- Identify whether the response variable is categorical or quantitative. - Count how many populations are being compared and identify the parameter being estimated.

1. The response is binary, and the parameter is a difference between two population proportions. 2. The samples are independent, so the appropriate procedure is a two-sample z-interval for a difference between two population proportions.

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

- Use population proportions for the parameter, not sample proportions. - Preserve the order requested in the question.

1. The requested population comparison is juniors minus seniors. 2. Therefore, the parameter is \(p_J-p_S\).

\(p_J-p_S\), the true junior attendance proportion minus the true senior attendance proportion.

- Distinguish the statistic computed from the sample from the unknown parameter being estimated. - Confidence intervals are designed to estimate population quantities.

1. The sample difference \(\hat p_1-\hat p_2\) is the point estimate used to construct the interval. 2. The interval itself estimates the population parameter \(p_1-p_2\).

It estimates \(p_1-p_2\).

- The condition can be met through random sampling or through randomized treatment assignment. - Focus on how the treatment groups were formed.

1. A randomized experiment satisfies the randomization condition when treatments are randomly assigned to experimental units. 2. The problem states that assignment was random, so the condition holds.

Yes, the randomization condition holds.

- What would a difference of \(0\) mean for the two neighborhoods? - How does the subtraction order affect the meaning of a positive or negative difference?

1. The interval lies entirely above \(0\). A difference of \(0\) would mean equal proportions, but \(0\) is not in the interval. 2. Every value in the interval is positive. Since the difference is Lakewood minus Oak Hill, the interval provides evidence that the proportion is higher in Lakewood.

A larger proportion in Lakewood. The \(95\%\) confidence interval is entirely above \(0\), so it supports a positive difference (Lakewood minus Oak Hill), not equal proportions.

- Check each sample against its own population size. - The interval condition requires both samples to satisfy the population-size restriction. - Do not combine the two population sizes into one denominator.

1. For population \(1\), \(0.10(1200)=120\), and \(80\le120\). 2. For population \(2\), \(0.10(700)=70\), but \(90>70\). 3. Because the condition must hold for both populations, it fails overall.

No. It fails for population \(2\) because \(90>70\).

- For an interval, use the observed successes and failures in each sample. - All four observed counts must meet the threshold. - One count below the threshold is enough for the condition to fail.

1. Sample \(1\) has \(18\) successes and \(42\) failures, both at least \(10\). 2. Sample \(2\) has \(9\) successes and \(41\) failures. 3. Because sample \(2\) has only \(9\) observed successes, the normality condition fails.

No. The condition fails because sample \(2\) has only \(9\) successes.

- Organize the check into randomization, population-size, and observed-count conditions. - For each sample, count both successes and failures. - The interval uses the separate observed counts rather than a pooled proportion.

1. Randomization holds because both samples are simple random samples and are independent. 2. The \(10\%\) condition is stated to hold for both populations. 3. The observed success and failure counts are \(44\), \(56\), \(63\), and \(57\), all at least \(10\). 4. Therefore, all conditions are satisfied.

All conditions for a two-sample z-interval are satisfied.

- Distinguish a randomized experiment from sampling without replacement. - Check the success and failure counts separately in both treatment groups. - Do not impose a population-size condition when the design does not require it.

1. Random assignment satisfies the randomization condition. 2. The \(10\%\) condition is not needed for a randomized experiment because sampling without replacement from a population is not the source of dependence. 3. The observed counts are \(52\) successes and \(28\) failures for treatment \(1\), and \(43\) successes and \(37\) failures for treatment \(2\); all are at least \(10\). 4. The needed conditions are satisfied.

Randomization and normality are satisfied; the \(10\%\) condition is unnecessary for this randomized experiment.

- Convert each success count to a sample proportion first. - Preserve the requested order when subtracting the two sample proportions. - The point estimate is a statistic, not yet a confidence interval.

1. \(\hat p_1=\frac{84}{120}=0.70\). 2. \(\hat p_2=\frac{65}{100}=0.65\). 3. The point estimate is \(0.70-0.65=0.05\).

\(0.05\).

- An interval standard error is unpooled. - Each sample contributes its own estimated variance term. - Add the variance terms before taking the square root.

1. For a confidence interval, use the two sample proportions separately. 2. \(SE=\sqrt{\frac{0.60(0.40)}{100}+\frac{0.60(0.40)}{140}}\). 3. Thus \(SE\approx0.0641\).

\(SE\approx0.0641\).

- The margin of error is the half-width of the interval. - Use the supplied critical value with the standard error.

1. Margin of error equals the critical value times the standard error. 2. \(ME=1.96(0.040)=0.0784\).

\(0.0784\).

- The point estimate comes from subtracting the two sample proportions. - The interval standard error uses the two sample proportions separately. - Keep the order consistent in both the statistic and its interpretation.

1. The point estimate is \(0.42-0.34=0.08\). 2. The standard error is \(\sqrt{\frac{0.42(0.58)}{200}+\frac{0.34(0.66)}{200}}\). 3. Thus \(SE\approx0.0484\).

Point estimate: \(0.08\). Standard error: \(0.0484\).

- A confidence interval is point estimate plus or minus margin of error. - The point estimate should be the midpoint of the two endpoints.

1. Subtract the margin of error from the point estimate: \(0.08-0.06=0.02\). 2. Add the margin of error to the point estimate: \(0.08+0.06=0.14\).

\((0.02,0.14)\).

- Refer to the two population proportions, not just the sample proportions. - Translate the positive endpoints into a directional comparison. - Include the confidence level and the context of the response variable.

1. The interval estimates the population difference \(p_1-p_2\), not a difference limited to the sampled employees. 2. We are \(95\%\) confident that company \(1\)'s true proportion is between \(0.03\) and \(0.11\) greater than company \(2\)'s true proportion.

We are \(95\%\) confident that the true proportion preferring a hybrid schedule is between \(3\) and \(11\) percentage points higher at company \(1\) than at company \(2\).

- Interpret the sign before translating the size of the difference. - A negative value of \(p_1-p_2\) means population \(1\) has the smaller proportion. - State the conclusion about population proportions in context.

1. Both endpoints are negative, so \(p_1\) is estimated to be smaller than \(p_2\). 2. The difference \(p_1-p_2\) is estimated to be between \(-0.14\) and \(-0.04\). 3. Equivalently, service \(1\)'s recommendation proportion is estimated to be between \(4\) and \(14\) percentage points lower than service \(2\)'s.

We are \(95\%\) confident that the true recommendation proportion for service \(1\) is between \(4\) and \(14\) percentage points lower than that for service \(2\).

- Think about repeating the entire sampling-and-interval procedure many times. - The population parameter is fixed; the interval endpoints change from sample to sample. - Interpret the percentage as a long-run capture rate for the method.

1. The confidence level describes the long-run performance of the interval-producing method. 2. In repeated sampling, approximately \(90\%\) of the intervals constructed by this method will contain the fixed population difference \(p_1-p_2\).

Over many repetitions of the sampling process, about \(90\%\) of the resulting intervals will capture the true value of \(p_1-p_2\).

- What value of \(p_1-p_2\) would mean the two population proportions are equal? - Ask whether each interval still allows that value. - For an interval that rules it out, consider what the sign of group \(1\) minus group \(2\) means.

1. Interval A lies entirely above \(0\), so it provides evidence that \(p_1>p_2\). 2. Interval B contains \(0\), so it does not provide convincing evidence of a difference. 3. Interval C lies entirely below \(0\), so it provides evidence that \(p_1<p_2\).

Intervals A and C provide evidence of a difference. A supports \(p_1>p_2\); C supports \(p_1<p_2\). B does not show a convincing difference.

- The point estimate lies halfway between the two endpoints. - The margin of error is the interval's half-width. - Use the scale on the horizontal axis rather than estimating from the segment length alone.

1. The endpoints shown are \(-0.04\) and \(0.12\). 2. The point estimate is the midpoint: \(\frac{-0.04+0.12}{2}=0.04\). 3. The margin of error is the distance from the midpoint to either endpoint: \(0.12-0.04=0.08\).

Point estimate: \(0.04\). Margin of error: \(0.08\).

- Think about what the interval is trying to capture when the confidence level rises. - The samples are unchanged, so ask which part of estimate \(\pm\) margin of error can change.

1. Higher confidence requires a larger critical value and a larger margin of error. 2. The \(99\%\) interval is wider, giving more possible values for the true difference and a higher long-run chance of capturing it. 3. Both intervals use the same sample difference as their midpoint, so the point estimate does not change.

The \(99\%\) interval is wider. It includes more possible values of the true difference, so it has a higher long-run capture rate. The point estimate stays the same; the margin of error increases.

- Ask what kind of quantity a confidence interval is estimating here. - How should the units of a difference between two proportions appear in a sentence?

1. Maya describes the proportion at one school, but the interval estimates the difference between the two population proportions. 2. We are \(95\%\) confident that the proportion who like lunch is between \(4\) and \(12\) percentage points higher at Pine High than at Westside High.

The interval does not estimate the proportion at Pine High alone. We are \(95\%\) confident that the proportion who like lunch is \(4\) to \(12\) percentage points higher at Pine High than at Westside High.

- Think of a confidence interval as a range of possible values for one fixed population difference. - For each sample, ask whether that fixed value could lie within its interval. - Why might intervals from new random samples cover different ranges?

1. Compare each horizontal interval with the dashed vertical line marking the fixed true difference. 2. Intervals A, C, E, and F cross the line; intervals B and D miss it. 3. A confidence-interval method can miss the true parameter on some samples, even if it has a high long-run capture rate.

A, C, E, and F contain the true difference; B and D do not. No, the method does not guarantee that every interval captures it.

- Try subtracting two easy proportions in both orders. What changes? - Think about how both ends of a range of possible differences would change when the order changes. - Does reversing the comparison change whether equality is among the possibilities?

1. Reversing the subtraction changes the sign of each possible difference. 2. Negate both endpoints and put them in increasing order: \((-0.02,0.08)\). 3. Both intervals include \(0\), so neither supports a clear direction for the population difference at the \(95\%\) confidence level.

Glenview minus Riverside: \((-0.02,0.08)\). The conclusion is unchanged: the interval does not show which neighborhood has the larger proportion.

- Convert the counts to sample proportions before computing the interval quantities. - Use an unpooled standard error for a confidence interval. - Multiply the standard error by the supplied critical value only after finding the standard error.

1. The sample proportions are \(\hat p_1=0.60\) and \(\hat p_2=0.50\), so the point estimate is \(0.10\). 2. \(SE=\sqrt{\frac{0.60(0.40)}{160}+\frac{0.50(0.50)}{150}}\approx0.0563\). 3. \(ME=1.645(0.056273\ldots)\approx0.0926\).

Point estimate: \(0.10\). Standard error: \(0.0563\). Margin of error: \(0.0926\).

- Begin with the difference in the two sample proportions. - The confidence-interval standard error is unpooled. - Form the two endpoints only after computing the margin of error.

1. \(\hat p_1=\frac{78}{120}=0.65\) and \(\hat p_2=\frac{60}{120}=0.50\), so the point estimate is \(0.15\). 2. \(SE=\sqrt{\frac{0.65(0.35)}{120}+\frac{0.50(0.50)}{120}}\approx0.0631\). 3. The margin of error is \(1.96(0.0630806\ldots)\approx0.1236\). 4. The interval is \(0.15\pm0.1236\), or approximately \((0.0264,0.2736)\).

\((0.0264,0.2736)\).

- Calculate the sample proportions before writing the standard error. - Use the confidence-level critical value supplied in the problem. - Check that the point estimate lies at the midpoint of your final interval.

1. \(\hat p_1=0.70\), \(\hat p_2=0.60\), and the point estimate is \(0.10\). 2. \(SE=\sqrt{\frac{0.70(0.30)}{150}+\frac{0.60(0.40)}{160}}\approx0.0539\). 3. The margin of error is \(1.645(0.0538516\ldots)\approx0.0886\). 4. The interval is approximately \((0.0114,0.1886)\).

\((0.0114,0.1886)\).

- Use the difference in the two sample proportions as the point estimate. - A higher confidence level uses a larger critical value and therefore a wider interval. - Keep the standard error unpooled for an interval.

1. The point estimate is \(0.38-0.31=0.07\). 2. \(SE=\sqrt{\frac{0.38(0.62)}{250}+\frac{0.31(0.69)}{300}}\approx0.0407\). 3. The margin of error is \(2.576(0.0406866\ldots)\approx0.1048\). 4. The interval is approximately \((-0.0348,0.1748)\).

\((-0.0348,0.1748)\).

- Find the point estimate before calculating the margin of error. - Each sample proportion has its own variance contribution. - A final interval entirely above \(0\) is possible even when the margin of error is fairly large.

1. The point estimate is \(0.56-0.44=0.12\). 2. The unpooled standard error is approximately \(0.0499\). 3. The margin of error is \(1.96(0.0498888\ldots)\approx0.0978\). 4. The interval is approximately \((0.0222,0.2178)\).

\((0.0222,0.2178)\).

- The requested order can make the point estimate negative. - Apply the margin of error symmetrically around that negative point estimate. - Check that both endpoints have the expected order from smaller to larger.

1. The point estimate is \(0.28-0.35=-0.07\). 2. The standard error is \(\sqrt{\frac{0.28(0.72)}{300}+\frac{0.35(0.65)}{280}}\approx0.0385\). 3. The margin of error is approximately \(1.645(0.0385292\ldots)=0.0634\). 4. The interval is approximately \((-0.1334,-0.0066)\).

\((-0.1334,-0.0066)\).

- Compute the numerical interval before comparing it with the displayed choices. - The correct graph must have midpoint \(0.10\). - Its half-width should equal the calculated margin of error.

1. The point estimate is \(0.62-0.52=0.10\). 2. \(SE=\sqrt{\frac{0.62(0.38)}{200}+\frac{0.52(0.48)}{200}}\approx0.0493\). 3. The margin of error is approximately \(1.96(0.0493)=0.0965\). 4. The interval is approximately \((0.0035,0.1965)\), which corresponds to interval B in the graph.

Interval B.

- Decide which quantity is fixed and which quantities would change under repeated sampling. - The confidence percentage describes the procedure that generated the interval. - Rewrite the conclusion using “we are confident” rather than assigning a post-data probability to the parameter.

1. In the usual frequentist interpretation, \(p_1-p_2\) is a fixed but unknown parameter after the populations are defined. 2. The \(95\%\) refers to the long-run success rate of the interval method, not to a probability assigned to the fixed parameter after this interval is computed. 3. A correct statement is that we are \(95\%\) confident that the interval from \(0.02\) to \(0.18\) contains the true difference \(p_1-p_2\).

The probability statement is not the standard frequentist interpretation. We are \(95\%\) confident that \((0.02,0.18)\) contains the true population difference \(p_1-p_2\).

- Both intervals use the same critical value, so compare how each margin depends on its standard error. - Use the point estimate to locate each interval relative to \(0\). - Consider the lower endpoints before deciding whether either interval includes \(0\).

1. Survey A has margin of error \(1.96(0.040)=0.0784\). 2. Survey B has margin of error \(1.96(0.025)=0.0490\). 3. Survey B has the smaller margin, so its interval is narrower. 4. The lower endpoints are \(0.06-0.0784=-0.0184\) for A and \(0.06-0.0490=0.0110\) for B. A includes \(0\); B does not.

Survey A: margin \(0.0784\), interval \((-0.0184,0.1384)\), which includes \(0\). Survey B: margin \(0.0490\), interval \((0.0110,0.1090)\), which excludes \(0\). B is narrower.

- Check both samples' designs and their success and failure counts before calculating. - What does each sample tell you about its own population proportion? - Interpret the two endpoints as differences between the libraries' population proportions, not as percentages of all adults combined.

1. Both samples are independent and random, and each is less than \(10\%\) of its population. The success and failure counts are \(126\), \(84\), \(84\), and \(116\), all at least \(10\). 2. The sample proportions are \(\hat p_R=\frac{126}{210}=0.60\) and \(\hat p_H=\frac{84}{200}=0.42\), so the point estimate is \(0.18\). 3. The unpooled standard error is \(SE=\sqrt{\frac{0.60(0.40)}{210}+\frac{0.42(0.58)}{200}}\approx0.04859\). 4. The margin of error is \(1.96(SE)\approx0.09523\). The interval is \(0.18\pm0.09523\approx(0.0848,0.2752)\). 5. We are \(95\%\) confident that the proportion of adult cardholders who borrowed an e-book is between \(8.48\) and \(27.52\) percentage points higher at Riverside than at Hillcrest.

Both samples are independent, random, and below \(10\%\) of their populations. Riverside has \(126\) e-book borrowers and \(84\) nonborrowers; Hillcrest has \(84\) borrowers and \(116\) nonborrowers. All four counts are at least \(10\). The \(95\%\) interval is \((0.0848,0.2752)\). We are \(95\%\) confident that Riverside's true e-book borrowing proportion is between \(8.48\) and \(27.52\) percentage points higher than Hillcrest's.

- Consider how random assignment changes the design conditions you need to check. - What information does each reminder group provide about its own attendance proportion? - What would a population difference of zero mean for the two reminders? Compare that possibility with the interval you calculate.

1. Random assignment supports a treatment comparison, and participants' attendance is assumed independent. The \(10\%\) sampling rule is not needed for random assignment. The observed success and failure counts are \(68\), \(32\), \(58\), and \(42\), all at least \(10\). 2. The sample difference is \(\hat p_T-\hat p_E=0.68-0.58=0.10\). 3. The unpooled standard error is \(SE=\sqrt{\frac{0.68(0.32)}{100}+\frac{0.58(0.42)}{100}}\approx0.06791\). 4. The interval is \(0.10\pm1.96(SE)\approx(-0.0331,0.2331)\). 5. The interval includes \(0\), so the experiment does not give convincing evidence that text reminders increase attendance. Random assignment supports causal inference for volunteers like these, but the data do not establish a positive effect.

Random assignment and the stated independence assumption apply. The attendance/nonattendance counts are \(68\)/\(32\) for text and \(58\)/\(42\) for email, all at least \(10\). The \(95\%\) confidence interval is \((-0.0331,0.2331)\). It includes \(0\), so the experiment does not give convincing evidence that texts cause higher attendance for volunteers like these.

- Identify every type of difference represented inside the interval: negative, zero, and positive. - “Not enough evidence of a difference” is not the same as “proof of equality.” - Confidence intervals support ranges of plausible parameter values rather than exact proofs.

1. The interval contains \(0\), so equality is a plausible value for the population difference. 2. The interval also contains negative and positive differences. 3. Therefore, the data do not provide convincing evidence of either direction, but including \(0\) does not prove exact equality.

Both statements are too strong. The interval does not provide convincing evidence that either proportion is larger, and it does not prove that the proportions are exactly equal.

- Compare the claim's threshold with every plausible value in the interval. - Evidence that a difference is positive is weaker than evidence that it exceeds a particular positive amount. - Use the lower endpoint to evaluate a “at least” claim.

1. Because the entire interval is above \(0\), the interval supports the conclusion that \(p_1>p_2\). 2. However, the interval contains values such as \(0.01\), \(0.02\), and \(0.04\), which are below \(0.05\). 3. Therefore, the interval does not support the stronger claim that the difference is at least \(0.05\).

The interval supports \(p_1>p_2\), but it does not support the claim that \(p_1-p_2\ge0.05\).

- Ask whether an interval-construction problem assumes the two population proportions are equal. - Compare the role of pooling in a null-hypothesis test with the role of estimation in a confidence interval. - A small numerical difference between methods does not make an unjustified method valid.

1. The sample proportions are \(0.60\) and \(0.48\), so the point estimate is \(0.12\). 2. A confidence interval does not assume \(p_1=p_2\), so its standard error is unpooled: \(SE=\sqrt{\frac{0.60(0.40)}{120}+\frac{0.48(0.52)}{100}}\approx0.0671\). 3. The correct interval is \(0.12\pm1.96(0.067052\ldots)\approx(-0.0114,0.2514)\). 4. The pooled proportion is \(\hat p_c=\frac{72+48}{120+100}\approx0.5455\), giving a pooled standard error of approximately \(0.0674\). 5. Chinedu's pooled interval is approximately \((-0.0121,0.2521)\). Although numerically close here, it is based on the wrong model for estimation.

The correct interval is approximately \((-0.0114,0.2514)\). The pooled interval is approximately \((-0.0121,0.2521)\), but pooling is inappropriate for a confidence interval.

- Intervals from the same data share a common midpoint but differ in width. - Higher confidence requires a wider interval when the data are fixed. - Use whether \(0\) is included to connect each interval to a claim about a difference.

1. Both intervals have midpoint \(0.06\), so they are consistent with the same point estimate. 2. Interval B is wider, so B is the \(99\%\) confidence interval and A is the \(90\%\) confidence interval. 3. Interval A lies entirely above \(0\), so at the \(90\%\) level it supports \(p_1>p_2\). 4. Interval B includes \(0\), so at the \(99\%\) level the data do not provide convincing evidence that \(p_1>p_2\). 5. Greater confidence requires a larger critical value and margin of error, which can change whether \(0\) is included.

A is the \(90\%\) interval and B is the \(99\%\) interval. A supports \(p_1>p_2\); B includes \(0\), so it does not support that conclusion at the higher confidence level.

- Write the lower endpoint as the point estimate minus a margin of error. - Work backward from the requirement that the lower endpoint be positive. - Check both proposed standard errors against the condition, not just against each other.

1. For the interval to lie entirely above \(0\), its lower endpoint must satisfy \(0.07-1.96(SE)>0\). 2. Rearranging gives \(SE<\frac{0.07}{1.96}\approx0.03571\). 3. An SE of \(0.050\) is too large, so its interval includes \(0\). An SE of \(0.030\) is small enough, so its interval lies entirely above \(0\), provided the point estimate stays at \(0.07\).

The standard error must be less than \(\frac{0.07}{1.96}\approx0.03571\). A survey with SE \(0.050\) does not meet the goal; one with SE \(0.030\) does, assuming the point estimate stays at \(0.07\).