probability hypothesis formula

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.

Hypothesis Test Example

Learn more about calculation of probability of type I and type II errors

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An important part of inferential statistics is hypothesis testing. As with learning anything related to mathematics, it is helpful to work through several examples. The following examines an example of a hypothesis test, and calculates the probability of type I and type II errors .

We will assume that the simple conditions hold. More specifically we will assume that we have a simple random sample from a population that is either normally distributed or has a large enough sample size that we can apply the central limit theorem . We will also assume that we know the population standard deviation.

Statement of the Problem

A bag of potato chips is packaged by weight. A total of nine bags are purchased, weighed and the mean weight of these nine bags is 10.5 ounces. Suppose that the standard deviation of the population of all such bags of chips is 0.6 ounces. The stated weight on all packages is 11 ounces. Set a level of significance at 0.01.

Does the sample support the hypothesis that true population mean is less than 11 ounces?

We have a lower tailed test . This is seen by the statement of our null and alternative hypotheses :

• H 0 : μ=11.
• H a : μ < 11.

The test statistic is calculated by the formula

z = ( x -bar - μ 0 )/(σ/√ n ) = (10.5 - 11)/(0.6/√ 9) = -0.5/0.2 = -2.5.

We now need to determine how likely this value of z is due to chance alone. By using a table of z -scores we see that the probability that z is less than or equal to -2.5 is 0.0062. Since this p-value is less than the significance level , we reject the null hypothesis and accept the alternative hypothesis. The mean weight of all bags of chips is less than 11 ounces.

What is the probability of a type I error?

A type I error occurs when we reject a null hypothesis that is true. The probability of such an error is equal to the significance level. In this case, we have a level of significance equal to 0.01, thus this is the probability of a type I error.

If the population mean is actually 10.75 ounces, what is the probability of a Type II error?

We begin by reformulating our decision rule in terms of the sample mean. For a significance level of 0.01, we reject the null hypothesis when z < -2.33. By plugging this value into the formula for the test statistics, we reject the null hypothesis when

( x -bar – 11)/(0.6/√ 9) < -2.33.

Equivalently we reject the null hypothesis when 11 – 2.33(0.2) > x -bar, or when x -bar is less than 10.534. We fail to reject the null hypothesis for x -bar greater than or equal to 10.534. If the true population mean is 10.75, then the probability that x -bar is greater than or equal to 10.534 is equivalent to the probability that z is greater than or equal to -0.22. This probability, which is the probability of a type II error, is equal to 0.587.

• The Difference Between Type I and Type II Errors in Hypothesis Testing
• An Example of a Hypothesis Test
• What Level of Alpha Determines Statistical Significance?
• How to Conduct a Hypothesis Test
• Type I and Type II Errors in Statistics
• Hypothesis Test for the Difference of Two Population Proportions
• What Is a P-Value?
• How to Calculate Percent Error
• What Is the Difference Between Alpha and P-Values?
• How to Do Hypothesis Tests With the Z.TEST Function in Excel
• How to Find Critical Values with a Chi-Square Table
• What Is ANOVA?
• Scientific Method Vocabulary Terms
• Chi-Square Goodness of Fit Test
• The Runs Test for Random Sequences
• Example of Two Sample T Test and Confidence Interval

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8.1: The Elements of Hypothesis Testing

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Learning Objectives

• To understand the logical framework of tests of hypotheses.
• To learn basic terminology connected with hypothesis testing.
• To learn fundamental facts about hypothesis testing.

Types of Hypotheses

A hypothesis about the value of a population parameter is an assertion about its value. As in the introductory example we will be concerned with testing the truth of two competing hypotheses, only one of which can be true.

Definition: null hypothesis and alternative hypothesis

• The null hypothesis , denoted \(H_0\), is the statement about the population parameter that is assumed to be true unless there is convincing evidence to the contrary.
• The alternative hypothesis , denoted \(H_a\), is a statement about the population parameter that is contradictory to the null hypothesis, and is accepted as true only if there is convincing evidence in favor of it.

Definition: statistical procedure

Hypothesis testing is a statistical procedure in which a choice is made between a null hypothesis and an alternative hypothesis based on information in a sample.

The end result of a hypotheses testing procedure is a choice of one of the following two possible conclusions:

• Reject \(H_0\) (and therefore accept \(H_a\)), or
• Fail to reject \(H_0\) (and therefore fail to accept \(H_a\)).

The null hypothesis typically represents the status quo, or what has historically been true. In the example of the respirators, we would believe the claim of the manufacturer unless there is reason not to do so, so the null hypotheses is \(H_0:\mu =75\). The alternative hypothesis in the example is the contradictory statement \(H_a:\mu <75\). The null hypothesis will always be an assertion containing an equals sign, but depending on the situation the alternative hypothesis can have any one of three forms: with the symbol \(<\), as in the example just discussed, with the symbol \(>\), or with the symbol \(\neq\). The following two examples illustrate the latter two cases.

Example \(\PageIndex{1}\)

A publisher of college textbooks claims that the average price of all hardbound college textbooks is \(\$127.50\). A student group believes that the actual mean is higher and wishes to test their belief. State the relevant null and alternative hypotheses.

The default option is to accept the publisher’s claim unless there is compelling evidence to the contrary. Thus the null hypothesis is \(H_0:\mu =127.50\). Since the student group thinks that the average textbook price is greater than the publisher’s figure, the alternative hypothesis in this situation is \(H_a:\mu >127.50\).

Example \(\PageIndex{2}\)

The recipe for a bakery item is designed to result in a product that contains \(8\) grams of fat per serving. The quality control department samples the product periodically to insure that the production process is working as designed. State the relevant null and alternative hypotheses.

The default option is to assume that the product contains the amount of fat it was formulated to contain unless there is compelling evidence to the contrary. Thus the null hypothesis is \(H_0:\mu =8.0\). Since to contain either more fat than desired or to contain less fat than desired are both an indication of a faulty production process, the alternative hypothesis in this situation is that the mean is different from \(8.0\), so \(H_a:\mu \neq 8.0\).

In Example \(\PageIndex{1}\), the textbook example, it might seem more natural that the publisher’s claim be that the average price is at most \(\$127.50\), not exactly \(\$127.50\). If the claim were made this way, then the null hypothesis would be \(H_0:\mu \leq 127.50\), and the value \(\$127.50\) given in the example would be the one that is least favorable to the publisher’s claim, the null hypothesis. It is always true that if the null hypothesis is retained for its least favorable value, then it is retained for every other value.

Thus in order to make the null and alternative hypotheses easy for the student to distinguish, in every example and problem in this text we will always present one of the two competing claims about the value of a parameter with an equality. The claim expressed with an equality is the null hypothesis. This is the same as always stating the null hypothesis in the least favorable light. So in the introductory example about the respirators, we stated the manufacturer’s claim as “the average is \(75\) minutes” instead of the perhaps more natural “the average is at least \(75\) minutes,” essentially reducing the presentation of the null hypothesis to its worst case.

The first step in hypothesis testing is to identify the null and alternative hypotheses.

The Logic of Hypothesis Testing

Although we will study hypothesis testing in situations other than for a single population mean (for example, for a population proportion instead of a mean or in comparing the means of two different populations), in this section the discussion will always be given in terms of a single population mean \(\mu\).

The null hypothesis always has the form \(H_0:\mu =\mu _0\) for a specific number \(\mu _0\) (in the respirator example \(\mu _0=75\), in the textbook example \(\mu _0=127.50\), and in the baked goods example \(\mu _0=8.0\)). Since the null hypothesis is accepted unless there is strong evidence to the contrary, the test procedure is based on the initial assumption that \(H_0\) is true. This point is so important that we will repeat it in a display:

The test procedure is based on the initial assumption that \(H_0\) is true.

The criterion for judging between \(H_0\) and \(H_a\) based on the sample data is: if the value of \(\overline{X}\) would be highly unlikely to occur if \(H_0\) were true, but favors the truth of \(H_a\), then we reject \(H_0\) in favor of \(H_a\). Otherwise we do not reject \(H_0\).

Supposing for now that \(\overline{X}\) follows a normal distribution, when the null hypothesis is true the density function for the sample mean \(\overline{X}\) must be as in Figure \(\PageIndex{1}\): a bell curve centered at \(\mu _0\). Thus if \(H_0\) is true then \(\overline{X}\) is likely to take a value near \(\mu _0\) and is unlikely to take values far away. Our decision procedure therefore reduces simply to:

• if \(H_a\) has the form \(H_a:\mu <\mu _0\) then reject \(H_0\) if \(\bar{x}\) is far to the left of \(\mu _0\);
• if \(H_a\) has the form \(H_a:\mu >\mu _0\) then reject \(H_0\) if \(\bar{x}\) is far to the right of \(\mu _0\);
• if \(H_a\) has the form \(H_a:\mu \neq \mu _0\) then reject \(H_0\) if \(\bar{x}\) is far away from \(\mu _0\) in either direction.

Think of the respirator example, for which the null hypothesis is \(H_0:\mu =75\), the claim that the average time air is delivered for all respirators is \(75\) minutes. If the sample mean is \(75\) or greater then we certainly would not reject \(H_0\) (since there is no issue with an emergency respirator delivering air even longer than claimed).

If the sample mean is slightly less than \(75\) then we would logically attribute the difference to sampling error and also not reject \(H_0\) either.

Values of the sample mean that are smaller and smaller are less and less likely to come from a population for which the population mean is \(75\). Thus if the sample mean is far less than \(75\), say around \(60\) minutes or less, then we would certainly reject \(H_0\), because we know that it is highly unlikely that the average of a sample would be so low if the population mean were \(75\). This is the rare event criterion for rejection: what we actually observed \((\overline{X}<60)\) would be so rare an event if \(\mu =75\) were true that we regard it as much more likely that the alternative hypothesis \(\mu <75\) holds.

In summary, to decide between \(H_0\) and \(H_a\) in this example we would select a “rejection region” of values sufficiently far to the left of \(75\), based on the rare event criterion, and reject \(H_0\) if the sample mean \(\overline{X}\) lies in the rejection region, but not reject \(H_0\) if it does not.

The Rejection Region

Each different form of the alternative hypothesis Ha has its own kind of rejection region:

• if (as in the respirator example) \(H_a\) has the form \(H_a:\mu <\mu _0\), we reject \(H_0\) if \(\bar{x}\) is far to the left of \(\mu _0\), that is, to the left of some number \(C\), so the rejection region has the form of an interval \((-\infty ,C]\);
• if (as in the textbook example) \(H_a\) has the form \(H_a:\mu >\mu _0\), we reject \(H_0\) if \(\bar{x}\) is far to the right of \(\mu _0\), that is, to the right of some number \(C\), so the rejection region has the form of an interval \([C,\infty )\);
• if (as in the baked good example) \(H_a\) has the form \(H_a:\mu \neq \mu _0\), we reject \(H_0\) if \(\bar{x}\) is far away from \(\mu _0\) in either direction, that is, either to the left of some number \(C\) or to the right of some other number \(C′\), so the rejection region has the form of the union of two intervals \((-\infty ,C]\cup [C',\infty )\).

The key issue in our line of reasoning is the question of how to determine the number \(C\) or numbers \(C\) and \(C′\), called the critical value or critical values of the statistic, that determine the rejection region.

Definition: critical values

The critical value or critical values of a test of hypotheses are the number or numbers that determine the rejection region.

Suppose the rejection region is a single interval, so we need to select a single number \(C\). Here is the procedure for doing so. We select a small probability, denoted \(\alpha\), say \(1\%\), which we take as our definition of “rare event:” an event is “rare” if its probability of occurrence is less than \(\alpha\). (In all the examples and problems in this text the value of \(\alpha\) will be given already.) The probability that \(\overline{X}\) takes a value in an interval is the area under its density curve and above that interval, so as shown in Figure \(\PageIndex{2}\) (drawn under the assumption that \(H_0\) is true, so that the curve centers at \(\mu _0\)) the critical value \(C\) is the value of \(\overline{X}\) that cuts off a tail area \(\alpha\) in the probability density curve of \(\overline{X}\). When the rejection region is in two pieces, that is, composed of two intervals, the total area above both of them must be \(\alpha\), so the area above each one is \(\alpha /2\), as also shown in Figure \(\PageIndex{2}\).

The number \(\alpha\) is the total area of a tail or a pair of tails.

Example \(\PageIndex{3}\)

In the context of Example \(\PageIndex{2}\), suppose that it is known that the population is normally distributed with standard deviation \(\alpha =0.15\) gram, and suppose that the test of hypotheses \(H_0:\mu =8.0\) versus \(H_a:\mu \neq 8.0\) will be performed with a sample of size \(5\). Construct the rejection region for the test for the choice \(\alpha =0.10\). Explain the decision procedure and interpret it.

If \(H_0\) is true then the sample mean \(\overline{X}\) is normally distributed with mean and standard deviation

\[\begin{align} \mu _{\overline{X}} &=\mu \nonumber \\[5pt] &=8.0 \nonumber \end{align} \nonumber \]

\[\begin{align} \sigma _{\overline{X}}&=\dfrac{\sigma}{\sqrt{n}} \nonumber \\[5pt] &= \dfrac{0.15}{\sqrt{5}} \nonumber\\[5pt] &=0.067 \nonumber \end{align} \nonumber \]

Since \(H_a\) contains the \(\neq\) symbol the rejection region will be in two pieces, each one corresponding to a tail of area \(\alpha /2=0.10/2=0.05\). From Figure 7.1.6, \(z_{0.05}=1.645\), so \(C\) and \(C′\) are \(1.645\) standard deviations of \(\overline{X}\) to the right and left of its mean \(8.0\):

\[C=8.0-(1.645)(0.067) = 7.89 \; \; \text{and}\; \; C'=8.0 + (1.645)(0.067) = 8.11 \nonumber \]

The result is shown in Figure \(\PageIndex{3}\). α = 0.1

The decision procedure is: take a sample of size \(5\) and compute the sample mean \(\bar{x}\). If \(\bar{x}\) is either \(7.89\) grams or less or \(8.11\) grams or more then reject the hypothesis that the average amount of fat in all servings of the product is \(8.0\) grams in favor of the alternative that it is different from \(8.0\) grams. Otherwise do not reject the hypothesis that the average amount is \(8.0\) grams.

The reasoning is that if the true average amount of fat per serving were \(8.0\) grams then there would be less than a \(10\%\) chance that a sample of size \(5\) would produce a mean of either \(7.89\) grams or less or \(8.11\) grams or more. Hence if that happened it would be more likely that the value \(8.0\) is incorrect (always assuming that the population standard deviation is \(0.15\) gram).

Because the rejection regions are computed based on areas in tails of distributions, as shown in Figure \(\PageIndex{2}\), hypothesis tests are classified according to the form of the alternative hypothesis in the following way.

Definitions: Test classifications

• If \(H_a\) has the form \(\mu \neq \mu _0\) the test is called a two-tailed test .
• If \(H_a\) has the form \(\mu < \mu _0\) the test is called a left-tailed test .
• If \(H_a\) has the form \(\mu > \mu _0\)the test is called a right-tailed test .

Each of the last two forms is also called a one-tailed test .

Two Types of Errors

The format of the testing procedure in general terms is to take a sample and use the information it contains to come to a decision about the two hypotheses. As stated before our decision will always be either

• reject the null hypothesis \(H_0\) in favor of the alternative \(H_a\) presented, or
• do not reject the null hypothesis \(H_0\) in favor of the alternative \(H_0\) presented.

There are four possible outcomes of hypothesis testing procedure, as shown in the following table:

As the table shows, there are two ways to be right and two ways to be wrong. Typically to reject \(H_0\) when it is actually true is a more serious error than to fail to reject it when it is false, so the former error is labeled “ Type I ” and the latter error “ Type II ”.

Definition: Type I and Type II errors

In a test of hypotheses:

• A Type I error is the decision to reject \(H_0\) when it is in fact true.
• A Type II error is the decision not to reject \(H_0\) when it is in fact not true.

Unless we perform a census we do not have certain knowledge, so we do not know whether our decision matches the true state of nature or if we have made an error. We reject \(H_0\) if what we observe would be a “rare” event if \(H_0\) were true. But rare events are not impossible: they occur with probability \(\alpha\). Thus when \(H_0\) is true, a rare event will be observed in the proportion \(\alpha\) of repeated similar tests, and \(H_0\) will be erroneously rejected in those tests. Thus \(\alpha\) is the probability that in following the testing procedure to decide between \(H_0\) and \(H_a\) we will make a Type I error.

Definition: level of significance

The number \(\alpha\) that is used to determine the rejection region is called the level of significance of the test. It is the probability that the test procedure will result in a Type I error .

The probability of making a Type II error is too complicated to discuss in a beginning text, so we will say no more about it than this: for a fixed sample size, choosing \(alpha\) smaller in order to reduce the chance of making a Type I error has the effect of increasing the chance of making a Type II error . The only way to simultaneously reduce the chances of making either kind of error is to increase the sample size.

Standardizing the Test Statistic

Hypotheses testing will be considered in a number of contexts, and great unification as well as simplification results when the relevant sample statistic is standardized by subtracting its mean from it and then dividing by its standard deviation. The resulting statistic is called a standardized test statistic . In every situation treated in this and the following two chapters the standardized test statistic will have either the standard normal distribution or Student’s \(t\)-distribution.

Definition: hypothesis test

A standardized test statistic for a hypothesis test is the statistic that is formed by subtracting from the statistic of interest its mean and dividing by its standard deviation.

For example, reviewing Example \(\PageIndex{3}\), if instead of working with the sample mean \(\overline{X}\) we instead work with the test statistic

\[\frac{\overline{X}-8.0}{0.067} \nonumber \]

then the distribution involved is standard normal and the critical values are just \(\pm z_{0.05}\). The extra work that was done to find that \(C=7.89\) and \(C′=8.11\) is eliminated. In every hypothesis test in this book the standardized test statistic will be governed by either the standard normal distribution or Student’s \(t\)-distribution. Information about rejection regions is summarized in the following tables:

Every instance of hypothesis testing discussed in this and the following two chapters will have a rejection region like one of the six forms tabulated in the tables above.

No matter what the context a test of hypotheses can always be performed by applying the following systematic procedure, which will be illustrated in the examples in the succeeding sections.

Systematic Hypothesis Testing Procedure: Critical Value Approach

• Identify the null and alternative hypotheses.
• Identify the relevant test statistic and its distribution.
• Compute from the data the value of the test statistic.
• Construct the rejection region.
• Compare the value computed in Step 3 to the rejection region constructed in Step 4 and make a decision. Formulate the decision in the context of the problem, if applicable.

The procedure that we have outlined in this section is called the “Critical Value Approach” to hypothesis testing to distinguish it from an alternative but equivalent approach that will be introduced at the end of Section 8.3.

Key Takeaway

• A test of hypotheses is a statistical process for deciding between two competing assertions about a population parameter.
• The testing procedure is formalized in a five-step procedure.

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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.

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8.4: Hypothesis Test Examples for Proportions

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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.

Full Hypothesis Test Examples

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.

Set up the hypothesis test:

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:

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

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

Interpretation of the \(p\text{-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%.

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

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.

Set up the Hypothesis Test:

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

Determine the distribution needed:

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 9.6.8.2

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 9.6.8.3

b. What is a success for this problem?

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

Exercise 9.6.8.4

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 9.6.8.5

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 9.6.8.6

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 \(\PageIndex{11}\)

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 9.6.11.

Figure 9.6.12.

• 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 9.6.13.

Figure 9.6.14.

• 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.

• Data from Amit Schitai. Director of Instructional Technology and Distance Learning. LBCC.
• Data from Bloomberg Businessweek . Available online at http://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).

Contributors and Attributions

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

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The P value, or calculated probability, is the probability of finding the observed, or more extreme, results when the null hypothesis (H 0 ) of a study question is true – the definition of ‘extreme’ depends on how the hypothesis is being tested. P is also described in terms of rejecting H 0 when it is actually true, however, it is not a direct probability of this state.

The null hypothesis is usually an hypothesis of "no difference" e.g. no difference between blood pressures in group A and group B. Define a null hypothesis for each study question clearly before the start of your study.

The only situation in which you should use a one sided P value is when a large change in an unexpected direction would have absolutely no relevance to your study. This situation is unusual; if you are in any doubt then use a two sided P value.

The term significance level (alpha) is used to refer to a pre-chosen probability and the term "P value" is used to indicate a probability that you calculate after a given study.

The alternative hypothesis (H 1 ) is the opposite of the null hypothesis; in plain language terms this is usually the hypothesis you set out to investigate. For example, question is "is there a significant (not due to chance) difference in blood pressures between groups A and B if we give group A the test drug and group B a sugar pill?" and alternative hypothesis is " there is a difference in blood pressures between groups A and B if we give group A the test drug and group B a sugar pill".

If your P value is less than the chosen significance level then you reject the null hypothesis i.e. accept that your sample gives reasonable evidence to support the alternative hypothesis. It does NOT imply a "meaningful" or "important" difference; that is for you to decide when considering the real-world relevance of your result.

The choice of significance level at which you reject H 0 is arbitrary. Conventionally the 5% (less than 1 in 20 chance of being wrong), 1% and 0.1% (P < 0.05, 0.01 and 0.001) levels have been used. These numbers can give a false sense of security.

In the ideal world, we would be able to define a "perfectly" random sample, the most appropriate test and one definitive conclusion. We simply cannot. What we can do is try to optimise all stages of our research to minimise sources of uncertainty. When presenting P values some groups find it helpful to use the asterisk rating system as well as quoting the P value:

P < 0.05 *

P < 0.01 **

P < 0.001

Most authors refer to statistically significant as P < 0.05 and statistically highly significant as P < 0.001 (less than one in a thousand chance of being wrong).

The asterisk system avoids the woolly term "significant". Please note, however, that many statisticians do not like the asterisk rating system when it is used without showing P values. As a rule of thumb, if you can quote an exact P value then do. You might also want to refer to a quoted exact P value as an asterisk in text narrative or tables of contrasts elsewhere in a report.

At this point, a word about error. Type I error is the false rejection of the null hypothesis and type II error is the false acceptance of the null hypothesis. As an aid memoir: think that our cynical society rejects before it accepts.

The significance level (alpha) is the probability of type I error. The power of a test is one minus the probability of type II error (beta). Power should be maximised when selecting statistical methods. If you want to estimate sample sizes then you must understand all of the terms mentioned here.

The following table shows the relationship between power and error in hypothesis testing:

If you are interested in further details of probability and sampling theory at this point then please refer to one of the general texts listed in the reference section .

You must understand confidence intervals if you intend to quote P values in reports and papers. Statistical referees of scientific journals expect authors to quote confidence intervals with greater prominence than P values.

Notes about Type I error :

• is the incorrect rejection of the null hypothesis
• maximum probability is set in advance as alpha
• is not affected by sample size as it is set in advance
• increases with the number of tests or end points (i.e. do 20 rejections of H 0 and 1 is likely to be wrongly significant for alpha = 0.05)

Notes about Type II error :

• is the incorrect acceptance of the null hypothesis
• probability is beta
• beta depends upon sample size and alpha
• can't be estimated except as a function of the true population effect
• beta gets smaller as the sample size gets larger
• beta gets smaller as the number of tests or end points increases

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7 Chapter 7: Introduction to Hypothesis Testing

alternative hypothesis

critical value

effect size

null hypothesis

probability value

rejection region

significance level

statistical power

statistical significance

test statistic

Type I error

Type II error

This chapter lays out the basic logic and process of hypothesis testing. We will perform z  tests, which use the z  score formula from Chapter 6 and data from a sample mean to make an inference about a population.

Logic and Purpose of Hypothesis Testing

A hypothesis is a prediction that is tested in a research study. The statistician R. A. Fisher explained the concept of hypothesis testing with a story of a lady tasting tea. Here we will present an example based on James Bond who insisted that martinis should be shaken rather than stirred. Let’s consider a hypothetical experiment to determine whether Mr. Bond can tell the difference between a shaken martini and a stirred martini. Suppose we gave Mr. Bond a series of 16 taste tests. In each test, we flipped a fair coin to determine whether to stir or shake the martini. Then we presented the martini to Mr. Bond and asked him to decide whether it was shaken or stirred. Let’s say Mr. Bond was correct on 13 of the 16 taste tests. Does this prove that Mr. Bond has at least some ability to tell whether the martini was shaken or stirred?

This result does not prove that he does; it could be he was just lucky and guessed right 13 out of 16 times. But how plausible is the explanation that he was just lucky? To assess its plausibility, we determine the probability that someone who was just guessing would be correct 13/16 times or more. This probability can be computed to be .0106. This is a pretty low probability, and therefore someone would have to be very lucky to be correct 13 or more times out of 16 if they were just guessing. So either Mr. Bond was very lucky, or he can tell whether the drink was shaken or stirred. The hypothesis that he was guessing is not proven false, but considerable doubt is cast on it. Therefore, there is strong evidence that Mr. Bond can tell whether a drink was shaken or stirred.

Let’s consider another example. The case study Physicians’ Reactions sought to determine whether physicians spend less time with obese patients. Physicians were sampled randomly and each was shown a chart of a patient complaining of a migraine headache. They were then asked to estimate how long they would spend with the patient. The charts were identical except that for half the charts, the patient was obese and for the other half, the patient was of average weight. The chart a particular physician viewed was determined randomly. Thirty-three physicians viewed charts of average-weight patients and 38 physicians viewed charts of obese patients.

The mean time physicians reported that they would spend with obese patients was 24.7 minutes as compared to a mean of 31.4 minutes for normal-weight patients. How might this difference between means have occurred? One possibility is that physicians were influenced by the weight of the patients. On the other hand, perhaps by chance, the physicians who viewed charts of the obese patients tend to see patients for less time than the other physicians. Random assignment of charts does not ensure that the groups will be equal in all respects other than the chart they viewed. In fact, it is certain the groups differed in many ways by chance. The two groups could not have exactly the same mean age (if measured precisely enough such as in days). Perhaps a physician’s age affects how long the physician sees patients. There are innumerable differences between the groups that could affect how long they view patients. With this in mind, is it plausible that these chance differences are responsible for the difference in times?

To assess the plausibility of the hypothesis that the difference in mean times is due to chance, we compute the probability of getting a difference as large or larger than the observed difference (31.4 − 24.7 = 6.7 minutes) if the difference were, in fact, due solely to chance. Using methods presented in later chapters, this probability can be computed to be .0057. Since this is such a low probability, we have confidence that the difference in times is due to the patient’s weight and is not due to chance.

The Probability Value

It is very important to understand precisely what the probability values mean. In the James Bond example, the computed probability of .0106 is the probability he would be correct on 13 or more taste tests (out of 16) if he were just guessing. It is easy to mistake this probability of .0106 as the probability he cannot tell the difference. This is not at all what it means.

The probability of .0106 is the probability of a certain outcome (13 or more out of 16) assuming a certain state of the world (James Bond was only guessing). It is not the probability that a state of the world is true. Although this might seem like a distinction without a difference, consider the following example. An animal trainer claims that a trained bird can determine whether or not numbers are evenly divisible by 7. In an experiment assessing this claim, the bird is given a series of 16 test trials. On each trial, a number is displayed on a screen and the bird pecks at one of two keys to indicate its choice. The numbers are chosen in such a way that the probability of any number being evenly divisible by 7 is .50. The bird is correct on 9/16 choices. We can compute that the probability of being correct nine or more times out of 16 if one is only guessing is .40. Since a bird who is only guessing would do this well 40% of the time, these data do not provide convincing evidence that the bird can tell the difference between the two types of numbers. As a scientist, you would be very skeptical that the bird had this ability. Would you conclude that there is a .40 probability that the bird can tell the difference? Certainly not! You would think the probability is much lower than .0001.

To reiterate, the probability value is the probability of an outcome (9/16 or better) and not the probability of a particular state of the world (the bird was only guessing). In statistics, it is conventional to refer to possible states of the world as hypotheses since they are hypothesized states of the world. Using this terminology, the probability value is the probability of an outcome given the hypothesis. It is not the probability of the hypothesis given the outcome.

This is not to say that we ignore the probability of the hypothesis. If the probability of the outcome given the hypothesis is sufficiently low, we have evidence that the hypothesis is false. However, we do not compute the probability that the hypothesis is false. In the James Bond example, the hypothesis is that he cannot tell the difference between shaken and stirred martinis. The probability value is low (.0106), thus providing evidence that he can tell the difference. However, we have not computed the probability that he can tell the difference.

The Null Hypothesis

The hypothesis that an apparent effect is due to chance is called the null hypothesis , written H 0 (“ H -naught”). In the Physicians’ Reactions example, the null hypothesis is that in the population of physicians, the mean time expected to be spent with obese patients is equal to the mean time expected to be spent with average-weight patients. This null hypothesis can be written as:

The null hypothesis in a correlational study of the relationship between high school grades and college grades would typically be that the population correlation is 0. This can be written as

Although the null hypothesis is usually that the value of a parameter is 0, there are occasions in which the null hypothesis is a value other than 0. For example, if we are working with mothers in the U.S. whose children are at risk of low birth weight, we can use 7.47 pounds, the average birth weight in the U.S., as our null value and test for differences against that.

For now, we will focus on testing a value of a single mean against what we expect from the population. Using birth weight as an example, our null hypothesis takes the form:

Keep in mind that the null hypothesis is typically the opposite of the researcher’s hypothesis. In the Physicians’ Reactions study, the researchers hypothesized that physicians would expect to spend less time with obese patients. The null hypothesis that the two types of patients are treated identically is put forward with the hope that it can be discredited and therefore rejected. If the null hypothesis were true, a difference as large as or larger than the sample difference of 6.7 minutes would be very unlikely to occur. Therefore, the researchers rejected the null hypothesis of no difference and concluded that in the population, physicians intend to spend less time with obese patients.

In general, the null hypothesis is the idea that nothing is going on: there is no effect of our treatment, no relationship between our variables, and no difference in our sample mean from what we expected about the population mean. This is always our baseline starting assumption, and it is what we seek to reject. If we are trying to treat depression, we want to find a difference in average symptoms between our treatment and control groups. If we are trying to predict job performance, we want to find a relationship between conscientiousness and evaluation scores. However, until we have evidence against it, we must use the null hypothesis as our starting point.

The Alternative Hypothesis

If the null hypothesis is rejected, then we will need some other explanation, which we call the alternative hypothesis, H A or H 1 . The alternative hypothesis is simply the reverse of the null hypothesis, and there are three options, depending on where we expect the difference to lie. Thus, our alternative hypothesis is the mathematical way of stating our research question. If we expect our obtained sample mean to be above or below the null hypothesis value, which we call a directional hypothesis, then our alternative hypothesis takes the form

based on the research question itself. We should only use a directional hypothesis if we have good reason, based on prior observations or research, to suspect a particular direction. When we do not know the direction, such as when we are entering a new area of research, we use a non-directional alternative:

We will set different criteria for rejecting the null hypothesis based on the directionality (greater than, less than, or not equal to) of the alternative. To understand why, we need to see where our criteria come from and how they relate to z  scores and distributions.

Critical Values, p Values, and Significance Level

The significance level is a threshold we set before collecting data in order to determine whether or not we should reject the null hypothesis. We set this value beforehand to avoid biasing ourselves by viewing our results and then determining what criteria we should use. If our data produce values that meet or exceed this threshold, then we have sufficient evidence to reject the null hypothesis; if not, we fail to reject the null (we never “accept” the null).

Figure 7.1. The rejection region for a one-tailed test. (“ Rejection Region for One-Tailed Test ” by Judy Schmitt is licensed under CC BY-NC-SA 4.0 .)

The rejection region is bounded by a specific z  value, as is any area under the curve. In hypothesis testing, the value corresponding to a specific rejection region is called the critical value , z crit  (“ z  crit”), or z * (hence the other name “critical region”). Finding the critical value works exactly the same as finding the z  score corresponding to any area under the curve as we did in Unit 1 . If we go to the normal table, we will find that the z  score corresponding to 5% of the area under the curve is equal to 1.645 ( z = 1.64 corresponds to .0505 and z = 1.65 corresponds to .0495, so .05 is exactly in between them) if we go to the right and −1.645 if we go to the left. The direction must be determined by your alternative hypothesis, and drawing and shading the distribution is helpful for keeping directionality straight.

Suppose, however, that we want to do a non-directional test. We need to put the critical region in both tails, but we don’t want to increase the overall size of the rejection region (for reasons we will see later). To do this, we simply split it in half so that an equal proportion of the area under the curve falls in each tail’s rejection region. For a = .05, this means 2.5% of the area is in each tail, which, based on the z  table, corresponds to critical values of z * = ±1.96. This is shown in Figure 7.2 .

Figure 7.2. Two-tailed rejection region. (“ Rejection Region for Two-Tailed Test ” by Judy Schmitt is licensed under CC BY-NC-SA 4.0 .)

Thus, any z  score falling outside ±1.96 (greater than 1.96 in absolute value) falls in the rejection region. When we use z  scores in this way, the obtained value of z (sometimes called z  obtained and abbreviated z obt ) is something known as a test statistic , which is simply an inferential statistic used to test a null hypothesis. The formula for our z  statistic has not changed:

Figure 7.3. Relationship between a , z obt , and p . (“ Relationship between alpha, z-obt, and p ” by Judy Schmitt is licensed under CC BY-NC-SA 4.0 .)

When the null hypothesis is rejected, the effect is said to have statistical significance , or be statistically significant. For example, in the Physicians’ Reactions case study, the probability value is .0057. Therefore, the effect of obesity is statistically significant and the null hypothesis that obesity makes no difference is rejected. It is important to keep in mind that statistical significance means only that the null hypothesis of exactly no effect is rejected; it does not mean that the effect is important, which is what “significant” usually means. When an effect is significant, you can have confidence the effect is not exactly zero. Finding that an effect is significant does not tell you about how large or important the effect is.

Do not confuse statistical significance with practical significance. A small effect can be highly significant if the sample size is large enough.

Why does the word “significant” in the phrase “statistically significant” mean something so different from other uses of the word? Interestingly, this is because the meaning of “significant” in everyday language has changed. It turns out that when the procedures for hypothesis testing were developed, something was “significant” if it signified something. Thus, finding that an effect is statistically significant signifies that the effect is real and not due to chance. Over the years, the meaning of “significant” changed, leading to the potential misinterpretation.

The Hypothesis Testing Process

A four-step procedure.

The process of testing hypotheses follows a simple four-step procedure. This process will be what we use for the remainder of the textbook and course, and although the hypothesis and statistics we use will change, this process will not.

Step 1: State the Hypotheses

Your hypotheses are the first thing you need to lay out. Otherwise, there is nothing to test! You have to state the null hypothesis (which is what we test) and the alternative hypothesis (which is what we expect). These should be stated mathematically as they were presented above and in words, explaining in normal English what each one means in terms of the research question.

Step 2: Find the Critical Values

Step 3: calculate the test statistic and effect size.

Once we have our hypotheses and the standards we use to test them, we can collect data and calculate our test statistic—in this case z . This step is where the vast majority of differences in future chapters will arise: different tests used for different data are calculated in different ways, but the way we use and interpret them remains the same. As part of this step, we will also calculate effect size to better quantify the magnitude of the difference between our groups. Although effect size is not considered part of hypothesis testing, reporting it as part of the results is approved convention.

Step 4: Make the Decision

Finally, once we have our obtained test statistic, we can compare it to our critical value and decide whether we should reject or fail to reject the null hypothesis. When we do this, we must interpret the decision in relation to our research question, stating what we concluded, what we based our conclusion on, and the specific statistics we obtained.

Example A Movie Popcorn

Our manager is looking for a difference in the mean weight of popcorn bags compared to the population mean of 8. We will need both a null and an alternative hypothesis written both mathematically and in words. We’ll always start with the null hypothesis:

In this case, we don’t know if the bags will be too full or not full enough, so we do a two-tailed alternative hypothesis that there is a difference.

Our critical values are based on two things: the directionality of the test and the level of significance. We decided in Step 1 that a two-tailed test is the appropriate directionality. We were given no information about the level of significance, so we assume that a = .05 is what we will use. As stated earlier in the chapter, the critical values for a two-tailed z  test at a = .05 are z * = ±1.96. This will be the criteria we use to test our hypothesis. We can now draw out our distribution, as shown in Figure 7.4 , so we can visualize the rejection region and make sure it makes sense.

Figure 7.4. Rejection region for z * = ±1.96. (“ Rejection Region z+-1.96 ” by Judy Schmitt is licensed under CC BY-NC-SA 4.0 .)

Now we come to our formal calculations. Let’s say that the manager collects data and finds that the average weight of this employee’s popcorn bags is M = 7.75 cups. We can now plug this value, along with the values presented in the original problem, into our equation for z :

So our test statistic is z = −2.50, which we can draw onto our rejection region distribution as shown in Figure 7.5 .

Figure 7.5. Test statistic location. (“ Test Statistic Location z-2.50 ” by Judy Schmitt is licensed under CC BY-NC-SA 4.0 .)

Effect Size

When we reject the null hypothesis, we are stating that the difference we found was statistically significant, but we have mentioned several times that this tells us nothing about practical significance. To get an idea of the actual size of what we found, we can compute a new statistic called an effect size. Effect size gives us an idea of how large, important, or meaningful a statistically significant effect is. For mean differences like we calculated here, our effect size is Cohen’s d :

This is very similar to our formula for z , but we no longer take into account the sample size (since overly large samples can make it too easy to reject the null). Cohen’s d is interpreted in units of standard deviations, just like z . For our example:

Cohen’s d is interpreted as small, moderate, or large. Specifically, d = 0.20 is small, d = 0.50 is moderate, and d = 0.80 is large. Obviously, values can fall in between these guidelines, so we should use our best judgment and the context of the problem to make our final interpretation of size. Our effect size happens to be exactly equal to one of these, so we say that there is a moderate effect.

Effect sizes are incredibly useful and provide important information and clarification that overcomes some of the weakness of hypothesis testing. Any time you perform a hypothesis test, whether statistically significant or not, you should always calculate and report effect size.

Looking at Figure 7.5 , we can see that our obtained z  statistic falls in the rejection region. We can also directly compare it to our critical value: in terms of absolute value, −2.50 > −1.96, so we reject the null hypothesis. We can now write our conclusion:

Reject H 0 . Based on the sample of 25 bags, we can conclude that the average popcorn bag from this employee is smaller ( M = 7.75 cups) than the average weight of popcorn bags at this movie theater, and the effect size was moderate, z = −2.50, p < .05, d = 0.50.

Example B Office Temperature

Let’s do another example to solidify our understanding. Let’s say that the office building you work in is supposed to be kept at 74 degrees Fahrenheit during the summer months but is allowed to vary by 1 degree in either direction. You suspect that, as a cost saving measure, the temperature was secretly set higher. You set up a formal way to test your hypothesis.

You start by laying out the null hypothesis:

Next you state the alternative hypothesis. You have reason to suspect a specific direction of change, so you make a one-tailed test:

You know that the most common level of significance is a  = .05, so you keep that the same and know that the critical value for a one-tailed z  test is z * = 1.645. To keep track of the directionality of the test and rejection region, you draw out your distribution as shown in Figure 7.6 .

Figure 7.6. Rejection region. (“ Rejection Region z1.645 ” by Judy Schmitt is licensed under CC BY-NC-SA 4.0 .)

Now that you have everything set up, you spend one week collecting temperature data:

This value falls so far into the tail that it cannot even be plotted on the distribution ( Figure 7.7 )! Because the result is significant, you also calculate an effect size:

The effect size you calculate is definitely large, meaning someone has some explaining to do!

Figure 7.7. Obtained z statistic. (“ Obtained z5.77 ” by Judy Schmitt is licensed under CC BY-NC-SA 4.0 .)

You compare your obtained z  statistic, z = 5.77, to the critical value, z * = 1.645, and find that z > z *. Therefore you reject the null hypothesis, concluding:

Reject H 0 . Based on 5 observations, the average temperature ( M = 76.6 degrees) is statistically significantly higher than it is supposed to be, and the effect size was large, z = 5.77, p < .05, d = 2.60.

Example C Different Significance Level

Finally, let’s take a look at an example phrased in generic terms, rather than in the context of a specific research question, to see the individual pieces one more time. This time, however, we will use a stricter significance level, a = .01, to test the hypothesis.

We will use 60 as an arbitrary null hypothesis value:

We will assume a two-tailed test:

We have seen the critical values for z  tests at a = .05 levels of significance several times. To find the values for a = .01, we will go to the Standard Normal Distribution Table and find the z  score cutting off .005 (.01 divided by 2 for a two-tailed test) of the area in the tail, which is z * = ±2.575. Notice that this cutoff is much higher than it was for a = .05. This is because we need much less of the area in the tail, so we need to go very far out to find the cutoff. As a result, this will require a much larger effect or much larger sample size in order to reject the null hypothesis.

We can now calculate our test statistic. We will use s = 10 as our known population standard deviation and the following data to calculate our sample mean:

The average of these scores is M = 60.40. From this we calculate our z  statistic as:

The Cohen’s d effect size calculation is:

Our obtained z  statistic, z = 0.13, is very small. It is much less than our critical value of 2.575. Thus, this time, we fail to reject the null hypothesis. Our conclusion would look something like:

Fail to reject H 0 . Based on the sample of 10 scores, we cannot conclude that there is an effect causing the mean ( M  = 60.40) to be statistically significantly different from 60.00, z = 0.13, p > .01, d = 0.04, and the effect size supports this interpretation.

Other Considerations in Hypothesis Testing

There are several other considerations we need to keep in mind when performing hypothesis testing.

Errors in Hypothesis Testing

In the Physicians’ Reactions case study, the probability value associated with the significance test is .0057. Therefore, the null hypothesis was rejected, and it was concluded that physicians intend to spend less time with obese patients. Despite the low probability value, it is possible that the null hypothesis of no true difference between obese and average-weight patients is true and that the large difference between sample means occurred by chance. If this is the case, then the conclusion that physicians intend to spend less time with obese patients is in error. This type of error is called a Type I error. More generally, a Type I error occurs when a significance test results in the rejection of a true null hypothesis.

The second type of error that can be made in significance testing is failing to reject a false null hypothesis. This kind of error is called a Type II error . Unlike a Type I error, a Type II error is not really an error. When a statistical test is not significant, it means that the data do not provide strong evidence that the null hypothesis is false. Lack of significance does not support the conclusion that the null hypothesis is true. Therefore, a researcher should not make the mistake of incorrectly concluding that the null hypothesis is true when a statistical test was not significant. Instead, the researcher should consider the test inconclusive. Contrast this with a Type I error in which the researcher erroneously concludes that the null hypothesis is false when, in fact, it is true.

A Type II error can only occur if the null hypothesis is false. If the null hypothesis is false, then the probability of a Type II error is called b (“beta”). The probability of correctly rejecting a false null hypothesis equals 1 − b and is called statistical power . Power is simply our ability to correctly detect an effect that exists. It is influenced by the size of the effect (larger effects are easier to detect), the significance level we set (making it easier to reject the null makes it easier to detect an effect, but increases the likelihood of a Type I error), and the sample size used (larger samples make it easier to reject the null).

Misconceptions in Hypothesis Testing

Misconceptions about significance testing are common. This section lists three important ones.

• Misconception: The probability value ( p value) is the probability that the null hypothesis is false. Proper interpretation: The probability value ( p value) is the probability of a result as extreme or more extreme given that the null hypothesis is true. It is the probability of the data given the null hypothesis. It is not the probability that the null hypothesis is false.
• Misconception: A low probability value indicates a large effect. Proper interpretation: A low probability value indicates that the sample outcome (or an outcome more extreme) would be very unlikely if the null hypothesis were true. A low probability value can occur with small effect sizes, particularly if the sample size is large.
• Misconception: A non-significant outcome means that the null hypothesis is probably true. Proper interpretation: A non-significant outcome means that the data do not conclusively demonstrate that the null hypothesis is false.
• In your own words, explain what the null hypothesis is.
• What are Type I and Type II errors?
• Why do we phrase null and alternative hypotheses with population parameters and not sample means?
• Why do we state our hypotheses and decision criteria before we collect our data?
• Why do you calculate an effect size?
• z = 1.99, two-tailed test at a = .05
• z = 0.34, z * = 1.645
• p = .03, a = .05
• p = .015, a = .01

Answers to Odd-Numbered Exercises

Your answer should include mention of the baseline assumption of no difference between the sample and the population.

Alpha is the significance level. It is the criterion we use when deciding to reject or fail to reject the null hypothesis, corresponding to a given proportion of the area under the normal distribution and a probability of finding extreme scores assuming the null hypothesis is true.

We always calculate an effect size to see if our research is practically meaningful or important. NHST (null hypothesis significance testing) is influenced by sample size but effect size is not; therefore, they provide complimentary information.

“ Null Hypothesis ” by Randall Munroe/xkcd.com is licensed under CC BY-NC 2.5 .)

Introduction to Statistics in the Psychological Sciences Copyright © 2021 by Linda R. Cote Ph.D.; Rupa G. Gordon Ph.D.; Chrislyn E. Randell Ph.D.; Judy Schmitt; and Helena Marvin is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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Statistics and probability

Unit 1: analyzing categorical data, unit 2: displaying and comparing quantitative data, unit 3: summarizing quantitative data, unit 4: modeling data distributions, unit 5: exploring bivariate numerical data, unit 6: study design, unit 7: probability, unit 8: counting, permutations, and combinations, unit 9: random variables, unit 10: sampling distributions, unit 11: confidence intervals, unit 12: significance tests (hypothesis testing), unit 13: two-sample inference for the difference between groups, unit 14: inference for categorical data (chi-square tests), unit 15: advanced regression (inference and transforming), unit 16: analysis of variance (anova).

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

p-value Calculator

What is p-value, how do i calculate p-value from test statistic, how to interpret p-value, how to use the p-value calculator to find p-value from test statistic, how do i find p-value from z-score, how do i find p-value from t, p-value from chi-square score (χ² score), p-value from f-score.

Welcome to our p-value calculator! You will never again have to wonder how to find the p-value, as here you can determine the one-sided and two-sided p-values from test statistics, following all the most popular distributions: normal, t-Student, chi-squared, and Snedecor's F.

P-values appear all over science, yet many people find the concept a bit intimidating. Don't worry – in this article, we will explain not only what the p-value is but also how to interpret p-values correctly . Have you ever been curious about how to calculate the p-value by hand? We provide you with all the necessary formulae as well!

🙋 If you want to revise some basics from statistics, our normal distribution calculator is an excellent place to start.

Formally, the p-value is the probability that the test statistic will produce values at least as extreme as the value it produced for your sample . It is crucial to remember that this probability is calculated under the assumption that the null hypothesis H 0 is true !

More intuitively, p-value answers the question:

Assuming that I live in a world where the null hypothesis holds, how probable is it that, for another sample, the test I'm performing will generate a value at least as extreme as the one I observed for the sample I already have?

It is the alternative hypothesis that determines what "extreme" actually means , so the p-value depends on the alternative hypothesis that you state: left-tailed, right-tailed, or two-tailed. In the formulas below, S stands for a test statistic, x for the value it produced for a given sample, and Pr(event | H 0 ) is the probability of an event, calculated under the assumption that H 0 is true:

Left-tailed test: p-value = Pr(S ≤ x | H 0 )

Right-tailed test: p-value = Pr(S ≥ x | H 0 )

Two-tailed test:

p-value = 2 × min{Pr(S ≤ x | H 0 ), Pr(S ≥ x | H 0 )}

(By min{a,b} , we denote the smaller number out of a and b .)

If the distribution of the test statistic under H 0 is symmetric about 0 , then: p-value = 2 × Pr(S ≥ |x| | H 0 )

or, equivalently: p-value = 2 × Pr(S ≤ -|x| | H 0 )

As a picture is worth a thousand words, let us illustrate these definitions. Here, we use the fact that the probability can be neatly depicted as the area under the density curve for a given distribution. We give two sets of pictures: one for a symmetric distribution and the other for a skewed (non-symmetric) distribution.

• Symmetric case: normal distribution:

• Non-symmetric case: chi-squared distribution:

In the last picture (two-tailed p-value for skewed distribution), the area of the left-hand side is equal to the area of the right-hand side.

To determine the p-value, you need to know the distribution of your test statistic under the assumption that the null hypothesis is true . Then, with the help of the cumulative distribution function ( cdf ) of this distribution, we can express the probability of the test statistics being at least as extreme as its value x for the sample:

Left-tailed test:

p-value = cdf(x) .

Right-tailed test:

p-value = 1 - cdf(x) .

p-value = 2 × min{cdf(x) , 1 - cdf(x)} .

If the distribution of the test statistic under H 0 is symmetric about 0 , then a two-sided p-value can be simplified to p-value = 2 × cdf(-|x|) , or, equivalently, as p-value = 2 - 2 × cdf(|x|) .

The probability distributions that are most widespread in hypothesis testing tend to have complicated cdf formulae, and finding the p-value by hand may not be possible. You'll likely need to resort to a computer or to a statistical table, where people have gathered approximate cdf values.

Well, you now know how to calculate the p-value, but… why do you need to calculate this number in the first place? In hypothesis testing, the p-value approach is an alternative to the critical value approach . Recall that the latter requires researchers to pre-set the significance level, α, which is the probability of rejecting the null hypothesis when it is true (so of type I error ). Once you have your p-value, you just need to compare it with any given α to quickly decide whether or not to reject the null hypothesis at that significance level, α. For details, check the next section, where we explain how to interpret p-values.

As we have mentioned above, the p-value is the answer to the following question:

What does that mean for you? Well, you've got two options:

• A high p-value means that your data is highly compatible with the null hypothesis; and
• A small p-value provides evidence against the null hypothesis , as it means that your result would be very improbable if the null hypothesis were true.

However, it may happen that the null hypothesis is true, but your sample is highly unusual! For example, imagine we studied the effect of a new drug and got a p-value of 0.03 . This means that in 3% of similar studies, random chance alone would still be able to produce the value of the test statistic that we obtained, or a value even more extreme, even if the drug had no effect at all!

The question "what is p-value" can also be answered as follows: p-value is the smallest level of significance at which the null hypothesis would be rejected. So, if you now want to make a decision on the null hypothesis at some significance level α , just compare your p-value with α :

• If p-value ≤ α , then you reject the null hypothesis and accept the alternative hypothesis; and
• If p-value ≥ α , then you don't have enough evidence to reject the null hypothesis.

Obviously, the fate of the null hypothesis depends on α . For instance, if the p-value was 0.03 , we would reject the null hypothesis at a significance level of 0.05 , but not at a level of 0.01 . That's why the significance level should be stated in advance and not adapted conveniently after the p-value has been established! A significance level of 0.05 is the most common value, but there's nothing magical about it. Here, you can see what too strong a faith in the 0.05 threshold can lead to. It's always best to report the p-value, and allow the reader to make their own conclusions.

Also, bear in mind that subject area expertise (and common reason) is crucial. Otherwise, mindlessly applying statistical principles, you can easily arrive at statistically significant, despite the conclusion being 100% untrue.

As our p-value calculator is here at your service, you no longer need to wonder how to find p-value from all those complicated test statistics! Here are the steps you need to follow:

Pick the alternative hypothesis : two-tailed, right-tailed, or left-tailed.

Tell us the distribution of your test statistic under the null hypothesis: is it N(0,1), t-Student, chi-squared, or Snedecor's F? If you are unsure, check the sections below, as they are devoted to these distributions.

If needed, specify the degrees of freedom of the test statistic's distribution.

Enter the value of test statistic computed for your data sample.

Our calculator determines the p-value from the test statistic and provides the decision to be made about the null hypothesis. The standard significance level is 0.05 by default.

Go to the advanced mode if you need to increase the precision with which the calculations are performed or change the significance level .

In terms of the cumulative distribution function (cdf) of the standard normal distribution, which is traditionally denoted by Φ , the p-value is given by the following formulae:

Left-tailed z-test:

p-value = Φ(Z score )

Right-tailed z-test:

p-value = 1 - Φ(Z score )

Two-tailed z-test:

p-value = 2 × Φ(−|Z score |)

p-value = 2 - 2 × Φ(|Z score |)

🙋 To learn more about Z-tests, head to Omni's Z-test calculator .

We use the Z-score if the test statistic approximately follows the standard normal distribution N(0,1) . Thanks to the central limit theorem, you can count on the approximation if you have a large sample (say at least 50 data points) and treat your distribution as normal.

A Z-test most often refers to testing the population mean , or the difference between two population means, in particular between two proportions. You can also find Z-tests in maximum likelihood estimations.

The p-value from the t-score is given by the following formulae, in which cdf t,d stands for the cumulative distribution function of the t-Student distribution with d degrees of freedom:

Left-tailed t-test:

p-value = cdf t,d (t score )

Right-tailed t-test:

p-value = 1 - cdf t,d (t score )

Two-tailed t-test:

p-value = 2 × cdf t,d (−|t score |)

p-value = 2 - 2 × cdf t,d (|t score |)

Use the t-score option if your test statistic follows the t-Student distribution . This distribution has a shape similar to N(0,1) (bell-shaped and symmetric) but has heavier tails – the exact shape depends on the parameter called the degrees of freedom . If the number of degrees of freedom is large (>30), which generically happens for large samples, the t-Student distribution is practically indistinguishable from the normal distribution N(0,1).

The most common t-tests are those for population means with an unknown population standard deviation, or for the difference between means of two populations , with either equal or unequal yet unknown population standard deviations. There's also a t-test for paired (dependent) samples .

🙋 To get more insights into t-statistics, we recommend using our t-test calculator .

Use the χ²-score option when performing a test in which the test statistic follows the χ²-distribution .

This distribution arises if, for example, you take the sum of squared variables, each following the normal distribution N(0,1). Remember to check the number of degrees of freedom of the χ²-distribution of your test statistic!

How to find the p-value from chi-square-score ? You can do it with the help of the following formulae, in which cdf χ²,d denotes the cumulative distribution function of the χ²-distribution with d degrees of freedom:

Left-tailed χ²-test:

p-value = cdf χ²,d (χ² score )

Right-tailed χ²-test:

p-value = 1 - cdf χ²,d (χ² score )

Remember that χ²-tests for goodness-of-fit and independence are right-tailed tests! (see below)

Two-tailed χ²-test:

p-value = 2 × min{cdf χ²,d (χ² score ), 1 - cdf χ²,d (χ² score )}

(By min{a,b} , we denote the smaller of the numbers a and b .)

The most popular tests which lead to a χ²-score are the following:

Testing whether the variance of normally distributed data has some pre-determined value. In this case, the test statistic has the χ²-distribution with n - 1 degrees of freedom, where n is the sample size. This can be a one-tailed or two-tailed test .

Goodness-of-fit test checks whether the empirical (sample) distribution agrees with some expected probability distribution. In this case, the test statistic follows the χ²-distribution with k - 1 degrees of freedom, where k is the number of classes into which the sample is divided. This is a right-tailed test .

Independence test is used to determine if there is a statistically significant relationship between two variables. In this case, its test statistic is based on the contingency table and follows the χ²-distribution with (r - 1)(c - 1) degrees of freedom, where r is the number of rows, and c is the number of columns in this contingency table. This also is a right-tailed test .

Finally, the F-score option should be used when you perform a test in which the test statistic follows the F-distribution , also known as the Fisher–Snedecor distribution. The exact shape of an F-distribution depends on two degrees of freedom .

To see where those degrees of freedom come from, consider the independent random variables X and Y , which both follow the χ²-distributions with d 1 and d 2 degrees of freedom, respectively. In that case, the ratio (X/d 1 )/(Y/d 2 ) follows the F-distribution, with (d 1 , d 2 ) -degrees of freedom. For this reason, the two parameters d 1 and d 2 are also called the numerator and denominator degrees of freedom .

The p-value from F-score is given by the following formulae, where we let cdf F,d1,d2 denote the cumulative distribution function of the F-distribution, with (d 1 , d 2 ) -degrees of freedom:

Left-tailed F-test:

p-value = cdf F,d1,d2 (F score )

Right-tailed F-test:

p-value = 1 - cdf F,d1,d2 (F score )

Two-tailed F-test:

p-value = 2 × min{cdf F,d1,d2 (F score ), 1 - cdf F,d1,d2 (F score )}

Below we list the most important tests that produce F-scores. All of them are right-tailed tests .

A test for the equality of variances in two normally distributed populations . Its test statistic follows the F-distribution with (n - 1, m - 1) -degrees of freedom, where n and m are the respective sample sizes.

ANOVA is used to test the equality of means in three or more groups that come from normally distributed populations with equal variances. We arrive at the F-distribution with (k - 1, n - k) -degrees of freedom, where k is the number of groups, and n is the total sample size (in all groups together).

A test for overall significance of regression analysis . The test statistic has an F-distribution with (k - 1, n - k) -degrees of freedom, where n is the sample size, and k is the number of variables (including the intercept).

With the presence of the linear relationship having been established in your data sample with the above test, you can calculate the coefficient of determination, R 2 , which indicates the strength of this relationship . You can do it by hand or use our coefficient of determination calculator .

A test to compare two nested regression models . The test statistic follows the F-distribution with (k 2 - k 1 , n - k 2 ) -degrees of freedom, where k 1 and k 2 are the numbers of variables in the smaller and bigger models, respectively, and n is the sample size.

You may notice that the F-test of an overall significance is a particular form of the F-test for comparing two nested models: it tests whether our model does significantly better than the model with no predictors (i.e., the intercept-only model).

Can p-value be negative?

No, the p-value cannot be negative. This is because probabilities cannot be negative, and the p-value is the probability of the test statistic satisfying certain conditions.

What does a high p-value mean?

A high p-value means that under the null hypothesis, there's a high probability that for another sample, the test statistic will generate a value at least as extreme as the one observed in the sample you already have. A high p-value doesn't allow you to reject the null hypothesis.

What does a low p-value mean?

A low p-value means that under the null hypothesis, there's little probability that for another sample, the test statistic will generate a value at least as extreme as the one observed for the sample you already have. A low p-value is evidence in favor of the alternative hypothesis – it allows you to reject the null hypothesis.

BMR - Harris-Benedict equation

Bertrand's box paradox, possible combinations.

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