- Identify the type of response variable and the number of populations. - The question asks for a test of a difference, not an interval estimate.

1. The response is binary, and the question compares two independent population proportions. 2. The appropriate procedure is a two-sample z-test for the difference between two population proportions.

A two-sample z-test for the difference between two population proportions.

- Use population proportions, not the observed sample proportions. - Preserve the stated subtraction order.

1. The test compares two population proportions. 2. In the requested order, the parameter is \(p_A-p_B\).

\(p_A-p_B\).

- Decide whether the goal is estimation or evaluation of a claim. - The response is binary and two independent populations are being compared.

1. The research question asks whether a directional claim about the population proportions is supported. 2. A hypothesis test is therefore the primary procedure. 3. The appropriate test is a two-sample z-test for the difference between two population proportions.

Use a two-sample z-test for the difference between two population proportions.

- Translate “equal proportions” into an equation. - Rewrite that equation as a difference between the two proportions.

1. Equality means \(p_1=p_2\). 2. Subtracting gives \(p_1-p_2=0\).

\(0\).

- Compare the p-value with the chosen cutoff for statistical significance. - Think about which decision is allowed when the sample evidence does not meet that cutoff.

1. The p-value \(0.12\) is larger than the significance level \(0.05\). 2. The result is not statistically significant at that level, so fail to reject \(H_0\).

Fail to reject \(H_0\).

- Identify the yes-or-no outcome. - Name both population groups clearly. - Preserve the proposed direction of the difference.

1. The variable is whether a member uses mobile deposit. 2. The two parameters are the population proportions for the two age groups. 3. The direction is that the proportion for members younger than \(40\) is greater than the proportion for members age \(40\) or older. 4. The population is all members of the credit union.

One valid question is: “Among credit-union members, is the proportion who use mobile deposit greater for members younger than \(40\) than for members age \(40\) or older?”

- The word “differ” does not specify a direction. - Write the hypotheses using population parameters rather than sample statistics. - Keep the same parameter expression in both hypotheses.

1. “No difference” is \(p_U-p_R=0\). 2. “Differ” requires a two-sided alternative, \(p_U-p_R\ne0\). 3. Thus \(H_0:p_U-p_R=0\) and \(H_a:p_U-p_R\ne0\).

\(H_0:p_U-p_R=0\); \(H_a:p_U-p_R\ne0\).

- Translate “higher” into the direction of the difference new minus old. - The null hypothesis uses equality. - State hypotheses about the population proportions, not the observed conversions.

1. The null hypothesis states no population difference: \(p_N-p_O=0\). 2. The claim “new is higher” gives the one-sided alternative \(p_N-p_O>0\).

\(H_0:p_N-p_O=0\); \(H_a:p_N-p_O>0\).

- Translate “lower” using the stated order line \(1\) minus line \(2\). - The null hypothesis represents no population difference. - Use population proportions in the hypotheses.

1. No difference is represented by \(p_1-p_2=0\). 2. “Line \(1\) is lower” means \(p_1-p_2<0\). 3. Therefore, \(H_0:p_1-p_2=0\) and \(H_a:p_1-p_2<0\).

\(H_0:p_1-p_2=0\); \(H_a:p_1-p_2<0\).

- Ask whether hypotheses are supposed to describe the samples already observed or the populations being studied. - Distinguish symbols with hats from population-parameter symbols. - Keep Eleni's two-sided direction while correcting the quantities being tested.

1. Hypotheses describe unknown population parameters, not random sample statistics. 2. The population parameter is \(p_1-p_2\). 3. The correct hypotheses are \(H_0:p_1-p_2=0\) and \(H_a:p_1-p_2\ne0\).

Eleni used sample statistics instead of population parameters. Correctly: \(H_0:p_1-p_2=0\); \(H_a:p_1-p_2\ne0\).

- Separate sample size from the way the data were collected. - Ask whether each member of the target populations had a random mechanism for selection. - A large sample can still be systematically biased.

1. The respondents were self-selected rather than obtained through independent random samples or randomized assignment. 2. Large sample sizes do not repair the lack of randomization. 3. Therefore, the randomization condition is not satisfied.

No. The samples are voluntary-response samples, not independent random samples or randomized treatment groups.

- Compare each sample size with \(10\%\) of its own population. - Both population-size checks must pass. - Do not use the combined sample size for this condition.

1. For population \(1\), \(0.10(900)=90\), and \(70\le90\). 2. For population \(2\), \(0.10(1200)=120\), but \(140>120\). 3. The condition fails because it must hold for both samples.

No. It fails for sample \(2\) because \(140>120\).

- For this hypothesis test, first combine the successes to estimate the common proportion under \(H_0\). - Use that pooled proportion to find expected successes and failures for each sample. - Check all four expected counts against the threshold.

1. The pooled proportion is \(\hat p_c=\frac{48+52}{80+120}=0.50\). 2. Under the null model, the expected successes and failures are \(80(0.50)=40\) and \(40\) for sample \(1\), and \(120(0.50)=60\) and \(60\) for sample \(2\). 3. All four expected counts are at least \(10\), so the normality condition holds.

Yes. The pooled expected counts are \(40\), \(40\), \(60\), and \(60\), all at least \(10\).

- Pool the observed successes before checking the test's expected counts. - Expected successes and failures are computed under the null model. - The condition fails if any required expected count is below the threshold.

1. The pooled proportion is \(\hat p_c=\frac{4+6}{30+40}=\frac{10}{70}\approx0.1429\). 2. The expected numbers of successes are \(30(0.1429)\approx4.29\) and \(40(0.1429)\approx5.71\). 3. These are below \(10\), so the pooled normality condition fails.

No. The pooled expected success counts are below \(10\) in both samples.

- Separate design conditions from numerical normality conditions. - Use the pooled proportion because the null hypothesis assumes equal success proportions. - Decide whether a population-size condition is relevant to random assignment.

1. Random assignment satisfies the randomization condition. 2. The \(10\%\) condition is unnecessary for this randomized experiment. 3. The pooled proportion is \(\hat p_c=\frac{90+75}{300}=0.55\). 4. Each group has expected successes \(150(0.55)=82.5\) and expected failures \(150(0.45)=67.5\), all at least \(10\). 5. Therefore, the needed conditions are satisfied.

Yes. Randomization and pooled normality are satisfied, and no \(10\%\) check is needed for the randomized experiment.

- Identify which part of the z statistic changes when the order of subtraction is reversed. - The pooled standard error uses the same two samples regardless of order. - Check whether the magnitude or only the sign should change.

1. Reversing the subtraction changes the observed difference from \(d\) to \(-d\). 2. The pooled standard error is unchanged because it does not depend on subtraction order. 3. Therefore, the z statistic changes sign from \(2.40\) to \(-2.40\).

\(z=-2.40\).

- The null difference is \(0\). - Use the pooled proportion already supplied to form the null standard error. - Standardize only after the pooled standard error is found.

1. The pooled standard error is \(\sqrt{0.40(0.60)(\frac{1}{100}+\frac{1}{150})}\). 2. This standard error is \(\sqrt{0.004}\approx0.06325\). 3. Therefore, \(z=\frac{0.10-0}{0.06325\ldots}\approx1.581\).

\(z\approx1.581\).

- Work backward from the z-score definition. - The null difference is \(0\), so no additional shift is needed after rescaling. - Multiply rather than divide when undoing the standardization.

1. Under the null hypothesis, the standardized statistic is \(z=\frac{(\hat p_1-\hat p_2)-0}{SE}\). 2. Rearranging gives \(\hat p_1-\hat p_2=z(SE)\). 3. Thus \(\hat p_1-\hat p_2=2.00(0.050)=0.10\).

\(0.10\).

- The direction of the alternative determines which tail is used. - A positive z statistic with a greater-than alternative uses the right tail. - The p-value is an area under the standard normal curve, not the z statistic itself.

1. Because the alternative is \(p_1>p_2\), the p-value is the standard normal area to the right of \(1.74\). 2. \(P(Z\ge1.74)\approx0.0409\).

The p-value is approximately \(0.0409\).

- Match the direction of the shaded tail to the greater-than alternative. - The boundary of the shaded region is the observed z statistic. - Use a standard normal table or calculator to find the area in that tail.

1. The greater-than alternative makes the p-value the area to the right of the observed z statistic. 2. Thus the p-value is \(P(Z\ge1.25)=1-\Phi(1.25)\approx0.1056\).

The p-value is approximately \(0.1056\).

- What would an equally extreme result with the opposite sign look like? - How could the bell curve's symmetry help you find the shaded area? - Read the marked z-score and work out what probability it represents in one tail.

1. For a two-sided test, the p-value includes both standard normal tails beyond \(-1.96\) and \(1.96\). 2. Each tail has area about \(0.0250\). 3. The total p-value is therefore about \(2(0.0250)=0.0500\).

The p-value is approximately \(0.0500\).

- Compare the p-value directly with the stated significance level. - The decision concerns the null hypothesis; the conclusion should address the alternative. - Keep the conclusion about population proportions, not just the sampled students.

1. Compare the p-value with the significance level: \(0.041<0.05\). 2. Reject \(H_0\). 3. There is statistically significant evidence that the population proportion of students who support a later start time is greater in district \(1\) than in district \(2\).

Reject \(H_0\). There is statistically significant evidence at the \(0.05\) level that the support proportion is greater in district \(1\) than in district \(2\).

- If the null hypothesis were true, would the return rate depend on which team a volunteer belongs to? - Think about what both groups' results would tell you in that situation.

1. The null hypothesis says the true return proportions at the warehouse and mobile pantry are equal. 2. Under that assumption, both samples provide information about the same unknown proportion, so we combine their counts to estimate a single common proportion when calculating the test standard error.

Under \(H_0\), both groups have the same population proportion. Combining their data estimates that common proportion for the test standard error.

- Begin by treating the null hypothesis as true, then think about outcomes from repeated samples. - Separate the probability of getting sample evidence from the probability that a hypothesis is true.

1. Assume the two schools have equal true support proportions. 2. If the study were repeated under that assumption, a sample difference at least as extreme as the one observed would occur about \(3\%\) of the time. 3. The p-value does not assign a \(3\%\) chance to the hypothesis being true.

If the population proportions were equal, sample results at least this unusual would occur about \(3\%\) of the time. This is not a \(3\%\) chance that the proportions are equal; the calculation assumes equality.

- Decide whether the shaded part represents unusually large or unusually small values of the difference. - Relate positive and negative values of the Ridgeview-minus-Harbor difference to the two population proportions. - In your explanation, distinguish what happened in the samples from the claim being tested about all students.

1. The shaded part is to the left of a negative z-score, so it represents evidence that the first proportion is smaller. 2. The matching alternative is \(H_a:p_R<p_H\). A difference in this tail means a smaller share of sampled Ridgeview students biked to school than sampled Harbor students. Such a result supports the claim that the Ridgeview population proportion is lower.

\(H_a:p_R<p_H\). The left tail represents unusually low Ridgeview-minus-Harbor sample differences: the proportion who bike is smaller in the Ridgeview sample than in the Harbor sample. This supports the claim that Ridgeview's population proportion is lower.

- How many sampled customers does each sample proportion represent? - How might a much larger sample affect a combined proportion? - Check whether the proposed estimate would describe all the sampled customers together.

1. The pooled estimate combines all successes and observations: \(\hat p_c=\frac{18+20}{30+70}=\frac{38}{100}=0.38\). 2. The sample proportions are \(\frac{18}{30}=0.60\) and \(\frac{20}{70}\approx0.2857\), whose simple average is about \(0.4429\). 3. That average gives equal weight to samples of different sizes; pooling counts each customer equally.

The pooled proportion is \(0.38\). The two sample sizes differ, so the pooled proportion weights each observation equally, rather than giving equal weight to the two samples.

- Compare each p-value with the same significance cutoff. - Look at the reported changes in attendance separately from the p-values. - Consider why stronger evidence does not always mean a larger observed change.

1. Springfield's p-value is \(0.01<0.05\), giving statistically significant evidence that reminders affect attendance there. Fairhaven's p-value is \(0.12>0.05\), so that study does not give convincing evidence of an effect. 2. The observed increase is larger in Fairhaven: \(8\) percentage points versus \(1\) percentage point in Springfield. 3. A p-value depends on the observed effect compared with its sampling uncertainty, not on the size of the effect alone. Different sample sizes or amounts of variation can make a small increase measured precisely and a larger increase uncertain.

Springfield gives statistically significant evidence of an effect at \(\alpha=0.05\). Fairhaven has the larger observed increase, but its p-value does not give convincing evidence of an effect. A p-value also depends on sampling uncertainty, not just on the increase.

- Imagine a z-score that is equally far from zero but has the opposite sign. - What does the symmetry of the standard normal curve tell you about the areas beyond those two z-scores? - Which probability needs to be compared with the significance level to make the decision?

1. A two-sided test includes equally extreme results in both tails, so the p-value is \(2(0.018)=0.036\). 2. Since \(0.036<0.05\), reject \(H_0\).

The p-value is \(0.036\); reject \(H_0\) at \(\alpha=0.05\).

- Ask whether the confidence interval assumes the two population proportions are equal. - For the test, consider what the null hypothesis says about the two populations. - Use the observed counts to form the proportion appropriate for each standard error.

1. The sample proportions are \(\hat p_E=\frac{84}{120}=0.70\) and \(\hat p_L=\frac{90}{180}=0.50\). 2. The confidence interval does not assume equal population proportions, so it uses the unpooled standard error: \(SE_{\text{CI}}=\sqrt{\frac{0.70(0.30)}{120}+\frac{0.50(0.50)}{180}}\approx0.0560\). 3. Under the equality null hypothesis, both samples estimate one proportion. The pooled estimate is \(\hat p_c=\frac{84+90}{120+180}=\frac{174}{300}=0.58\). 4. The test uses \(SE_{\text{test}}=\sqrt{0.58(0.42)(\frac{1}{120}+\frac{1}{180})}\approx0.0582\). 5. Putri's pooled standard error is appropriate for the equality test but not for the confidence interval.

No. The confidence interval uses the unpooled standard error, \(0.0560\), while the equality test uses the pooled standard error, \(0.0582\). Pooling is justified by the test's null hypothesis, not by the confidence interval.

- Pool successes and total observations across both samples first. - The pooled standard error represents the null model of a common population proportion. - Keep the two sample-size reciprocals separate inside the final factor.

1. The pooled proportion is \(\hat p_c=\frac{78+63}{120+140}=\frac{141}{260}\approx0.5423\). 2. The null standard error is \(\sqrt{\hat p_c(1-\hat p_c)(\frac{1}{120}+\frac{1}{140})}\). 3. Substitution gives \(SE\approx0.0620\).

\(\hat p_c\approx0.5423\) and \(SE\approx0.0620\).

- Find the observed difference in sample proportions first. - Under the null hypothesis, estimate the common proportion by pooling the successes. - Standardize the observed difference using the pooled standard error.

1. \(\hat p_1=0.56\), \(\hat p_2=0.44\), and \(\hat p_1-\hat p_2=0.12\). 2. The pooled proportion is \(\hat p_c=\frac{112+88}{400}=0.50\). 3. The pooled standard error is \(\sqrt{0.50(0.50)(\frac{1}{200}+\frac{1}{200})}=0.05\). 4. Thus \(z=\frac{0.12-0}{0.05}=2.40\).

\(z=2.40\).

- The alternative direction affects the p-value later, not the formula for the z statistic. - Pool the successes because the null hypothesis assumes equal population proportions. - Standardize the observed sample difference relative to the null value \(0\).

1. The sample proportions are \(0.50\) and \(0.40\), so the observed difference is \(0.10\). 2. The pooled proportion is \(\frac{75+60}{300}=0.45\). 3. The pooled standard error is \(\sqrt{0.45(0.55)(\frac{1}{150}+\frac{1}{150})}\approx0.05745\). 4. Therefore, \(z=\frac{0.10}{0.05745\ldots}\approx1.741\).

\(z\approx1.741\).

- Compute both sample proportions before pooling the success counts. - The pooled proportion is used only in the null standard error. - Divide the observed difference by the pooled standard error after centering at the null difference.

1. The sample proportions are \(0.54\) and \(0.48\), so the observed difference is \(0.06\). 2. The pooled proportion is \(\hat p_c=\frac{54+72}{250}=0.5040\). 3. The pooled standard error is \(\sqrt{0.504(0.496)(\frac{1}{100}+\frac{1}{150})}\approx0.0645\). 4. The test statistic is \(z=\frac{0.06}{0.0645477\ldots}\approx0.9295\).

\(\hat p_c=0.5040\), \(SE\approx0.0645\), and \(z\approx0.9295\).

- The not-equal alternative makes this a two-sided test. - Use the magnitude of the observed z statistic to identify equally extreme values in both tails. - Find one tail area and account for its symmetric counterpart.

1. A two-sided alternative requires probability in both tails at least as extreme as \(|z|=2.10\). 2. One tail has area \(P(Z\le-2.10)\approx0.01786\). 3. Doubling gives \(2(0.01786\ldots)\approx0.0357\).

The p-value is approximately \(0.0357\).

- Begin the interpretation by assuming the null hypothesis is true. - Describe a probability about sample results, not a probability that either hypothesis is true. - Use the greater-than alternative to identify what “at least as extreme” means here.

1. A p-value is interpreted under the assumption that the null hypothesis is true. 2. If the two population recommendation proportions are equal, the probability of obtaining a sample difference \(\hat p_1-\hat p_2\) at least as large as the one observed, from random sampling alone, is about \(0.032\).

If the two population recommendation proportions are equal, there is about a \(3.2\%\) chance of obtaining a sample difference in favor of chain \(1\) at least as large as the observed difference from random sampling alone.

- First compare the p-value with \(\alpha\). - Distinguish “fail to reject” from proving the null hypothesis. - Phrase the final statement in terms of the evidence for the alternative hypothesis.

1. Since \(0.081>0.05\), fail to reject \(H_0\). 2. Failing to reject a null hypothesis does not establish that it is true, so “accept \(H_0\)” and “the proportions are equal” are too strong. 3. The correct conclusion is that the data do not provide statistically significant evidence at the \(0.05\) level that the two population proportions differ.

Fail to reject \(H_0\). The data do not provide statistically significant evidence that the population proportions differ; they do not prove that the proportions are equal.

- Compare the same p-value separately with each significance level. - A smaller significance level demands stronger evidence before rejection. - Separate the numerical evidence from the rule used to make a decision from that evidence.

1. At \(\alpha=0.05\), \(0.032<0.05\), so reject \(H_0\). 2. At \(\alpha=0.01\), \(0.032>0.01\), so fail to reject \(H_0\). 3. The p-value summarizes the same evidence from the data in both cases. Changing \(\alpha\) changes the threshold required for rejection, so the decision can change without any change in the data.

Reject \(H_0\) at \(\alpha=0.05\), but fail to reject \(H_0\) at \(\alpha=0.01\). The evidence is unchanged; only the rejection threshold changes.

- Use the number and location of shaded tails to identify the alternative direction. - Read the z boundary from the graph before finding the standard normal tail area. - Compare the resulting tail area with the stated significance level.

1. Only the right tail is shaded, so the graph represents the greater-than alternative \(H_a:p_1>p_2\). 2. The labeled boundary is \(z=2.05\), so the p-value is \(P(Z\ge2.05)\approx0.0202\). 3. Since \(0.0202<0.05\), reject \(H_0\).

The graph represents \(H_a:p_1>p_2\). The p-value is approximately \(0.0202\), so reject \(H_0\) at \(\alpha=0.05\).

- Ask what assumption about \(p_1\) and \(p_2\) is imposed by the null hypothesis. - A confidence interval estimates a difference without first setting that difference equal to zero. - Connect pooling to the model assumed while calculating the test statistic.

1. A confidence interval estimates the unknown difference without assuming that \(p_1=p_2\), so it estimates each population proportion separately. 2. Its standard error is \(\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}}\). 3. The hypothesis test is carried out under \(H_0:p_1=p_2\), so the null model treats the two population proportions as one common value. 4. Therefore the test uses the pooled estimate \(\hat p_c=\frac{x_1+x_2}{n_1+n_2}\) and standard error \(\sqrt{\hat p_c(1-\hat p_c)(\frac{1}{n_1}+\frac{1}{n_2})}\).

The confidence interval uses the unpooled standard error based on \(\hat p_1\) and \(\hat p_2\) separately. The test uses a pooled standard error because \(H_0\) assumes a common population proportion.

- Ask what the null hypothesis assumes about the two population proportions. - Decide how that null assumption changes the way the sampling variability should be estimated. - After finding the null-model spread, standardize the observed difference and check the requested precision.

1. A test of \(H_0:p_1=p_2\) assumes one common population proportion under the null, so its null standard error must use a pooled estimate. 2. The pooled proportion is \(\hat p_c=\frac{62+60}{100+120}=\frac{122}{220}\approx0.5545\). 3. The pooled standard error is \(\sqrt{0.5545(1-0.5545)(\frac{1}{100}+\frac{1}{120})}\approx0.06730\). 4. The observed difference is \(0.62-0.50=0.12\), so \(z=\frac{0.12}{0.067296\ldots}\approx1.783\).

Daniela's error is using the confidence-interval, unpooled standard error in a null-hypothesis test. The correct pooled test statistic is \(z\approx1.783\).

- The shaded side of the standard normal curve tells you the direction of the alternative. - Read the observed z statistic from the labeled boundary of the shaded tail. - Compare the resulting tail area with \(0.05\) only after identifying the correct direction.

1. The shaded region is in the left tail, so the alternative is \(H_a:p_1<p_2\). 2. The graph labels the observed statistic as \(z=-1.65\). 3. The left-tail area is \(P(Z\le-1.65)\approx0.0495\). 4. Since \(0.0495<0.05\), reject \(H_0\).

The alternative is \(H_a:p_1<p_2\). The p-value is approximately \(0.0495\), so reject \(H_0\) at \(\alpha=0.05\).

- Use the p-value and significance level to make the test decision first. - Random sampling and random assignment support different kinds of conclusions. - Keep the inferential claim about the two population proportions measured by the survey.

1. Since \(0.006<0.05\), reject \(H_0:p_A=p_B\). 2. There is statistically significant evidence that the population proportions of adults who support the proposal differ between the two cities. 3. Because the samples were random, the inference can be generalized to the adult populations represented by the sampling processes. 4. The study does not impose a treatment, so the result does not establish that living in one city causes a person's opinion.

Reject \(H_0\). There is significant evidence that the two citywide support proportions differ. The random sampling supports generalization to the sampled populations, but the comparison does not establish causation.

- Compare the standard-error formulas used for estimation and for testing equality. - Ask what extra assumption is imposed when the null hypothesis sets the two population proportions equal. - Decide whether two procedures with different standard errors must always give exactly matching boundaries.

1. The interval excludes \(0\), so the confidence interval itself supports a nonzero population difference at the \(95\%\) confidence level. 2. However, the usual two-sample z-interval uses an unpooled standard error based on \(\hat p_1\) and \(\hat p_2\) separately. 3. The standard z-test of \(H_0:p_1=p_2\) uses a pooled standard error based on the common proportion assumed under the null hypothesis. 4. Because those standard errors are not generally identical, the two procedures are not exact mathematical inverses. The pooled test statistic and p-value must be computed from the sample data to guarantee the test decision.

No. Excluding \(0\) supports a difference through the confidence-interval procedure, but it does not by itself guarantee the decision of the standard pooled two-sample z-test because the interval and test use different standard errors.

- Check whether the hypotheses describe population parameters or observed sample statistics. - Ask what the equality assumption under the null changes about the standard-error model. - Separate the numerical decision from what a two-sided alternative allows you to conclude about direction and certainty. - Recheck each stage of the test against the stated null and alternative before comparing the p-value with \(\alpha\).

1. Hypotheses must be about the population parameters: \(H_0:p_1=p_2\) and \(H_a:p_1\ne p_2\), not about the sample proportions. 2. Under \(H_0\), use a pooled estimate rather than the unpooled confidence-interval standard error. The pooled proportion is \(\hat p_c=\frac{66+45}{100+90}=\frac{111}{190}\approx0.5842\). 3. The pooled standard error is \(\sqrt{\frac{111}{190}(1-\frac{111}{190})(\frac{1}{100}+\frac{1}{90})}\approx0.07161\). 4. The observed difference is \(0.66-0.50=0.16\), so \(z=\frac{0.16}{0.0716105\ldots}\approx2.234\). 5. For a two-sided test, the p-value is \(2P(Z\ge2.234\ldots)\approx0.0255\). 6. Since \(0.0255<0.05\), reject \(H_0\). The data provide statistically significant evidence that the two population proportions differ. The conclusion is not certainty, and a two-sided alternative does not state a pre-specified greater-than direction.

Corrections include using population parameters in the hypotheses, pooling for the null standard error, and replacing “accept” and “definitely” with an evidence-based conclusion. The correct results are \(z\approx2.234\) and p-value \(\approx0.0255\). Reject \(H_0\); there is significant evidence that \(p_1\ne p_2\).

- Define the two population proportions before writing the hypotheses. - If the two population proportions were equal, what shared quantity would both samples help estimate? - Think about expected renewals and nonrenewals in each group under that assumption. - Use the claim being tested to choose the relevant tail, then state what your decision says about the two plans.

1. Let \(p_B\) and \(p_P\) be the population renewal proportions for Basic and Premium. Test \(H_0:p_B=p_P\) against \(H_a:p_B>p_P\). 2. The samples are random and independent, and each is less than \(10\%\) of its population. Under the null hypothesis, \(\hat p_c=\frac{117+96}{360}=\frac{213}{360}\approx0.5917\). Each group has \(180(\frac{213}{360})=106.5\) expected successes and \(180(\frac{147}{360})=73.5\) expected failures, so the normal condition is satisfied. 3. The sample proportions are \(\hat p_B=\frac{117}{180}=0.65\) and \(\hat p_P=\frac{96}{180}\approx0.5333\), giving an observed difference of about \(0.1167\). 4. The pooled standard error is \(\sqrt{\frac{213}{360}(1-\frac{213}{360})(\frac{1}{180}+\frac{1}{180})}\approx0.05181\). 5. The test statistic is \(z=\frac{0.65-0.5333\ldots}{0.0518113\ldots}\approx2.252\). 6. For the greater-than alternative, the p-value is \(P(Z\ge2.252\ldots)\approx0.0122\). 7. Since \(0.0122<0.05\), reject \(H_0\). There is statistically significant evidence that the population renewal proportion is greater for Basic than for Premium.

The samples are independent and random and meet the \(10\%\) rule. Under the null, each plan has \(106.5\) expected renewals and \(73.5\) expected nonrenewals, all at least \(10\). With \(H_0:p_B=p_P\) and \(H_a:p_B>p_P\), \(z\approx2.252\) and the p-value is approximately \(0.0122\). Reject \(H_0\); there is significant evidence that Basic has the greater population renewal proportion.

- Define the two population proportions and keep Pinebrook minus Riverton in the same order throughout. - Before calculating, check the random sampling, population sizes, and expected yes/no counts under the null. - Under the claim that the population proportions are equal, what rate would the two samples be estimating? - Choose the tail that matches the alternative, then interpret the decision as evidence about population proportions.

1. Let \(p_P\) and \(p_R\) be the population support proportions. Test \(H_0:p_P=p_R\) against \(H_a:p_P<p_R\). 2. The samples are independently and randomly chosen, and each is below \(10\%\) of its city's adult population. With \(\hat p_c=\frac{65+85}{150+150}=0.50\), each sample has \(75\) expected yes responses and \(75\) expected no responses, meeting the large-counts condition. 3. The observed proportions are \(\hat p_P=\frac{65}{150}\) and \(\hat p_R=\frac{85}{150}\). The pooled standard error is \(SE=\sqrt{0.50(0.50)(\frac{1}{150}+\frac{1}{150})}\approx0.0577\). 4. The test statistic is \(z=\frac{\frac{65}{150}-\frac{85}{150}}{\sqrt{0.50(0.50)(\frac{1}{150}+\frac{1}{150})}}\approx-2.3094\). The left-tail p-value is approximately \(0.0105\). 5. Since \(0.0105<0.05\), reject \(H_0\). The samples give convincing evidence that a lower proportion of adults in Pinebrook supports protected bike lanes than in Riverton. The survey does not show why support differs.

The independent random samples meet the \(10\%\) rule. With pooled proportion \(0.50\), each sample has \(75\) expected yes responses and \(75\) expected no responses. For \(H_0:p_P=p_R\) and \(H_a:p_P<p_R\), \(z\approx-2.3094\) and the p-value is approximately \(0.0105\). Reject \(H_0\); the data support lower population support in Pinebrook than in Riverton.

- The question asks whether the reminders differ, without choosing a direction first. What does that mean for the alternative and p-value? - Random assignment and random sampling are different study designs. Decide which conditions apply here. - Under equality, the two groups' attendance rates are estimates of what common quantity? - Keep the observed percentage-point difference separate from its statistical significance and from any cost decision.

1. Let \(p_P\) and \(p_S\) be attendance proportions under personalized and standard reminders. Test \(H_0:p_P=p_S\) against \(H_a:p_P\ne p_S\). 2. Random assignment makes the treatment groups comparable, and individual attendance is assumed independent. The pooled estimate is \(\hat p_c=\frac{610+560}{2000}=0.585\). Under the null, each group expects \(585\) attendees and \(415\) nonattendees, so the large-counts condition is met. The \(10\%\) sampling rule is not needed for random assignment. 3. The observed difference is \(0.610-0.560=0.050\). The pooled standard error is \(SE=\sqrt{0.585(0.415)(\frac{1}{1000}+\frac{1}{1000})}\approx0.0220\). 4. The test statistic is \(z=\frac{0.050}{\sqrt{0.585(0.415)(\frac{1}{1000}+\frac{1}{1000})}}\approx2.2691\). The two-sided p-value is approximately \(0.0233\). 5. Since \(0.0233<0.05\), reject \(H_0\). There is evidence that reminder type affects attendance for these volunteers. The observed difference is \(5\) percentage points, but the test alone does not show whether that improvement is worth the extra cost or would carry over to other groups.

Random assignment and the stated independence assumption apply. Under \(H_0:p_P=p_S\), the pooled expected counts in each group are \(585\) attendees and \(415\) nonattendees. For \(H_a:p_P\ne p_S\), \(z\approx2.2691\) and the two-sided p-value is approximately \(0.0233\). Reject \(H_0\). Reminder type appears to affect attendance among these volunteers; whether a \(5\)-percentage-point increase is worthwhile requires more information, and the result may not generalize to everyone.