

The probability that, if the null hypothesis were true, the result found in the sample would occur.

How low the p value must be before the sample result is considered unlikely in null hypothesis testing.

When there is less than a 5% chance of a result as extreme as the sample result occurring and the null hypothesis is rejected.

Research Methods in Psychology - 2nd Canadian Edition Copyright © 2015 by Paul C. Price, Rajiv Jhangiani, & I-Chant A. Chiang is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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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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AP®︎/College Statistics

Course: ap®︎/college statistics   >   unit 10.

• Idea behind hypothesis testing

Examples of null and alternative hypotheses

• Writing null and alternative hypotheses
• P-values and significance tests
• Comparing P-values to different significance levels
• Estimating a P-value from a simulation
• Estimating P-values from simulations
• Using P-values to make conclusions

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Video transcript

Null Hypothesis Examples

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In statistical analysis, the null hypothesis assumes there is no meaningful relationship between two variables. Testing the null hypothesis can tell you whether your results are due to the effect of manipulating ​a dependent variable or due to chance. It's often used in conjunction with an alternative hypothesis, which assumes there is, in fact, a relationship between two variables.

The null hypothesis is among the easiest hypothesis to test using statistical analysis, making it perhaps the most valuable hypothesis for the scientific method. By evaluating a null hypothesis in addition to another hypothesis, researchers can support their conclusions with a higher level of confidence. Below are examples of how you might formulate a null hypothesis to fit certain questions.

What Is the Null Hypothesis?

The null hypothesis states there is no relationship between the measured phenomenon (the dependent variable ) and the independent variable , which is the variable an experimenter typically controls or changes. You do not​ need to believe that the null hypothesis is true to test it. On the contrary, you will likely suspect there is a relationship between a set of variables. One way to prove that this is the case is to reject the null hypothesis. Rejecting a hypothesis does not mean an experiment was "bad" or that it didn't produce results. In fact, it is often one of the first steps toward further inquiry.

To distinguish it from other hypotheses , the null hypothesis is written as ​ H 0  (which is read as “H-nought,” "H-null," or "H-zero"). A significance test is used to determine the likelihood that the results supporting the null hypothesis are not due to chance. A confidence level of 95% or 99% is common. Keep in mind, even if the confidence level is high, there is still a small chance the null hypothesis is not true, perhaps because the experimenter did not account for a critical factor or because of chance. This is one reason why it's important to repeat experiments.

Examples of the Null Hypothesis

To write a null hypothesis, first start by asking a question. Rephrase that question in a form that assumes no relationship between the variables. In other words, assume a treatment has no effect. Write your hypothesis in a way that reflects this.

Other Types of Hypotheses

In addition to the null hypothesis, the alternative hypothesis is also a staple in traditional significance tests . It's essentially the opposite of the null hypothesis because it assumes the claim in question is true. For the first item in the table above, for example, an alternative hypothesis might be "Age does have an effect on mathematical ability."

Key Takeaways

• In hypothesis testing, the null hypothesis assumes no relationship between two variables, providing a baseline for statistical analysis.
• Rejecting the null hypothesis suggests there is evidence of a relationship between variables.
• By formulating a null hypothesis, researchers can systematically test assumptions and draw more reliable conclusions from their experiments.
• Difference Between Independent and Dependent Variables
• Examples of Independent and Dependent Variables
• What Is a Hypothesis? (Science)
• Definition of a Hypothesis
• What 'Fail to Reject' Means in a Hypothesis Test
• Null Hypothesis Definition and Examples
• Scientific Method Vocabulary Terms
• Null Hypothesis and Alternative Hypothesis
• Hypothesis Test for the Difference of Two Population Proportions
• How to Conduct a Hypothesis Test
• What Is a P-Value?
• Hypothesis Test Example
• What Are the Elements of a Good Hypothesis?
• What Is the Difference Between Alpha and P-Values?
• Example of a Chi-Square Goodness of Fit Test
• Understanding Path Analysis

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11.3: Goodness-of-Fit Test

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In this type of hypothesis test, you determine whether the data "fit" a particular distribution or not. For example, you may suspect your unknown data fit a binomial distribution. You use a chi-square test (meaning the distribution for the hypothesis test is chi-square) to determine if there is a fit or not. The null and the alternative hypotheses for this test may be written in sentences or may be stated as equations or inequalities.

The test statistic for a goodness-of-fit test is:

\[\sum_k \frac{(O - E)^{2}}{E}\]

• \(O =\) observed values (data)
• \(E =\) expected values (from theory)
• \(k =\) the number of different data cells or categories

The observed values are the data values and the expected values are the values you would expect to get if the null hypothesis were true. There are \(n\) terms of the form \(\frac{(O - E)^{2}}{E}\).

The number of degrees of freedom is \(df = (\text{number of categories} - 1)\).

The goodness-of-fit test is almost always right-tailed. If the observed values and the corresponding expected values are not close to each other, then the test statistic can get very large and will be way out in the right tail of the chi-square curve.

The expected value for each cell needs to be at least five in order for you to use this test.

Example 11.3.1

Absenteeism of college students from math classes is a major concern to math instructors because missing class appears to increase the drop rate. Suppose that a study was done to determine if the actual student absenteeism rate follows faculty perception. The faculty expected that a group of 100 students would miss class according to the table below.

A random survey across all mathematics courses was then done to determine the actual number (observed) of absences in a course. The chart in the table below displays the results of that survey.

Determine the null and alternative hypotheses needed to conduct a goodness-of-fit test.

• \(H_{0}\) : Student absenteeism fits faculty perception.
• \(H_{a}\) : Student absenteeism does not fit faculty perception.

Exercise 11.3.1.1

a. Can you use the information as it appears in the charts to conduct the goodness-of-fit test?

a. No. Notice that the expected number of absences for the "12+" entry is less than five (it is two). Combine that group with the "9–11" group to create new tables where the number of students for each entry are at least five. The new results are in the table below.

Exercise 11.3.1.2

b. What is the number of degrees of freedom (\(df\))?

b. There are four "cells" or categories in each of the new tables.

\(df = \text{number of cells} - 1 = 4 - 1 = 3\)

Exercise 11.3.1

A factory manager needs to understand how many products are defective versus how many are produced.  The number of expected defects is listed in the table below.

A random sample was taken to determine the actual number of defects. The table below shows the results of the survey.

State the null and alternative hypotheses needed to conduct a goodness-of-fit test, and state the degrees of freedom.

\(H_{0}\):The number of defects fits expectations.

\(H_{a}\):The number of defects does not fit expectations.

Example 11.3.2

Employers want to know which days of the week employees are absent in a five-day work week. Most employers would like to believe that employees are absent equally during the week. Suppose a random sample of 60 managers were asked on which day of the week they had the highest number of employee absences. The results were distributed as in the table below. For the population of employees, do the days for the highest number of absences occur with equal frequencies during a five-day work week? Test at a 5% significance level.

The null and alternative hypotheses are:

• \(H_{0}\): The absent days occur with equal frequencies, that is, they fit a uniform distribution.
• \(H_{a}\): The absent days occur with unequal frequencies, that is, they do not fit a uniform distribution.

If the absent days occur with equal frequencies, then, out of 60 absent days (the total in the sample: \(15 + 12 + 9 + 9 + 15 = 60\)), there would be 12 absences on Monday, 12 on Tuesday, 12 on Wednesday, 12 on Thursday, and 12 on Friday. These numbers are the expected (\(E\)) values. The values in the table are the observed (\(O\)) values or data.

This time, calculate the \(\chi^{2}\) test statistic by hand. Make a chart with the following headings and fill in the columns:

• Expected (\(E\)) values \((12, 12, 12, 12, 12)\)
• Observed (\(O\)) values \((15, 12, 9, 9, 15)\)
• \((O – E)\)
• \((O – E)^{2}\)
• \(\frac{(O - E)^{2}}{E}\)

Now add (sum) the last column. The sum is three. This is the \(\chi^{2}\) test statistic.

To find the p -value, calculate \(P(\chi^{2} > 3)\). This test is right-tailed. (Use a computer or calculator to find the p -value. You should get \(p\text{-value} = 0.5578\).)

The \(dfs\) are the \(\text{number of cells} - 1 = 5 - 1 = 4\)

Press 2nd DISTR . Arrow down to \(\chi^{2}\)cdf . Press ENTER . Enter (3,10^99,4) . Rounded to four decimal places, you should see 0.5578, which is the \(p\text{-value}\).

Next, complete a graph like the following one with the proper labeling and shading. (You should shade the right tail.)

The decision is not to reject the null hypothesis.

Conclusion: At a 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the absent days do not occur with equal frequencies.

TI-83+ and some TI-84 calculators do not have a special program for the test statistic for the goodness-of-fit test. The next example Example has the calculator instructions. The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF . To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list. Press STAT TESTS and Chi2 GOF . Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or draw . Make sure you clear any lists before you start. To Clear Lists in the calculators: Go into STAT EDIT and arrow up to the list name area of the particular list. Press CLEAR and then arrow down. The list will be cleared. Alternatively, you can press STAT and press 4 (for ClrList ). Enter the list name and press ENTER .

 \(\chi^{2}\) Goodness of fit Calculator

Type in the values from the observed and expected sets separated by commas, for example, 2,4,5,8,11,2.  Then hit Calculate and the test statistic, \(\chi^{2}\), and the p-value, p, will be shown.

Exercise 11.3.2

Teachers want to know which night each week their students are doing most of their homework. Most teachers think that students do homework equally throughout the week. Suppose a random sample of 49 students were asked on which night of the week they did the most homework. The results were distributed as in the table below.

From the population of students, do the nights for the highest number of students doing the majority of their homework occur with equal frequencies during a week? What type of hypothesis test should you use?

\(p\text{-value} = 0.6093\)

We decline to reject the null hypothesis. There is not enough evidence to support that students do not do the majority of their homework equally throughout the week.

Example 11.3.3

One study indicates that the number of televisions that American families have is distributed (this is the given distribution for the American population) as in the table below.

The table contains expected (\(E\)) percents.

A random sample of 600 families in the far western United States resulted in the data in the table below.

The table contains observed (\(O\)) frequency values.

Exercise 11.3.3.1

At the 1% significance level, does it appear that the distribution "number of televisions" of far western United States families is different from the distribution for the American population as a whole?

This problem asks you to test whether the far western United States families distribution fits the distribution of the American families. This test is always right-tailed.

The first table contains expected percentages. To get expected ( E ) frequencies, multiply the percentage by 600. The expected frequencies are shown in the table below.

Therefore, the expected frequencies are 60, 96, 330, 66, and 48. In the TI calculators, you can let the calculator do the math. For example, instead of 60, enter \(0.10*600\).

\(H_{0}\): The "number of televisions" distribution of far western United States families is the same as the "number of televisions" distribution of the American population.

\(H_{a}\): The "number of televisions" distribution of far western United States families is different from the "number of televisions" distribution of the American population.

Distribution for the test: \(\chi^{2}_{4}\) where \(df = (\text{the number of cells}) - 1 = 5 - 1 = 4\).

Note 11.3.3.1

\(df \neq 600 - 1\)

Calculate the test statistic: \(\chi^{2} = 29.65\)

Probability statement: \(p\text{-value} = P(\chi^{2} > 29.65) = 0.000006\)

Compare α and the p -value:

\(\alpha = 0.01\)

\(p\text{-value} = 0.000006\)

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

This means you reject the belief that the distribution for the far western states is the same as that of the American population as a whole.

Conclusion: At the 1% significance level, from the data, there is sufficient evidence to conclude that the "number of televisions" distribution for the far western United States is different from the "number of televisions" distribution for the American population as a whole.

Press STAT and ENTER . Make sure to clear lists L1 , L2 , and L3 if they have data in them (see the note at the end of Example ). Into L1 , put the observed frequencies 66 , 119 , 349 , 60 , 15 . Into L2 , put the expected frequencies .10*600, .16*600 , .55*600 , .11*600 , .08*600 . Arrow over to list L3 and up to the name area "L3" . Enter (L1-L2)^2/L2 and ENTER . Press 2nd QUIT . Press 2nd LIST and arrow over to MATH . Press 5 . You should see "sum" (Enter L3) . Rounded to 2 decimal places, you should see 29.65 . Press 2nd DISTR . Press 7 or Arrow down to 7:χ2cdf and press ENTER . Enter (29.65,1E99,4) . Rounded to four places, you should see 5.77E-6 = .000006 (rounded to six decimal places), which is the p-value.

The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF . To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list. Press STAT TESTS and Chi2 GOF . Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or draw . Make sure you clear any lists before you start.

Exercise 11.3.3

The expected percentage of the number of pets students have in their homes is distributed (this is the given distribution for the student population of the United States) as in the table below.

A random sample of 1,000 students from the Eastern United States resulted in the data in the table below.

At the 1% significance level, does it appear that the distribution “number of pets” of students in the Eastern United States is different from the distribution for the United States student population as a whole? What is the \(p\text{-value}\)?

\(p\text{-value} = 0.0036\)

We reject the null hypothesis that the distributions are the same. There is sufficient evidence to conclude that the distribution “number of pets” of students in the Eastern United States is different from the distribution for the United States student population as a whole.

Example 11.3.4

Suppose you flip two coins 100 times. The results are 20 HH , 27 HT , 30 TH , and 23 TT . Are the coins fair? Test at a 5% significance level.

This problem can be set up as a goodness-of-fit problem. The sample space for flipping two fair coins is \({HH, HT, TH, TT}\). Out of 100 flips, you would expect 25 HH , 25 HT , 25 TH , and 25 TT . This is the expected distribution. The question, "Are the coins fair?" is the same as saying, "Does the distribution of the coins (\(20 HH, 27 HT, 30 TH, 23 TT\)) fit the expected distribution?"

Random Variable: Let \(X =\) the number of heads in one flip of the two coins. \(X\) takes on the values 0, 1, 2. (There are 0, 1, or 2 heads in the flip of two coins.) Therefore, the number of cells is three . Since \(X =\) the number of heads, the observed frequencies are 20 (for two heads), 57 (for one head), and 23 (for zero heads or both tails). The expected frequencies are 25 (for two heads), 50 (for one head), and 25 (for zero heads or both tails). This test is right-tailed.

\(H_{0}\): The coins are fair.

\(H_{a}\): The coins are not fair.

Distribution for the test: \(\chi^{2}_{2}\) where \(df = 3 - 1 = 2\).

Calculate the test statistic: \(\chi^{2} = 2.14\)

Probability statement: \(p\text{-value} = P(\chi^{2} > 2.14) = 0.3430\)

\(\alpha = 0.05\)

\(p\text{-value} = 0.3430\)

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

Conclusion: There is insufficient evidence to conclude that the coins are not fair.

Press STAT and ENTER . Make sure you clear lists L1 , L2 , and L3 if they have data in them. Into L1 , put the observed frequencies 20 , 57 , 23 . Into L2 , put the expected frequencies 25 , 50 , 25 . Arrow over to list L3 and up to the name area "L3" . Enter (L1-L2)^2/L2 and ENTER . Press 2nd QUIT . Press 2nd LIST and arrow over to MATH . Press 5 . You should see "sum" . Enter L3 . Rounded to two decimal places, you should see 2.14 . Press 2nd DISTR . Arrow down to 7:χ2cdf (or press 7 ). Press ENTER . Enter 2.14,1E99,2) . Rounded to four places, you should see .3430 , which is the p-value.

Exercise 11.3.4

Students in a social studies class hypothesize that the literacy rates across the world for every region are 82%. The table below shows the actual literacy rates across the world broken down by region. What are the test statistic and the degrees of freedom?

\(\chi^{2} \text{test statistic} = 26.38\)

Press STAT and ENTER . Make sure you clear lists L1, L2, and L3 if they have data in them. Into L1, put the observed frequencies 99, 99.5, 67.3, 62.5, 91, 93.8, 61.9, 91.9, 84.5, 66.4 . Into L2 , put the expected frequencies 82, 82, 82, 82, 82, 82, 82, 82, 82, 82 . Arrow over to list L3 and up to the name area "L3" . Enter (L1-L2)^2/L2 and ENTER . Press 2nd QUIT . Press 2nd LIST and arrow over to MATH . Press 5 . You should see "sum" . Enter L3 . Rounded to two decimal places, you should see 26.38 . Press 2nd DISTR . Arrow down to 7:χ2cdf (or press 7 ). Press ENTER . Enter 26.38,1E99,9) . Rounded to four places, you should see .0018 , which is the p -value.

• Data from the U.S. Census Bureau
• Data from the College Board. Available online at http://www.collegeboard.com .
• Data from the U.S. Census Bureau, Current Population Reports.
• Ma, Y., E.R. Bertone, E.J. Stanek III, G.W. Reed, J.R. Hebert, N.L. Cohen, P.A. Merriam, I.S. Ockene, “Association between Eating Patterns and Obesity in a Free-living US Adult Population.” American Journal of Epidemiology volume 158, no. 1, pages 85-92.
• Ogden, Cynthia L., Margaret D. Carroll, Brian K. Kit, Katherine M. Flegal, “Prevalence of Obesity in the United States, 2009–2010.” NCHS Data Brief no. 82, January 2012. Available online at http://www.cdc.gov/nchs/data/databriefs/db82.pdf (accessed May 24, 2013).
• Stevens, Barbara J., “Multi-family and Commercial Solid Waste and Recycling Survey.” Arlington Count, VA. Available online at http://www.arlingtonva.us/department.../file84429.pdf (accessed May 24,2013).

Chapter Review

To assess whether a data set fits a specific distribution, you can apply the goodness-of-fit hypothesis test that uses the chi-square distribution. The null hypothesis for this test states that the data come from the assumed distribution. The test compares observed values against the values you would expect to have if your data followed the assumed distribution. The test is almost always right-tailed. Each observation or cell category must have an expected value of at least five.

Formula Review

\(\sum_k \frac{(O - E)^{2}}{E}\) goodness-of-fit test statistic where:

\(O\): observed values

\(E\): expected value

\(k\): number of different data cells or categories

\(df = k - 1\) degrees of freedom

Determine the appropriate test to be used in the next three exercises.

Exercise 11.3.5

An archeologist is calculating the distribution of the frequency of the number of artifacts she finds in a dig site. Based on previous digs, the archeologist creates an expected distribution broken down by grid sections in the dig site. Once the site has been fully excavated, she compares the actual number of artifacts found in each grid section to see if her expectation was accurate.

Exercise 11.3.6

An economist is deriving a model to predict outcomes on the stock market. He creates a list of expected points on the stock market index for the next two weeks. At the close of each day’s trading, he records the actual points on the index. He wants to see how well his model matched what actually happened.

a goodness-of-fit test

Exercise 11.3.7

A personal trainer is putting together a weight-lifting program for her clients. For a 90-day program, she expects each client to lift a specific maximum weight each week. As she goes along, she records the actual maximum weights her clients lifted. She wants to know how well her expectations met with what was observed.

Use the following information to answer the next five exercises: A teacher predicts that the distribution of grades on the final exam will be and they are recorded in the table below.

The actual distribution for a class of 20 is in the table below.

Exercise 11.3.8

\(df =\) ______

Exercise 11.3.9

State the null and alternative hypotheses.

Exercise 11.3.10

\(\chi^{2} \text{test statistic} =\) ______

Exercise 11.3.11

\(p\text{-value} =\) ______

Exercise 11.3.12

At the 5% significance level, what can you conclude?

We decline to reject the null hypothesis. There is not enough evidence to suggest that the observed test scores are significantly different from the expected test scores.

Use the following information to answer the next nine exercises: The following data are real. The cumulative number of AIDS cases reported for Santa Clara County is broken down by ethnicity as in the table below.

The percentage of each ethnic group in Santa Clara County is as in the table below.

Exercise 11.3.13

If the ethnicities of AIDS victims followed the ethnicities of the total county population, fill in the expected number of cases per ethnic group.

Perform a goodness-of-fit test to determine whether the occurrence of AIDS cases follows the ethnicities of the general population of Santa Clara County.

Exercise 11.3.14

\(H_{0}\): _______

\(H_{0}\): the distribution of AIDS cases follows the ethnicities of the general population of Santa Clara County.

Exercise 11.3.15

\(H_{a}\): _______

Exercise 11.3.16

Is this a right-tailed, left-tailed, or two-tailed test?

right-tailed

Exercise 11.3.17

degrees of freedom = _______

Exercise 11.3.18

\(\chi^{2} \text{test statistic}\) = _______

Exercise 11.3.19

\(p\text{-value} =\) _______

Exercise 11.3.20

Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade in the region corresponding to the \(p\text{-value}\).

Let \(\alpha = 0.05\)

Decision: ________________

Reason for the Decision: ________________

Conclusion (write out in complete sentences): ________________

Graph: Check student’s solution.

Decision: Reject the null hypothesis.

Reason for the Decision: \(p\text{-value} < \alpha\)

Conclusion (write out in complete sentences): The make-up of AIDS cases does not fit the ethnicities of the general population of Santa Clara County.

Exercise 11.3.21

Does it appear that the pattern of AIDS cases in Santa Clara County corresponds to the distribution of ethnic groups in this county? Why or why not?

Statistics Made Easy

How to Write Hypothesis Test Conclusions (With Examples)

A   hypothesis test is used to test whether or not some hypothesis about a population parameter is true.

To perform a hypothesis test in the real world, researchers obtain a random sample from the population and perform a hypothesis test on the sample data, using a null and alternative hypothesis:

• Null Hypothesis (H 0 ): The sample data occurs purely from chance.
• Alternative Hypothesis (H A ): The sample data is influenced by some non-random cause.

If the p-value of the hypothesis test is less than some significance level (e.g. α = .05), then we reject the null hypothesis .

Otherwise, if the p-value is not less than some significance level then we fail to reject the null hypothesis .

When writing the conclusion of a hypothesis test, we typically include:

• Whether we reject or fail to reject the null hypothesis.
• The significance level.
• A short explanation in the context of the hypothesis test.

For example, we would write:

We reject the null hypothesis at the 5% significance level.   There is sufficient evidence to support the claim that…

Or, we would write:

We fail to reject the null hypothesis at the 5% significance level.   There is not sufficient evidence to support the claim that…

The following examples show how to write a hypothesis test conclusion in both scenarios.

Example 1: Reject the Null Hypothesis Conclusion

Suppose a biologist believes that a certain fertilizer will cause plants to grow more during a one-month period than they normally do, which is currently 20 inches. To test this, she applies the fertilizer to each of the plants in her laboratory for one month.

She then performs a hypothesis test at a 5% significance level using the following hypotheses:

• H 0 : μ = 20 inches (the fertilizer will have no effect on the mean plant growth)
• H A : μ > 20 inches (the fertilizer will cause mean plant growth to increase)

Suppose the p-value of the test turns out to be 0.002.

Here is how she would report the results of the hypothesis test:

We reject the null hypothesis at the 5% significance level.   There is sufficient evidence to support the claim that this particular fertilizer causes plants to grow more during a one-month period than they normally do.

Example 2: Fail to Reject the Null Hypothesis Conclusion

Suppose the manager of a manufacturing plant wants to test whether or not some new method changes the number of defective widgets produced per month, which is currently 250. To test this, he measures the mean number of defective widgets produced before and after using the new method for one month.

He performs a hypothesis test at a 10% significance level using the following hypotheses:

• H 0 : μ after = μ before (the mean number of defective widgets is the same before and after using the new method)
• H A : μ after ≠ μ before (the mean number of defective widgets produced is different before and after using the new method)

Suppose the p-value of the test turns out to be 0.27.

Here is how he would report the results of the hypothesis test:

We fail to reject the null hypothesis at the 10% significance level.   There is not sufficient evidence to support the claim that the new method leads to a change in the number of defective widgets produced per month.

Additional Resources

The following tutorials provide additional information about hypothesis testing:

Introduction to Hypothesis Testing 4 Examples of Hypothesis Testing in Real Life How to Write a Null Hypothesis

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Hey there. My name is Zach Bobbitt. I have a Masters of Science degree in Applied Statistics and I’ve worked on machine learning algorithms for professional businesses in both healthcare and retail. I’m passionate about statistics, machine learning, and data visualization and I created Statology to be a resource for both students and teachers alike.  My goal with this site is to help you learn statistics through using simple terms, plenty of real-world examples, and helpful illustrations.

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S.3.2 hypothesis testing (p-value approach).

The P -value approach involves determining "likely" or "unlikely" by determining the probability — assuming the null hypothesis was true — of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed. If the P -value is small, say less than (or equal to) \(\alpha\), then it is "unlikely." And, if the P -value is large, say more than \(\alpha\), then it is "likely."

If the P -value is less than (or equal to) \(\alpha\), then the null hypothesis is rejected in favor of the alternative hypothesis. And, if the P -value is greater than \(\alpha\), then the null hypothesis is not rejected.

Specifically, the four steps involved in using the P -value approach to conducting any hypothesis test are:

• Specify the null and alternative hypotheses.
• Using the sample data and assuming the null hypothesis is true, calculate the value of the test statistic. Again, to conduct the hypothesis test for the population mean μ , we use the t -statistic \(t^*=\frac{\bar{x}-\mu}{s/\sqrt{n}}\) which follows a t -distribution with n - 1 degrees of freedom.
• Using the known distribution of the test statistic, calculate the P -value : "If the null hypothesis is true, what is the probability that we'd observe a more extreme test statistic in the direction of the alternative hypothesis than we did?" (Note how this question is equivalent to the question answered in criminal trials: "If the defendant is innocent, what is the chance that we'd observe such extreme criminal evidence?")
• Set the significance level, \(\alpha\), the probability of making a Type I error to be small — 0.01, 0.05, or 0.10. Compare the P -value to \(\alpha\). If the P -value is less than (or equal to) \(\alpha\), reject the null hypothesis in favor of the alternative hypothesis. If the P -value is greater than \(\alpha\), do not reject the null hypothesis.

Example S.3.2.1

Mean gpa section  .

In our example concerning the mean grade point average, suppose that our random sample of n = 15 students majoring in mathematics yields a test statistic t * equaling 2.5. Since n = 15, our test statistic t * has n - 1 = 14 degrees of freedom. Also, suppose we set our significance level α at 0.05 so that we have only a 5% chance of making a Type I error.

Right Tailed

The P -value for conducting the right-tailed test H 0 : μ = 3 versus H A : μ > 3 is the probability that we would observe a test statistic greater than t * = 2.5 if the population mean \(\mu\) really were 3. Recall that probability equals the area under the probability curve. The P -value is therefore the area under a t n - 1 = t 14 curve and to the right of the test statistic t * = 2.5. It can be shown using statistical software that the P -value is 0.0127. The graph depicts this visually.

The P -value, 0.0127, tells us it is "unlikely" that we would observe such an extreme test statistic t * in the direction of H A if the null hypothesis were true. Therefore, our initial assumption that the null hypothesis is true must be incorrect. That is, since the P -value, 0.0127, is less than \(\alpha\) = 0.05, we reject the null hypothesis H 0 : μ = 3 in favor of the alternative hypothesis H A : μ > 3.

Note that we would not reject H 0 : μ = 3 in favor of H A : μ > 3 if we lowered our willingness to make a Type I error to \(\alpha\) = 0.01 instead, as the P -value, 0.0127, is then greater than \(\alpha\) = 0.01.

Left Tailed

In our example concerning the mean grade point average, suppose that our random sample of n = 15 students majoring in mathematics yields a test statistic t * instead of equaling -2.5. The P -value for conducting the left-tailed test H 0 : μ = 3 versus H A : μ < 3 is the probability that we would observe a test statistic less than t * = -2.5 if the population mean μ really were 3. The P -value is therefore the area under a t n - 1 = t 14 curve and to the left of the test statistic t* = -2.5. It can be shown using statistical software that the P -value is 0.0127. The graph depicts this visually.

The P -value, 0.0127, tells us it is "unlikely" that we would observe such an extreme test statistic t * in the direction of H A if the null hypothesis were true. Therefore, our initial assumption that the null hypothesis is true must be incorrect. That is, since the P -value, 0.0127, is less than α = 0.05, we reject the null hypothesis H 0 : μ = 3 in favor of the alternative hypothesis H A : μ < 3.

Note that we would not reject H 0 : μ = 3 in favor of H A : μ < 3 if we lowered our willingness to make a Type I error to α = 0.01 instead, as the P -value, 0.0127, is then greater than \(\alpha\) = 0.01.

In our example concerning the mean grade point average, suppose again that our random sample of n = 15 students majoring in mathematics yields a test statistic t * instead of equaling -2.5. The P -value for conducting the two-tailed test H 0 : μ = 3 versus H A : μ ≠ 3 is the probability that we would observe a test statistic less than -2.5 or greater than 2.5 if the population mean μ really was 3. That is, the two-tailed test requires taking into account the possibility that the test statistic could fall into either tail (hence the name "two-tailed" test). The P -value is, therefore, the area under a t n - 1 = t 14 curve to the left of -2.5 and to the right of 2.5. It can be shown using statistical software that the P -value is 0.0127 + 0.0127, or 0.0254. The graph depicts this visually.

Note that the P -value for a two-tailed test is always two times the P -value for either of the one-tailed tests. The P -value, 0.0254, tells us it is "unlikely" that we would observe such an extreme test statistic t * in the direction of H A if the null hypothesis were true. Therefore, our initial assumption that the null hypothesis is true must be incorrect. That is, since the P -value, 0.0254, is less than α = 0.05, we reject the null hypothesis H 0 : μ = 3 in favor of the alternative hypothesis H A : μ ≠ 3.

Note that we would not reject H 0 : μ = 3 in favor of H A : μ ≠ 3 if we lowered our willingness to make a Type I error to α = 0.01 instead, as the P -value, 0.0254, is then greater than \(\alpha\) = 0.01.

Now that we have reviewed the critical value and P -value approach procedures for each of the three possible hypotheses, let's look at three new examples — one of a right-tailed test, one of a left-tailed test, and one of a two-tailed test.

The good news is that, whenever possible, we will take advantage of the test statistics and P -values reported in statistical software, such as Minitab, to conduct our hypothesis tests in this course.