Two-proportion z interval and z test practice problems

Q: Why does the test pool the two samples but the interval does not?

A test starts by assuming H₀: p₁ = p₂ is true. If the two proportions really are equal, then every observation in both samples estimates the same number, so the best estimate combines them into p-hat_c = (X₁ + X₂)/(n₁ + n₂) and uses it in both terms of the standard error. An interval makes no such assumption. It is estimating the difference, not testing whether the difference is 0, so it has nothing to pool toward and uses each sample's own proportion.

Q: Does the order of subtraction change my conclusion?

It changes every sign but not the verdict. Switching from p₁ - p₂ to p₂ - p₁ swaps and negates the interval endpoints and flips the sign of z, while the p-value and whether 0 sits inside the interval stay exactly the same. State your order in the first step and keep it, because a conclusion that says the wrong group is ahead loses the point even when the arithmetic is right.

By Jude Wallis · Published July 29, 2026

This set has 8 problems on comparing two proportions: two-sample z intervals for a difference, two-proportion z tests with the pooled standard error, Large Counts checks on both samples, and reading an interval that does or does not contain 0. Solve each on paper, then open the steps.

AP Statistics: Unit 3 (topics 3.10 Constructing a Confidence Interval for the Difference Between Two Population Proportions, 3.11 Justifying a Claim Based on a Confidence Interval for the Difference Between Two Population Proportions, 3.12 Setting Up a Test for the Difference Between Two Population Proportions, 3.13 Carrying Out a Test for the Difference Between Two Population Proportions). These problems sit at the end of Unit 3 of the Fall 2026 AP Statistics course, where the one-proportion procedures of topics 3.3 through 3.7 extend to comparing two independent groups, and Unit 3 carries 15 to 25% of the multiple-choice section.

What these problems build

These 8 problems build the two Unit 3 procedures for comparing two groups on a categorical variable: the two-sample z interval for $p_1 - p_2$ and the two-proportion z test of $H_0: p_1 = p_2$ . Both start from the same point estimate, the difference in sample proportions $\hat{p}_1 - \hat{p}_2$ (read 'p-hat one minus p-hat two'), where each $\hat{p}$ is that sample's successes divided by its own sample size $n$ .

The interval takes the form $(\hat{p}_1 - \hat{p}_2) \pm z^*SE$ , with $z^* = 1.645$ at 90% confidence, $z^* = 1.96$ at 95%, and $z^* = 2.576$ at 99%. The test standardizes the same difference against the 0 that $H_0$ claims:

z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{SE}

What changes between the two is the standard error, which is the subject of the next section. Work each problem with the four-step State, Plan, Do, Conclude structure before opening the solution, and fix your order of subtraction in the first step. Define the difference as $p_1 - p_2$ and keep that order to the last sentence, because reversing it flips the sign of every number you report.

One note on p-values. A t-table forces you to trap a p-value between two columns, but the standard normal table the exam supplies hands you the area directly. These solutions round $z$ to two decimal places, look the area up on the z table, and report it, so the digits below are table values; software working from the unrounded $z$ will differ slightly in the last place. More sets are on the practice page.

Pooled for the test, unpooled for the interval

This is the most common lost point in two-proportion inference, and it comes down to one question: do you have a null hypothesis to lean on?

An interval does not. It estimates each group's variability from that group's own sample proportion, and the two pieces stay separate:

SE = \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}

A test begins by assuming $H_0: p_1 = p_2$ is true. Under that assumption the two groups share one common proportion, and the best estimate of it throws every observation into one pile:

\hat{p}_c = \frac{X_1 + X_2}{n_1 + n_2}

where $X_1$ and $X_2$ are the two success counts. That single pooled proportion then goes into both terms:

SE = \sqrt{\hat{p}_c(1-\hat{p}_c)\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}

Two warnings come with it. Pool counts, not proportions: $\hat{p}_c$ is combined successes over combined sample sizes, and it equals the plain average of $\hat{p}_1$ and $\hat{p}_2$ only when $n_1 = n_2$ . And because the two formulas return slightly different numbers, an interval and a test on the same data can land on opposite sides of a borderline cutoff, so report whichever procedure the question asked for. When a confidence interval and a p-value disagree works through that case. One-proportion vs two-proportion z test sets the two up side by side, and the proportion z test calculator checks your arithmetic.

Checking Large Counts on both samples

Large Counts needs four numbers to clear 10, not two, and which four depends on the procedure.

For an interval, use each sample's observed counts: $X_1$ , $n_1 - X_1$ , $X_2$ , and $n_2 - X_2$ .
For a test, use the expected counts under $H_0$ , built from the pooled proportion: $n_1\hat{p}_c$ , $n_1(1-\hat{p}_c)$ , $n_2\hat{p}_c$ , and $n_2(1-\hat{p}_c)$ .

A sample of 500 does not rescue a sample of 30, and a healthy success count does not rescue a thin failure count. Problem 5 is a case where both procedures are off the table, and the right answer there is to name the failed condition and stop.

The rest of the checklist:

Random. Two independent random samples, or one group of subjects randomly assigned to two treatments. The second kind supports a cause and effect conclusion; the first does not.
10%. Needed only when you sample without replacement, so it applies to the survey problems here and not to the randomized experiments. Check each sample against its own population: $n_1 \le 0.10N_1$ and $n_2 \le 0.10N_2$ .
Independent groups. The two samples must not be the same subjects measured twice. Paired data call for a different procedure.

Why 10 successes and 10 failures explains where the threshold comes from, and what a two-proportion interval containing 0 means covers the conclusion wording problems 1 and 6 ask for. The full sequence sits in Unit 3.

Frequently asked questions

Why does the test pool the two samples but the interval does not?

A test starts by assuming $H_0: p_1 = p_2$ is true. If the two proportions really are equal, then every observation in both samples estimates the same number, so the best estimate combines them into $\hat{p}_c = \frac{X_1 + X_2}{n_1 + n_2}$ and uses it in both terms of the standard error. An interval makes no such assumption. It is estimating the difference, not testing whether the difference is 0, so it has nothing to pool toward and uses each sample's own proportion.

Do I check Large Counts with observed counts or expected counts?

Observed counts for an interval, expected counts for a test. An interval checks $X_1$ , $n_1 - X_1$ , $X_2$ , and $n_2 - X_2$ against 10. A test checks $n_1\hat{p}_c$ , $n_1(1-\hat{p}_c)$ , $n_2\hat{p}_c$ , and $n_2(1-\hat{p}_c)$ , all built from the pooled proportion. Either way you need all four numbers to reach 10, and a large sample in one group does not cover for a small count in the other.

My interval for the difference contains 0. What do I write?

Write that 0 is a plausible value for the difference, so there is not convincing evidence that the two proportions differ. Do not write that the proportions are equal or that there is no difference. The interval supports every value between its endpoints, and 0 is only one of them. Name the confidence level, both endpoints, and both populations in context.

Does the order of subtraction change my conclusion?

It changes every sign but not the verdict. Switching from $p_1 - p_2$ to $p_2 - p_1$ swaps and negates the interval endpoints and flips the sign of $z$ , while the p-value and whether 0 sits inside the interval stay exactly the same. State your order in the first step and keep it, because a conclusion that says the wrong group is ahead loses the point even when the arithmetic is right.

Problem 1

A city compares recycling participation in two neighborhoods. An independent random sample of 250 households in Neighborhood A contains 130 that recycle. An independent random sample of 200 households in Neighborhood B contains 88 that recycle. Construct and interpret a 95% confidence interval for $p_A - p_B$ , the difference in the true proportions of households that recycle.

Show the worked solution

State. Let $p_A$ be the true proportion of all Neighborhood A households that recycle and $p_B$ the same proportion for Neighborhood B. Estimate $p_A - p_B$ with a 95% confidence interval.
Plan. Two-sample z interval for a difference of proportions. Random: two independent random samples. 10%: 250 must be at most 10% of Neighborhood A and 200 at most 10% of Neighborhood B, so assume the neighborhoods hold at least 2,500 and 2,000 households. Large Counts: use the observed counts, $130$ recyclers and $120$ non-recyclers in sample A, $88$ and $112$ in sample B, all four at least 10.
Find the sample proportions and the point estimate. $\hat{p}_A = \frac{130}{250} = 0.52$ and $\hat{p}_B = \frac{88}{200} = 0.44$ , so $\hat{p}_A - \hat{p}_B = 0.52 - 0.44 = 0.08$ .
Find the standard error. An interval uses each sample's own proportion, with no pooling. $\frac{0.52(0.48)}{250} = \frac{0.2496}{250} = 0.0009984$ and $\frac{0.44(0.56)}{200} = \frac{0.2464}{200} = 0.001232$ , so $SE = \sqrt{0.0009984 + 0.001232} = \sqrt{0.0022304} = 0.04723$ .
Find the margin of error. For 95% confidence, $z^* = 1.96$ , so $ME = 1.96(0.04723) = 0.0926$ .
Build the interval. $0.08 \pm 0.0926$ gives $(-0.0126, 0.1726)$ .
Conclude. We are 95% confident that the interval from $-0.0126$ to $0.1726$ captures the true difference in the proportions of households that recycle, Neighborhood A minus Neighborhood B. Because 0 lies inside the interval, it is plausible that the two neighborhoods recycle at the same rate, so there is not convincing evidence of a difference. That is not the same as saying the rates are equal: the interval also supports a Neighborhood A lead as large as about 17 percentage points.

$\hat{p}_A - \hat{p}_B = 0.08$ , $SE \approx 0.04723$ , $ME \approx 0.0926$ , interval $(-0.0126, 0.1726)$ . The interval contains 0, so a difference of 0 is a plausible value and there is not convincing evidence that the recycling rates differ.

Problem 2

A company randomly assigns 400 new employees to one of two versions of a safety training video. Of the 180 who watch Version A, 117 pass the certification test on the first try. Of the 220 who watch Version B, 121 pass on the first try. At $\alpha = 0.05$ , is there convincing evidence that the two versions produce different first-try pass rates?

Show the worked solution

State. Let $p_A$ and $p_B$ be the true first-try pass proportions for Version A and Version B. $H_0: p_A = p_B$ versus $H_a: p_A \ne p_B$ , with $\alpha = 0.05$ .
Plan. Two-proportion z test. Random: the employees were randomly assigned to a version, so the two groups are independent. Large Counts: a test checks expected counts from the pooled proportion $\hat{p}_c = \frac{117 + 121}{180 + 220} = \frac{238}{400} = 0.595$ . Those counts are $180(0.595) = 107.1$ , $180(0.405) = 72.9$ , $220(0.595) = 130.9$ , and $220(0.405) = 89.1$ , all at least 10.
Find the sample proportions. $\hat{p}_A = \frac{117}{180} = 0.65$ and $\hat{p}_B = \frac{121}{220} = 0.55$ , a difference of $0.10$ .
Find the pooled standard error. $\frac{1}{180} + \frac{1}{220} = \frac{11}{1980} + \frac{9}{1980} = \frac{20}{1980} = \frac{1}{99}$ , and $0.595(0.405) = 0.240975$ , so $SE = \sqrt{\frac{0.240975}{99}} = \sqrt{0.0024341} = 0.04934$ .
Find the test statistic. $z = \frac{0.65 - 0.55}{0.04934} = \frac{0.10}{0.04934} = 2.03$ .
Find the p-value. Two-sided, so p-value $= 2 \times P(Z > 2.03) = 2(1 - 0.9788) = 2(0.0212) = 0.0424$ .
Conclude. Since $0.0424 < 0.05$ , reject $H_0$ . There is convincing evidence that the two versions of the training video produce different first-try pass rates. Because employees were randomly assigned to a version, the difference can be attributed to the video for the employees in this experiment.

$\hat{p}_c = 0.595$ , $SE \approx 0.04934$ , $z \approx 2.03$ , two-sided p-value $\approx 0.0424$ . Reject $H_0$ at $\alpha = 0.05$ ; convincing evidence the two video versions give different first-try pass rates.

Problem 3

A traffic study watches a random sample of 400 vehicles approaching a stop sign at Intersection A and records 240 that come to a complete stop. An independent random sample of 400 vehicles at Intersection B has 200 that come to a complete stop. Construct a 90% confidence interval for $p_A - p_B$ and say what it shows about which intersection has the higher stopping rate.

Show the worked solution

State. Let $p_A$ and $p_B$ be the true proportions of vehicles that come to a complete stop at Intersection A and Intersection B. Estimate $p_A - p_B$ with a 90% confidence interval.
Plan. Two-sample z interval. Random: two independent random samples. 10%: 400 vehicles is less than 10% of all vehicles passing each intersection. Large Counts: the observed counts are $240$ stopping and $160$ not at A, $200$ and $200$ at B, all four at least 10.
Find the sample proportions and the point estimate. $\hat{p}_A = \frac{240}{400} = 0.60$ and $\hat{p}_B = \frac{200}{400} = 0.50$ , so the point estimate is $0.60 - 0.50 = 0.10$ .
Find the standard error. $\frac{0.60(0.40)}{400} = \frac{0.24}{400} = 0.0006$ and $\frac{0.50(0.50)}{400} = \frac{0.25}{400} = 0.000625$ , so $SE = \sqrt{0.0006 + 0.000625} = \sqrt{0.001225} = 0.035$ exactly.
Find the margin of error. For 90% confidence, $z^* = 1.645$ , so $ME = 1.645(0.035) = 0.0576$ .
Build the interval. $0.10 \pm 0.0576$ gives $(0.0424, 0.1576)$ .
Conclude. We are 90% confident that the interval from $0.0424$ to $0.1576$ captures the true difference in the proportions of vehicles that come to a complete stop, Intersection A minus Intersection B. Both endpoints are positive, so 0 is not a plausible value and every plausible difference points the same way. There is convincing evidence that Intersection A has the higher stopping rate, by somewhere between about 4 and 16 percentage points.

Point estimate $0.10$ , $SE = 0.035$ , $ME \approx 0.0576$ , interval $(0.0424, 0.1576)$ . Both endpoints are positive, so 0 is not plausible and there is convincing evidence Intersection A has the higher stopping rate.

Problem 4

A financial aid office runs an experiment on 600 seniors. It randomly assigns 240 to receive weekly text reminders about the federal aid application and 360 to receive no reminders. By the deadline, 168 of the text group and 216 of the no-reminder group have filed. At $\alpha = 0.05$ , is there convincing evidence that the reminders raise the filing rate?

Show the worked solution

State. Let $p_1$ be the true filing proportion under text reminders and $p_2$ the true filing proportion with no reminders. $H_0: p_1 = p_2$ versus $H_a: p_1 > p_2$ , with $\alpha = 0.05$ .
Plan. Two-proportion z test. Random: the 600 seniors were randomly assigned to the two conditions, so the groups are independent. No 10% check is needed, because this is a randomized experiment rather than sampling without replacement. Large Counts: $\hat{p}_c = \frac{168 + 216}{240 + 360} = \frac{384}{600} = 0.64$ , giving expected counts $240(0.64) = 153.6$ , $240(0.36) = 86.4$ , $360(0.64) = 230.4$ , and $360(0.36) = 129.6$ , all at least 10.
Find the sample proportions. $\hat{p}_1 = \frac{168}{240} = 0.70$ and $\hat{p}_2 = \frac{216}{360} = 0.60$ , a difference of $0.10$ .
Find the pooled standard error. $\frac{1}{240} + \frac{1}{360} = \frac{3}{720} + \frac{2}{720} = \frac{5}{720} = \frac{1}{144}$ , and $0.64(0.36) = 0.2304$ , so $SE = \sqrt{\frac{0.2304}{144}} = \sqrt{0.0016} = 0.04$ exactly.
Find the test statistic. $z = \frac{0.70 - 0.60}{0.04} = \frac{0.10}{0.04} = 2.50$ .
Find the p-value. Right-tailed, so p-value $= P(Z > 2.50) = 1 - 0.9938 = 0.0062$ .
Conclude. Since $0.0062 < 0.05$ , reject $H_0$ . There is convincing evidence that weekly text reminders raise the true proportion of seniors who file the aid application. Because the seniors were randomly assigned, this supports a cause and effect conclusion for the seniors in this study.

$\hat{p}_c = 0.64$ , $SE = 0.04$ , $z = 2.50$ , one-sided p-value $\approx 0.0062$ . Reject $H_0$ at $\alpha = 0.05$ ; convincing evidence the reminders raise the filing rate, and random assignment supports a causal reading.

Problem 5

A quality engineer compares defect rates on two production lines. A random sample of 40 units from Line A contains 3 defective units. An independent random sample of 45 units from Line B contains 1 defective unit. (a) Can the engineer construct a 95% confidence interval for $p_A - p_B$ ? (b) Can the engineer carry out a two-proportion z test? (c) If the combined defect rate stayed the same, how large would each sample have to be, using equal sample sizes, for the test's Large Counts condition to hold?

Show the worked solution

(a) Check Large Counts the way an interval checks it, with each sample's own observed counts. Line A gives $3$ defective and $40 - 3 = 37$ non-defective. Line B gives $1$ defective and $45 - 1 = 44$ non-defective. The defect counts $3$ and $1$ are both below 10, so the condition fails and the two-sample z interval is not appropriate for these data.
(b) Check Large Counts the way a test checks it, with expected counts from the pooled proportion. $\hat{p}_c = \frac{3 + 1}{40 + 45} = \frac{4}{85} = 0.0471$ , so $1 - \hat{p}_c = 0.9529$ . The four expected counts are $40(0.0471) = 1.88$ , $40(0.9529) = 38.12$ , $45(0.0471) = 2.12$ , and $45(0.9529) = 42.88$ . Two of the four fall below 10, so this condition fails as well and the two-proportion z test is not appropriate either.
(b) Say why in one line. Both procedures approximate the sampling distribution of $\hat{p}_1 - \hat{p}_2$ with a normal curve, and success counts of 1 and 3 leave that distribution too skewed for the approximation to hold. Naming the failed condition and stopping is the correct response; running the procedure anyway produces a z statistic the normal model does not support.
(c) Solve for the sample size. Hold the combined defect rate at $\hat{p}_c = \frac{4}{85}$ and let $n$ be the common sample size. The binding requirement is the defect count, $n\hat{p}_c \ge 10$ , so $n \ge \frac{10}{4/85} = \frac{850}{4} = 212.5$ .
(c) Round the right way. Sample sizes are whole numbers and $212$ would give $212\left(\frac{4}{85}\right) = 9.98$ , just short, so round up: each line needs at least $213$ units, which gives $213\left(\frac{4}{85}\right) = 10.02$ . Check the other side, $213(0.9529) \approx 203$ , which clears 10 with room to spare.

(a) No: the observed defect counts $3$ and $1$ are below 10. (b) No: with $\hat{p}_c = \frac{4}{85} \approx 0.0471$ the expected defect counts are $1.88$ and $2.12$ , both below 10. Name the failed condition and stop. (c) At least 213 units per line, since $n\left(\frac{4}{85}\right) \ge 10$ needs $n \ge 212.5$ .

Problem 6

A university surveys an independent random sample of 150 students who eat at Dining Hall A and finds 96 who rate the food good or better. An independent random sample of 150 students at Dining Hall B has 84 who rate it good or better. (a) Construct a 99% confidence interval for $p_A - p_B$ . (b) Use that interval to decide the test of $H_0: p_A = p_B$ against $H_a: p_A \ne p_B$ at $\alpha = 0.01$ , and check the decision by running the test.

Show the worked solution

(a) State. Let $p_A$ and $p_B$ be the true proportions of students at each dining hall who rate the food good or better. Estimate $p_A - p_B$ with a 99% confidence interval.
(a) Plan. Two-sample z interval. Random: two independent random samples. 10%: 150 must be at most 10% of the students who eat at each hall, so assume each hall serves at least 1,500 students. Large Counts: the observed counts are $96$ and $54$ at Hall A, $84$ and $66$ at Hall B, all four at least 10.
(a) Sample proportions and point estimate. $\hat{p}_A = \frac{96}{150} = 0.64$ and $\hat{p}_B = \frac{84}{150} = 0.56$ , so the point estimate is $0.64 - 0.56 = 0.08$ .
(a) Standard error. $\frac{0.64(0.36)}{150} = \frac{0.2304}{150} = 0.001536$ and $\frac{0.56(0.44)}{150} = \frac{0.2464}{150} = 0.0016427$ , so $SE = \sqrt{0.001536 + 0.0016427} = \sqrt{0.0031787} = 0.05638$ .
(a) Margin of error and interval. For 99% confidence, $z^* = 2.576$ , so $ME = 2.576(0.05638) = 0.1452$ and the interval is $0.08 \pm 0.1452$ , or $(-0.0652, 0.2252)$ . We are 99% confident this interval captures the true difference $p_A - p_B$ .
(b) Read the interval as a decision. A 99% confidence interval usually agrees with a two-sided test at $\alpha = 1 - 0.99 = 0.01$ , though the interval's unpooled standard error and the test's pooled one can send the two verdicts opposite ways on a borderline problem. Because 0 lies inside $(-0.0652, 0.2252)$ , fail to reject $H_0$ . There is not convincing evidence that the two dining halls differ in the true proportion of students who rate the food good or better. Note the wording: 0 being plausible does not make the proportions equal, and a Hall A lead near $0.22$ is equally plausible.
(b) Check by running the test. Pooling gives $\hat{p}_c = \frac{96 + 84}{150 + 150} = \frac{180}{300} = 0.60$ , with expected counts $150(0.60) = 90$ and $150(0.40) = 60$ in each group, all at least 10. Then $SE = \sqrt{0.60(0.40)\left(\frac{1}{150} + \frac{1}{150}\right)} = \sqrt{\frac{0.24 \times 2}{150}} = \sqrt{\frac{0.48}{150}} = \sqrt{0.0032} = 0.05657$ , so $z = \frac{0.08}{0.05657} = 1.41$ and the two-sided p-value is $2(1 - 0.9207) = 2(0.0793) = 0.1586$ . Since $0.1586 > 0.01$ , fail to reject $H_0$ : the same verdict, reached through a slightly different standard error.

(a) Point estimate $0.08$ , $SE \approx 0.05638$ , $ME \approx 0.1452$ , interval $(-0.0652, 0.2252)$ . (b) The interval contains 0, so fail to reject $H_0$ at $\alpha = 0.01$ ; the test agrees, with $\hat{p}_c = 0.60$ , $z \approx 1.41$ , and a two-sided p-value of about $0.1586$ .

Problem 7

A university randomly assigns 1,000 students taking an online statistics course to one of two formats: 500 to a self-paced version and 500 to a cohort-based version. By the end of the term, 305 self-paced students and 260 cohort-based students have completed the course. (a) Construct a 95% confidence interval for $p_1 - p_2$ , the difference in true completion proportions, self-paced minus cohort-based. (b) Carry out a two-proportion z test at $\alpha = 0.05$ . (c) Explain why the standard error in part (a) differs from the one in part (b).

Show the worked solution

(a) State and plan. Let $p_1$ and $p_2$ be the true completion proportions for the self-paced and cohort-based formats. Two-sample z interval for $p_1 - p_2$ . Random: the 1,000 students were randomly assigned, so the groups are independent, and no 10% check applies to a randomized experiment. Large Counts: the observed counts are $305$ completions and $195$ non-completions in the self-paced group, $260$ and $240$ in the cohort group, all four at least 10.
(a) Point estimate and standard error. $\hat{p}_1 = \frac{305}{500} = 0.61$ and $\hat{p}_2 = \frac{260}{500} = 0.52$ , so the point estimate is $0.09$ . Unpooled, $\frac{0.61(0.39)}{500} = \frac{0.2379}{500} = 0.0004758$ and $\frac{0.52(0.48)}{500} = \frac{0.2496}{500} = 0.0004992$ , so $SE = \sqrt{0.000975} = 0.03122$ .
(a) Interval and interpretation. $z^* = 1.96$ , so $ME = 1.96(0.03122) = 0.0612$ and the interval is $0.09 \pm 0.0612$ , or $(0.0288, 0.1512)$ . We are 95% confident this interval captures the true difference in completion proportions. Both endpoints are positive, so 0 is not plausible and the self-paced format has the higher completion rate.
(b) State. $H_0: p_1 = p_2$ versus $H_a: p_1 \ne p_2$ , with $\alpha = 0.05$ .
(b) Plan and pooled standard error. Large Counts for a test uses $\hat{p}_c = \frac{305 + 260}{500 + 500} = \frac{565}{1000} = 0.565$ , giving expected counts $500(0.565) = 282.5$ and $500(0.435) = 217.5$ in each group, all at least 10. Then $0.565(0.435) = 0.245775$ and $\frac{1}{500} + \frac{1}{500} = 0.004$ , so $SE = \sqrt{0.245775 \times 0.004} = \sqrt{0.0009831} = 0.03135$ .
(b) Test statistic, p-value, and conclusion. $z = \frac{0.09}{0.03135} = 2.87$ , so the two-sided p-value is $2 \times P(Z > 2.87) = 2(1 - 0.9979) = 2(0.0021) = 0.0042$ . Since $0.0042 < 0.05$ , reject $H_0$ . There is convincing evidence that the two formats produce different completion rates, and because students were randomly assigned, the format is a plausible cause.
(c) Explain the gap. The interval has no null hypothesis to lean on, so it measures each group's variability with that group's own proportion, $0.61$ and $0.52$ . The test assumes $H_0: p_1 = p_2$ is true, and under that assumption both groups share one proportion, best estimated by combining the samples into $\hat{p}_c = 0.565$ . That is why the interval used $SE = 0.03122$ while the test used $SE = 0.03135$ . The two agreed here, but on a borderline problem they can disagree, so give the question the procedure it asked for.

(a) Point estimate $0.09$ , $SE \approx 0.03122$ , interval $(0.0288, 0.1512)$ , which excludes 0. (b) $\hat{p}_c = 0.565$ , $SE \approx 0.03135$ , $z \approx 2.87$ , two-sided p-value $\approx 0.0042$ ; reject $H_0$ . (c) The interval uses each sample's own proportion; the test pools because $H_0$ says the proportions are equal.

Problem 8

A state agency compares how small businesses file their taxes in two regions. An independent random sample of 420 businesses in Region 1 has 273 that filed online. An independent random sample of 380 businesses in Region 2 has 209 that filed online. (a) Construct and interpret a 95% confidence interval for $p_1 - p_2$ . (b) Carry out a test of $H_0: p_1 = p_2$ against $H_a: p_1 \ne p_2$ at $\alpha = 0.05$ . (c) Interpret the p-value in context. (d) Name the error the decision in part (b) could be and describe it in context.

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(a) State and plan. Let $p_1$ and $p_2$ be the true proportions of small businesses that file online in Region 1 and Region 2. Two-sample z interval for $p_1 - p_2$ . Random: two independent random samples. 10%: 420 must be at most 10% of Region 1's businesses and 380 at most 10% of Region 2's, so assume the regions hold at least 4,200 and 3,800 small businesses. Large Counts: the observed counts are $273$ online and $147$ not online in Region 1, $209$ and $171$ in Region 2, all four at least 10.
(a) Point estimate and standard error. $\hat{p}_1 = \frac{273}{420} = 0.65$ and $\hat{p}_2 = \frac{209}{380} = 0.55$ , so the point estimate is $0.10$ . Unpooled, $\frac{0.65(0.35)}{420} = \frac{0.2275}{420} = 0.0005417$ and $\frac{0.55(0.45)}{380} = \frac{0.2475}{380} = 0.0006513$ , so $SE = \sqrt{0.0005417 + 0.0006513} = \sqrt{0.0011930} = 0.03454$ .
(a) Interval and interpretation. $z^* = 1.96$ , so $ME = 1.96(0.03454) = 0.0677$ and the interval is $0.10 \pm 0.0677$ , or $(0.0323, 0.1677)$ . We are 95% confident that the interval from $0.0323$ to $0.1677$ captures the true difference in the proportions of small businesses that file online, Region 1 minus Region 2. Both endpoints are positive, so 0 is not plausible and Region 1 has the higher online filing rate, by roughly 3 to 17 percentage points.
(b) Plan and pooled standard error. Two-proportion z test with $H_0: p_1 = p_2$ against $H_a: p_1 \ne p_2$ at $\alpha = 0.05$ . Large Counts for a test uses $\hat{p}_c = \frac{273 + 209}{420 + 380} = \frac{482}{800} = 0.6025$ , giving expected counts $420(0.6025) = 253.05$ , $420(0.3975) = 166.95$ , $380(0.6025) = 228.95$ , and $380(0.3975) = 151.05$ , all at least 10. Then $0.6025(0.3975) = 0.23949375$ and $\frac{1}{420} + \frac{1}{380} = 0.00501253$ , so $SE = \sqrt{0.23949375 \times 0.00501253} = \sqrt{0.00120047} = 0.03465$ .
(b) Test statistic, p-value, and conclusion. $z = \frac{0.65 - 0.55}{0.03465} = \frac{0.10}{0.03465} = 2.89$ , so the two-sided p-value is $2 \times P(Z > 2.89) = 2(1 - 0.9981) = 2(0.0019) = 0.0038$ . Since $0.0038 < 0.05$ , reject $H_0$ . There is convincing evidence that the true proportions of small businesses filing online differ between the two regions. These are observational samples rather than a randomized experiment, so the conclusion is about a difference in rates and not about what causes it.
(c) Interpret the p-value. If the two regions truly had the same online filing rate, there would be about a $0.38\%$ chance of drawing two samples of these sizes whose sample proportions differ by at least $0.10$ in either direction from sampling variability alone.
(d) Name the error. A Type I error is the only one available, because a Type I error can accompany a rejection and a Type II error can accompany only a failure to reject. Here it would mean concluding that the two regions have different true online filing rates when in fact the rates are identical. Its long-run rate is the significance level you chose, $\alpha = 0.05$ .

(a) Point estimate $0.10$ , $SE \approx 0.03454$ , interval $(0.0323, 0.1677)$ , which excludes 0. (b) $\hat{p}_c = 0.6025$ , $SE \approx 0.03465$ , $z \approx 2.89$ , two-sided p-value $\approx 0.0038$ ; reject $H_0$ . (c) About a $0.38\%$ chance of a gap this large if the true rates were equal. (d) A Type I error: calling the regions different when they are the same.