- First distinguish the full interval width from the margin of error. - The margin of error measures the distance from the center to either endpoint. - Use the two endpoints to find that half-width.

1. The total interval width is \(0.62-0.54=0.08\). 2. The margin of error is half the width, so \(E=\frac{0.08}{2}=0.04\).

The margin of error is \(0.04\), or \(4\) percentage points.

- Focus on what changes when the confidence level rises. - The sample information is unchanged, so the estimated standard error does not change. - Think about the multiplier needed to capture a larger central percentage of a normal distribution.

1. A higher confidence level requires a larger critical value. 2. With the sample size and sample proportion unchanged, the estimated standard error stays the same. 3. Therefore, the larger critical value makes the margin of error increase.

The margin of error increases because a higher confidence level uses a larger critical value while the estimated standard error is unchanged.

- Read the two endpoints from the diagram. - How does the distance from the center to an endpoint compare with the full interval width? - Convert a proportion difference to percentage points.

1. The interval runs from \(0.38\) to \(0.54\). 2. Its half-width is \(\frac{0.54-0.38}{2}=0.08\), or \(8\) percentage points.

\(8\) percentage points.

- Compare the original and new sample sizes. - How does the sample size enter the formula for the margin of error?

1. The sample size grows by a factor of \(4\). 2. Margin of error varies approximately as \(1/\sqrt{n}\), so it is halved: \(6\div2=3\) percentage points.

About \(3\) percentage points.

- How would the interval have to change to capture a greater proportion of possible sample results? - Compare the critical values for the two confidence levels.

1. Greater confidence requires covering more of the sampling distribution. 2. With the same sample and method, the \(99\%\) interval has a larger critical value and therefore a larger margin of error.

The \(99\%\) interval; greater confidence means a wider interval when the sample is unchanged.

- Compare the lengths of the colored intervals. - How is the distance from the center to an endpoint related to the full width?

1. Interval a) runs from \(0.38\) to \(0.62\), so its margin of error is \(\frac{0.62-0.38}{2}=0.12\). 2. Interval b) runs from \(0.44\) to \(0.56\), giving \(\frac{0.56-0.44}{2}=0.06\). 3. Interval b) has the smaller margin of error.

Interval b), with margin of error \(0.06\), or \(6\) percentage points.

- Use the pilot result rather than assuming the two outcomes are equally likely. - Think about which quantity must be isolated when planning the study. - A sample-size answer must be rounded in the direction that preserves the desired precision.

1. Use the pilot estimate \(\hat p=0.62\), the \(95\%\) critical value \(z^*=1.96\), and the target margin of error \(0.03\). 2. Solving the margin-of-error relationship for \(n\) gives \(n=\frac{(1.96)^2(0.62)(0.38)}{(0.03)^2}\approx 1005.65\). 3. Round up to ensure the target margin of error is met, giving \(n=1006\).

The agency should sample at least \(1006\) riders.

- Convert the audit count to a sample proportion first. - Estimate the sampling variability from the observed proportion. - Distinguish a proportion such as \(0.03\) from \(3\) percentage points.

1. The sample proportion is \(\hat p=\frac{24}{300}=0.08\). 2. The sample is random, \(300\le 0.10(5000)=500\), and the observed counts are \(24\) and \(276\), both at least \(10\). 3. The standard error is \(\sqrt{\frac{0.08(0.92)}{300}}\approx 0.0157\). 4. The margin of error is \(1.96(0.0157)\approx 0.0307\). 5. In percentage-point form, the margin of error is approximately \(3.07\) percentage points.

The margin of error is approximately \(0.0307\), or \(3.07\) percentage points.

- The estimated standard error is fixed, so only the confidence multiplier changes. - Compute each candidate margin before making the selection. - Choose the greatest confidence level that stays within the precision limit.

1. At \(90\%\), the margin of error is \(1.645(0.020)=0.0329\). 2. At \(95\%\), the margin of error is \(1.96(0.020)=0.0392\). 3. At \(99\%\), the margin of error is \(2.576(0.020)=0.0515\). 4. The \(95\%\) margin of error is at most \(0.040\), but the \(99\%\) margin is not. Therefore, \(95\%\) is the highest allowable confidence level.

a) \(90\%: 0.0329\); \(95\%: 0.0392\); \(99\%: 0.0515\). b) \(95\%\).

- Hold the confidence level and anticipated population composition fixed. - Compare the old and new sample sizes through their square roots. - Distinguish square-root scaling from direct proportionality.

1. With the confidence level and anticipated proportion fixed, margin of error is proportional to \(1/\sqrt{n}\). 2. Changing the sample size from \(1600\) to \(800\) multiplies the margin of error by \(\sqrt{\frac{1600}{800}}=\sqrt{2}\approx 1.414\). 3. The new margin of error is \(0.018\sqrt{2}\approx 0.02546\). 4. The margin of error changes with the square root of sample size, not in direct inverse proportion to sample size.

a) It increases by a factor of \(\sqrt{2}\approx 1.414\). b) Approximately \(0.0255\). c) Margin of error is proportional to the reciprocal of the square root of sample size, so halving \(n\) multiplies it by \(\sqrt{2}\), not \(2\).

- Convert a proportion difference to percentage points by changing its unit, not its meaning. - For a relative comparison, compare the margin with the estimate itself. - Distinguish an absolute change in a proportion from a percent change relative to that proportion.

1. A margin of error of \(0.02\) equals \(2\) percentage points. 2. Relative to the estimate, the margin is \(\frac{0.02}{0.08}=0.25\), or \(25\%\) of the point estimate. 3. The phrase “\(2\%\)” could mean an absolute change of \(2\) percentage points or a relative change equal to \(2\%\) of the estimate. These are different quantities.

a) \(2\) percentage points. b) \(25\%\) of the point estimate. c) It can refer either to an absolute percentage-point margin or to a relative percent of the estimate.

- Consider what aspect of uncertainty sample size addresses. - Identify how people entered the sample and whether that mechanism could be selective. - Separate narrowness of an interval from whether its center targets the correct population value.

1. A large sample size reduces the estimated standard error, so the formula-based margin of error can be small. 2. Voluntary respondents may differ systematically from customers who ignore the pop-up. Because the sample was not randomly selected from all customers, selection bias can shift the estimate away from the population proportion. 3. Precision refers to small random sampling variability, reflected by a narrow interval. Validity requires a design that supports inference to the target population. A precise interval can be centered at a biased value.

a) The large sample size reduces the formula-based standard error and margin of error. b) Self-selection can bias the sample, so the interval may be centered away from the true customer proportion. c) The interval can be precise but not valid for the intended population.

- Convert total tail probability into central confidence. - Split the remaining tail probability equally between the two sides. - Use the resulting cutoff with the estimated variability.

1. The confidence level is \(1-0.03=0.97\), or \(97\%\). 2. Each tail has area \(0.015\), so the upper cumulative area is \(0.985\). This gives \(z^*\approx 2.170\). 3. The margin of error is \(2.170(0.020)\approx 0.04340\).

a) \(97\%\). b) \(z^*\approx 2.170\). c) Approximately \(0.04340\).

- Convert the verbal precision requirement into the same scale used for proportions. - Use the planning value that gives the largest possible variability when no estimate is available. - Check the final rounding direction against the phrase “at most.”

1. Three percentage points must be entered in proportion form as \(E=0.03\), not \(E=3\). 2. With the conservative planning value \(p=0.50\), \(n=\frac{(1.96)^2(0.25)}{(0.03)^2}\approx1067.11\). 3. Round up to ensure the margin of error is no greater than the target, giving \(n=1068\).

The correct input is \(E=0.03\). The conservative minimum sample size is \(1068\).

- Separate the desired central area from the cutoff measured in standard-deviation units. - Compute the estimated standard error only once, then compare the two multipliers. - Check whether the proposed critical value is a plausible standard normal cutoff for a wide central interval.

1. The estimated standard error is \(\sqrt{\frac{(0.40)(0.60)}{400}}\approx0.02449\). 2. Farah's margin of error is \(0.95(0.02449)\approx0.02327\). 3. The correct critical value is \(1.96\), giving \(1.96(0.02449)\approx0.04801\). 4. The confidence level is a central probability, while the critical value is a standard normal cutoff. They are not the same number.

a) Approximately \(0.02327\). b) Approximately \(0.04801\). c) Farah confused the \(95\%\) central confidence level with the standard normal critical value \(1.96\).

- Separate random sample-to-sample variation from systematic errors in who enters the sample. - Ask which assumptions are required before a margin of error has its usual interpretation. - Consider whether increasing the number of self-selected respondents would fix the source of bias.

1. A margin of error describes the estimated sampling variability of the sample proportion under the assumptions of the confidence-interval procedure. 2. It does not measure voluntary-response bias or undercoverage because those errors come from how the sample was obtained, not from random sample-to-sample variation. 3. The reported \(2.1\)-percentage-point margin of error therefore cannot make the survey representative of the intended population.

The margin of error is intended to measure random sampling uncertainty around the sample proportion. It does not account for voluntary-response bias or the undercoverage of adults without internet access, so the reported precision does not make the survey results representative.

- Use the number of observations that actually contributes to each estimate. - Compare how the uncertainty changes when the effective sample size becomes much smaller. - Do not assume one reported precision applies automatically to every subgroup.

1. For the full sample, the margin of error is \(1.96\sqrt{\frac{(0.50)(0.50)}{1000}}\approx 0.03099\). 2. For the subgroup, the margin of error is \(1.96\sqrt{\frac{(0.50)(0.50)}{240}}\approx 0.06326\). 3. The subgroup margin of error is about \(2.04\) times the full-sample margin because the subgroup estimate is based on far fewer observations.

a) Approximately \(0.0310\), or \(3.10\) percentage points. b) Approximately \(0.0633\), or \(6.33\) percentage points. c) The subgroup estimate uses only \(240\) observations, so its sampling variability is much larger than that of the full sample.

- Write the estimator of the benchmark difference in terms of the sample proportion. - Ask whether subtracting a fixed number changes random spread. - Separate a change in the estimate’s center from a change in its uncertainty.

1. The estimated standard error of \(\hat p\) is \(\sqrt{\frac{(0.67)(0.33)}{500}}\). 2. The \(95\%\) margin of error is \(1.96\sqrt{\frac{(0.67)(0.33)}{500}}\approx 0.04122\). 3. The estimator of \(d\) is \(\hat d=\hat p-0.60\). Subtracting a fixed constant changes the center but not the sampling variability. 4. Therefore, the margin of error for \(d\) is also approximately \(0.04122\).

a) Approximately \(0.04122\). b) Approximately \(0.04122\). c) Subtracting the fixed benchmark \(0.60\) shifts the estimate but does not change its standard error, so the margin of error is unchanged.

- Preserve the original count information as long as possible. - Perform the two calculations with all other inputs held fixed. - Compare the results only after carrying sufficient precision through both computations.

1. The exact sample proportion is \(\hat p=\frac{137}{260}\approx 0.526923\). 2. Using the exact value, the margin of error is \(1.96\sqrt{\frac{\left(\frac{137}{260}\right)\left(\frac{123}{260}\right)}{260}}\approx 0.0606888\). 3. Using \(0.53\), the margin of error is \(1.96\sqrt{\frac{(0.53)(0.47)}{260}}\approx 0.0606675\). 4. The rounded-input result is smaller by approximately \(0.0000213\). 5. The margin of error based on the exact count should be reported because intermediate rounding is unnecessary and can introduce avoidable error.

a) Approximately \(0.0606888\). b) Approximately \(0.0606675\). c) The rounded-input result is about \(0.0000213\) smaller. Report the calculation based on \(\frac{137}{260}\), rounding only the final result.

- Compare the reported estimate with each interval endpoint. - Which quantity describes variation across random samples, and which describes the interval's size?

1. The distance from the estimate to an endpoint is \(0.63-0.60=0.03\), so the margin of error is \(0.030\). 2. The standard error of \(0.015\) measures estimated sample-to-sample variation; the margin of error gives the interval's half-width, here twice the standard error.

Margin of error: \(0.030\). The standard error \(0.015\) measures sampling variation, while the margin gives the distance from the estimate to an interval endpoint.

- What does the critical value tell you about distance measured in standard errors? - Does your result represent an interval width or a distance from its center?

1. Margin of error equals critical value times standard error. 2. \(ME=1.645(0.018)=0.02961\approx0.0296\).

\(ME\approx0.0296\).

- Begin with the usual margin-of-error equation for a proportion. - Rearrange it to place sample size on one side. - Think about which direction to round when the goal is a maximum allowed error.

1. Use \(ME=z^*\sqrt{\frac{p(1-p)}{n}}\). 2. Solving for \(n\) gives \(n\ge\frac{1.645^2(0.50)(0.50)}{0.10^2}=67.650625\). 3. Round up to get at least \(68\) responses.

\(n=68\) responses.

- Compare the desired precision with the current precision as a ratio. - Sample size changes with the square of that ratio. - Round upward so the new survey is not less precise than requested.

1. With the other factors fixed, margin of error is approximately proportional to \(1/\sqrt n\). 2. Therefore, \(n_2=n_1\left(\frac{0.050}{0.035}\right)^2=196\left(\frac{0.050}{0.035}\right)^2\approx 400.00\). 3. Rounding up gives a required sample size of \(400\).

The organization should use at least \(400\) observations.

- Use the prior estimate only in the first planning calculation. - For the second calculation, choose the proportion that gives the largest uncertainty. - Compare the variability terms to explain the difference in required sizes.

1. Using \(0.20\), \(n=\frac{(1.96)^2(0.20)(0.80)}{(0.04)^2}\approx 384.16\), so round up to \(385\). 2. With no prior estimate, use \(0.50\): \(n=\frac{(1.96)^2(0.50)(0.50)}{(0.04)^2}\approx 600.25\), so round up to \(601\). 3. The conservative choice \(0.50\) maximizes \(p(1-p)\), producing the largest required sample size.

a) \(385\). b) \(601\). c) Using \(0.50\) assumes the greatest possible variability.

- Recover the center and half-width of the original interval. - Use the square-root relationship between sample size and margin of error. - Keep the new interval centered at the assumed unchanged sample proportion.

1. The original margin of error is \(\frac{0.54-0.42}{2}=0.06\), and the center is \(0.48\). 2. Quadrupling the sample size halves the margin of error, so the new margin of error is approximately \(0.03\). 3. The new interval is approximately \(0.48\pm0.03\), or \((0.45, 0.51)\). 4. This approximation assumes the same confidence level, a similar sample proportion, and comparable random-sampling conditions.

a) Approximately \(0.03\). b) Approximately \((0.45, 0.51)\). c) The confidence level, sample proportion, and sampling design are assumed to remain comparable.

- Separate the formula’s direct inputs from conditions that justify using it. - Compare the planned sample with each population rather than comparing populations alone. - Think about when removing one person noticeably changes the remaining population.

1. The statement is generally incorrect. For large populations, the margin of error depends mainly on the confidence level, sample proportion, and sample size, not directly on population size. 2. Once the sample is no more than \(10\%\) of the population, both populations can use essentially the same sample size for the same target margin of error. 3. Population size matters when the proposed sample exceeds \(10\%\) of the population, because the usual independence approximation no longer applies.

a) No. b) Once the \(10\%\) condition is met, the same sample size can give about the same margin of error in both populations. c) Population size matters when sampling more than \(10\%\) without replacement.

- Separate the effect of the confidence multiplier from the effect of sample size. - For the first two parts, the estimated standard error stays fixed. - For the last part, compare how precision changes when the multiplier and sample size both change.

1. The \(95\%\) margin of error is \(1.96(0.0192)=0.037632\approx 0.0376\). 2. The \(99\%\) margin of error is \(2.576(0.0192)=0.0494592\approx 0.0495\). 3. With the anticipated proportion unchanged, margin of error is proportional to \(z^*/\sqrt{n}\). Thus \(n_{\text{new}}=625\left(\frac{2.576}{1.96}\right)^2\approx 1079.59\). 4. Rounding up gives a minimum sample size of \(1080\).

a) \(0.0376\). b) \(0.0495\). c) \(1080\).

- When no reasonable estimate is available, plan for the population proportion that creates the greatest uncertainty. - A sample-size requirement must be rounded in the direction that guarantees the target precision. - Compare the available responses with the requirement, then evaluate the precision under the actual limit.

1. With no prior estimate, use \(\hat p=0.50\) for the largest possible variability. 2. The required sample size is \(n=\frac{1.96^2(0.50)(0.50)}{0.025^2}=1536.64\), which rounds up to \(1537\). 3. Because \(900<1537\), the response budget is not sufficient for the desired margin of error. 4. With \(n=900\), the conservative margin of error is \(1.96\sqrt{\frac{0.25}{900}}\approx 0.03267\).

a) \(1537\) responses. b) No. A maximum of \(900\) responses is not enough. c) Approximately \(0.0327\), or \(3.27\) percentage points.

- First use the sample data to quantify the estimated variability. - Compare the reported margin with that variability to recover the multiplier. - Match the multiplier to a familiar standard confidence level.

1. The sample proportion is \(\hat p=\frac{210}{500}=0.42\). 2. The estimated standard error is \(\sqrt{\frac{0.42(0.58)}{500}}\approx 0.02207\). 3. The critical value is \(z^*=\frac{0.0433}{0.02207}\approx 1.962\). 4. This is essentially \(1.96\), the critical value for a \(95\%\) confidence interval.

a) Approximately \(0.02207\). b) \(z^*\approx 1.962\). c) \(95\%\) confidence.

- Separate an interval's total width from the distance between its center and one endpoint. - Recover the underlying variability before changing the confidence level. - The same sample keeps the estimated standard error fixed while the confidence multiplier changes.

1. The margin of error is half the interval width: \(\frac{0.084}{2}=0.042\). 2. Using \(z^*=1.645\) for \(90\%\) confidence, the estimated standard error is \(\frac{0.042}{1.645}\approx 0.02553\). 3. With the same sample, the \(95\%\) margin of error is \(1.96(0.02553)\approx 0.05004\). 4. The new total width is \(2(0.05004)\approx 0.1001\).

a) \(0.042\). b) Approximately \(0.02553\). c) Approximately \(0.1001\).

- Identify the parts of the margin of error that are common to both designs. - Compare the confidence multiplier after accounting for the square root of sample size. - Precision depends on both confidence level and sample size, not sample size alone.

1. The common factor \(\sqrt{p(1-p)}\) cancels when the margins of error are compared. 2. Design A has proportional factor \(\frac{2.576}{\sqrt{600}}\approx 0.10516\). Design B has proportional factor \(\frac{1.96}{\sqrt{400}}=0.09800\). 3. Because \(0.09800<0.10516\), Design B has the smaller margin of error. 4. The ratio is \(\frac{0.10516}{0.09800}\approx 1.073\), so Design A's margin of error is about \(7.3\%\) larger. 5. Design A's higher confidence level requires a substantially larger critical value, more than offsetting its larger sample size.

a) Design B. b) Approximately \(1.073\). c) The lower confidence level uses a smaller critical value, which more than offsets Design B's smaller sample size.

- Within the allowed range, identify which proportion is closest to an even split between outcomes. - Use the value that produces the greatest anticipated variability. - Round the final requirement in the direction that preserves the precision target.

1. Over the interval from \(0.30\) to \(0.40\), the quantity \(p(1-p)\) is largest at \(p=0.40\), the value closer to \(0.50\). 2. The required sample size is \(n=\frac{1.96^2(0.40)(0.60)}{0.035^2}\approx 752.64\). 3. Rounding up gives \(n=753\). 4. Using \(0.30\) would assume smaller variability and could underestimate the sample size needed if the true proportion is closer to \(0.40\).

a) Use \(0.40\). b) \(753\). c) The value \(0.30\) gives a smaller estimated variance and therefore a smaller, less protective sample-size requirement.

- Consider what happens to precision when the sample size is reduced below the calculated requirement. - Check the proposed whole number directly against the target. - Choose the smallest integer that still satisfies the precision condition.

1. A minimum sample size must be rounded up because any smaller whole number falls below the calculated requirement. 2. For \(n=384\), the conservative margin of error is \(1.96\sqrt{\frac{0.25}{384}}\approx0.050010\), which is slightly greater than \(0.050\). 3. The correct minimum is \(n=385\). 4. For \(n=385\), the conservative margin of error is \(1.96\sqrt{\frac{0.25}{385}}\approx0.049945\), which meets the target.

a) Round up because the requirement is a minimum. b) Approximately \(0.050010\), which is too large. c) \(385\); its conservative margin of error is approximately \(0.049945\).

- Separate invitations from completed surveys. - Apply the response rate to find the expected count. - What changes in the margin-of-error formula when the number of completed surveys changes?

1. Solve \(0.60m=1200\): \(m=2000\) invitations. 2. With \(1500\) invitations, expect \(0.60(1500)=900\) completed responses. 3. Since margin of error scales as \(1/\sqrt{n}\), the factor is \(\sqrt{\frac{1200}{900}}=\sqrt{\frac{4}{3}}\approx1.155\).

a) \(2000\) invitations. b) \(900\) completed responses. c) About \(1.155\) times as large, an increase of about \(15.5\%\).

- Distinguish the full distance between endpoints from the distance between the center and one endpoint. - Use the conservative variability choice because no estimate is available. - Examine how the precision target enters the sample-size calculation to compare the two plans.

1. The margin of error is half the total width, so the correct target is \(\frac{0.060}{2}=0.030\). 2. The correct conservative sample size is \(n=\frac{1.96^2(0.25)}{0.030^2}\approx1067.11\), which rounds up to \(1068\). 3. Using \(0.060\) incorrectly gives \(n=\frac{1.96^2(0.25)}{0.060^2}\approx266.78\), which rounds up to \(267\). 4. Doubling the allowed margin divides the required sample size by approximately \(2^2=4\), so the mistaken plan is about one fourth as large.

a) \(0.030\). b) \(1068\). c) \(267\); using twice the correct margin reduces the calculated sample size by a factor of about \(4\).

- Determine which split between two outcomes creates the greatest uncertainty. - Use the same sample size and confidence multiplier in both calculations. - Compare the variability of an even split with that of a rare outcome.

1. The largest value of \(\hat p(1-\hat p)\) occurs at \(\hat p=0.50\). 2. The maximum margin of error is \(1.96\sqrt{\frac{0.25}{1000}}\approx 0.03099\). 3. At \(\hat p=0.01\), the margin of error is \(1.96\sqrt{\frac{0.01(0.99)}{1000}}\approx 0.00617\). 4. A proportion near \(0.50\) has much greater binomial variability than a proportion near \(0\) or \(1\).

a) Approximately \(0.0310\). b) Approximately \(0.00617\). c) Estimated sampling variability depends on \(\hat p(1-\hat p)\), which is largest near \(0.50\) and much smaller near a boundary.

- Compare each new sample size with the original one through a square-root ratio. - Compute the two improvements as differences in margins of error. - Relate the unequal improvements to the shape of the reciprocal square-root relationship.

1. At \(n=600\), the margin of error is \(0.049\sqrt{\frac{400}{600}}\approx 0.04001\). 2. At \(n=800\), the margin of error is \(0.049\sqrt{\frac{400}{800}}\approx 0.03465\). 3. The first \(200\) added responses reduce the margin by about \(0.04900-0.04001=0.00899\). 4. The next \(200\) reduce it by about \(0.04001-0.03465=0.00536\). 5. Equal additions produce diminishing improvements because margin of error depends on \(1/\sqrt{n}\), which flattens as \(n\) grows.

a) Approximately \(0.04001\). b) Approximately \(0.03465\). c) The first reduction is about \(0.00899\), while the second is about \(0.00536\); precision improves with diminishing returns.

- First determine the sample size required by the precision goal. - Then translate the sampling-fraction condition into a minimum population size. - Check the proposed finite population against that requirement.

1. The conservative sample-size calculation is \(n=\frac{1.96^2(0.25)}{0.030^2}\approx 1067.11\), so the minimum sample size is \(1068\). 2. To satisfy the \(10\%\) condition, the population size must be at least \(10(1068)=10{,}680\). 3. If the population size is \(8000\), then \(1068>0.10(8000)=800\), so the sample is too large a fraction of the population for the usual independence-based interval calculation.

a) \(1068\). b) \(10{,}680\). c) The \(10\%\) condition fails because \(1068>800\).

- First find the sample size without the purchasing restriction. - Apply both the precision requirement and the block-size requirement. - Verify the final allowed sample size by recomputing its margin of error.

1. The unconstrained requirement is \(n=\frac{(1.96)^2(0.25)}{(0.025)^2}=1536.64\). 2. The next whole-number sample would be \(1537\), but purchases must be multiples of \(50\). 3. The smallest allowed sample size at least this large is \(1550\). 4. Its planned margin of error is \(1.96\sqrt{\frac{0.25}{1550}}\approx 0.02489\), which meets the target.

The organization should purchase \(1550\) responses. The planned margin of error is approximately \(0.02489\).

- Carry out the planning calculation separately for the two critical values. - Apply the whole-number requirement after each calculation. - Recheck the larger plan using the intended confidence level rather than the rounded shortcut.

1. With \(z^*=1.96\), \(n=\frac{(1.96)^2(0.25)}{(0.040)^2}=600.25\), so the minimum is \(601\). 2. With \(z^*=2.00\), \(n=\frac{(2.00)^2(0.25)}{(0.040)^2}=625\). 3. The rounded critical value requires \(625-601=24\) extra observations. 4. With \(n=625\) and \(z^*=1.96\), the planned margin of error is \(1.96\sqrt{\frac{0.25}{625}}=0.0392\).

a) Using \(1.96\): \(601\); using \(2.00\): \(625\). b) \(24\) extra observations. c) The actual planned margin of error is \(0.0392\).

- Convert the allowed error in people into an allowed error in the population proportion. - Use a planning value that protects against the greatest possible variability. - After finding the sample size, compare it with the finite population size.

1. A count uncertainty of \(1200\) out of \(80{,}000\) corresponds to a proportion margin of error of \(E=\frac{1200}{80{,}000}=0.015\). 2. Use \(z^*=1.645\) for \(90\%\) confidence and the conservative planning value \(p=0.50\). 3. The required sample size is \(n=\frac{(1.645)^2(0.25)}{(0.015)^2}\approx 3006.69\). 4. Round up to \(3007\). This is less than \(10\%\) of the voter population, so the independence condition is also satisfied.

The county should sample at least \(3007\) registered voters.

- Express both margins using the same estimated standard error. - Focus on the part of the difference that still depends on the sample proportion. - Identify where a proportion and its complement have the largest product.

1. The difference is \((1.96-1.645)\sqrt{\frac{\hat p(1-\hat p)}{400}}\). 2. This expression is largest when \(\hat p(1-\hat p)\) is largest, which occurs at \(\hat p=0.50\). 3. The maximum difference is \((1.96-1.645)\sqrt{\frac{0.25}{400}}=0.007875\).

The difference is largest at \(\hat p=0.50\), and the maximum difference is \(0.007875\), or about \(0.788\) percentage points.

- Convert the fixed budget into a completed-response count for each method. - Use the same planning proportion and confidence level so the designs can be compared fairly. - Smaller margin of error means greater statistical precision, assuming the sampling quality is equal.

1. The phone design can obtain \(\frac{6000}{12}=500\) responses. Its conservative margin of error is \(1.96\sqrt{\frac{0.25}{500}}\approx 0.04383\). 2. The online design can obtain \(\frac{6000}{5}=1200\) responses. Its conservative margin of error is \(1.96\sqrt{\frac{0.25}{1200}}\approx 0.02829\). 3. The online design has the smaller margin of error and is therefore more precise under the stated assumption that both designs are equally valid random samples.

Phone survey: margin of error \(\approx 0.04383\). Online survey: margin of error \(\approx 0.02829\). The online design is more precise.

- Convert the stated odds into a proportion before planning the sample. - Use that estimate together with its complement in the variability calculation. - Check which whole-number rounding direction preserves the maximum-error requirement.

1. Odds of \(3\) to \(2\) correspond to a planning proportion of \(p=\frac{3}{3+2}=0.60\). 2. The complementary proportion is \(0.40\). 3. The required sample size is \(n=\frac{(1.96)^2(0.60)(0.40)}{(0.025)^2}\approx 1475.17\). 4. Round up to \(1476\) so the planned margin of error does not exceed \(0.025\).

The minimum required sample size is \(1476\).

- Read each interval’s center and endpoint distances from the display. - Keep the confidence level and observed proportion fixed. - Use the inverse square relationship between margin of error and sample size.

1. From the display, Interval A has margin of error \(0.05\), and Interval B has margin of error \(0.03\). 2. With the same confidence level and sample proportion, sample size is inversely proportional to the square of the margin of error. 3. The ratio is \(\frac{n_B}{n_A}\approx\left(\frac{0.05}{0.03}\right)^2=\frac{25}{9}\approx2.78\). 4. Study B likely used about \(2.78\) times the sample size of Study A.

a) Study B. b) Approximately \(n_B/n_A=2.78\). c) Its smaller margin of error implies a larger sample, with sample size scaling inversely with the square of margin of error.

- Identify the unit that was randomly selected and the unit that supplied responses. - Consider shared influences within each selected group. - Think about whether correlated observations provide as much information as independent ones.

1. Students were sampled in clusters, not independently across the district. Responses within a classroom may be correlated. 2. Students in the same classroom may share the same teacher, assignments, reminders, or school resources, making their reading-program participation more similar. 3. Treating correlated observations as independent can underestimate the true sampling variability, so the reported margin of error may be too small.

a) The students are clustered within only \(8\) classrooms, so their responses may not be independent. b) Shared classroom experiences can make responses similar. c) The usual calculation may underestimate the margin of error.

- Treat the reported half-width as the product of a confidence multiplier and an estimated spread. - Work backward from the spread to the sample size. - Consider how rounding affects inverse calculations.

1. Use \(0.04=1.96\sqrt{\frac{0.50(0.50)}{n}}\). 2. Solving gives \(n=\frac{(1.96)^2(0.25)}{(0.04)^2}\approx 600.25\), so the sample size was about \(600\). 3. If \(0.04\) is a rounded margin of error, several nearby integer sample sizes could produce the displayed value.

a) About \(600\) observations. b) Rounding the margin of error prevents recovering a unique exact sample size.

- Set the old and new precision expressions equal while holding the anticipated proportion fixed. - Solve for the new standard normal cutoff. - Translate that cutoff into the central area between its negative and positive values.

1. To keep the margin of error unchanged, set \(\frac{z^*_{\text{new}}}{\sqrt{600}}=\frac{1.96}{\sqrt{900}}\). 2. Thus \(z^*_{\text{new}}=1.96\sqrt{\frac{600}{900}}\approx 1.6003\). 3. The corresponding central probability is \(2P(Z\le 1.6003)-1\approx 0.8905\), or about \(89.1\%\) confidence. 4. Preserving precision with fewer responses requires accepting a lower confidence level.

a) \(z^*\approx 1.600\). b) Approximately \(89.1\%\) confidence. c) The confidence level must decrease to offset the larger sampling variability from the smaller sample.

- Write the precision requirement as an inequality involving the unknown sample proportion. - Find the boundary values first, then determine which side of those values satisfies the requirement. - Use the symmetry of a proportion and its complement to check the endpoints.

1. Require \(1.96\sqrt{\frac{\hat p(1-\hat p)}{500}}\ge 0.040\). 2. Squaring and rearranging gives \(\hat p(1-\hat p)\ge 500\left(\frac{0.040}{1.96}\right)^2\approx 0.208247\). 3. The boundary equation \(\hat p^2-\hat p+0.208247=0\) has solutions \(\hat p\approx 0.295663\) and \(\hat p\approx 0.704337\). 4. The product \(\hat p(1-\hat p)\) is largest near \(0.5\), so the inequality holds between the two boundary values.

The estimated margin of error is at least \(0.040\) when approximately \(0.296\le \hat p\le 0.704\).

- Read both published endpoints from the display. - Determine the full interval of original values that could round to each displayed endpoint. - Convert the possible interval widths into possible margins of error.

1. Reading the display, the published endpoints are \(0.412\) and \(0.488\). 2. An endpoint displayed as \(0.412\) could come from an unrounded lower endpoint in \([0.4115, 0.4125)\). 3. An endpoint displayed as \(0.488\) could come from an unrounded upper endpoint in \([0.4875, 0.4885)\). 4. The margin of error is half the distance between the endpoints. 5. The smallest possible endpoint distance is greater than \(0.4875-0.4125=0.0750\), and the largest is less than \(0.4885-0.4115=0.0770\). 6. Therefore, the unrounded margin of error satisfies \(0.0375<E<0.0385\).

The unrounded margin of error must satisfy \(0.0375<E<0.0385\).

- Convert the rounded point estimate into an interval, then restrict it to sample proportions obtainable from a whole-number count out of \(500\). - Determine how estimated variability changes across the attainable values. - Evaluate the margin of error at the smallest and largest attainable sample proportions.

1. Rounding to \(0.30\) means \(0.295\le \hat p<0.305\). 2. Because \(\hat p=x/500\) for a whole-number success count, the possible counts are \(148\), \(149\), \(150\), \(151\), and \(152\). Thus, the possible sample proportions range from \(0.296\) through \(0.304\) in increments of \(0.002\). 3. Over these values, which are below \(0.50\), the product \(\hat p(1-\hat p)\) and the margin of error increase as \(\hat p\) increases. 4. At \(\hat p=0.296\), \(E=1.96\sqrt{\frac{(0.296)(0.704)}{500}}\approx 0.04001\). At \(\hat p=0.304\), \(E=1.96\sqrt{\frac{(0.304)(0.696)}{500}}\approx 0.04032\).

The possible margin of error ranges from approximately \(0.04001\) to \(0.04032\), inclusive.

- Treat the precision calculation and the approximation conditions as separate requirements. - Check the expected count of the rare outcome after obtaining the initial sample size. - The final plan must meet whichever requirement demands the larger sample.

1. The precision calculation gives \(n=\frac{(1.96)^2(0.01)(0.99)}{(0.020)^2}\approx 95.08\), so it suggests \(96\). 2. At \(n=96\), the expected success count is \(96(0.01)=0.96\), which is below \(10\). The large-count condition fails. 3. To have at least \(10\) expected successes, require \(n(0.01)\ge 10\), so \(n\ge 1000\). The expected failure count is then \(990\). 4. At \(n=1000\), the planned margin of error is \(1.96\sqrt{\frac{(0.01)(0.99)}{1000}}\approx 0.00617\), so the precision requirement is also met.

a) \(96\). b) It fails because the expected success count is only \(0.96\). c) The minimum sample size satisfying both requirements is \(1000\).

- Consider how large the product involving the sample proportion can be. - A larger value of that product permits a larger sample size for the same reported uncertainty. - Check that the maximizing proportion corresponds to a possible whole-number success count.

1. The margin-of-error equation is \(0.049=1.96\sqrt{\frac{\hat p(1-\hat p)}{n}}\). 2. For any sample proportion, \(\hat p(1-\hat p)\le 0.25\), with equality at \(\hat p=0.50\). 3. Therefore, \(n\le \frac{(1.96)^2(0.25)}{(0.049)^2}=400\). 4. With \(n=400\) and \(\hat p=0.50\), the success count is \(200\), and the margin of error is exactly \(1.96\sqrt{\frac{(0.50)(0.50)}{400}}=0.049\).

The largest possible sample size is \(400\). It is achieved with \(\hat p=0.50\), corresponding to \(200\) successes.