7.3 A Population Proportion

Step 1 — Read off the target. The margin of error is \(EBP = 0.05\), and 90% confidence gives \(\alpha = 0.10\), \(\frac{\alpha}{2} = 0.05\), so \(z_{0.05} = 1.645\).

Step 2 — Use the worst-case proportion. No survey has been run yet, so there is no \(p'\). Take \(p' = q' = 0.5\), which makes \(p'q' = 0.25\), the largest that product can be.

Step 3 — Substitute.

Step 4 — Round up. A sample size must be a whole number, and rounding down would miss the target margin of error.

Answer: survey a minimum of 271 women.

Step 1 — See what changes. Raising the confidence level above 90% shrinks \(\alpha\), which pushes \(z_{\frac{\alpha}{2}}\) further out into the tail. A higher confidence level always means a larger critical value.

Step 2 — Trace it through the sample-size formula. In \(n = \frac{\left(z_{\frac{\alpha}{2}}\right)^{2}(p'q')}{EBP^{2}}\), the critical value sits in the numerator and is squared. If \(z\) goes up and the target error bound stays at 0.05, \(n\) must go up too.

Answer: the sample size needed would increase. As the confidence level increases, \(\alpha\) decreases and \(z_{\frac{\alpha}{2}}\) increases, so to hold the same error bound the sample has to get bigger.

Step 1 — Identify the successes. A "success" here is a household where the woman makes the majority of the purchasing decisions. The survey found 120 of those.

Step 2 — Identify the sample size. They surveyed 200 households.

Step 3 — Compute the sample proportion.

Answer: a) \(X = 120\); b) \(n = 200\); c) \(p' = 0.6\).

Step 1 — Define \(X\). \(X\) counts things, so it is a number of households.

Step 2 — Define \(P'\). \(P'\) is that count divided by the sample size, so it is a proportion of the sample.

Answer: \(X\) is the number of "successes" — households in the sample where the woman makes the majority of the purchasing decisions. \(P'\) is the proportion of households in the sample where the woman makes the majority of the purchasing decisions.

Step 1 — Recognize the parameter. The question is about a percent of households, not an average of measurements, so this is a proportion problem.

Step 2 — Check the condition. \(np' = 200(0.6) = 120 > 5\) and \(nq' = 200(0.4) = 80 > 5\), so the normal approximation to the binomial is allowed.

Step 3 — Write the distribution.

Answer: use the normal distribution for proportions, \(P' \sim N\!\left(0.6, \sqrt{\frac{(0.6)(0.4)}{200}}\right)\).

Step 1 — Collect \(p'\) and \(q'\). \(p' = 0.6\) and \(q' = 0.4\), with \(n = 200\).

Step 2 — Find the critical value. \(CL = 0.95\), so \(\alpha = 0.05\), \(\frac{\alpha}{2} = 0.025\), and \(z_{0.025} = 1.96\).

Step 3 — Compute the error bound.

Step 4 — Build the interval.

Step 5 — Sketch the graph. Draw a normal curve centred at 0.6. Mark 0.5321 and 0.6679 on the horizontal axis, shade the region between them, and label that shaded middle area 0.95, with 0.025 in each tail.

Answer: the 95% confidence interval is \((0.5321, 0.6679)\), with error bound \(EBP = 0.0679\). We estimate with 95% confidence that between 53.21% and 66.79% of all households have the woman making the majority of purchasing decisions.

Step 1 — Ask who can be reached. An email survey only reaches households with an email address the company already has, so households without email — or without an address on that list — have no chance of being selected. That is not a random sample of all households.

Step 2 — Ask who responds. Even among those reached, people choose whether to reply. The households that bother to answer a survey about purchasing decisions are unlikely to be a fair cross-section of the ones that ignore it.

Answer: two difficulties are (1) coverage — households with no email address, or none on the company's list, can never be selected, so the frame is not the whole population; and (2) non-response bias — the households that choose to reply may differ systematically from those that do not.

Step 1 — Identify the group being studied. The question is about executives, so the relevant sample size is the 160 executives, not the 1,050 adults.

Step 2 — Define both variables. \(X\) is a count; \(P'\) is that count over 160.

Answer: \(X\) is the number of "successes" — executives in the sample who prefer trucks. \(P'\) is the proportion of the sampled executives who prefer trucks.

Step 1 — Recognize the parameter. We want a percent of executives, so this is a proportion problem.

Step 2 — Check the condition. With \(n = 160\) and \(p' = 0.26\): \(np' = 41.6 > 5\) and \(nq' = 118.4 > 5\).

Step 3 — Write the distribution.

Answer: use the normal distribution for proportions, \(P' \sim N\!\left(0.26, \sqrt{\frac{(0.26)(0.74)}{160}}\right)\).

Step 1 — Collect \(p'\) and \(q'\). \(p' = 0.26\), \(q' = 0.74\), \(n = 160\).

Step 2 — Find the critical value. \(CL = 0.95\) gives \(z_{0.025} = 1.96\).

Step 3 — Compute the error bound.

Step 4 — Build the interval.

Step 5 — Sketch the graph. A normal curve centred at 0.26, shaded between 0.192 and 0.328, with 0.95 in the middle and 0.025 in each tail.

A note on the calculator. 1-PropZInt needs a whole number for \(x\), and 26% of 160 is 41.6. Entering \(x = 42\) and \(n = 160\) gives \(p' = 0.2625\) and the interval \((0.1943, 0.3307)\) — the same answer to within the rounding forced by that whole-number requirement.

Answer: the 95% confidence interval is \((0.192, 0.328)\), with \(EBP = 0.068\). We estimate with 95% confidence that between 19.2% and 32.8% of all executives prefer trucks.

Step 1 — Look at where the sample size sits. The error bound is \(EBP = z_{\frac{\alpha}{2}}\sqrt{\frac{p'q'}{n}}\), and \(n\) is in the denominator under a square root.

Step 2 — Increase it. A larger \(n\) makes the whole square root smaller, which makes the error bound smaller, without touching the confidence level.

Answer: increase the sample size. (Lowering the confidence level would also shrink the interval, but it buys the narrower interval with less certainty rather than with more information.)

Step 1 — Name what the number is. The \(\pm 2\%\) is the error bound — the distance the survey adds to and subtracts from its sample proportion to build the interval.

Step 2 — Say what it claims. It says the true population proportion is estimated to lie within 2 percentage points of the reported sample percent, at the survey's stated confidence level.

Answer: the sampling error means the true population proportion can be up to 2 percentage points above or below the sample proportion the survey reported.

Step 1 — \(X\) counts. In a proportion problem \(X\) is always the number of successes in the sample.

Answer: \(X\) is the number of voters in the sample of 1,200 who said the economy is the most significant issue in the upcoming election.

Step 1 — \(P'\) is a proportion. It is \(X\) divided by the sample size.

Answer: \(P'\) is the proportion of the 1,200 voters sampled who said the economy is the most important issue in the upcoming election.

Step 1 — Check the condition. With \(n = 1{,}200\) and \(p' = 0.65\): \(np' = 780 > 5\) and \(nq' = 420 > 5\).

Step 2 — Write the distribution.

Answer: use the normal distribution for proportions, \(P' \sim N\!\left(0.65, \sqrt{\frac{(0.65)(0.35)}{1{,}200}}\right)\).

Step 1 — Collect \(p'\) and \(q'\). \(p' = 0.65\), \(q' = 0.35\), \(n = 1{,}200\).

Step 2 — Find the critical value. \(CL = 0.90\), so \(\alpha = 0.10\), \(\frac{\alpha}{2} = 0.05\), and \(z_{0.05} = 1.645\).

Step 3 — Compute the error bound.

Step 4 — Build the interval.

Answer: the 90% confidence interval is \((0.62735, 0.67265)\), with \(EBP = 0.02265\).

Step 1 — See what a higher confidence level changes. At 95%, \(z_{0.025} = 1.96\) replaces \(z_{0.05} = 1.645\). The sample data do not change, so \(p'\), \(q'\), and \(n\) are all the same.

Step 2 — Recompute the error bound.

Step 3 — Compare. The 95% interval is \((0.623, 0.677)\), wider than the 90% interval \((0.627, 0.673)\).

Answer: the confidence interval would get wider. Buying more confidence in the same data always costs precision.

Answer: the number of girls, ages 8 to 12, in the 5 P.M. Monday night beginning ice-skating class.

Step 1 — \(X\) counts the successes. A "success" here is a child in the class who is a girl.

Answer: \(X\) is the number of girls, ages 8 to 12, among the 80 children in the selected beginning ice-skating class.

Step 1 — Count the successes. The class has 64 girls, so \(x = 64\).

Step 2 — Count the whole sample. The class has 64 girls and 16 boys, so \(n = 64 + 16 = 80\).

Step 3 — Divide.

Answer: a) \(x = 64\); b) \(n = 80\); c) \(p' = 0.8\).

Step 1 — Identify the model. Each of the 80 children is either a girl (success) or not, so \(X\) is binomial with \(n = 80\) trials.

Step 2 — Use \(p'\) as the estimate of \(p\). We do not know the true \(p\), so we estimate it with \(p' = 0.8\).

Answer: \(X \sim B(80,\ 0.8)\).

Step 1 — Build \(P'\). Define \(P' = \frac{X}{n} = \frac{X}{80}\), the proportion of girls in the sample.

Step 2 — Say what it estimates. A sample proportion is a point estimate of the corresponding population proportion.

Answer: \(p'\) is estimating \(p\), the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet.

Answer: \(P'\) is the proportion of girls, ages 8 to 12, in the sample of 80 children taken from the beginning ice-skating classes at the Ice Chalet.

Step 1 — State the distribution. With \(p' = 0.8\) and \(n = 80\),

Step 2 — Find the critical value for 92% confidence. \(\alpha = 1 - 0.92 = 0.08\) and \(\frac{\alpha}{2} = 0.04\), so \(z_{0.04} = 1.751\).

Step 3 — Compute the error bound.

Step 4 — Build the interval.

Answer: \(P' \sim N\!\left(0.8, \sqrt{\frac{(0.8)(0.2)}{80}}\right)\), and the 92% confidence interval is \((0.72171, 0.87829)\).

Step 1 — Subtract the confidence level from 1. The confidence level is 92%, so the area left over for the two tails is \(\alpha = 1 - 0.92 = 0.08\).

Answer: 0.08.

Step 1 — Split the leftover area evenly. The normal curve is symmetric, so the two tails share \(\alpha\) equally: \(\frac{\alpha}{2} = \frac{0.08}{2}\).

Answer: 0.04 in each tail.

Step 1 — Recall the error bound from Problem 7.3.24.

Step 2 — Subtract and add.

Answer: a) lower limit \(= 0.72171\); b) upper limit \(= 0.87829\); c) error bound \(= 0.07829\).

Step 1 — Read the limits off Problem 7.3.27 and round to two decimal places.

Answer: \((0.72,\ 0.88)\).

Step 1 — Label the centre. The peak of the curve sits at the sample proportion, \(p' = 0.8\).

Step 2 — Label the boundaries. The lower limit is 0.72 and the upper limit is 0.88 (from Problem 7.3.28).

Step 3 — Label the areas. The shaded middle region carries the confidence level, 0.92. Each unshaded tail carries \(\frac{\alpha}{2} = 0.04\).

Answer: centre \(p' = 0.8\); lower bound 0.72; upper bound 0.88; middle area \(C.L. = 0.92\); each tail area \(\frac{\alpha}{2} = 0.04\).

Step 1 — State the confidence, the parameter, and the range. A correct interpretation names all three and describes the population, not the sample.

Answer: with 92% confidence, we estimate the proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet to be between 72% and 88%.

Step 1 — Change only \(n\). Keep \(p' = 0.8\) and \(z_{0.04} = 1.751\), and raise \(n\) from 80 to 100.

Step 2 — Recompute.

Step 3 — Compare. The old error bound was 0.07829; the new one is 0.070.

Answer: the error bound would become smaller. The sample size is in the denominator under a square root, so a larger \(n\) shrinks the standard deviation of \(P'\) and tightens the interval.

Step 1 — Change only the confidence level. At 98%, \(\alpha = 0.02\) and \(\frac{\alpha}{2} = 0.01\), so \(z_{0.01} = 2.326\), up from 1.751.

Step 2 — Recompute.

Answer: the error bound would increase, from 0.078 to about 0.104. Raising the confidence level pushes the critical value further into the tail, which widens the interval.

Step 1 — Look at the sample-size formula.

Step 2 — Shrink the error bound. EBP sits in the denominator and is squared, so dividing the target error bound in half multiplies \(n\) by four.

Answer: a smaller allowable error bound means a smaller denominator in the sample-size formula, so \(n\) must grow. Demanding more precision at the same confidence level always costs more data.

Part a — Step 1: read off the target. \(EBP = 0.03\), and 95% confidence gives \(z_{0.025} = 1.96\).

Part a — Step 2: use the worst case. With no prior survey, \(p' = q' = 0.5\), so \(p'q' = 0.25\).

Part a — Step 3: substitute and round up.

Part b — See what changes. A confidence level above 95% pushes \(z_{\frac{\alpha}{2}}\) further out, and the critical value is squared in the numerator.

Answer: a) survey a minimum of 1,068 drivers; b) the sample size would need to increase, since the critical value increases as the confidence level increases.

Part a — Read off the counts. \(x = 320\) drivers claimed they always buckle up out of \(n = 400\), so \(p' = \frac{320}{400} = 0.8\).

Part b — Define the variables. \(X\) is the number of drivers in the sample who claim they always buckle up. \(P'\) is the proportion of the sampled drivers who claim they always buckle up.

Part c — Pick the distribution. This is a proportion, and \(np' = 320 > 5\) with \(nq' = 80 > 5\), so the normal approximation to the binomial applies:

Part d — Build the interval. \(CL = 0.95\) gives \(z_{0.025} = 1.96\).

Sketch a normal curve centred at 0.8, shade between 0.7608 and 0.8392, and label the shaded area 0.95 with 0.025 in each tail.

Part e — Telephone difficulties. People without a phone, or with an unlisted or mobile-only number, may never be reachable; many of those reached will refuse or hang up, and refusers may differ from responders; and people tend to over-report socially approved behaviour like buckling up when a live person is asking.

Answer: a) i. \(x = 320\), ii. \(n = 400\), iii. \(p' = 0.8\); c) normal for proportions, \(N(0.8, 0.02)\); d) i. \((0.7608, 0.8392)\), iii. \(EBP = 0.0392\).

Part a — Define the variables. \(X\) is the number of people in the sample who feel the president is doing an acceptable job. \(P'\) is the proportion of the sampled people who feel the president is doing an acceptable job.

Part b — Pick the distribution. This is a proportion, with \(n = 1{,}200\) and \(p' = 0.61\). Since \(np' = 732 > 5\) and \(nq' = 468 > 5\),

Part c — Build the interval. \(CL = 0.90\) gives \(z_{0.05} = 1.645\).

Sketch a normal curve centred at 0.61, shaded between 0.587 and 0.633, with area 0.90 in the middle and 0.05 in each tail.

Answer: c) i. the 90% confidence interval is \((0.59, 0.63)\) to two decimal places; iii. \(EBP \approx 0.02\).

Part a — Define the variables. \(X\) is the number of Black adults in the sample who said they would welcome a White person into their families. \(P'\) is the proportion of the sampled Black adults who said they would welcome a White person into their families.

Part b — Pick the distribution. The relevant sample is the 323 adults who identified as Black, not all 1,709, and \(p' = 0.86\). Since \(np' = 277.8 > 5\) and \(nq' = 45.2 > 5\),

Part c — Build the interval. \(CL = 0.95\) gives \(z_{0.025} = 1.96\).

Sketch a normal curve centred at 0.86, shaded between 0.822 and 0.898, labelled 0.95 in the middle and 0.025 per tail.

Answer: c) i. the 95% confidence interval is \((0.822, 0.898)\); iii. \(EBP = 0.038\). We estimate with 95% confidence that between 82.2% and 89.8% of all Black adults would welcome a White person into their families.

Part a — Three intervals, same sample size. All three use the 254 adults who identified as Asian, with \(z_{0.025} = 1.96\).

i. White person: \(p' = 0.77\), \(EBP = (1.96)\sqrt{\frac{(0.77)(0.23)}{254}} = 0.052\), giving \((0.72, 0.82)\).

ii. Hispanic/Latino person: \(p' = 0.71\), \(EBP = (1.96)\sqrt{\frac{(0.71)(0.29)}{254}} = 0.056\), giving \((0.65, 0.77)\).

iii. Black person: \(p' = 0.66\), \(EBP = (1.96)\sqrt{\frac{(0.66)(0.34)}{254}} = 0.058\), giving \((0.60, 0.72)\).

Part b — Look for overlap. Compare the endpoints: \((0.72, 0.82)\) and \((0.65, 0.77)\) share the whole stretch from 0.72 to 0.77. \((0.65, 0.77)\) and \((0.60, 0.72)\) share the stretch from 0.65 to 0.72. Intervals i and iii only meet at 0.72 — carried to four decimals they are \((0.7182, 0.8218)\) and \((0.6017, 0.7183)\), so they touch across a width of about one ten-thousandth, which is no meaningful overlap at all.

Part c — What overlap means. When two intervals overlap, the same value of the true proportion is plausible for both groups, so the data do not establish a real difference.

Part d — What no overlap means. When two intervals do not overlap, no single value is plausible for both, so the difference between the two true proportions is significant.

Answer: a) i. \((0.72, 0.82)\), ii. \((0.65, 0.77)\), iii. \((0.60, 0.72)\); b) yes — intervals i and ii overlap, and ii and iii overlap; c) there does not appear to be a significant difference between the proportion of Asian adults who would welcome a White person and the proportion who would welcome a Hispanic/Latino person; d) there is a significant difference between the proportion who would welcome a White person and the proportion who would welcome a Black person.

Part a — Define the variables. \(X\) is the number of people over 50 in the study who ran and died during the eight-year period. \(P'\) is the proportion of the sampled 50-Plus Fitness Association members who ran and died during that period.

Part b — Pick the distribution. With \(n = 451\) and \(p' = 0.015\), check the condition: \(np' = 6.765 > 5\) and \(nq' = 444.2 > 5\). It clears the bar, though not by much — with a proportion this small, a slightly smaller study would not have qualified.

Part c — Build the interval. \(CL = 0.97\), so \(\alpha = 0.03\), \(\frac{\alpha}{2} = 0.015\), and \(z_{0.015} = 2.17\).

Sketch a normal curve centred at 0.015, shaded between 0.0026 and 0.0274, with 0.97 in the middle and 0.015 in each tail.

Part d — Interpret the confidence level. It describes the method, not this one interval: if the same study were repeated many times and an interval built the same way each time, 97% of those intervals would contain the true proportion.

Answer: c) i. the 97% confidence interval is \((0.0026, 0.0274)\); iii. \(EBP = 0.0124\). We estimate with 97% confidence that between 0.26% and 2.74% of all people over 50 who run died in the same eight-year period.

Part a — Define the variables. \(X\) is the number of adult Americans in the sample who feel that crime is the main problem facing the country. \(P'\) is the proportion of the sampled adult Americans who feel that crime is the main problem.

Part b — Pick the distribution. We are estimating a proportion, with \(p' = 0.20\) and \(n = 1{,}000\). Since \(np' = 200 > 5\) and \(nq' = 800 > 5\),

Part c — Build the interval. \(CL = 0.95\) gives \(z_{0.025} = 1.96\).

Rounded to two decimals, \((0.18, 0.22)\). Sketch a normal curve centred at 0.2 with that region shaded.

Part d — Lower the sampling error. Increase the sample size; \(n\) is under a square root in the denominator of the error bound.

Part e — What \(\pm 3\%\) represents. It is the maximum error bound the pollsters are reporting. They are saying their estimate of the percentage of adult Americans who feel crime is the main problem is off by no more than 3 percentage points, so the true percentage is estimated to be between 17% and 23%.

Answer: c) i. the 95% confidence interval is \((0.18, 0.22)\); iii. \(EBP \approx 0.02\); d) increase the sample size.

Part a — Define the variables. \(X\) is the number of adult Americans in the sample who are worried a lot about the quality of education in our schools. \(P'\) is the proportion of the sampled adult Americans who are worried a lot about it.

Part b — Pick the distribution. A proportion again, with \(p' = 0.63\) and \(n = 1{,}000\). Since \(np' = 630 > 5\) and \(nq' = 370 > 5\),

Part c — Build the interval. \(CL = 0.95\) gives \(z_{0.025} = 1.96\).

Sketch a normal curve centred at 0.63, shaded between 0.60 and 0.66.

Part d — What \(\pm 3\%\) represents. It is the maximum error bound the pollsters report: their estimate is claimed to be within 3 percentage points of the true percentage, so between 60% and 66% of adult Americans are worried a lot about school quality.

Answer: c) i. the 95% confidence interval is \((0.60, 0.66)\); iii. \(EBP \approx 0.03\).

Step 1 — Recall what a point estimate is. The point estimate for a population proportion is the sample proportion \(p' = \frac{x}{n}\).

Step 2 — Compute it.

Step 3 — Rule out the others. 0.90 is the confidence level, 1.27 is not a proportion at all, and 400 is the count \(x\), not the proportion.

Answer: c. 0.79.

Step 1 — Collect the pieces. \(p' = \frac{400}{506} = 0.7905\), \(q' = 0.2095\), \(n = 506\), and \(CL = 0.90\) gives \(z_{0.05} = 1.645\).

Step 2 — Compute the error bound.

Step 3 — Build the interval.

Answer: a. (0.761, 0.820).

Step 1 — Take the error bound from Problem 7.3.43.

Step 2 — Match it to the choices. That rounds to 0.030.

Answer: d. 0.030.

Step 1 — Find \(p'\) and \(q'\). 173 of 511 homes met the minimum recommendations.

Step 2 — Find the critical value. \(CL = 0.90\) gives \(z_{0.05} = 1.645\).

Step 3 — Compute the error bound.

Step 4 — Build the interval.

Answer: c. (0.3041, 0.3730).

Step 1 — Read the question carefully. It asks about the homes that did NOT meet the recommendations, which is 338 of them, not 173.

Step 2 — Divide.

Step 3 — Rule out the others. 0.3386 is the proportion that DID meet the recommendations, and 173 and 338 are counts, not proportions.

Answer: a. 0.6614.

Part a — The estimated proportion. Gallup reported that 76% of those surveyed believe they will keep working past retirement age, so \(p' = 0.76\).

Part b — The sample size. \(n = 1{,}005\).

Part c — Identify CL and \(\alpha\). The confidence level is \(CL = 0.95\), so \(\alpha = 1 - 0.95 = 0.05\) and \(\frac{\alpha}{2} = 0.025\), giving \(z_{0.025} = 1.96\).

Part d — Calculate the error bound.

Part e — Compare to Gallup's \(\pm 3\%\). Our calculation gives about 2.6%, slightly less than the 3% Gallup reported. Gallup likely rounded up to a conservative whole number, and polling organizations often quote a margin of error computed at the worst case \(p' = 0.5\) — which here would give \((1.96)\sqrt{\frac{0.25}{1{,}005}} = 0.0309\), about 3%. Reported margins also sometimes include an allowance for design effects that our formula ignores.

Part f — Create the interval.

Part g — Explaining it on the news. She should avoid the phrase "95% chance". A workable version: "About three-quarters of workers told Gallup they expect to keep working past retirement age. Because this is a survey of about a thousand people rather than everyone, the real figure for all U.S. workers is probably between 73% and 79%. If Gallup ran this same poll over and over, 95% of the ranges they reported this way would contain the true figure."

Answer: a) 0.76; b) 1,005; c) \(CL = 0.95\), \(\alpha = 0.05\); d) \(EBP = 0.0264\); f) \((0.734, 0.786)\).

Part a — Recover the point estimate and error bound. The interval runs from 0.49 to 0.55. The point estimate is the midpoint, and the error bound is the distance from the midpoint to either endpoint.

Part b — Can we conclude more than half? The interval \((0.49, 0.55)\) contains values at and below 0.50, so a true proportion of one half or less is still plausible.

Part c — Build a 75% interval. \(CL = 0.75\), so \(\alpha = 0.25\) and \(\frac{\alpha}{2} = 0.125\), giving \(z_{0.125} = 1.150\) (the area to the left of it is \(1 - 0.125 = 0.875\)).

Part d — Can we conclude at least half at 75%? This interval's lower endpoint, 0.502, sits above 0.50, so every plausible value in it is more than half.

Answer: a) \(p' = 0.52\), \(EBP = 0.03\); b) no — the 95% interval includes values at or below 0.50, so it is possible that less than half of the population believes this; c) \((0.502, 0.538)\); d) yes — this interval does not reach below 0.50, so at 75% confidence we can conclude at least half of all American adults believe this. Note that the weaker claim comes with weaker confidence: narrowing the interval by dropping to 75% is what made the conclusion possible.

Part a — Build the 99% interval. With \(x = 80\) and \(n = 571\),

Check the condition: \(np' = 80 > 5\) and \(nq' = 491 > 5\). For \(CL = 0.99\), \(\alpha = 0.01\), \(\frac{\alpha}{2} = 0.005\), and \(z_{0.005} = 2.576\).

Part b — What the margin of error does not cover. The error bound only accounts for the randomness of which people ended up in the sample. It says nothing about people who never pick up an automated call, or who have no landline; about respondents who under-report an admission of illegal activity to an automated voice; about how the question was worded; or about who hangs up partway through.

Part c — Dropping to 90%. Lowering the confidence level shrinks the critical value from 2.576 to 1.645, so the interval gets narrower and stays centred on 0.1401.

Answer: a) the 99% confidence interval is \((0.103, 0.177)\); b) non-response, coverage of the automated calling frame, and under-reporting of an illegal behaviour are all uncovered by the margin of error; c) the interval would become narrower.

Step 1 — Read off the target. The margin of error is \(EBP = 0.05\), and 95% confidence gives \(\alpha = 0.05\), \(\frac{\alpha}{2} = 0.025\), and \(z_{0.025} = 1.96\).

Step 2 — Use the worst-case proportion. You have no prior estimate of the turnout proportion on your campus, so take \(p' = q' = 0.5\) and \(p'q' = 0.25\).

Step 3 — Substitute.

Step 4 — Round up.

Answer: you must interview 385 students.

Step 1 — Apply the plus four adjustment. With \(x = 9\) and \(n = 48\), use \(x = 9 + 2 = 11\) and \(n = 48 + 4 = 52\). The sample is small and the confidence level is above 90%, which is exactly when this method is recommended.

Step 2 — Find \(p'\) and \(q'\).

Step 3 — Find the critical value. \(CL = 0.97\), so \(\alpha = 0.03\), \(\frac{\alpha}{2} = 0.015\), and \(z_{0.015} = 2.17\).

Step 4 — Compute the error bound.

Step 5 — Build the interval.

Answer: the 97% plus four confidence interval is \((0.089, 0.334)\). We are 97% confident that between 8.9% and 33.4% of all American adults believe a terrorist attack in their community is likely or very likely. The interval is very wide because only 48 people were polled — with a sample this small, the poll narrows the answer down to somewhere between "fewer than one in ten" and "a third", which is not a precise finding.