When do I use a t-test instead of a z-test for a mean?

Use a t-test whenever you do not know the true population standard deviation and must estimate it with the sample standard deviation s — which is essentially always in AP Statistics. The z-test for a mean requires a known , a situation that almost never appears on the exam, so the one-sample t-test is your default procedure for a mean.

What degrees of freedom do I use for a one-sample t-test?

Use df = n - 1, where n is the sample size. For example, a sample of 25 observations uses 24 degrees of freedom. This determines which t-distribution you use to find the p-value.

How do I decide between a one-sided and two-sided alternative?

Read the research question before looking at the data. If it asks whether the mean 'increased,' 'is greater than,' or 'is less than' a value, use a one-sided alternative in that direction. If it asks whether the mean 'differs,' 'changed,' or 'is not equal to,' use a two-sided alternative. Never pick the direction based on which way your sample happened to come out.

What exactly does the p-value mean in a t-test?

The p-value is the probability of getting a sample mean as extreme as, or more extreme than, the one you observed, assuming the null hypothesis is true. A small p-value means your data would be surprising under H_0, which is why it counts as evidence against the null. It is not the probability that H_0 is true.

AP-STATS-7.5-7.6

U7.5 One-Sample t-Test for a Mean

Master the one-sample t-test for a population mean: set up hypotheses, check conditions, compute the t-statistic and p-value, and conclude in context.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U7.5 One-Sample t-Test for a Mean, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start U7.5 One-Sample t-Test for a Mean free

What this lesson covers

You have already built a confidence interval for a mean using the

t

-distribution. Now you will use that same machinery to answer a yes-or-no question: is there convincing evidence that a population mean differs from some claimed value? That is the job of the one-sample

t

-test.

This lesson walks through the full four-step significance test the exam rewards: state hypotheses, check the conditions, calculate the test statistic and

p

-value, and write a conclusion tied to the context. Nail these steps and you will pick up nearly all the points on any mean-testing FRQ, because the AP rubric grades the process almost as much as the final answer.

Setting Up the Hypotheses

Every significance test starts with two competing claims about a parameter, here the true population mean

\mu

. The null hypothesis always states no effect or no difference:

H_0: \mu = \mu_0

, where

\mu_0

is the specific value from the problem (a manufacturer's claim, a historical average, a target). The alternative hypothesis expresses what you are trying to find evidence for, and it takes one of three forms:

Alternative	When to use it	Wording clue
$H_a: \mu > \mu_0$	one-sided, upper	"greater than," "increased"
$H_a: \mu < \mu_0$	one-sided, lower	"less than," "decreased"
$H_a: \mu \neq \mu_0$	two-sided	"differs," "changed"

Three rules save easy points. First, hypotheses are about the population mean

\mu

, never the sample mean

\bar{x}

— writing

H_0: \bar{x} = 12

loses credit. Second, define

\mu

in words in context ("

\mu

= the true mean battery life, in hours, of all batteries"). Third, decide the direction from the research question, not from your data. Choosing a one-sided alternative after seeing which way the sample came out is a serious error. If the question only asks whether the mean "differs" or "changed," use the two-sided form.

Checking the Conditions

Before you trust any

t

-procedure, verify three conditions and state them explicitly with evidence from the problem.

The Random condition requires that the data come from a random sample or randomized experiment. Quote the design: "The problem states a random sample of 30 students was selected." This lets you generalize to the population.

The Independence (10% condition) requires that individual observations are independent. When sampling without replacement, check that the sample is at most 10% of the population:

n \leq 0.10N

. State it: "It is reasonable to assume there are more than 300 students in the population."

The Normal/Large Sample condition protects the sampling distribution of

\bar{x}

. It is met if the population is stated to be approximately normal, OR the sample size is large (

n \geq 30

) by the Central Limit Theorem, OR — for small samples — a graph of the data (dotplot, boxplot, histogram) shows no strong skew and no outliers. On the exam, if

n < 30

and no distribution is given, you must reference the provided graph.

A common misconception: the Normal condition is about the shape of the population or the sampling distribution, not the sample itself. Also, never skip stating conditions to save time; the FRQ rubric usually has a dedicated point for correctly verifying all three.

Computing the t-Statistic and p-Value

Because we estimate the population standard deviation with the sample standard deviation

s

, we standardize using the

t

-distribution rather than the normal. The test statistic measures how many standard errors the observed mean sits from the hypothesized mean:

t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}

The quantity

\frac{s}{\sqrt{n}}

is the standard error of the mean. The relevant

t

-distribution has

df = n - 1

degrees of freedom. A larger

|t|

means the data are farther from

H_0

and give stronger evidence against it.

The

p

-value is the probability of getting a test statistic at least as extreme as the observed

t

, assuming

H_0

is true. Find it from the

t

-distribution with

n-1

degrees of freedom:

Alternative	p-value
$\mu > \mu_0$	area to the right of $t$
$\mu < \mu_0$	area to the left of $t$
$\mu \neq \mu_0$	$2\times$ area beyond $\|t\|$

On a calculator, tcdf or T-Test gives the

p

-value directly. If using a table, you can only bracket the

p

-value between two columns — that is acceptable for the FRQ as long as you report the interval, e.g.

0.02 < p < 0.05

Making a Conclusion in Context

The conclusion compares your

p

-value to the significance level

\alpha

(use

\alpha = 0.05

unless the problem specifies otherwise) and always links back to the alternative hypothesis in context. Use a two-part template.

If

p \leq \alpha

: "Because the

p

-value of ___ is less than

\alpha =

___, we reject

H_0

. There is convincing evidence that [statement of

H_a

in context]."

If

p > \alpha

: "Because the

p

-value of ___ is greater than

\alpha =

___, we fail to reject

H_0

. There is not convincing evidence that [statement of

H_a

in context]."

Two phrasings cost points on the exam. Never say you "accept

H_0

" — failing to reject means the data are consistent with

H_0

, not that

H_0

is proven true. And never conclude about the sample; the conclusion is a claim about the population mean

\mu

.

Remember the interpretation of the

p

-value itself, which readers frequently ask you to state: it is the probability, computed assuming

H_0

is true, of observing a sample mean as extreme as or more extreme than the one obtained. A small

p

-value means such data would be surprising if

H_0

were true, which is why it counts as evidence against

H_0

Key terms

Null hypothesis ( $H_0$ ).: The default claim of no effect, stated as $\mu = \mu_0$ , that the test assumes true when computing the p-value.
Alternative hypothesis ( $H_a$ ).: The claim you seek evidence for, written as $\mu > \mu_0$ , $\mu < \mu_0$ , or $\mu \neq \mu_0$ .
Standard error of the mean.: An estimate of the variability of the sample mean, equal to $\frac{s}{\sqrt{n}}$ , used in the denominator of the t-statistic.
t-statistic.: The standardized distance of $\bar{x}$ from $\mu_0$ : $t = \frac{\bar{x}-\mu_0}{s/\sqrt{n}}$ , evaluated with $n-1$ degrees of freedom.
Degrees of freedom.: The parameter $df = n - 1$ that selects the specific t-distribution for a one-sample test.
p-value.: The probability, assuming $H_0$ is true, of getting a test statistic at least as extreme as the observed value.
Significance level ( $\alpha$ ).: The threshold, often 0.05, against which the p-value is compared to decide whether to reject $H_0$ .

Worked example

A cereal company claims its boxes contain a mean of 18 ounces of cereal. A consumer group suspects the true mean is less. They randomly sample 25 boxes and find

\bar{x} = 17.7

ounces with

s = 0.6

ounces. A dotplot of the 25 weights shows no strong skew or outliers. Test the consumer group's claim at

\alpha = 0.05

Step 1 — Hypotheses. Let

\mu

= the true mean weight, in ounces, of all cereal boxes. Because the group suspects the mean is less than the claim, use a one-sided test:

H_0: \mu = 18

versus

H_a: \mu < 18

.

Step 2 — Conditions. Random: the 25 boxes were randomly sampled. Independence: it is reasonable that the company produces more than

10 \times 25 = 250

boxes, so the 10% condition holds. Normal:

n = 25 < 30

, but the dotplot shows no strong skew and no outliers, so the sampling distribution of

\bar{x}

is approximately normal. All conditions are met, so a one-sample

t

-test is appropriate.

Step 3 — Test statistic and p-value. The standard error is

\frac{s}{\sqrt{n}} = \frac{0.6}{\sqrt{25}} = 0.12

. Then

t = \frac{17.7 - 18}{0.12} = \frac{-0.3}{0.12} = -2.5

with

df = 25 - 1 = 24

. For a lower-tailed test, the

p

-value is the area to the left of

-2.5

p \approx 0.0098

.

Step 4 — Conclusion. Because the

p

-value of about

0.0098

is less than

\alpha = 0.05

, we reject

H_0

. There is convincing evidence that the true mean weight of the cereal boxes is less than 18 ounces.

Practice questions

A researcher runs a one-sample t-test with

H_a: \mu \neq 50

, a sample of

n = 16

, and computes

t = 2.10

. Which of the following correctly describes how to find the p-value?

Find the area to the right of $t = 2.10$ using 16 degrees of freedom
Find the area to the right of $t = 2.10$ using 15 degrees of freedom
Double the area to the right of $t = 2.10$ using 15 degrees of freedom
Double the area to the right of $t = 2.10$ using 16 degrees of freedom

Answer: Double the area to the right of $t = 2.10$ using 15 degrees of freedom

The degrees of freedom for a one-sample t-test are

n - 1 = 16 - 1 = 15

, not

n

. Because the alternative is two-sided (

\mu \neq 50

), the p-value is the combined area in both tails, which equals twice the area beyond

|t| = 2.10

A quality inspector tests whether the mean fill volume of a bottling machine differs from the target of 500 mL. From a random sample of 40 bottles she finds

\bar{x} = 502.3

mL and

s = 5

mL. State the hypotheses, compute the test statistic, and describe how you would reach a conclusion at

\alpha = 0.05

Answer: $H_0: \mu = 500$ , $H_a: \mu \neq 500$ ; $t \approx 2.91$ with 39 df; two-sided p-value $\approx 0.006$ , so reject $H_0$ .

Let

\mu

be the true mean fill volume. Since the question asks whether the mean 'differs,' the alternative is two-sided. The standard error is

\frac{5}{\sqrt{40}} \approx 0.7906

, so

t = \frac{502.3 - 500}{0.7906} \approx 2.91

with

df = 39

. The two-sided p-value is about

0.006

, which is less than

0.05

, so you would reject

H_0

and conclude there is convincing evidence the mean fill volume differs from 500 mL. With

n = 40 \geq 30

the Normal condition is satisfied by the Central Limit Theorem.

A test produces a p-value of 0.18 at

\alpha = 0.05

. A student writes: 'Since

p > \alpha

, we accept the null hypothesis and conclude the population mean equals

\mu_0

.' Identify and correct the error.

Answer: You never accept $H_0$ ; you fail to reject it, concluding only that there is not convincing evidence for $H_a$ .

Failing to reject the null does not prove it true — the data are simply consistent with it, and the true mean could still differ by an amount too small to detect. The correct wording is: 'Because

p = 0.18 > 0.05

, we fail to reject

H_0

. There is not convincing evidence that the population mean differs from

\mu_0

FAQ

When do I use a t-test instead of a z-test for a mean?: Use a t-test whenever you do not know the true population standard deviation $\sigma$ and must estimate it with the sample standard deviation $s$ — which is essentially always in AP Statistics. The z-test for a mean requires a known $\sigma$ , a situation that almost never appears on the exam, so the one-sample t-test is your default procedure for a mean.
What degrees of freedom do I use for a one-sample t-test?: Use $df = n - 1$ , where $n$ is the sample size. For example, a sample of 25 observations uses 24 degrees of freedom. This determines which t-distribution you use to find the p-value.
How do I decide between a one-sided and two-sided alternative?: Read the research question before looking at the data. If it asks whether the mean 'increased,' 'is greater than,' or 'is less than' a value, use a one-sided alternative in that direction. If it asks whether the mean 'differs,' 'changed,' or 'is not equal to,' use a two-sided alternative. Never pick the direction based on which way your sample happened to come out.
What exactly does the p-value mean in a t-test?: The p-value is the probability of getting a sample mean as extreme as, or more extreme than, the one you observed, assuming the null hypothesis is true. A small p-value means your data would be surprising under $H_0$ , which is why it counts as evidence against the null. It is not the probability that $H_0$ is true.

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