AP Statistics Topic 4.7: Constructing a Confidence Interval for the Difference Between Two Population Means

01 · Start here

The concept in 60 seconds

$2$ sample averages give a point estimate of the difference between $2$ population averages. Because different samples would produce different estimates, report an interval to describe the uncertainty.

$\text{Two-sample }t\text{-interval}=\text{difference in sample means}\pm\text{margin of error}$

The difference in sample means is $\bar x_{1}-\bar x_{2}$ . The margin of error is a $t$ critical value times the estimated standard error of that difference.

The central idea: Center the interval on the observed sample difference. Extend it in both directions by an amount that reflects sample variability, sample sizes and the chosen confidence level.

This method estimates $\mu_{1}-\mu_{2}$ for independent groups when the population standard deviations are unknown. The interval concerns a difference between averages, not the difference between every pair of individuals.

02 · One situation, deeper understanding

How much longer does service A take?

A business wants to estimate the difference in population mean delivery times between services A and B. In this study, the population means and standard deviations are unknown. $2$ independent SRSs give these sample summaries:

Observed sample summaries, in minutes
Group	Sample mean and $\operatorname{SD}$	Sample and population sizes
1: Service A	$\bar x_{1}=32$ ; $s_{1}=8$	$n_{1}=40$ from $N_{1}=2{,}000$
2: Service B	$\bar x_{2}=28$ ; $s_{2}=6$	$n_{2}=36$ from $N_{2}=1{,}800$

Let $\mu_{A}-\mu_{B}$ be the difference in mean delivery time, in minutes, for the $2$ defined delivery populations, using $\mathrm A-\mathrm B$ . A positive difference means A takes longer on average.

The point estimate is $32-28=4 \,\mathrm{min}$ . A $95\%$ unpooled two-sample $t$ -interval will be approximately $(0.785, 7.215)\,\mathrm{min}$ .

Pause and predict: Why not report exactly $4 \,\mathrm{min}$ as the population difference? $4$ is the observed sample difference. The interval allows for sampling variation around that estimate.

We will justify the procedure, calculate its standard error and margin of error, and connect the endpoints to the original question.

03 · Understand the language

Key ideas and notation

Parameter

$\mu_{1}-\mu_{2}$ is the difference between $2$ population means. Define both populations, the quantitative response and the subtraction order.

“A-minus-B mean delivery time” is more precise than “the difference.”

Point estimate

$\bar x_{1}-\bar x_{2}$ is the sample-based estimate at the center of the interval.

For the deliveries, it is $4 \,\mathrm{min}$ .

Standard error

$\operatorname{SE}$ estimates how much the difference between sample means varies across repeated independent sample pairs.

Use each sample’s $\operatorname{SD}$ with its own sample size.

Margin of error

$\operatorname{ME}=t^* \times \operatorname{SE}$ is the distance from the point estimate to either endpoint.

The full interval width is $2 \times \operatorname{ME}$ .

Confidence level and critical value

The chosen confidence level, such as $95\%$ , determines the central area used to find $t^*$ in a $t$ distribution with appropriate degrees of freedom. The confidence level describes the long-run success rate of the interval method under its conditions.

$t^*$ is a positive multiplier for interval construction. It is different from a test statistic calculated to test a null claim.

04 · Match the procedure to the design

Choose the right method

Ask whether the response is quantitative and whether the $2$ groups are independent. Check that you are estimating a difference between population means and that population standard deviations are unknown.

Different data designs call for different mean intervals

1

group $1$ population mean

Use a one-sample $t$ -interval for $\mu$ when appropriate. There is $1$ sample mean to estimate $1$ mean.

Matched observationsMean paired difference

Form $1$ within-pair difference per complete pair. Use a one-sample $t$ -interval on the differences.

Independent groupsDifference of population means

Use a two-sample $t$ -interval for $\mu_{1}-\mu_{2}$ , with separate sample means, standard deviations and sizes.

The study design chooses the method. Equal sample sizes do not create matched pairs, and $2$ different group labels do not automatically establish independence.

For the delivery study, the samples are independently selected from $2$ service populations, so the independent two-sample method fits the design.

Use the unpooled procedure here: Estimate each group’s contribution to variability separately. The formula in this lesson does not assume equal population variances. On the calculator, choose Pooled: No.

05 · Justify the interval

Check the conditions

A confidence interval needs more than $6$ summary statistics. Explain why the collection method and data support the procedure.

Randomization

2

independent random samples, or an appropriate randomized experiment with separate treatment groups.

IndependenceThe groups are independent. Within each sample, observations should be independent or sufficiently close to independent for the sampling approximation.

$10\%$ for each sampleWhen sampling without replacement:

n_1\le0.10N_1

and

n_2\le0.10N_2

Sample data / shapeBoth

n_{1}

and

n_{2}

are at least

30

, or suitable normal-population information supports the procedure. If either sample is smaller than

30

, examine both sample distributions for strong skewness and outliers.

Apply the checks to the deliveries

Randomization: The study gives $2$ independently selected SRSs.
$10\%$ : $40\le 0.10(2{,}000)=200$ and $36\le 0.10(1{,}800)=180$ .
Sample sizes: $40\ge 30$ and $36\ge 30$ support the course’s large-sample condition for the two-sample $t$ procedure.

These separate $10\%$ checks support treating observations within each sample as approximately independent. They do not make sampling without replacement exactly independent.

Small samples and experiments

When a sample is small, inspect the shape rather than adding the $2$ sample sizes together. Under the course’s small-sample check, both sample distributions should be free from strong skewness and outliers; approximately normal population models also support the method.

For a randomized experiment, explain the random assignment of treatments. The sampling-fraction $10\%$ condition is not required merely for treatment assignment; a separate finite-population sampling stage still needs its own check. The design and sample-data checks still matter.

If a check fails: State the limitation. A large sample does not remove volunteer bias, and a calculator cannot repair a paired design treated as independent.

06 · Build it step by step

Calculate the interval

$(\bar x_1-\bar x_2)\pm t^*\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

Step 1: find the point estimate

Use the declared order: $\bar x_{A}-\bar x_{B}=32-28=4 \,\mathrm{min}$ .

Step 2: calculate the standard error

$\operatorname{SE}=\sqrt{\frac{8^{2}}{40}+\frac{6^{2}}{36}}=\sqrt{1.6+1}=\sqrt{2.6}\approx 1.612452 \,\mathrm{min}$

Add the estimated variances of the separate sample means, then take the square root. The subtraction in the estimate does not mean subtracting standard deviations.

Step 3: find $t^*$ and the margin of error

For a $95\%$ unpooled interval, technology gives $\mathrm{df}\approx 71.752722$ and $t^*\approx 1.993581$ . Thus:

$\operatorname{ME}=t^* \times \operatorname{SE}\approx 3.215 \,\mathrm{min}$

Step 4: calculate both endpoints

Using the rounded margin of error, the interval is approximately $4\pm3.215$ , or $(0.785, 7.215)\,\mathrm{min}$ .

\text{Point estimate}\pm\text{margin of error}

The dot marks the observed A-minus-B sample mean difference. Each endpoint is $1$ margin of error from the estimate. The interval estimates a population mean difference, not a range of individual delivery times.

A concise interpretation: We are $95\%$ confident that service A’s population mean delivery time is between about $0.785$ and $7.215 \,\mathrm{min}$ longer than service B’s for the defined delivery populations.

Rounding: Keep full precision for $\operatorname{SE}$ and $t^*$ while calculating. Round the final endpoints and include units. Using rounded displayed values can change the last decimal slightly.

07 · Choose the multiplier

$t$ critical values, degrees of freedom and precision

Population standard deviations are unknown, so the $\operatorname{SE}$ uses sample standard deviations. A $t$ critical value accounts for this additional uncertainty. For confidence level $C$ , expressed as a proportion with $0\lt C\lt1$ , $t^*$ leaves central area $C$ under the chosen $t$ curve and area $\frac{1-C}{2}$ in each tail.

For $95\%$ confidence, each tail has area $0.025$ . If you already know $\mathrm{df}$ , the inverse- $t$ input is $\operatorname{invT}(0.975,\operatorname{df})$ .

Why might $\mathrm{df}$ be a decimal?

The usual unpooled two-sample procedure uses Welch’s approximate degrees of freedom, calculated from both sample sizes and standard deviations. Fractional $\mathrm{df}$ is normal. Use the technology’s value without forcing it to $n_{1}+n_{2}-2$ .

The unpooled $\mathrm{df}$ lies between the smaller of $n_{1}-1$ and $n_{2}-1$ and $n_{1}+n_{2}-2$ . For the deliveries, those bounds are $35$ and $74$ ; $\mathrm{df}\approx 71.752722$ is within them.

Optional: where the technology’s

\mathrm{df}

comes from

Let $a=\frac{s_1^2}{n_1}$ and $b=\frac{s_2^2}{n_2}$ . Welch’s approximation is $\operatorname{df}=\frac{(a+b)^2}{\frac{a^2}{n_1-1}+\frac{b^2}{n_2-1}}$ .

For the deliveries, $a=1.6$ and $b=1$ , giving $\mathrm{df}\approx 71.752722$ . You can use a two-sample interval command to obtain this automatically.

If you are using a printed $t$ table

A conservative classroom option is $\operatorname{df}=\min(n_1-1,n_2-1)$ . For the deliveries, $\mathrm{df}=35$ gives $t^*\approx 2.030108$ and a $95\%$ interval of approximately $(0.727, 7.273)\,\mathrm{min}$ .

This interval is slightly wider than the technology-based unpooled interval because the conservative $\mathrm{df}$ gives a larger $t^*$ . State which method you used. If your table lacks the selected row, follow the table’s instructions; using an available lower $\mathrm{df}$ preserves a conservative critical value.

What makes an interval wider or narrower?

Higher confidence: a larger $t^*$ and wider interval for the same data.
More variability: larger sample standard deviations increase $\operatorname{SE}$ and width, all else equal.
Larger samples: reduce contributions $\frac{s_{1}^{2}}{n_{1}}$ and $\frac{s_{2}^{2}}{n_{2}}$ , improving precision when variability stays similar.

If both sample sizes are multiplied by $4$ and both sample standard deviations remain the same, $\operatorname{SE}$ becomes $\frac12$ of its original value. Under the usual unpooled procedure, $\mathrm{df}$ and $t^{* [Truncated]}$

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Constructing a Confidence Interval for the Difference Between Two Population Means

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

The concept in 60 seconds

How much longer does service A take?