AP Statistics Topic 4.5: Carrying Out a Test for a Population Mean or Population Mean Difference

01 · Start here

The concept in 60 seconds

A sample mean usually differs from a claimed population mean just because samples vary. A significance test asks whether the observed result is unusually far from the claim, after allowing for that variation.

The $t$ -statistic measures the difference in estimated standard-error units. The $p$ -value measures how unusual the statistic would be if the null hypothesis were true. The alternative hypothesis tells us which results count as more extreme.

The essential sequence: Check the test setup → calculate $t$ and $\mathrm{df}$ → find $p$ in the correct tail or tails → compare $p$ with $\alpha$ → conclude about the population in context.

For paired data, first turn each complete pair into $1$ difference. Then carry out a one-sample $t$ -test on those differences. The sample size is the number of pairs.

02 · One situation, deeper understanding

Are the bottles averaging $500\,\mathrm{mL}$ ?

A bottling company labels a production lot as averaging $500 \,\mathrm{mL}$ per bottle. An inspector takes an SRS of $25$ bottles from a lot of $10{,}000$ . The sample mean is $501.2 \,\mathrm{mL}$ and the sample standard deviation is $7.5 \,\mathrm{mL}$ . The fill-volume population is approximately normal; its standard deviation is unknown.

The question is whether the lot’s mean differs from $500 \,\mathrm{mL}$ . Let $\mu$ be the mean fill volume of all bottles in this lot. The inspector sets $\alpha =0.05$ before analyzing the sample.

$H_0:\mu=500\,\mathrm{mL}$
$H_a:\mu\ne500\,\mathrm{mL}$

Use a one-sample $t$ -test. The sample is random, and $25\le 0.10(10{,}000)=1{,}000$ supports treating observations as approximately independent when sampling without replacement. The approximately normal population supports the $t$ procedure with $n=25$ .

Pause and predict: Is a sample mean $1.2 \,\mathrm{mL}$ above the claim necessarily convincing evidence? We need to compare that difference with the estimated variability of sample means, rather than judging $1.2$ by itself.

You will find $t=0.8$ and $p\approx 0.43156$ . That $p$ -value is much larger than $0.05$ , so this sample does not provide sufficient evidence that the lot’s population mean differs from $500 \,\mathrm{mL}$ .

03 · Understand the language

Key ideas and notation

Null benchmark

$\mu_0$ is the population mean specified by $H_0$ . For paired differences, use $\mu_{d,0}$ , often $0$ for no average change.

Bottle example: $\mu_0=500 \,\mathrm{mL}$ .

Sample summaries

$\bar{x}$ and $s$ are the sample mean and sample $\operatorname{SD}$ . For differences, use $\bar d$ and $s_d$ .

$s$ describes individual values; $\frac{s}{\sqrt{n}}$ estimates variability of sample means.

$t$ and $\mathrm{df}$

$t$ is the standardized test statistic. Its sign shows whether the sample mean is above or below the null benchmark.

For these one-sample procedures, $\mathrm{df}=n-1$ .

$p$ and $\alpha$

$p$ is the calculated tail probability under the null model. $\alpha$ is the significance level chosen for the decision.

Compare these probabilities on the same scale: $0.05$ means $5\%$ .

Paired mean difference

$\mu_d$ is the population mean of the differences formed in a stated order. If $d=\text{before}-\text{after}$ , a positive difference means the after value is lower.

Each pair supplies $1$ observation to the difference sample. $2$ separate groups with no matching require a different procedure; equal group sizes alone do not make data paired.

04 · Measure the distance

Calculate the test statistic

Start with the same idea for both procedures:

$t=\frac{\text{sample estimate}-\text{null value}}{\text{estimated standard error}}$

1

population mean $t=\frac{\bar x-\mu_0}{s/\sqrt n}$

Estimated standard error: $\operatorname{SE}=\frac{s}{\sqrt{n}}$

Degrees of freedom: $\mathrm{df}=n-1$

Population mean of paired differences $t=\frac{\bar d-\mu_{d,0}}{s_d/\sqrt n}$

Estimated standard error: $\operatorname{SE}=\frac{s_d}{\sqrt{n}}$

$n$ counts pairs; $\mathrm{df}=n-1$ .

Apply the formula to the bottles

Estimate variability: $\operatorname{SE}=\frac{7.5}{\sqrt{25}}=1.5 \,\mathrm{mL}$ .
Find the observed difference: $501.2-500=1.2 \,\mathrm{mL}$ .
Standardize: $t=\frac{1.2}{1.5}=0.8$ .
Choose the reference distribution: $\mathrm{df}=25-1=24$ .

The sample mean is $0.8$ estimated standard errors above the null mean. The units cancel, so $t$ has no measurement units.

Keep the denominator straight: Use the sample $\operatorname{SD}$ divided by $\sqrt{n}$ , not $s$ alone. For pairs, calculate the $\operatorname{SD}$ of the differences; subtracting the $2$ original standard deviations does not give $s_d$ .

Use full precision for intermediate calculations. Round the reported $t$ -statistic and $p$ -value at the end. If $s=0$ , the usual $t$ formula is undefined; do not divide by $0$ or force a standard test result.

05 · Let the question choose the area

Choose the correct $p$ -value tail

Choose the alternative from the investigative question before looking at the result. Then use that alternative to select the area under the $t$ curve. In the table, $T$ represents a statistic from the null $t$ distribution and $t_{\mathrm{obs}}$ is the observed statistic.

The alternative determines the tail or tails
Alternative	Count as more extreme	$p$ -value
$H_a:\mu\lt\mu_0$	Statistics at or below the observed $t$ .	$P(T\le t_{\mathrm{obs}})$ — left tail
$H_a:\mu\gt\mu_0$	Statistics at or above the observed $t$ .	$P(T\ge t_{\mathrm{obs}})$ — right tail
$H_a:\mu\ne\mu_0$	Statistics at least as far from $0$ in either direction.	$P(\lvert T\rvert\ge\lvert t_{\mathrm{obs}}\rvert)=2P(T\ge\lvert t_{\mathrm{obs}}\rvert)$

The same tail rules apply to $\mu_d$ . Remember that the meaning of “greater” or “less” depends on your difference order.

One question, one correct choice of tails

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Carrying Out a Test for a Population Mean or Population Mean Difference

By the end of this lesson, you should be able to:

The concept in 60 seconds

Are the bottles averaging $500\,\mathrm{mL}$ ?