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What Is P-Value?

Understanding p-value.

• P-Value in Hypothesis Testing

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P-Value: What It Is, How to Calculate It, and Why It Matters

Yarilet Perez is an experienced multimedia journalist and fact-checker with a Master of Science in Journalism. She has worked in multiple cities covering breaking news, politics, education, and more. Her expertise is in personal finance and investing, and real estate.

In statistics, a p-value is defined as a number that indicates how likely you are to obtain a value that is at least equal to or more than the actual observation if the null hypothesis is correct.

The p-value serves as an alternative to rejection points to provide the smallest level of significance at which the null hypothesis would be rejected. A smaller p-value means stronger evidence in favor of the alternative hypothesis.

P-value is often used to promote credibility for studies or reports by government agencies. For example, the U.S. Census Bureau stipulates that any analysis with a p-value greater than 0.10 must be accompanied by a statement that the difference is not statistically different from zero. The Census Bureau also has standards in place stipulating which p-values are acceptable for various publications.

Key Takeaways

• A p-value is a statistical measurement used to validate a hypothesis against observed data.
• A p-value measures the probability of obtaining the observed results, assuming that the null hypothesis is true.
• The lower the p-value, the greater the statistical significance of the observed difference.
• A p-value of 0.05 or lower is generally considered statistically significant.
• P-value can serve as an alternative to—or in addition to—preselected confidence levels for hypothesis testing.

Jessica Olah / Investopedia

P-values are usually found using p-value tables or spreadsheets/statistical software. These calculations are based on the assumed or known probability distribution of the specific statistic tested. The sample size, which determines the reliability of the observed data, directly influences the accuracy of the p-value calculation. he p-value approach to hypothesis testing uses the calculated he p-value approach to hypothesis testing uses the calculated P-values are calculated from the deviation between the observed value and a chosen reference value, given the probability distribution of the statistic, with a greater difference between the two values corresponding to a lower p-value.

Mathematically, the p-value is calculated using integral calculus from the area under the probability distribution curve for all values of statistics that are at least as far from the reference value as the observed value is, relative to the total area under the probability distribution curve. Standard deviations, which quantify the dispersion of data points from the mean, are instrumental in this calculation.

The calculation for a p-value varies based on the type of test performed. The three test types describe the location on the probability distribution curve: lower-tailed test, upper-tailed test, or two-tailed test . In each case, the degrees of freedom play a crucial role in determining the shape of the distribution and thus, the calculation of the p-value.

In a nutshell, the greater the difference between two observed values, the less likely it is that the difference is due to simple random chance, and this is reflected by a lower p-value.

The P-Value Approach to Hypothesis Testing

The p-value approach to hypothesis testing uses the calculated probability to determine whether there is evidence to reject the null hypothesis. This determination relies heavily on the test statistic, which summarizes the information from the sample relevant to the hypothesis being tested. The null hypothesis, also known as the conjecture, is the initial claim about a population (or data-generating process). The alternative hypothesis states whether the population parameter differs from the value of the population parameter stated in the conjecture.

In practice, the significance level is stated in advance to determine how small the p-value must be to reject the null hypothesis. Because different researchers use different levels of significance when examining a question, a reader may sometimes have difficulty comparing results from two different tests. P-values provide a solution to this problem.

Even a low p-value is not necessarily proof of statistical significance, since there is still a possibility that the observed data are the result of chance. Only repeated experiments or studies can confirm if a relationship is statistically significant.

For example, suppose a study comparing returns from two particular assets was undertaken by different researchers who used the same data but different significance levels. The researchers might come to opposite conclusions regarding whether the assets differ.

If one researcher used a confidence level of 90% and the other required a confidence level of 95% to reject the null hypothesis, and if the p-value of the observed difference between the two returns was 0.08 (corresponding to a confidence level of 92%), then the first researcher would find that the two assets have a difference that is statistically significant , while the second would find no statistically significant difference between the returns.

To avoid this problem, the researchers could report the p-value of the hypothesis test and allow readers to interpret the statistical significance themselves. This is called a p-value approach to hypothesis testing. Independent observers could note the p-value and decide for themselves whether that represents a statistically significant difference or not.

Example of P-Value

An investor claims that their investment portfolio’s performance is equivalent to that of the Standard & Poor’s (S&P) 500 Index . To determine this, the investor conducts a two-tailed test.

The null hypothesis states that the portfolio’s returns are equivalent to the S&P 500’s returns over a specified period, while the alternative hypothesis states that the portfolio’s returns and the S&P 500’s returns are not equivalent—if the investor conducted a one-tailed test , the alternative hypothesis would state that the portfolio’s returns are either less than or greater than the S&P 500’s returns.

The p-value hypothesis test does not necessarily make use of a preselected confidence level at which the investor should reset the null hypothesis that the returns are equivalent. Instead, it provides a measure of how much evidence there is to reject the null hypothesis. The smaller the p-value, the greater the evidence against the null hypothesis.

Thus, if the investor finds that the p-value is 0.001, there is strong evidence against the null hypothesis, and the investor can confidently conclude that the portfolio’s returns and the S&P 500’s returns are not equivalent.

Although this does not provide an exact threshold as to when the investor should accept or reject the null hypothesis, it does have another very practical advantage. P-value hypothesis testing offers a direct way to compare the relative confidence that the investor can have when choosing among multiple different types of investments or portfolios relative to a benchmark such as the S&P 500.

For example, for two portfolios, A and B, whose performance differs from the S&P 500 with p-values of 0.10 and 0.01, respectively, the investor can be much more confident that portfolio B, with a lower p-value, will actually show consistently different results.

Is a 0.05 P-Value Significant?

A p-value less than 0.05 is typically considered to be statistically significant, in which case the null hypothesis should be rejected. A p-value greater than 0.05 means that deviation from the null hypothesis is not statistically significant, and the null hypothesis is not rejected.

What Does a P-Value of 0.001 Mean?

A p-value of 0.001 indicates that if the null hypothesis tested were indeed true, then there would be a one-in-1,000 chance of observing results at least as extreme. This leads the observer to reject the null hypothesis because either a highly rare data result has been observed or the null hypothesis is incorrect.

How Can You Use P-Value to Compare 2 Different Results of a Hypothesis Test?

If you have two different results, one with a p-value of 0.04 and one with a p-value of 0.06, the result with a p-value of 0.04 will be considered more statistically significant than the p-value of 0.06. Beyond this simplified example, you could compare a 0.04 p-value to a 0.001 p-value. Both are statistically significant, but the 0.001 example provides an even stronger case against the null hypothesis than the 0.04.

The p-value is used to measure the significance of observational data. When researchers identify an apparent relationship between two variables, there is always a possibility that this correlation might be a coincidence. A p-value calculation helps determine if the observed relationship could arise as a result of chance.

U.S. Census Bureau. “ Statistical Quality Standard E1: Analyzing Data .”

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In Statistics, the researcher checks the significance of the observed result, which is known as test static . For this test, a hypothesis test is also utilized. The P-value  or probability value concept is used everywhere in statistical analysis. It determines the statistical significance and the measure of significance testing. In this article, let us discuss its definition, formula, table, interpretation and how to use P-value to find the significance level etc. in detail.

Table of Contents:

P-value Definition

The P-value is known as the probability value. It is defined as the probability of getting a result that is either the same or more extreme than the actual observations. The P-value is known as the level of marginal significance within the hypothesis testing that represents the probability of occurrence of the given event. The P-value is used as an alternative to the rejection point to provide the least significance at which the null hypothesis would be rejected. If the P-value is small, then there is stronger evidence in favour of the alternative hypothesis.

P-value Table

The P-value table shows the hypothesis interpretations:

Generally, the level of statistical significance is often expressed in p-value and the range between 0 and 1. The smaller the p-value, the stronger the evidence and hence, the result should be statistically significant. Hence, the rejection of the null hypothesis is highly possible, as the p-value becomes smaller.

Let us look at an example to better comprehend the concept of P-value.

Let’s say a researcher flips a coin ten times with the null hypothesis that it is fair. The total number of heads is the test statistic, which is two-tailed. Assume the researcher notices alternating heads and tails on each flip (HTHTHTHTHT). As this is the predicted number of heads, the test statistic is 5 and the p-value is 1 (totally unexceptional).

Assume that the test statistic for this research was the “number of alternations” (i.e., the number of times H followed T or T followed H), which is two-tailed once again. This would result in a test statistic of 9, which is extremely high and has a p-value of 1/2 8 = 1/256, or roughly 0.0039. This would be regarded as extremely significant, much beyond the 0.05 level. These findings suggest that the data set is exceedingly improbable to have happened by random in terms of one test statistic, yet they do not imply that the coin is biased towards heads or tails.

The data have a high p-value according to the first test statistic, indicating that the number of heads observed is not impossible. The data have a low p-value according to the second test statistic, indicating that the pattern of flips observed is extremely unlikely. There is no “alternative hypothesis,” (therefore only the null hypothesis can be rejected), and such evidence could have a variety of explanations – the data could be falsified, or the coin could have been flipped by a magician who purposefully swapped outcomes.

This example shows that the p-value is entirely dependent on the test statistic used and that p-values can only be used to reject a null hypothesis, not to explore an alternate hypothesis.

P-value Formula

We Know that P-value is a statistical measure, that helps to determine whether the hypothesis is correct or not. P-value is a number that lies between 0 and 1. The level of significance(α) is a predefined threshold that should be set by the researcher. It is generally fixed as 0.05. The formula for the calculation for P-value is

Step 1: Find out the test static Z is

P0 = assumed population proportion in the null hypothesis

N = sample size

Step 2: Look at the Z-table to find the corresponding level of P from the z value obtained.

P-Value Example

An example to find the P-value is given here.

Question: A statistician wants to test the hypothesis H 0 : μ = 120 using the alternative hypothesis Hα: μ > 120 and assuming that α = 0.05. For that, he took the sample values as

n =40, σ = 32.17 and x̄ = 105.37. Determine the conclusion for this hypothesis?

We know that,

Now substitute the given values

Now, using the test static formula, we get

t = (105.37 – 120) / 5.0865

Therefore, t = -2.8762

Using the Z-Score table , we can find the value of P(t>-2.8762)

From the table, we get

P (t<-2.8762) = P(t>2.8762) = 0.003

If P(t>-2.8762) =1- 0.003 =0.997

P- value =0.997 > 0.05

Therefore, from the conclusion, if p>0.05, the null hypothesis is accepted or fails to reject.

Hence, the conclusion is “fails to reject H 0. ”

Frequently Asked Questions on P-Value

What is meant by p-value.

The p-value is defined as the probability of obtaining the result at least as extreme as the observed result of a statistical hypothesis test, assuming that the null hypothesis is true.

What does a smaller P-value represent?

The smaller the p-value, the greater the statistical significance of the observed difference, which results in the rejection of the null hypothesis in favour of alternative hypotheses.

What does the p-value greater than 0.05 represent?

If the p-value is greater than 0.05, then the result is not statistically significant.

Can the p-value be greater than 1?

P-value means probability value, which tells you the probability of achieving the result under a certain hypothesis. Since it is a probability, its value ranges between 0 and 1, and it cannot exceed 1.

What does the p-value less than 0.05 represent?

If the p-value is less than 0.05, then the result is statistically significant, and hence we can reject the null hypothesis in favour of the alternative hypothesis.

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7.4.1 - Hypothesis Testing

Five step hypothesis testing procedure.

In the remaining lessons, we will use the following five step hypothesis testing procedure. This is slightly different from the five step procedure that we used when conducting randomization tests. 

• Check assumptions and write hypotheses.  The assumptions will vary depending on the test. In this lesson we'll be confirming that the sampling distribution is approximately normal by visually examining the randomization distribution. In later lessons you'll learn more objective assumptions. The null and alternative hypotheses will always be written in terms of population parameters; the null hypothesis will always contain the equality (i.e., \(=\)).
• Calculate the test statistic.  Here, we'll be using the formula below for the general form of the test statistic.
• Determine the p-value.  The p-value is the area under the standard normal distribution that is more extreme than the test statistic in the direction of the alternative hypothesis.
• Make a decision.  If \(p \leq \alpha\) reject the null hypothesis. If \(p>\alpha\) fail to reject the null hypothesis.
• State a "real world" conclusion.  Based on your decision in step 4, write a conclusion in terms of the original research question.

General Form of a Test Statistic

When using a standard normal distribution (i.e., z distribution), the test statistic is the standardized value that is the boundary of the p-value. Recall the formula for a z score: \(z=\frac{x-\overline x}{s}\). The formula for a test statistic will be similar. When conducting a hypothesis test the sampling distribution will be centered on the null parameter and the standard deviation is known as the standard error.

This formula puts our observed sample statistic on a standard scale (e.g., z distribution). A z score tells us where a score lies on a normal distribution in standard deviation units. The test statistic tells us where our sample statistic falls on the sampling distribution in standard error units.

7.4.1.1 - Video Example: Mean Body Temperature

Research question:  Is the mean body temperature in the population different from 98.6° Fahrenheit?

7.4.1.2 - Video Example: Correlation Between Printer Price and PPM

Research question:  Is there a positive correlation in the population between the price of an ink jet printer and how many pages per minute (ppm) it prints?

7.4.1.3 - Example: Proportion NFL Coin Toss Wins

Research question:  Is the proportion of NFL overtime coin tosses that are won different from 0.50?

StatKey was used to construct a randomization distribution:

Step 1: Check assumptions and write hypotheses

From the given StatKey output, the randomization distribution is approximately normal.

\(H_0\colon p=0.50\)

\(H_a\colon p \ne 0.50\)

Step 2: Calculate the test statistic

\(test\;statistic=\dfrac{sample\;statistic-null\;parameter}{standard\;error}\)

The sample statistic is the proportion in the original sample, 0.561. The null parameter is 0.50. And, the standard error is 0.024.

\(test\;statistic=\dfrac{0.561-0.50}{0.024}=\dfrac{0.061}{0.024}=2.542\)

Step 3: Determine the p value

The p value will be the area on the z distribution that is more extreme than the test statistic of 2.542, in the direction of the alternative hypothesis. This is a two-tailed test:

The p value is the area in the left and right tails combined: \(p=0.0055110+0.0055110=0.011022\)

Step 4: Make a decision

The p value (0.011022) is less than the standard 0.05 alpha level, therefore we reject the null hypothesis.

Step 5: State a "real world" conclusion

There is evidence that the proportion of all NFL overtime coin tosses that are won is different from 0.50

7.4.1.4 - Example: Proportion of Women Students

Research question : Are more than 50% of all World Campus STAT 200 students women?

Data were collected from a representative sample of 501 World Campus STAT 200 students. In that sample, 284 students were women and 217 were not women. 

StatKey was used to construct a sampling distribution using randomization methods:

Because this randomization distribution is approximately normal, we can find the p value by computing a standardized test statistic and using the z distribution.

The assumption here is that the sampling distribution is approximately normal. From the given StatKey output, the randomization distribution is approximately normal. 

\(H_0\colon p=0.50\) \(H_a\colon p>0.50\)

2. Calculate the test statistic

\(test\;statistic=\dfrac{sample\;statistic-hypothesized\;parameter}{standard\;error}\)

The sample statistic is \(\widehat p = 284/501 = 0.567\).

The hypothesized parameter is the value from the hypotheses: \(p_0=0.50\).

The standard error on the randomization distribution above is 0.022.

\(test\;statistic=\dfrac{0.567-0.50}{0.022}=3.045\)

3. Determine the p value

We can find the p value by constructing a standard normal distribution and finding the area under the curve that is more extreme than our observed test statistic of 3.045, in the direction of the alternative hypothesis. In other words, \(P(z>3.045)\):

Our p value is 0.0011634

4. Make a decision

Our p value is less than or equal to the standard 0.05 alpha level, therefore we reject the null hypothesis.

5. State a "real world" conclusion

There is evidence that the proportion of all World Campus STAT 200 students who are women is greater than 0.50.

7.4.1.5 - Example: Mean Quiz Score

Research question:  Is the mean quiz score different from 14 in the population?

\(H_0\colon \mu = 14\)

\(H_a\colon \mu \ne 14\)

The sample statistic is the mean in the original sample, 13.746 points. The null parameter is 14 points. And, the standard error, 0.142, can be found on the StatKey output.

\(test\;statistic=\dfrac{13.746-14}{0.142}=\dfrac{-0.254}{0.142}=-1.789\)

The p value will be the area on the z distribution that is more extreme than the test statistic of -1.789, in the direction of the alternative hypothesis:

This was a two-tailed test. The p value is the area in the left and right tails combined: \(p=0.0368074+0.0368074=0.0736148\)

The p value (0.0736148) is greater than the standard 0.05 alpha level, therefore we fail to reject the null hypothesis.

There is not enough evidence to state that the mean quiz score in the population is different from 14 points. 

7.4.1.6 - Example: Difference in Mean Commute Times

Research question:  Do the mean commute times in Atlanta and St. Louis differ in the population? 

 From the given StatKey output, the randomization distribution is approximately normal.

\(H_0: \mu_1-\mu_2=0\)

\(H_a: \mu_1 - \mu_2 \ne 0\)

Step 2: Compute the test statistic

\(test\;statistic=\dfrac{sample\;statistic - null \; parameter}{standard \;error}\)

The observed sample statistic is \(\overline x _1 - \overline x _2 = 7.14\). The null parameter is 0. And, the standard error, from the StatKey output, is 1.136.

\(test\;statistic=\dfrac{7.14-0}{1.136}=6.285\)

The p value will be the area on the z distribution that is more extreme than the test statistic of 6.285, in the direction of the alternative hypothesis:

This was a two-tailed test. The area in the two tailed combined is 0.000000. Theoretically, the p value cannot be 0 because there is always some chance that a Type I error was committed. This p value would be written as p < 0.001.

The p value is smaller than the standard 0.05 alpha level, therefore we reject the null hypothesis. 

There is evidence that the mean commute times in Atlanta and St. Louis are different in the population. 

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AP®︎/College Statistics

Course: ap®︎/college statistics   >   unit 10.

• Idea behind hypothesis testing
• Examples of null and alternative hypotheses
• Writing null and alternative hypotheses

P-values and significance tests

• Comparing P-values to different significance levels
• Estimating a P-value from a simulation
• Estimating P-values from simulations
• Using P-values to make conclusions

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Video transcript

P-value Formula 

The P-value formula is short for probability value. P-value defines the probability of getting a result that is either the same or more extreme than the other actual observations. The P-value represents the probability of occurrence of the given event. The P-value formula is used as an alternative to the rejection point to provide the least significance for which the null hypothesis would be rejected. The smaller the P-value, the stronger is the evidence in favor of the alternative hypothesis given observed frequency and expected frequency. 

What is P-value Formula?

P-value is an important statistical measure, that helps to determine whether the hypothesis is correct or not. P-value always only lies between 0 and 1. The level of significance(α) is a predefined threshold that should be set by the researcher. It is generally fixed as 0.05. The formula for the calculation for P-value is:

Step 1: Find out the test static Z is

\(Z = \frac{\hat{p}-p 0}{\sqrt{\frac{p 0(1-p 0)}{n}}}\)

• \(\hat{p}=\)Sample Proportion
• \(\mathrm{P0}=\) assumed population proportion in the null hypothesis
• N = sample size

Step 2: Look at the Z-table to find the corresponding level of P from the z value obtained.

P-value Formula

The formula to calculate the P-value is:

\(\hat{p}=\)Sample Proportion \(\mathrm{P0}=\) assumed population proportion in the null hypothesis

P-value Table

The below-mentioned P-value table helps in determining the hypothesis according to the p-value. 

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Examples Using P-value Formula 

Example 1: A statistician is testing the hypothesis H0: μ = 120 using the approach of alternative hypothesis Hα: μ > 120 and assuming that α = 0.05. The sample values that he took are as n =40, σ = 32.17 and x̄ = 105.37. What is the conclusion for this hypothesis?

We know that, \(\sigma_{\bar{x}}=\dfrac{\sigma}{\sqrt{n}}\) Now substitute the given values \(\sigma_{\bar{x}}=\dfrac{32.17}{\sqrt{40}}=5.0865\)

As per the test static formula, we get

t = (105.37 – 120) / 5.0865

Therefore, t = -2.8762

Using the Z-Score table, finding the value of P(t > -2.8762)

P (t < -2.8762) = P(t > 2.8762) = 0.003

If P(t > -2.8762) =1 - 0.003 =0.997

P- value =0.997 > 0.05

As the value of p > 0.05, the null hypothesis is accepted.

Therefore, the null hypothesis is accepted.

Example 2: P-value is 0.3105. If the level of significance is 5%, find if we can reject the null hypothesis.

Solution: Looking at the P-value table, the p-value of 0.3105 is greater than the level of significance of 0.05 (5%), we fail to reject the null hypothesis.

Example 3: P-value is 0.0219. If the level of significance is 5%, find if we can reject the null hypothesis.

Solution: Looking at the P-value table,   the p-value of 0.0219 is less than the level of significance of 0.05, we reject the null hypothesis.

FAQs on P-value Formula 

What is meant by p-value formula.

The P-value formula is short for probability value. P-value defines the probability of getting a result that is either the same or more extreme than the other actual observations. The P-value represents the probability of occurrence of the given event. The formula to calculate the p-value is: \(Z = \frac{\hat{p}-p 0}{\sqrt{\frac{p 0(1-p 0)}{n}}}\)

What is the Formula to Calculate the P-value?

What is the p-value formula table .

The P-value formula table is: 

Using the P-value Formula Table, Check if the Hypothesis is Rejected or not when the P-value is 0.354 with 5% Level of Significance.

Looking at the table, the p-value of 0.354 is greater than the level of significance of 0.05 (5%), we fail to reject the null hypothesis.

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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] .

t-test Calculator

Table of contents

Welcome to our t-test calculator! Here you can not only easily perform one-sample t-tests , but also two-sample t-tests , as well as paired t-tests .

Do you prefer to find the p-value from t-test, or would you rather find the t-test critical values? Well, this t-test calculator can do both! 😊

What does a t-test tell you? Take a look at the text below, where we explain what actually gets tested when various types of t-tests are performed. Also, we explain when to use t-tests (in particular, whether to use the z-test vs. t-test) and what assumptions your data should satisfy for the results of a t-test to be valid. If you've ever wanted to know how to do a t-test by hand, we provide the necessary t-test formula, as well as tell you how to determine the number of degrees of freedom in a t-test.

When to use a t-test?

A t-test is one of the most popular statistical tests for location , i.e., it deals with the population(s) mean value(s).

There are different types of t-tests that you can perform:

• A one-sample t-test;
• A two-sample t-test; and
• A paired t-test.

In the next section , we explain when to use which. Remember that a t-test can only be used for one or two groups . If you need to compare three (or more) means, use the analysis of variance ( ANOVA ) method.

The t-test is a parametric test, meaning that your data has to fulfill some assumptions :

• The data points are independent; AND
• The data, at least approximately, follow a normal distribution .

If your sample doesn't fit these assumptions, you can resort to nonparametric alternatives. Visit our Mann–Whitney U test calculator or the Wilcoxon rank-sum test calculator to learn more. Other possibilities include the Wilcoxon signed-rank test or the sign test.

Which t-test?

Your choice of t-test depends on whether you are studying one group or two groups:

One sample t-test

Choose the one-sample t-test to check if the mean of a population is equal to some pre-set hypothesized value .

The average volume of a drink sold in 0.33 l cans — is it really equal to 330 ml?

The average weight of people from a specific city — is it different from the national average?

Two-sample t-test

Choose the two-sample t-test to check if the difference between the means of two populations is equal to some pre-determined value when the two samples have been chosen independently of each other.

In particular, you can use this test to check whether the two groups are different from one another .

The average difference in weight gain in two groups of people: one group was on a high-carb diet and the other on a high-fat diet.

The average difference in the results of a math test from students at two different universities.

This test is sometimes referred to as an independent samples t-test , or an unpaired samples t-test .

Paired t-test

A paired t-test is used to investigate the change in the mean of a population before and after some experimental intervention , based on a paired sample, i.e., when each subject has been measured twice: before and after treatment.

In particular, you can use this test to check whether, on average, the treatment has had any effect on the population .

The change in student test performance before and after taking a course.

The change in blood pressure in patients before and after administering some drug.

How to do a t-test?

So, you've decided which t-test to perform. These next steps will tell you how to calculate the p-value from t-test or its critical values, and then which decision to make about the null hypothesis.

Decide on the alternative hypothesis :

Use a two-tailed t-test if you only care whether the population's mean (or, in the case of two populations, the difference between the populations' means) agrees or disagrees with the pre-set value.

Use a one-tailed t-test if you want to test whether this mean (or difference in means) is greater/less than the pre-set value.

Compute your T-score value :

Formulas for the test statistic in t-tests include the sample size , as well as its mean and standard deviation . The exact formula depends on the t-test type — check the sections dedicated to each particular test for more details.

Determine the degrees of freedom for the t-test:

The degrees of freedom are the number of observations in a sample that are free to vary as we estimate statistical parameters. In the simplest case, the number of degrees of freedom equals your sample size minus the number of parameters you need to estimate . Again, the exact formula depends on the t-test you want to perform — check the sections below for details.

The degrees of freedom are essential, as they determine the distribution followed by your T-score (under the null hypothesis). If there are d degrees of freedom, then the distribution of the test statistics is the t-Student distribution with d degrees of freedom . This distribution has a shape similar to N(0,1) (bell-shaped and symmetric) but has heavier tails . If the number of degrees of freedom is large (>30), which generically happens for large samples, the t-Student distribution is practically indistinguishable from N(0,1).

💡 The t-Student distribution owes its name to William Sealy Gosset, who, in 1908, published his paper on the t-test under the pseudonym "Student". Gosset worked at the famous Guinness Brewery in Dublin, Ireland, and devised the t-test as an economical way to monitor the quality of beer. Cheers! 🍺🍺🍺

p-value from t-test

Recall that the p-value is the probability (calculated under the assumption that the null hypothesis is true) that the test statistic will produce values at least as extreme as the T-score produced for your sample . As probabilities correspond to areas under the density function, p-value from t-test can be nicely illustrated with the help of the following pictures:

The following formulae say how to calculate p-value from t-test. By cdf t,d we denote the cumulative distribution function of the t-Student distribution with d degrees of freedom:

p-value from left-tailed t-test:

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

p-value from right-tailed t-test:

p-value = 1 − cdf t,d (t score )

p-value from two-tailed t-test:

p-value = 2 × cdf t,d (−|t score |)

or, equivalently: p-value = 2 − 2 × cdf t,d (|t score |)

However, the cdf of the t-distribution is given by a somewhat complicated formula. To find the p-value by hand, you would need to resort to statistical tables, where approximate cdf values are collected, or to specialized statistical software. Fortunately, our t-test calculator determines the p-value from t-test for you in the blink of an eye!

t-test critical values

Recall, that in the critical values approach to hypothesis testing, you need to set a significance level, α, before computing the critical values , which in turn give rise to critical regions (a.k.a. rejection regions).

Formulas for critical values employ the quantile function of t-distribution, i.e., the inverse of the cdf :

Critical value for left-tailed t-test: cdf t,d -1 (α)

critical region:

(-∞, cdf t,d -1 (α)]

Critical value for right-tailed t-test: cdf t,d -1 (1-α)

[cdf t,d -1 (1-α), ∞)

Critical values for two-tailed t-test: ±cdf t,d -1 (1-α/2)

(-∞, -cdf t,d -1 (1-α/2)] ∪ [cdf t,d -1 (1-α/2), ∞)

To decide the fate of the null hypothesis, just check if your T-score lies within the critical region:

If your T-score belongs to the critical region , reject the null hypothesis and accept the alternative hypothesis.

If your T-score is outside the critical region , then you don't have enough evidence to reject the null hypothesis.

How to use our t-test calculator

Choose the type of t-test you wish to perform:

A one-sample t-test (to test the mean of a single group against a hypothesized mean);

A two-sample t-test (to compare the means for two groups); or

A paired t-test (to check how the mean from the same group changes after some intervention).

Two-tailed;

Left-tailed; or

Right-tailed.

This t-test calculator allows you to use either the p-value approach or the critical regions approach to hypothesis testing!

Enter your T-score and the number of degrees of freedom . If you don't know them, provide some data about your sample(s): sample size, mean, and standard deviation, and our t-test calculator will compute the T-score and degrees of freedom for you .

Once all the parameters are present, the p-value, or critical region, will immediately appear underneath the t-test calculator, along with an interpretation!

One-sample t-test

The null hypothesis is that the population mean is equal to some value μ 0 \mu_0 μ 0 ​ .

The alternative hypothesis is that the population mean is:

• different from μ 0 \mu_0 μ 0 ​ ;
• smaller than μ 0 \mu_0 μ 0 ​ ; or
• greater than μ 0 \mu_0 μ 0 ​ .

One-sample t-test formula :

• μ 0 \mu_0 μ 0 ​ — Mean postulated in the null hypothesis;
• n n n — Sample size;
• x ˉ \bar{x} x ˉ — Sample mean; and
• s s s — Sample standard deviation.

Number of degrees of freedom in t-test (one-sample) = n − 1 n-1 n − 1 .

The null hypothesis is that the actual difference between these groups' means, μ 1 \mu_1 μ 1 ​ , and μ 2 \mu_2 μ 2 ​ , is equal to some pre-set value, Δ \Delta Δ .

The alternative hypothesis is that the difference μ 1 − μ 2 \mu_1 - \mu_2 μ 1 ​ − μ 2 ​ is:

• Different from Δ \Delta Δ ;
• Smaller than Δ \Delta Δ ; or
• Greater than Δ \Delta Δ .

In particular, if this pre-determined difference is zero ( Δ = 0 \Delta = 0 Δ = 0 ):

The null hypothesis is that the population means are equal.

The alternate hypothesis is that the population means are:

• μ 1 \mu_1 μ 1 ​ and μ 2 \mu_2 μ 2 ​ are different from one another;
• μ 1 \mu_1 μ 1 ​ is smaller than μ 2 \mu_2 μ 2 ​ ; and
• μ 1 \mu_1 μ 1 ​ is greater than μ 2 \mu_2 μ 2 ​ .

Formally, to perform a t-test, we should additionally assume that the variances of the two populations are equal (this assumption is called the homogeneity of variance ).

There is a version of a t-test that can be applied without the assumption of homogeneity of variance: it is called a Welch's t-test . For your convenience, we describe both versions.

Two-sample t-test if variances are equal

Use this test if you know that the two populations' variances are the same (or very similar).

Two-sample t-test formula (with equal variances) :

where s p s_p s p ​ is the so-called pooled standard deviation , which we compute as:

• Δ \Delta Δ — Mean difference postulated in the null hypothesis;
• n 1 n_1 n 1 ​ — First sample size;
• x ˉ 1 \bar{x}_1 x ˉ 1 ​ — Mean for the first sample;
• s 1 s_1 s 1 ​ — Standard deviation in the first sample;
• n 2 n_2 n 2 ​ — Second sample size;
• x ˉ 2 \bar{x}_2 x ˉ 2 ​ — Mean for the second sample; and
• s 2 s_2 s 2 ​ — Standard deviation in the second sample.

Number of degrees of freedom in t-test (two samples, equal variances) = n 1 + n 2 − 2 n_1 + n_2 - 2 n 1 ​ + n 2 ​ − 2 .

Two-sample t-test if variances are unequal (Welch's t-test)

Use this test if the variances of your populations are different.

Two-sample Welch's t-test formula if variances are unequal:

• s 1 s_1 s 1 ​ — Standard deviation in the first sample;
• s 2 s_2 s 2 ​ — Standard deviation in the second sample.

The number of degrees of freedom in a Welch's t-test (two-sample t-test with unequal variances) is very difficult to count. We can approximate it with the help of the following Satterthwaite formula :

Alternatively, you can take the smaller of n 1 − 1 n_1 - 1 n 1 ​ − 1 and n 2 − 1 n_2 - 1 n 2 ​ − 1 as a conservative estimate for the number of degrees of freedom.

🔎 The Satterthwaite formula for the degrees of freedom can be rewritten as a scaled weighted harmonic mean of the degrees of freedom of the respective samples: n 1 − 1 n_1 - 1 n 1 ​ − 1 and n 2 − 1 n_2 - 1 n 2 ​ − 1 , and the weights are proportional to the standard deviations of the corresponding samples.

As we commonly perform a paired t-test when we have data about the same subjects measured twice (before and after some treatment), let us adopt the convention of referring to the samples as the pre-group and post-group.

The null hypothesis is that the true difference between the means of pre- and post-populations is equal to some pre-set value, Δ \Delta Δ .

The alternative hypothesis is that the actual difference between these means is:

Typically, this pre-determined difference is zero. We can then reformulate the hypotheses as follows:

The null hypothesis is that the pre- and post-means are the same, i.e., the treatment has no impact on the population .

The alternative hypothesis:

• The pre- and post-means are different from one another (treatment has some effect);
• The pre-mean is smaller than the post-mean (treatment increases the result); or
• The pre-mean is greater than the post-mean (treatment decreases the result).

Paired t-test formula

In fact, a paired t-test is technically the same as a one-sample t-test! Let us see why it is so. Let x 1 , . . . , x n x_1, ... , x_n x 1 ​ , ... , x n ​ be the pre observations and y 1 , . . . , y n y_1, ... , y_n y 1 ​ , ... , y n ​ the respective post observations. That is, x i , y i x_i, y_i x i ​ , y i ​ are the before and after measurements of the i -th subject.

For each subject, compute the difference, d i : = x i − y i d_i := x_i - y_i d i ​ := x i ​ − y i ​ . All that happens next is just a one-sample t-test performed on the sample of differences d 1 , . . . , d n d_1, ... , d_n d 1 ​ , ... , d n ​ . Take a look at the formula for the T-score :

Δ \Delta Δ — Mean difference postulated in the null hypothesis;

n n n — Size of the sample of differences, i.e., the number of pairs;

x ˉ \bar{x} x ˉ — Mean of the sample of differences; and

s s s  — Standard deviation of the sample of differences.

Number of degrees of freedom in t-test (paired): n − 1 n - 1 n − 1

t-test vs Z-test

We use a Z-test when we want to test the population mean of a normally distributed dataset, which has a known population variance . If the number of degrees of freedom is large, then the t-Student distribution is very close to N(0,1).

Hence, if there are many data points (at least 30), you may swap a t-test for a Z-test, and the results will be almost identical. However, for small samples with unknown variance, remember to use the t-test because, in such cases, the t-Student distribution differs significantly from the N(0,1)!

🙋 Have you concluded you need to perform the z-test? Head straight to our z-test calculator !

What is a t-test?

A t-test is a widely used statistical test that analyzes the means of one or two groups of data. For instance, a t-test is performed on medical data to determine whether a new drug really helps.

What are different types of t-tests?

Different types of t-tests are:

• One-sample t-test;
• Two-sample t-test; and
• Paired t-test.

How to find the t value in a one sample t-test?

To find the t-value:

• Subtract the null hypothesis mean from the sample mean value.
• Divide the difference by the standard deviation of the sample.
• Multiply the resultant with the square root of the sample size.

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ol{padding-top:0px;}.css-4okk7a ul:not(:first-child),.css-4okk7a ol:not(:first-child){padding-top:4px;} Test setup

Choose test type

t-test for the population mean, μ, based on one independent sample . Null hypothesis H 0 : μ = μ 0  

Alternative hypothesis H 1

Test details

Significance level α

The probability that we reject a true H 0 (type I error).

Degrees of freedom

Calculated as sample size minus one.

#ez_toc_widget_sticky--1 .ez-toc-widget-sticky-container ul.ez-toc-widget-sticky-list li.active{ background-color: #ededed; } Steps To Follow

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Peek through the statistical keyhole to discover the power of the P-value. It’s the probability that study results are due to chance if the null hypothesis holds true—essential for determining statistical significance. In this article, we will learn 2 quick methods on how to calculate p value in Microsoft Excel .

Key Takeaways:

• P-Value: The Statistical Key – It unveils the likelihood of chance influencing study outcomes if the null hypothesis were accurate.
• Data Preparation is Crucial – Before diving into P-value calculations, ensure your data is pristine, organized, and free from errors.
• Simple Methods for P-Value Calculation – Excel offers user-friendly functions like T.TEST and Data Analysis ToolPak for swift P-value determination.
• Interpreting P-Values – A P-value below 0.05 suggests significant findings, while above indicates results may be due to chance.

Table of Contents

Unlocking the Mystery of the P-Value in Excel

What is a p-value.

Imagine you’re peering through a keyhole at the secret workings of statistical analysis . Right there, in the center of it all, is the P-value—an intriguing figure that holds the power to unlock the truth behind your data. In statistics, the P-value represents the probability that the results of your study could occur by chance if the null hypothesis, which is usually a statement of ‘no effect’ or ‘no difference’, were true.

It’s essentially the key that statisticians use to determine whether or not they’ve stumbled upon something that could be statistically significant.

The Role of P-Value in Hypothesis Testing

When you’re delving into hypothesis testing, the P-value acts like a trusty sidekick, helping you to navigate the tricky terrain of statistical significance. Think of it as a guide that tells you whether the observational differences or relationships uncovered in your data are likely to hold true for the wider population, or if they might just be the result of random chance.

A smaller P-value, one that dips below the agreed-upon threshold (often 0.05), whispers a hint that your findings may challenge the status quo, suggesting that the null hypothesis might not hold up. On the flip side, a larger P-value, one that looms above this benchmark, suggests that the null hypothesis cannot be dismissed—the results could simply be a fluke of chance.

Setting the Stage for P-Value Calculation

Preparing your data for analysis.

Before you get to unravel the secrets of the P-value, it’s crucial that you set the stage right by preparing your data. This means you should start off by arranging your experimental results into two clean columns, each representing different conditions or groups for comparison. Make sure that your data is neatly formatted without any stray values that could throw off your calculations. Remove duplicates , correct any entry errors, and ensure that all data points are consistent in their units of measurement.

This meticulous preparation clears the path for a smoother analysis and more reliable insights, setting a rock-solid foundation for your P-value calculations.

The Roadmap to Calculating P-Value

Method 1: using the insert function button.

If you’re looking to calculate the P-value in Excel without getting too technical, then using the Insert Function button is like taking the scenic route on a data analysis road trip. Here’s what you need to do:

STEP 1: Park your cursor in the cell where you want the P-value to appear.

STEP 2: Click the “fx” button next to the formula bar, bringing up the Insert Function dialog box.

STEP 3: Now, type “T.TEST” into the search bar. When it pops up in the list, select it to open the door to calculate the P-value.

STEP 4: Next, you’ll be greeted by the Function Arguments dialog box. This is where you’ll input your data ranges for both groups you’re comparing into “Array1” and “Array2,” select whether you’re conducting a one-tailed or two-tailed test, and choose the type of t-test you’re running.

STEP 5: After hitting “OK,” Excel does the legwork and delivers the P-value right where you need it.

By using this friendly function, Excel handles the complex calculations in the background while you comfortably navigate through the intuitive interface.

Method 2: Delving into Analysis ToolPAk

To calculate the P-value using this method, follow the steps below:

STEP 1: Before we start, go to the Data Tab in Excel Ribbon and look for Data Analysis. If not found, follow the steps below to install it. Otherwise, you can directly go to STEP 5.

STEP 2: Go to File > Options.

STEP 3: In the dialog box, go to Add-ins in the left pane. Select Excel Add-ins from the dropdown and click on Go.

STEP 4: Select Analysis ToolPak and Click OK.

STEP 5: Go to Data Tab > Data Analysis.

STEP 6: In the Data Analysis dialog box, select ‘t-Test: Paired Two Sample for Means’ and click ‘OK’.

STEP 7: In the ‘t-Test: Paired Two Sample for Means’ dialogue box, select –

• Select the first data set as Variable Range 1.
• Select the second data set as Variable Range 2.
• Select Output Range.

The p-value will be displayed in the t-test result.

Specific Paths for Different Tests

Deciphering one-tailed p-value calculations.

Engage your statistical gears, because when you’re calculating a one-tailed P-value, you’re taking a more directed approach. You’re essentially investigating if one dataset is significantly greater than or less than the other, not just different. Excel simplifies this process through its functions, allowing you to focus on whether your data leans significantly in one direction.

As you unpack your one-tailed results, remember that a P-value greater than your alpha level (commonly set at 0.05) signals that you shouldn’t reject the null hypothesis. Your findings fall within the realm of normal range, with no significant slant. However, if your P-value cruises below that threshold, it’s waving a flag that the alternative hypothesis deserves some spotlight—that there’s a significant tilt in your data.

Navigating Through Two-Tailed P-Value Computations

Navigating through two-tailed P-value computations is like setting sail in open waters—you need to be prepared for anything, as you’re testing for any significant difference without specifying a direction. In Excel, achieving this balanced approach is a matter of adjusting your sails—simply replace the tails argument with 2 when using the T.TEST function or the T Distribution functions.

In this wider sea of possibilities, a small P-value, again compared against the alpha level of 0.05 or your chosen criterion, hints at the presence of a significant difference. But unlike a one-tailed test, direction doesn’t factor in. The key is the magnitude of the difference—whether one dataset is either much greater or much less—it’s the difference itself that’s significant here.

So batten down the hatches and make sure your data is shipshape. When done correctly, two-tailed tests provide a robust analysis of your datasets, ensuring you capture any significant differences, wherever they may lie.

Interpreting Your Results

What does your p-value mean.

Your P-value is like a secret message about your data waiting to be decoded. A small P-value, typically less than 0.05, is a whisper of a revelation—it hints strongly that your results might not be due to just luck or random chance, instead pointing to a real effect or difference. In contrast, a P-value that’s larger tells you to hold your horses—there’s not enough evidence to declare that what you’re observing is anything more than a statistical mirage.

Think of your P-value as a gauge on your dashboard, with 0.05 being a critical threshold. If the needle is to the left, switch on your “Eureka!” alert. To the right, and it’s time to keep the champagne on ice, as further investigation or new data might be necessary to draw firm conclusions.

Making Informed Decisions Based on P-Values

Making decisions based on P-values is akin to using a compass on a data-driven expedition. It doesn’t point you to an X-marks-the-spot treasure chest directly, but it helps you avoid wandering aimlessly or falling into the quicksand of faulty conclusions.

With a P-value in hand, you can make decisions with greater confidence. If the P-value sails below the 0.05 threshold, it’s a strong wind in your sails towards rejecting the null hypothesis and considering your alternative hypothesis credible. On the flip side, if the P-value is above 0.05, it’s wise to consider that your hypothesis might need revisiting, or that more data could be needed to clarify the waters.

However, it’s essential to weigh your P-value against the context of your study, including the effect size and confidence intervals. Sailing toward the truth involves considering all instruments at your disposal—not just the P-value compass.

FAQs on Calculating P-Value in Excel

What is p-value in excel.

The p-value in Excel is a probability score that tells you the strength of your evidence against the null hypothesis. It’s a way to measure if the data at hand could very well be a fluke or if there’s a statistically significant difference or association. In simpler terms, the p-value helps you understand whether your findings are due to actual effects or mere random chance.

How do you find the p-value in Excel?

To find the p-value in Excel, you’d typically use built-in functions like T.TEST for t-tests or the Data Analysis tool after installing the Analysis ToolPak. These Excel features help you calculate the p-value by inputting your data range, selecting tail type, and choosing the test type, guiding you swiftly towards the statistical significances in your data.

Is it necessary to have a strong statistics background to calculate p-values in Excel?

Not at all! While a solid grip on statistics can be beneficial, Excel’s user-friendly interface and functions greatly simplify the process of calculating p-values. You can perform these calculations effectively even with a basic understanding of statistical principles, thanks to Excel’s guided tools and wide array of resources available for support.

How can one-tailed and two-tailed tests affect the p-value in Excel?

In Excel, a one-tailed test evaluates if a dataset is significantly greater or less than another, influencing the p-value by focusing on a single direction of interest . Meanwhile, a two-tailed test checks for any significant difference regardless of direction, effectively doubling the critical area and can result in a larger p-value. Choosing between the two hinges on your hypothesis specifics, directly impacting the p-value and your conclusions.

Can Excel handle all types of hypothesis tests for p-value calculations?

Excel boasts a robust suite of statistical functions, able to tackle a wide range of hypothesis tests for p-value calculations. However, it may not be the best fit for extremely specialized or complex analyses, where dedicated statistical software would offer more nuanced control and options. For most standard tests, though, Excel has you covered.

John Michaloudis

John Michaloudis is a former accountant and finance analyst at General Electric, a Microsoft MVP since 2020, an Amazon #1 bestselling author of 4 Microsoft Excel books and teacher of Microsoft Excel & Office over at his flagship Academy Online Course .

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