How to read t-test, z, chi-square calculator output

Q: What is the lowercase p on a calculator test screen?

It is the p-value, the probability of a result at least as extreme as your sample if the null hypothesis were true. It is not a proportion. On a proportion test the sample proportion appears on a separate line labeled p-hat (p-hat), so read the two lines carefully.

Q: What does a p-value like 2.5E-4 mean?

That is scientific notation for 0.00025. The E-4 tells you to move the decimal point four places to the left, which gives a very small p-value, not a large one. A tiny p-value like this is strong evidence against the null hypothesis.

Q: Is writing only the calculator command enough on a free-response question?

As general advice, no. A bare command such as the menu name leaves out the statistic, the p-value, and the conclusion in context that show your reasoning, so the safer practice is to write those out yourself. Treat this as guidance for clear communication rather than a fixed scoring rule.

Q: How do I find degrees of freedom if the screen does not show them?

For a one-sample t-test, degrees of freedom equal n - 1, the sample size minus one. For a chi-square test for homogeneity or independence, they equal (r - 1)(c - 1) for a table with r rows and c columns. A proportion z-test has no degrees of freedom.

By Jude Wallis · Published July 28, 2026

On a graphing calculator, test output shows the test statistic (t, z, or chi-square), the p-value, degrees of freedom where relevant, and your sample statistics echoed back. Copy the statistic, the p-value, and a conclusion in context into your answer, not the calculator command.

This guide supports reading calculator output across the inference procedures of the Fall 2026 AP Statistics course, mainly the proportion and chi-square tests in Unit 3 and the mean tests in Unit 4, where a graphing calculator with statistical capabilities is expected.

How to read t-test calculator output

When you run an inference test, a graphing calculator returns a short results screen. Learning how to read t test calculator output means recognizing four kinds of numbers on that screen: the test statistic, the p-value, the degrees of freedom, and the sample statistics the calculator echoes back from your data.

For a test about a mean, the test statistic is labeled $t$ (the t statistic), a standardized measure of how far your sample mean sits from the value in the null hypothesis. The p-value is labeled $p$ , the probability of a result at least as extreme as yours if the null hypothesis were true. The remaining lines repeat your inputs so you can confirm the machine used the data you meant to enter.

This page keeps the walkthrough device-generic, because menu paths differ across calculators and the exam expects a graphing calculator with statistical capabilities rather than one specific model. Field names vary a little too, so focus on what each number means, not on its exact label.

The fields on a results screen, one by one

Here is the generic anatomy of calculator output in AP Statistics, using a one-sample t-test as the model. Your screen will show most of these lines, though the order and the exact labels depend on the device.

The alternative hypothesis, echoed at the top (for example $\mu > 20$ , read as the population mean $\mu$ (mu) is greater than 20). This confirms you set a one-sided or two-sided test the way you intended.
The test statistic, $t$ for a mean or $z$ for a proportion (the z statistic, a standardized score). Report it to two or three decimals.
The p-value, labeled $p$ . Watch for scientific notation: a display of $2.5\text{E-}4$ means $0.00025$ , a very small p-value, not a large one.
Degrees of freedom, labeled $df$ , which appear for t procedures and chi-square procedures. For a one-sample t-test, $df = n - 1$ .
The sample statistics, echoed from your data: the sample mean $\bar{x}$ (x-bar), the sample standard deviation written $\text{Sx}$ or $s$ , and the sample size $n$ . Checking these against your data is the fastest way to catch a typo in your entry.

One label causes constant confusion. Lowercase $p$ on the output is the p-value. It is not a proportion. The sample proportion, when you run a proportion test, is written $\hat{p}$ (p-hat) on its own separate line.

How output fields map to State, Plan, Do, and Conclude

A written inference answer is commonly organized into four steps: State, Plan, Do, and Conclude. Each calculator field lands in a specific step, so the output works like a checklist for the middle of your response.

State: name the parameter and write the hypotheses in terms of that parameter, such as $H_0: \mu = 20$ and $H_a: \mu > 20$ . The calculator does not do this for you, though the echoed alternative confirms your setup.
Plan: name the procedure (a one-sample t-test) and check the conditions. No output field goes here, but naming the test tells the reader which statistic to expect.
Do: this is where the output lives. Copy the test statistic, the degrees of freedom, and the p-value. The echoed sample statistics support this step.
Conclude: compare the p-value to your significance level $\alpha$ (alpha, the threshold you set, often 0.05) and state a decision in the context of the question.

The AP course framework describes the same sequence for a hypothesis test: identify the parameter and hypotheses, identify a procedure and check conditions, calculate a test statistic and p-value, and give a conclusion in context justified by linking the p-value to the significance level. See null vs alternative hypothesis if the State step still feels shaky.

What to copy into your answer, and what not to

The output is a source of numbers, not a finished answer. A good habit is to copy three things into the Do and Conclude steps: the test statistic, the p-value (with degrees of freedom for a t or chi-square test), and a sentence of interpretation in context.

What generally does not communicate your reasoning is a bare calculator command. Writing only the name of a menu function, or only "T-Test" with nothing beside it, leaves out the parameter, the statistic, the p-value, and the conclusion a reader needs to follow your work. Treat that as advice rather than a fixed scoring rule, but the safe practice is to show the numbers and the reasoning in your own writing.

The AP framework is explicit that hypotheses must be stated in terms of population parameters, not sample statistics, and that a decision should carry a numerical justification, for example rejecting the null hypothesis because the p-value is at or below 0.05. The calculator hands you the p-value; the justification and the context are yours to write.

How to read z-test and chi-square output

A proportion z-test screen follows the same pattern with different labels. The test statistic is $z$ , the p-value is $p$ , the sample proportion is $\hat{p}$ , and the sample size is $n$ . There are no degrees of freedom for a z procedure. A two-proportion test simply echoes two sample proportions and two sample sizes.

A chi-square test for homogeneity or independence reports the chi-square statistic $\chi^2$ (chi-square), the p-value $p$ , and the degrees of freedom $df$ . For a table with $r$ rows and $c$ columns of categories, $df = (r - 1)(c - 1)$ . Many calculators also store a matrix of expected counts, which you can display to check the condition that the expected counts are large enough.

The reading habit does not change across procedures. Find the statistic, find the p-value, note the degrees of freedom if the procedure has them, and confirm the echoed sample values match your data before you write anything down. For help choosing the procedure in the first place, see which statistical test to use.

How this fits the Fall 2026 AP course

On the AP Statistics course that takes effect in Fall 2026, with the first exam in May 2027, a graphing calculator with statistical capabilities is expected for both the multiple-choice and free-response sections. Formulas and tables are provided for both sections, so your job is to choose the right procedure and read its output correctly, not to memorize critical values.

Test output shows up across the inference units: proportion z-tests and chi-square tests for homogeneity or independence in Unit 3, Inference for Categorical Data, and t-tests for means in Unit 4, Inference for Quantitative Data. Free-response Question 3 is an inference question, a hypothesis test or a confidence interval, where reading and reporting output cleanly matters most.

For the full structure of a written response, see the AP Statistics FRQ guide. To check a p-value you have read off the screen, use the p-value calculator.

Reading a one-sample t-test screen

A quality check on battery lifetimes, in hours, for a sample of 6 batteries gives the data 20, 22, 19, 23, 21, 21. The claim is that the mean lifetime is 20 hours, and you test whether the true mean is greater than 20. The calculator's t-test screen shows: alternative $\mu > 20$ ; $t = 1.732051$ ; $p = 0.0719$ ; $df = 5$ ; $\bar{x} = 21$ ; $\text{Sx} = 1.414214$ ; $n = 6$ . Read the output and say what you would write.

Confirm the echoed sample mean matches your data. Sample mean $\bar{x} = \frac{20 + 22 + 19 + 23 + 21 + 21}{6} = \frac{126}{6} = 21$ hours, which matches the screen.
Confirm the test statistic. The standard error is $\frac{\text{Sx}}{\sqrt{n}} = \frac{1.414214}{\sqrt{6}} = \frac{1.414214}{2.449490} = 0.577350$ . Then $t = \frac{\bar{x} - \mu_0}{SE} = \frac{21 - 20}{0.577350} = 1.732051$ , matching the $t$ on the screen (here $\mu_0 = 20$ is the hypothesized mean).
Read the degrees of freedom: $df = n - 1 = 6 - 1 = 5$ , matching the screen.
Read the p-value directly: $p = 0.0719$ . Compare it to a significance level of $\alpha = 0.05$ . Since $0.0719 > 0.05$ , you fail to reject the null hypothesis.
Write it up. Do: a one-sample t-test gives $t = 1.73$ with $df = 5$ and a p-value of $0.0719$ . Conclude: because $0.0719 > 0.05$ , there is not convincing evidence that the mean battery lifetime exceeds 20 hours.

Report $t \approx 1.73$ , $df = 5$ , and a p-value of about $0.0719$ , then state a conclusion in context: at $\alpha = 0.05$ you fail to reject $H_0$ , so the data do not give convincing evidence that the mean lifetime is more than 20 hours. Do not write only the calculator command.

Reading a one-proportion z-test screen

In a poll of 400 randomly selected voters, 220 support a measure. You test whether the population proportion of support differs from 0.5. The one-proportion z-test screen shows: alternative prop is not equal to 0.5; $z = 2$ ; $p = 0.0455$ ; $\hat{p} = 0.55$ ; $n = 400$ . Read the output.

Confirm the sample proportion. $\hat{p} = \frac{220}{400} = 0.55$ , matching the screen. Note that $\hat{p} = 0.55$ is the sample proportion, while the lowercase $p = 0.0455$ is the p-value; they sit on different lines.
Confirm the test statistic. Using the hypothesized proportion $p_0 = 0.5$ , the standard error is $\sqrt{\frac{p_0(1 - p_0)}{n}} = \sqrt{\frac{0.5 \times 0.5}{400}} = \sqrt{\frac{0.25}{400}} = \sqrt{0.000625} = 0.025$ . Then $z = \frac{\hat{p} - p_0}{0.025} = \frac{0.55 - 0.5}{0.025} = \frac{0.05}{0.025} = 2$ , matching the screen.
A z procedure has no degrees of freedom, so there is no $df$ line to read.
Read the p-value: $p = 0.0455$ . This is a two-sided test, so the calculator has already doubled the one-tail area; from a z-table, $P(Z > 2) = 0.02275$ and $2 \times 0.02275 = 0.0455$ .
Compare to $\alpha = 0.05$ . Since $0.0455 < 0.05$ , you reject the null hypothesis.

Report $z = 2.00$ and a p-value of about $0.0455$ , with sample proportion $\hat{p} = 0.55$ . At $\alpha = 0.05$ you reject $H_0$ because $0.0455 < 0.05$ , so there is convincing evidence that the proportion of support differs from 0.5.

Reading a chi-square test screen

A chi-square test for homogeneity compares two groups across two categories. The observed counts are 30 and 20 in group 1, and 20 and 30 in group 2. The calculator's chi-square screen shows: $\chi^2 = 4$ ; $p = 0.0455$ ; $df = 1$ . Read the output and confirm the numbers.

Find the expected counts. Each row totals 50, each column totals 50, and the grand total is 100, so every expected count is $\frac{(\text{row total})(\text{column total})}{\text{grand total}} = \frac{50 \times 50}{100} = 25$ .
Confirm the chi-square statistic: $\chi^2 = \sum \frac{(\text{Observed} - \text{Expected})^2}{\text{Expected}} = \frac{(30 - 25)^2 + (20 - 25)^2 + (20 - 25)^2 + (30 - 25)^2}{25} = \frac{25 + 25 + 25+ 25}{25} = \frac{100}{25} = 4$ , matching the screen.
Confirm the degrees of freedom for a table with $r = 2$ rows and $c = 2$ columns: $df = (r - 1)(c - 1) = (2 - 1)(2 - 1) = 1$ , matching the screen.
Read the p-value directly: $p = 0.0455$ . Compare it to $\alpha = 0.05$ ; since $0.0455 < 0.05$ , you reject the null hypothesis of homogeneity.

Report $\chi^2 = 4.00$ with $df = 1$ and a p-value of $0.0455$ . At $\alpha = 0.05$ you reject $H_0$ , so there is convincing evidence that the distribution of the category differs between the two groups.

Frequently asked questions

What is the lowercase p on a calculator test screen?

It is the p-value, the probability of a result at least as extreme as your sample if the null hypothesis were true. It is not a proportion. On a proportion test the sample proportion appears on a separate line labeled $\hat{p}$ (p-hat), so read the two lines carefully.

What does a p-value like 2.5E-4 mean?

That is scientific notation for $0.00025$ . The E-4 tells you to move the decimal point four places to the left, which gives a very small p-value, not a large one. A tiny p-value like this is strong evidence against the null hypothesis.

Is writing only the calculator command enough on a free-response question?

As general advice, no. A bare command such as the menu name leaves out the statistic, the p-value, and the conclusion in context that show your reasoning, so the safer practice is to write those out yourself. Treat this as guidance for clear communication rather than a fixed scoring rule.

How do I find degrees of freedom if the screen does not show them?

For a one-sample t-test, degrees of freedom equal $n - 1$ , the sample size minus one. For a chi-square test for homogeneity or independence, they equal $(r - 1)(c - 1)$ for a table with $r$ rows and $c$ columns. A proportion z-test has no degrees of freedom.