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- Distinguish the full population from the people actually sampled. - The population proportion is a parameter; the sample proportion is calculated from the sample. - Use the number of sampled supporters over the total sample size.

1. The symbol \(p\) represents the unknown population proportion of all registered voters who support the bond measure. 2. The sample proportion is \(\hat p=\frac{118}{200}=0.59\).

a) \(p\) is the proportion of all registered voters who support the bond measure. b) \(\hat p=0.59\).

- Changing the sample size does not change the population proportion being estimated. - Think about how sample size affects sampling variability. - Compare the two sample sizes before worrying about exact arithmetic.

1. For either sample size, the sampling distribution of \(\hat p\) is centered at the population proportion, so both centers are \(0.40\). 2. The standard deviation of \(\hat p\) is proportional to \(\frac{1}{\sqrt n}\). Increasing the sample size from \(100\) to \(400\) makes the spread smaller.

Both sampling distributions are centered at \(0.40\). Sample B has the smaller spread because it uses the larger sample size.

- Use the population proportion and its complement to describe sampling spread. - Keep the sample size fixed while comparing the two populations. - Compare the products that determine the variance.

1. For Population A, \(\sigma_{\hat p_A}=\sqrt{\frac{0.10(0.90)}{100}}=0.03\). 2. For Population B, \(\sigma_{\hat p_B}=\sqrt{\frac{0.50(0.50)}{100}}=0.05\). 3. The second sampling distribution is more variable because \(p(1-p)\) is larger at \(p=0.50\) than at \(p=0.10\).

a) \(\sigma_{\hat p_A}=0.03\) and \(\sigma_{\hat p_B}=0.05\). b) \(\hat p_B\) is more variable because \(p(1-p)\) is larger for Population B.

- Relate the support spacing to a one-person change in the sample count. - Use the center of the sampling distribution to identify the population quantity. - Treat the sample size and population proportion as separate features.

1. Consecutive sample proportions differ by \(1/n\). 2. Since \(1/n=0.02\), the sample size is \(n=50\). 3. For this sampling design, the mean of \(\hat p\) is the population proportion, so \(p=0.64\).

a) \(n=50\). b) \(p=0.64\).

- Keep the sample size fixed and compare the part of the variability expression that changes. - Proportions equally far from \(0.50\) have a useful symmetry. - Check which population gives the most balanced split between the two outcomes.

1. For \(p=0.10\), \(\sigma_{\hat p}=\sqrt{\frac{0.10(0.90)}{300}}\approx 0.0173\). 2. For \(p=0.50\), \(\sigma_{\hat p}=\sqrt{\frac{0.50(0.50)}{300}}\approx 0.0289\). 3. For \(p=0.90\), \(\sigma_{\hat p}=\sqrt{\frac{0.90(0.10)}{300}}\approx 0.0173\). 4. The \(p=0.50\) distribution is most variable because \(p(1-p)\) is largest at \(p=0.50\). 5. The values \(0.10(0.90)\) and \(0.90(0.10)\) are equal, so the two standard deviations match.

a) Approximately \(0.0173\), \(0.0289\), and \(0.0173\), respectively. b) The distribution with \(p=0.50\) is most variable. c) The products \(p(1-p)\) are equal for complementary proportions \(0.10\) and \(0.90\).

- Recall what unbiasedness says about the long-run average of a sample statistic. - Compare the reported center directly with the population parameter. - Think about setup or coding mistakes that could shift every simulated result.

1. The sample proportion is an unbiased estimator of the population proportion, so its sampling distribution should be centered at \(p\). 2. The expected center is \(0.52\), not \(0.47\). 3. A difference of \(0.05\) after \(5000\) simulations strongly suggests that the simulation used the wrong success probability, coded successes incorrectly, or did not generate the intended random samples.

a) No. b) The expected center is \(0.52\). c) The simulation likely used an incorrect probability or coded the response categories incorrectly.

- Use the relationship between the sampling-distribution center and the population parameter. - Check whether the proposed spread reflects averaging across many observations. - A sample statistic should become less variable as sample size grows.

1. The mean of \(\hat p\) equals the population proportion, so \(p=0.73\). 2. The standard deviation is \(\sqrt{\frac{0.73(0.27)}{300}}\approx0.0256\). 3. Finn omitted division by the sample size inside the square root. Sampling variability decreases as sample size increases.

a) \(p=0.73\). b) Approximately \(0.0256\). c) Finn omitted the sample-size factor that reduces the variability of a sample proportion.

- Separate the rule for the center from the rule for the spread. - Check each proposed number against the population proportion and sample size. - Use expected outcome counts to assess the shape.

1. The mean of the sampling distribution is \(\mu_{\hat p}=p=0.36\). 2. Its standard deviation is \(\sigma_{\hat p}=\sqrt{\frac{0.36(0.64)}{225}}=0.032\). Therefore, Student A is correct. 3. Student B used the complementary population proportion as the mean. Student C doubled the correct standard deviation. 4. The expected counts are \(225(0.36)=81\) and \(225(0.64)=144\), both at least \(10\), so an approximately normal shape is reasonable.

a) Student A. b) Student B used \(1-p\) as the mean; Student C used twice the correct standard deviation. c) Yes. The expected counts are \(81\) and \(144\).

- Identify who controls whether an observation enters the sample. - Separate random sampling variability from systematic selection effects. - Consider what design would give each population member a known chance to be selected.

1. Visitors choose whether to respond, so respondents can differ systematically from nonrespondents. The resulting sample proportion need not be centered at the population proportion. 2. A larger voluntary-response sample can reduce random variability around the biased center, but it does not remove the selection bias. 3. A random sample of site visitors, selected from an appropriate sampling frame and invited to respond, would support the usual sampling-distribution model if the other conditions are met.

a) Voluntary response can create selection bias, so \(E(\hat p)=p\) is not justified. b) A larger sample does not remove systematic bias from self-selection. c) Use a random sample of site visitors from a suitable sampling frame.

- Treat the two outcomes as complementary categories. - Compare the variability expressions after swapping the two population proportions. - Use the fact that every sampled member belongs to exactly one of the two categories.

1. For \(\hat p\), the mean is \(0.27\), and the standard deviation is \(\sqrt{\frac{0.27(0.73)}{300}}\approx 0.02563\). 2. Because the complementary population proportion is \(q=0.73\), the mean of \(\hat q\) is \(0.73\). 3. The standard deviation of \(\hat q\) is \(\sqrt{\frac{0.73(0.27)}{300}}\approx 0.02563\), the same as for \(\hat p\). 4. In every sample, \(\hat q=1-\hat p\), so the two sample proportions always sum to \(1\) and move in opposite directions.

a) Mean \(0.27\); standard deviation approximately \(0.02563\). b) Mean \(0.73\); standard deviation approximately \(0.02563\). c) \(\hat p+\hat q=1\) for every sample.

- Use the population proportion and sample size to quantify sampling spread. - Build the requested interval directly around the center. - Translate the endpoints into standard-deviation units before finding probability.

1. The standard deviation is \(\sigma_{\hat p}=\sqrt{\frac{0.47(0.53)}{500}}\approx 0.02232\). 2. The one-standard-deviation interval is \(0.47\pm 0.02232\approx(0.4477, 0.4923)\). 3. Under a normal model, the probability of being within one standard deviation of the mean is \(P(-1\le Z\le 1)\approx 0.6827\). 4. Standardizing any normal distribution converts an interval one standard deviation from its mean into the same interval \([-1, 1]\) on the standard normal scale.

a) Approximately \(0.02232\). b) Approximately \((0.4477, 0.4923)\). c) Approximately \(0.6827\). d) Standardization maps every such interval to \([-1, 1]\) on the standard normal distribution.

- Compare how the first selected item affects the probabilities on later selections in each design. - Check the sample size against the finite population size for the design without replacement. - Once independence is established, identify the center and spread of the sample proportion.

1. Sampling with replacement makes the draws independent, so the usual formula applies to design a). 2. For design b), \(30\) is more than \(10\%\) of \(120\), so the usual independence condition for sampling without replacement is not met. 3. Under design a), the mean is \(\mu_{\hat p}=p=0.40\). 4. The standard deviation is \(\sigma_{\hat p}=\sqrt{\frac{(0.40)(0.60)}{30}}\approx 0.08944\).

The usual formula is justified for sampling with replacement. For that design, \(\mu_{\hat p}=0.40\) and \(\sigma_{\hat p}\approx 0.08944\).

- Distinguish a measure of squared spread from a measure on the original scale. - Check whether the expression being used gives variance or standard deviation. - Use the units of the random variable as a reasonableness check.

1. The value \(\frac{(0.40)(0.60)}{150}=0.0016\) is the variance of \(\hat p\), not its standard deviation. 2. The standard deviation is the square root of the variance: \(\sqrt{0.0016}=0.040\). 3. Variance is measured in squared proportion units, while standard deviation is measured on the same scale as \(\hat p\).

The variance is \(0.0016\), and the standard deviation is \(0.040\). Khaled forgot to take the square root.

- Work directly with the sampling distribution of the sample proportion. - Check the large-count condition before using a normal approximation. - Use \(\sqrt{p(1-p)/n}\) for the standard deviation of \(\hat p\). - Compare the observed sample proportion with the predicted interval.

1. Under the claim, \(p=0.85,\qquad n=600.\) The large-count values are \(np=510,\qquad n(1-p)=90,\) so the sampling distribution of \(\hat p\) is approximately normal. 2. Its mean is \(\mu_{\hat p}=p=0.85,\) and its standard deviation is \(\sigma_{\hat p} =\sqrt{\frac{p(1-p)}{n}} =\sqrt{\frac{0.85(0.15)}{600}} \approx0.01458.\) 3. A central \(99\%\) interval uses \(z^*\approx2.576\): \(0.85\pm2.576(0.01458) \approx[0.8125,0.8875].\) 4. The observed sample proportion is \(\hat p=\frac{486}{600}=0.81.\) Since \(0.81<0.8125\), the observed proportion lies just below the central \(99\%\) interval under the claim.

a) Approximately normal because \(np=510\) and \(n(1-p)=90\). Mean \(0.85\); standard deviation approximately \(0.01458\). b) Approximately \([0.8125,0.8875]\). c) \(\hat p=0.81\), which lies below the interval; the observed response rate is unusually low relative to the \(85\%\) claim at this prediction level.

- Use the center of the sampling distribution to identify the population proportion. - Square the stated sampling standard deviation to obtain the variance. - Solve the sampling-variance relationship for the sample size.

1. The mean of the sampling distribution is the population proportion, so \(p=0.35\). 2. The variance is \(0.002275\), and \(\operatorname{Var}(\hat p)=\frac{p(1-p)}{n}\). 3. Thus, \(0.002275=\frac{0.35(0.65)}{n}\). 4. Since \(0.35(0.65)=0.2275\), \(n=\frac{0.2275}{0.002275}=100\).

\(p=0.35\) and \(n=100\).

- Focus on the part of the sampling variance that changes with the population proportion. - Consider where a proportion and its complement are most balanced. - Use the fixed sample size after identifying the maximizing proportion.

1. The sampling variance is \(\frac{p(1-p)}{64}\). 2. The product \(p(1-p)\) is largest at \(p=0.50\), where it equals \(0.25\). 3. The maximum standard deviation is \(\sqrt{\frac{0.25}{64}}=0.0625\).

a) \(p=0.50\). b) \(0.0625\).

- List the possible counts of the characteristic in the sample. - Count samples using the available members of each type. - Convert counts to proportions before finding the center.

1. The sample can contain \(0,1,\) or \(2\) members with the characteristic, so \(\hat p\) can be \(0,0.5,1\). 2. The probabilities are \(\frac{\binom{3}{0}\binom{2}{2}}{\binom{5}{2}}=0.10\), \(\frac{\binom{3}{1}\binom{2}{1}}{\binom{5}{2}}=0.60\), and \(\frac{\binom{3}{2}\binom{2}{0}}{\binom{5}{2}}=0.30\). 3. The sampling-distribution mean is \(0(0.10)+0.5(0.60)+1(0.30)=0.60\), equal to the population proportion \(3/5\).

a) <table> <tr><th>\(\hat p\)</th><th>\(0\)</th><th>\(0.5\)</th><th>\(1\)</th></tr> <tr><th>Probability</th><td>\(0.10\)</td><td>\(0.60\)</td><td>\(0.30\)</td></tr> </table> b) The mean is \(0.60\), equal to the population proportion.

- Start with the usual sample-proportion spread. - Account for the substantial fraction of the population being sampled. - Interpret why sampling without replacement reduces remaining uncertainty.

1. Ignoring the finite population gives \(\sqrt{\frac{0.40(0.60)}{100}}=\sqrt{0.0024}\approx0.0490\). 2. The finite-population correction factor is \(\sqrt{\frac{500-100}{500-1}}=\sqrt{\frac{400}{499}}\). 3. The corrected standard deviation is \(\sqrt{0.0024}\sqrt{\frac{400}{499}}\approx0.0439\). 4. The corrected value is smaller because sampling without replacement removes a substantial fraction of the population from further selection.

a) \(\sqrt{0.0024}\sqrt{\frac{400}{499}}\approx0.0439\). b) Ignoring the correction gives \(\sqrt{0.0024}\approx0.0490\), which is larger.

- Connect the sample proportion to the number of labeled items in a sample. - Determine how much the proportion changes when the count changes by one. - Consider whether both possible outcome counts are expected to occur often enough for symmetry.

1. The center is \(\mu_{\hat p}=p=0.02\). 2. The standard deviation is \(\sigma_{\hat p}=\sqrt{\frac{0.02(0.98)}{100}}=0.014\). 3. Because the sample count can be \(0,1,2,3,\ldots\), the first four proportions are \(0,0.01,0.02,0.03\). 4. The expected number with the label is only \(100(0.02)=2\), so the distribution is strongly right-skewed rather than approximately normal.

a) Mean \(0.02\); standard deviation \(0.014\). b) \(0,0.01,0.02,0.03\). c) No. The expected labeled count is only \(2\), so the sampling distribution is strongly right-skewed.

- Express the second sample proportion using the first one. - A complement changes the center but not the amount of sample-to-sample variation. - Think about what happens to one statistic whenever the other increases.

1. The complementary population proportion is \(q=1-0.35=0.65\). 2. The sampling-distribution mean is \(\mu_{\hat q}=0.65\). 3. Its standard deviation is \(\sqrt{\frac{0.65(0.35)}{200}}\approx0.0337\). 4. In every sample, \(\hat q=1-\hat p\), so the two statistics move in exactly opposite directions and have the same standard deviation.

a) Mean \(0.65\); standard deviation \(\sqrt{\frac{0.65(0.35)}{200}}\approx0.0337\). b) \(\hat q=1-\hat p\) in every sample, so their sampling distributions are mirror images and the statistics are perfectly negatively related.

- The target population proportion determines both centers. - Compare the sample sizes inside the square-root expression for spread. - One additional success changes a sample proportion by the reciprocal of the sample size.

1. Both sampling distributions have mean \(0.30\). 2. For \(n=50\), the standard deviation is \(\sqrt{\frac{0.30(0.70)}{50}}\approx0.0648\), and possible proportions are spaced by \(\frac{1}{50}=0.02\). 3. For \(n=200\), the standard deviation is \(\sqrt{\frac{0.30(0.70)}{200}}\approx0.0324\), and possible proportions are spaced by \(\frac{1}{200}=0.005\). 4. Quadrupling the sample size halves the spread and makes the support four times finer.

For \(n=50\): mean \(0.30\), standard deviation \(\sqrt{\frac{0.30(0.70)}{50}}\approx0.0648\), spacing \(0.02\). For \(n=200\): mean \(0.30\), standard deviation \(\sqrt{\frac{0.30(0.70)}{200}}\approx0.0324\), spacing \(0.005\).

- Separate the population proportion from the statistic computed from a sample. - Check both the independence requirement and whether the expected counts are large enough. - Convert each endpoint to its relative position in the sampling distribution before finding the middle probability.

1. The mean is \(\mu_{\hat p}=p=0.38\). 2. The \(10\%\) condition holds because \(200 \le 0.10(5000)=500\). Also, \(np=76\) and \(n(1-p)=124\), both at least \(10\). 3. The standard deviation is \(\sigma_{\hat p}=\sqrt{\frac{0.38(0.62)}{200}}\approx 0.0343\). 4. The standardized endpoints are \(z\approx -1.75\) and \(z\approx 1.75\). 5. Therefore, \(P(0.32 \le \hat p \le 0.44)\approx 0.9196\).

a) \(\mu_{\hat p}=0.38\) and \(\sigma_{\hat p}\approx 0.0343\). b) The randomization, \(10\%\), and large-count conditions are satisfied. c) \(P(0.32 \le \hat p \le 0.44)\approx 0.9196\).

- Estimate the center of the displayed distribution before checking the sample size. - Compare the expected variability for the remaining sample-size choices. - A larger sample should produce a tighter distribution of sample proportions.

1. The displayed distribution is centered near \(0.40\), so the population proportion must be \(0.40\). This eliminates C. 2. For A, the standard deviation is \(\sqrt{\frac{0.40(0.60)}{100}}\approx 0.0490\). 3. For B, the standard deviation is \(\sqrt{\frac{0.40(0.60)}{400}}\approx 0.0245\). 4. The displayed distribution has a spread consistent with about \(0.0245\), so it matches B.

B. The center identifies \(p=0.40\), and the displayed spread is consistent with the standard deviation \(0.0245\) for \(n=400\).

- Convert the count condition into a condition on the sample proportion. - Verify the shape before using a normal model. - Locate the cutoff below the center and find the corresponding lower-tail area.

1. The sample is random, and \(250\le 0.10(10{,}000)=1000\), so the observations are approximately independent. 2. The mean is \(\mu_{\hat p}=0.72\), and the standard deviation is \(\sigma_{\hat p}=\sqrt{\frac{0.72(0.28)}{250}}\approx 0.0284\). 3. The expected counts are \(180\) renewals and \(70\) nonrenewals, so a normal approximation is appropriate. 4. Fewer than \(165\) renewals corresponds to \(\hat p<\frac{165}{250}=0.66\). 5. The standardized cutoff is \(z\approx -2.11\), so \(P(\hat p<0.66)\approx 0.0173\).

a) Approximately normal with mean \(0.72\) and standard deviation \(0.0284\); the randomization, \(10\%\), and large-count conditions are satisfied. b) The probability is approximately \(0.0173\).

- Translate the count range into a range of sample proportions. - Verify that the sampling distribution can be treated as approximately normal. - Find the area between the two standardized boundaries.

1. Convert the count bounds to proportions: \(\frac{160}{320}=0.500\) and \(\frac{190}{320}=0.59375\). 2. The sampling distribution has mean \(0.55\) and standard deviation \(\sqrt{\frac{0.55(0.45)}{320}}\approx 0.0278\). 3. The sample is random, \(320\le 0.10(5000)=500\), and the expected counts are \(176\) and \(144\), so a normal approximation is appropriate. 4. Standardizing the two bounds and finding the middle area gives \(P(0.500\le \hat p\le 0.59375)\approx 0.9061\).

\(P(160\le X\le 190)\approx 0.9061\).

- Describe the center and spread before locating the requested percentile. - The requested cutoff is above the mean because it captures most of the distribution to its left. - Translate a standard-normal percentile back to the sample-proportion scale.

1. The samples are random, \(225\le 0.10(5000)=500\), and the expected counts are \(225(0.48)=108\) and \(225(0.52)=117\), so a normal approximation is appropriate. 2. The sampling distribution has mean \(0.48\) and standard deviation \(\sqrt{\frac{0.48(0.52)}{225}}\approx 0.0333\). 3. The \(90\)th percentile of the standard normal distribution is \(z\approx 1.282\). 4. Therefore, \(c=0.48+1.282(0.0333)\approx 0.5227\).

\(c\approx 0.5227\). About \(90\%\) of sample proportions are at or below this value.

- Write a separate minimum-count requirement for each outcome. - The smaller population proportion will usually be the binding condition. - Apply the population-size condition only after finding the sample size.

1. The expected success count requires \(0.04n\ge10\), so \(n\ge250\). 2. The expected failure count requires \(0.96n\ge10\), which is already satisfied when \(n=250\). 3. Thus, the smallest sample size is \(n=250\). 4. The \(10\%\) condition requires \(N\ge10n=2500\). 5. The rare outcome determines the minimum because its expected count reaches \(10\) last.

a) \(n=250\). b) \(N\ge 2500\). c) The expected success count \(np\) determines the minimum because \(p=0.04\) is the rarer outcome.

- Measure the observed statistic in units of the sampling-distribution standard deviation. - A percentile is the area at or below the observed value. - The final probability is the complementary upper-tail area.

1. The sample is random, \(100\le0.10(2000)=200\), and the expected counts are \(55\) and \(45\), so a normal approximation is appropriate. 2. The sampling distribution has standard deviation \(\sqrt{\frac{0.55(0.45)}{100}}\approx 0.0497\). 3. The standardized value is \(z=\frac{0.61-0.55}{0.0497}\approx 1.21\). 4. The cumulative probability is approximately \(0.8861\), so \(0.61\) is about the \(88.6\)th percentile. 5. The upper-tail probability is \(1-0.8861\approx 0.1139\).

a) \(z\approx 1.21\). b) About the \(88.6\)th percentile. c) \(P(\hat p\ge0.61)\approx 0.1139\).

- Model the long-run behavior of the sample proportion, not individual renewals. - The cutoff is above the center, so use the upper tail. - Translate the final area into a proportion of repeated samples.

1. The samples are random, and \(600\le0.10(10{,}000)=1000\), so the observations are approximately independent. 2. The mean is \(0.40\), and the standard deviation is \(\sqrt{\frac{0.40(0.60)}{600}}=0.0200\). 3. The expected counts are \(240\) and \(360\), so a normal approximation is appropriate. 4. The standardized value for \(0.45\) is \(z=\frac{0.45-0.40}{0.0200}=2.50\). 5. The upper-tail probability is \(P(\hat p\ge0.45)\approx 0.0062\). 6. In repeated random samples of size \(600\), about \(0.62\%\) would have a sample renewal proportion of at least \(0.45\).

a) Mean \(0.40\); standard deviation \(0.0200\). b) Approximately \(0.0062\). c) About \(0.62\%\) of repeated samples would produce \(\hat p\ge0.45\).

- Compare the spreads before calculating the interval probabilities. - The interval around the population proportion is the same for both samples. - A tighter distribution places more probability near its center.

1. Both sample sizes satisfy the \(10\%\) condition because \(200\le0.10(5000)=500\). 2. For \(n=50\), the standard deviation is \(\sqrt{\frac{0.70(0.30)}{50}}\approx 0.0648\). 3. For \(n=200\), the standard deviation is approximately \(0.0324\). 4. The large-count condition holds for both sample sizes. 5. For \(n=50\), \(P(0.65\le\hat p\le0.75)\approx 0.5596\). 6. For \(n=200\), the probability is approximately \(0.8772\). 7. The larger sample produces a narrower sampling distribution, so a fixed interval around \(p\) captures more of it.

a) Approximately \(0.0648\) and \(0.0324\). b) Approximately \(0.5596\) for \(n=50\) and \(0.8772\) for \(n=200\). c) The larger sample has less sampling variability.

- First find the probability for one random sample. - The cutoff is above the population proportion, so use an upper-tail area. - Convert the probability to an expected count across all simulations.

1. The sampling distribution has mean \(0.40\) and standard deviation \(\sqrt{\frac{0.40(0.60)}{100}}\approx 0.0490\). 2. The large-count condition is satisfied. 3. The standardized cutoff for \(0.50\) is \(z\approx 2.04\), so \(P(\hat p\ge0.50)\approx 0.0206\). 4. The expected number among \(500\) simulations is \(500(0.0206)\approx 10.3\), or about \(10\) samples.

About \(10\) of the \(500\) samples.

- Check the expected count for the rare outcome. - Consider the boundary that sample proportions cannot cross. - Think about where most samples will fall and which direction unusual larger counts can extend.

1. The mean is \(0.02\), and the standard deviation is \(\sqrt{\frac{0.02(0.98)}{100}}\approx 0.0140\). 2. The expected counts are \(2\) successes and \(98\) failures, so the large-count condition fails. 3. Most samples will have very few successes, placing \(\hat p\) near \(0\), while occasional samples with more successes extend the distribution to the right.

a) Mean \(0.02\); standard deviation approximately \(0.0140\). b) The condition fails because \(np=2<10\). c) The lower boundary at \(0\) and the rarity of successes concentrate values near \(0\) with a longer right tail.

- Connect each possible proportion to an integer number of heads. - Determine how much the proportion changes when the count changes by one. - Approximate normality concerns the overall shape, not whether every value on the number line is possible.

1. The mean is \(0.50\), and the standard deviation is \(\sqrt{\frac{0.50(0.50)}{40}}\approx 0.0791\). 2. Since the number of heads changes by whole numbers, consecutive sample proportions differ by \(\frac{1}{40}=0.025\). 3. The expected counts are \(20\) heads and \(20\) tails, both at least \(10\). The discrete probability masses therefore follow an approximately bell-shaped pattern that can be modeled by a normal distribution.

a) Mean \(0.50\); standard deviation approximately \(0.0791\). b) The spacing is \(0.025\). c) The large expected counts make the discrete distribution approximately bell-shaped.

- Calculate the actual spread of the sample proportion before judging the stated distance. - Express the allowed difference in standard-deviation units. - Account for deviations in both directions from the population proportion.

1. The sample is random, \(260\le0.10(5000)=500\), and the expected counts are \(260(0.65)=169\) renewals and \(260(0.35)=91\) nonrenewals, so a normal approximation is appropriate. 2. The sampling distribution has mean \(0.65\) and standard deviation \(\sqrt{\frac{(0.65)(0.35)}{260}}\approx 0.02958\). 3. A difference of \(0.05\) is \(\frac{0.05}{0.02958}\approx 1.69\) standard deviations, not about one standard deviation. 4. Therefore, \(P(|\hat p-0.65|\ge 0.05)\approx 2P(Z\ge 1.69)\approx 0.0910\). 5. The analyst’s estimate of \(0.32\) is too large because it uses an incorrect standardized distance.

The analyst’s claim is incorrect. The probability is approximately \(0.0910\), because a difference of \(0.05\) is about \(1.69\) standard deviations from the sampling-distribution mean.

- Connect the sample proportion to the number of pages with the rare outcome. - For no pages to require correction, every selected page must have the complementary outcome. - Check whether both expected outcome counts are large enough before using a bell-shaped model.

1. The mean is \(\mu_{\hat p}=p=0.015\). 2. The standard deviation is \(\sigma_{\hat p}=\sqrt{\frac{0.015(0.985)}{80}}\approx 0.01359\). 3. The event \(\hat p=0\) means that none of the \(80\) pages require correction. Thus \(P(\hat p=0)=(0.985)^{80}\approx 0.2985\). 4. The expected success count is \(80(0.015)=1.2\), which is below \(10\). Therefore, the sampling distribution is not approximately normal and has substantial probability at \(0\).

a) Mean \(0.015\); standard deviation approximately \(0.01359\). b) \(P(\hat p=0)\approx 0.2985\). c) The expected number of pages requiring correction is only \(1.2\), so the large-count condition fails.

- The center of a random-sample proportion does not depend on the population size. - Compare each sample size with one tenth of its population. - Use the standard spread calculation only where approximate independence is supported.

1. For both populations, the sample proportion is unbiased, so each sampling distribution has mean \(0.52\). 2. For Population A, \(120>0.10(900)=90\), so the \(10\%\) condition fails. For Population B, \(120\le 0.10(5000)=500\), so the condition is satisfied. 3. For Population B, \(\sigma_{\hat p}=\sqrt{\frac{0.52(0.48)}{120}}\approx 0.04561\). 4. Sampling a large fraction of Population A creates stronger dependence among selections, so the usual independence-based formula overstates the spread unless a finite-population adjustment is made.

a) Both means are \(0.52\). b) Population B. c) Approximately \(0.04561\). d) Population A fails the \(10\%\) condition because \(120>90\).

- Convert the proportion cutoff into a count and respect that counts must be whole numbers. - Check the expected numbers in both outcome categories. - Standardize the proportion cutoff using the center and spread of its sampling distribution.

1. Because \(0.437(160)=69.92\), the smallest possible count is \(70\), which gives \(\hat p=\frac{70}{160}=0.4375\). Thus, the event \(\hat p\ge0.437\) is the same as \(\hat p\ge0.4375\). 2. The sample is random, \(160\le0.10(5000)=500\), and the expected counts are \(160(0.38)=60.8\) and \(160(0.62)=99.2\), both at least \(10\). 3. The sampling distribution has mean \(0.38\) and standard deviation \(\sqrt{\frac{0.38(0.62)}{160}}\approx 0.03837\). 4. The standardized attainable cutoff is \(z=\frac{0.4375-0.38}{0.03837}\approx 1.498\). 5. Therefore, \(P(\hat p\ge 0.437)\approx P(Z\ge 1.498)\approx 0.0670\).

a) \(70\) members. b) Yes; the randomization, \(10\%\), and large-count conditions are satisfied. c) Approximately \(0.0670\).

- Start with the possible whole-number success counts in a sample of two. - Convert each count to a proportion and use the corresponding binomial probability. - Compute the weighted average of the possible sample proportions.

1. With two observations, the possible success counts are \(0\), \(1\), and \(2\), so the possible sample proportions are \(0\), \(0.5\), and \(1\). 2. The probabilities are \(P(\hat p=0)=(0.70)^2=0.49\), \(P(\hat p=0.5)=2(0.30)(0.70)=0.42\), and \(P(\hat p=1)=(0.30)^2=0.09\). 3. The mean is \(0(0.49)+0.5(0.42)+1(0.09)=0.30\), equal to the population proportion. 4. The distribution has only three possible values, and the expected success count is \(2(0.30)=0.60\), far below \(10\).

a) \(0\), \(0.5\), and \(1\). b) \(0.49\), \(0.42\), and \(0.09\), respectively. c) The mean is \(0.30\). d) The distribution is highly discrete and fails the large-count condition.

- Relate sampling spread to the square root of sample size. - A smaller spread corresponds to more observations when the population is unchanged. - Form a ratio so the common population term disappears.

1. For a fixed population proportion, \(\sigma_{\hat p}\) is proportional to \(1/\sqrt{n}\). Therefore, the smaller standard deviation for Distribution B comes from the larger sample size. 2. The ratio satisfies \(\frac{n_B}{n_A}=\left(\frac{0.030}{0.020}\right)^2=2.25\). 3. The common factor \(p(1-p)\) appears in both variance expressions and cancels when the ratio is formed.

a) Distribution B. b) \(\frac{n_B}{n_A}=2.25\). c) The common population-variability factor cancels in the comparison.

- Check both expected outcome counts, not only the larger one. - Compare the boundary of the possible proportion scale with the proposed normal model. - Use the model's center and spread to see how much area it places beyond that boundary.

1. The mean is \(\mu_{\hat p}=0.95\), and the standard deviation is \(\sigma_{\hat p}=\sqrt{\frac{0.95(0.05)}{40}}\approx 0.03446\). 2. The expected counts are \(40(0.95)=38\) and \(40(0.05)=2\). Because the expected failure count is below \(10\), the large-count condition fails. 3. Under the inappropriate normal model, \(z=\frac{1-0.95}{0.03446}\approx 1.451\), giving \(P(\hat p>1)\approx 0.0734\). 4. Assigning positive probability to proportions above \(1\) shows that the normal model is a poor approximation in this case.

a) Mean \(0.95\); standard deviation approximately \(0.03446\). b) The condition fails because the expected failure count is only \(2\). c) About \(0.0734\); the impossible probability exposes the inadequacy of the normal approximation.

- Translate the exact sample proportion into an exact success count. - Use the probability of that whole-number count rather than the population proportion itself. - Distinguish the expected value of a statistic from the probability that it equals that value.

1. The event \(\hat p=0.50\) means exactly \(10\) successes in \(20\) trials. 2. Therefore, \(P(\hat p=0.50)=\binom{20}{10}(0.50)^{10}(0.50)^{10}=\frac{\binom{20}{10}}{2^{20}}\approx 0.1762\). 3. The value \(0.50\) is the population success probability and the mean of \(\hat p\), not the probability that the random statistic equals its mean exactly. 4. The sampling-distribution mean is \(E(\hat p)=p=0.50\).

a) Approximately \(0.1762\). b) The parameter \(0.50\) is the center of the distribution, not the probability of one exact sample-proportion value. c) \(0.50\).

- Express every possible sample proportion as a whole-number count divided by the sample size. - Locate the two counts nearest the expected count. - Use symmetry of the success probability to compare their probabilities.

1. The mean of \(\hat p\) is \(0.50\), but a sample proportion must equal \(x/21\) for a whole-number success count \(x\). Because \(21(0.50)=10.5\) is not a whole number, \(\hat p\) cannot equal \(0.50\). 2. The nearest success counts are \(10\) and \(11\), giving \(\hat p=\frac{10}{21}\) and \(\hat p=\frac{11}{21}\). 3. By symmetry, the probabilities are equal. Each is \(\binom{21}{10}(0.50)^{21}=\frac{\binom{21}{10}}{2^{21}}\approx 0.1682\).

a) A success count of \(10.5\) is impossible. b) \(\frac{10}{21}\) and \(\frac{11}{21}\). c) Each has probability approximately \(0.1682\).

- Determine which split between the two outcomes maximizes population variability. - Use that maximum to place an upper bound on sampling spread. - Compare the claimed value with the bound.

1. The variance is \(\frac{p(1-p)}{100}\), and the product \(p(1-p)\) is largest at \(p=0.50\), where it equals \(0.25\). 2. The largest possible standard deviation is \(\sqrt{\frac{0.25}{100}}=0.050\). 3. Because \(0.052>0.050\), the reported standard deviation is impossible under the stated sampling model.

a) \(0.050\). b) No. A standard deviation of \(0.052\) is impossible for \(n=100\). c) \(p=0.50\).

- Relate one additional success to the resulting change in the sample proportion. - Once the sample size is known, identify the sampling distribution’s center and spread. - Check both expected outcome counts before using a normal shape.

1. Adjacent sample proportions differ by \(\frac{1}{n}\), so \(\frac{1}{n}=0.004\) and \(n=250\). 2. The mean is \(\mu_{\hat p}=p=0.28\). 3. The standard deviation is \(\sigma_{\hat p}=\sqrt{\frac{(0.28)(0.72)}{250}}\approx 0.02840\). 4. The expected counts are \(np=70\) and \(n(1-p)=180\), both at least \(10\), so the normal model is reasonable.

a) \(n=250\). b) \(\mu_{\hat p}=0.28\) and \(\sigma_{\hat p}\approx 0.02840\). c) Yes. The expected counts are \(70\) and \(180\).

- Distinguish a parameter that centers repeated samples from one statistic produced by one sample. - Use the population proportion when describing the theoretical sampling distribution. - Measure the observed statistic's distance from the distribution center in standard-deviation units.

1. The mean of the sampling distribution is the population proportion, \(\mu_{\hat p}=0.42\). The value \(0.448\) is one observed sample proportion. 2. The standard deviation is \(\sigma_{\hat p}=\sqrt{\frac{(0.42)(0.58)}{250}}\approx0.03122\). 3. The standardized value is \(z=\frac{0.448-0.42}{0.03122}\approx0.897\).

The sampling-distribution mean is \(0.42\), not \(0.448\). Its standard deviation is approximately \(0.03122\), and the observed sample proportion has \(z\approx0.897\).

- Translate the bounds on the sample proportion into allowable whole-number success counts. - For the approximation, extend the count interval by half a unit at each endpoint. - Use the count mean and standard deviation when standardizing the corrected boundaries. - Compare the two final probabilities using their difference.

1. The event \(0.45\le\hat p\le0.55\) is equivalent to \(18\le X\le22\), where \(X\sim\operatorname{Bin}(40,0.50)\). 2. The exact binomial probability is \(\sum_{k=18}^{22}\binom{40}{k}(0.50)^{40}\approx0.57041\). 3. For the continuity correction, use \(17.5\le Y\le22.5\), where \(Y\) is normal with mean \(20\) and standard deviation \(\sqrt{10}\). 4. The corresponding standardized bounds are approximately \(-0.7906\) and \(0.7906\), giving a normal approximation of \(0.57080\). 5. The approximation exceeds the exact probability by \(0.5708047\ldots-0.5704095\ldots=0.0003952\ldots\approx0.00040\) using the requested rounded values.

a) \(\sum_{k=18}^{22}\binom{40}{k}(0.50)^{40}\approx0.57041\). b) \(0.57080\). c) The approximation is higher by approximately \(0.00040\).

- Write one expected-count requirement for each of the two outcomes. - Solve both inequalities for the same population proportion. - Keep only values that satisfy the two requirements simultaneously.

1. The large-count condition requires \(np\ge 10\) and \(n(1-p)\ge 10\). 2. From \(80p\ge 10\), \(p\ge 0.125\). 3. From \(80(1-p)\ge 10\), \(1-p\ge 0.125\), so \(p\le 0.875\). 4. Both conditions hold when \(0.125\le p\le 0.875\).

The normal large-count condition holds for \(0.125\le p\le 0.875\).

- Reexpress the condition as a statement about how far the sample proportion is from the population proportion. - Identify the center and spread of that error across repeated samples. - Standardize both symmetric error limits before finding the central area.

1. The sampling error \(\hat p-p\) has mean \(0\). 2. Its standard deviation is the standard deviation of \(\hat p\): \(\sqrt{\frac{(0.25)(0.75)}{300}}=0.025\). 3. The error bounds standardize to \(z=\frac{\pm 0.04}{0.025}=\pm 1.60\). 4. The probability between these bounds is \(P(-1.60\le Z\le 1.60)\approx 0.8904\).

The probability is approximately \(0.8904\), or \(89.04\%\).

- Translate “fewer than half” into the largest allowable whole-number success count. - Use a boundary halfway between adjacent counts when moving to a continuous model. - Standardize the corrected boundary using the count mean and standard deviation.

1. Fewer than half of \(200\) means at most \(99\) successes. 2. The success count has mean \(np=200(0.58)=116\) and standard deviation \(\sqrt{200(0.58)(0.42)}\approx6.9800\). 3. With a continuity correction, use the boundary \(99.5\). 4. The standardized boundary is \(z=\frac{99.5-116}{6.979971\ldots}\approx-2.364\). 5. The lower-tail probability is approximately \(0.00904\).

Approximately \(0.00904\), or \(0.904\%\).

- Ask whether repeated executions of the stated procedure can produce different samples. - Describe the full set of possible values of the sample proportion. - Check the sampling-fraction condition behind the usual variability formula.

1. Every census contains the same \(80\) featured objects out of \(200\), so \(\hat p=\frac{80}{200}=0.40\) every time. 2. The sampling distribution is concentrated entirely at \(0.40\), with mean \(0.40\) and standard deviation \(0\). 3. The usual formula assumes independent sampling or a small sampling fraction when sampling without replacement. 4. A census samples \(100\%\) of the finite population, so there is no sampling variability; the nonzero value \(0.03464\) ignores this dependence.

a) \(\hat p=0.40\) with probability \(1\), so its standard deviation is \(0\). b) The usual formula is not applicable to a census because the sample is the entire finite population and has no sampling uncertainty.

- Treat the center and shape of a sampling distribution as separate properties. - Check the expected counts before deciding on a normal approximation. - An estimator’s bias is determined by its long-run mean, not by symmetry.

1. The mean is \(\mu_{\hat p}=p=0.06\), so \(\hat p\) is unbiased for \(p\). 2. The standard deviation is \(\sigma_{\hat p}=\sqrt{\frac{(0.06)(0.94)}{100}}\approx 0.02375\). 3. The expected success count is \(np=6\), which is below \(10\), so the large-count condition for a normal shape is not met. 4. Unbiasedness concerns the distribution’s center, while normality concerns its shape. A skewed distribution can still have mean \(0.06\).

The mean is \(0.06\), and the standard deviation is approximately \(0.02375\). The estimator is unbiased because its mean equals \(p\), even though the small expected success count makes the distribution nonnormal.

- View the estimated total as a constant multiple of the sample proportion. - Transform both the center and the spread by the same constant. - Translate the allowed error in account counts into a distance from the sampling-distribution mean.

1. The estimator is unbiased, so \(\mu_{\hat T}=40{,}000(0.08)=3200\). 2. The sampling conditions hold because \(500\le 0.10(40{,}000)\), \(500(0.08)=40\), and \(500(0.92)=460\). 3. The standard deviation is \(\sigma_{\hat T}=40{,}000\sqrt{\frac{(0.08)(0.92)}{500}}\approx 485.30\). 4. Being within \(2000\) accounts corresponds to being within \(\frac{2000}{485.30}\approx 4.121\) standard deviations of the mean. 5. The normal approximation gives \(P(|\hat T-3200|\le 2000)\approx 0.99996\).

a) \(\mu_{\hat T}=3200\) accounts and \(\sigma_{\hat T}\approx 485.30\) accounts. b) The probability is approximately \(0.99996\).

- Rewrite the weighted expression in terms of the total number of successes across both samples. - Use the combined number of independent observations when finding the spread. - Standardize the cutoff relative to the pooled sampling distribution.

1. The pooled statistic is the total number of successes in the two samples divided by the combined sample size \(500\). 2. Because both samples come independently from the same population, \(\mu_{\hat p_{\text{pool}}}=0.40\). 3. The combined expected counts are \(500(0.40)=200\) successes and \(500(0.60)=300\) failures, so a normal approximation is appropriate. 4. The standard deviation is \(\sigma_{\hat p_{\text{pool}}}=\sqrt{\frac{(0.40)(0.60)}{500}}\approx 0.02191\). 5. The standardized cutoff is \(z=\frac{0.44-0.40}{0.02191}\approx 1.826\). 6. Therefore, \(P(\hat p_{\text{pool}}>0.44)\approx 0.0339\).

a) \(\mu_{\hat p_{\text{pool}}}=0.40\) and \(\sigma_{\hat p_{\text{pool}}}\approx 0.02191\). b) \(P(\hat p_{\text{pool}}>0.44)\approx 0.0339\).

- Use the defining relationship between the population proportion and the center of the sampling distribution. - Treat the reported variance as a clue about sample size. - Once the distribution is identified, locate the cutoff relative to its center and spread.

1. Because \(\mu_{\hat p}=p\), the population proportion is \(p=0.65\). 2. Since \(\sigma_{\hat p}^2=\frac{p(1-p)}{n}\), solve \(0.000455=\frac{0.65(0.35)}{n}\), giving \(n=500\). 3. The standard deviation is \(\sigma_{\hat p}=\sqrt{0.000455}\approx 0.0213\). 4. The standardized value for \(0.69\) is \(z\approx 1.88\), so \(P(\hat p>0.69)\approx 0.0304\).

a) \(p=0.65\). b) \(n=500\). c) \(P(\hat p>0.69)\approx 0.0304\).

- Translate the central coverage statement into a number of standard deviations around the mean. - Express the desired half-width in terms of the sampling-distribution spread. - Round upward so the coverage target is not weakened.

1. For approximately \(95\%\) of values to lie within \(0.04\), set \(1.96\sigma_{\hat p}\le0.04\). 2. With \(p=0.50\), \(1.96\sqrt{\frac{0.25}{n}}\le0.04\). 3. Solving gives \(n\ge\frac{(1.96)^2(0.25)}{(0.04)^2}\approx 600.25\). 4. Round up to \(n=601\).

The minimum sample size is \(601\).

- Read the two marked percentile values and use their symmetry to locate the center. - Relate each percentile’s distance from the center to a standard normal percentile. - Once the spread is known, connect it to the population proportion and sample size.

1. Reading the display gives the fifth percentile \(0.3766\) and the ninety-fifth percentile \(0.4634\). For a normal distribution, these percentiles are symmetric about the mean. Thus \(p=\mu_{\hat p}=\frac{0.3766+0.4634}{2}=0.4200\). 2. Each percentile is approximately \(1.645\) standard deviations from the mean. Therefore, \(\sigma_{\hat p}=\frac{0.4634-0.3766}{2(1.645)}\approx 0.02638\). 3. Using \(\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}\), solve for \(n\): \(n=\frac{0.42(0.58)}{(0.02638)^2}\approx 349.97\). 4. The estimated sample size is \(350\).

a) \(p\approx 0.420\). b) \(\sigma_{\hat p}\approx 0.02638\). c) \(n\approx 350\).

- Square the spread relationship to isolate a product involving the population proportion. - Expect a pair of complementary answers because the two outcomes can be interchanged. - The center of each sampling distribution follows directly from its population proportion.

1. The standard deviation condition gives \(0.020=\sqrt{\frac{p(1-p)}{400}}\), so \(p(1-p)=400(0.020)^2=0.16\). 2. Solving \(p-p^2=0.16\) gives \(p^2-p+0.16=0\). 3. Thus \(p=\frac{1\pm\sqrt{1-0.64}}{2}=\frac{1\pm 0.60}{2}\), so \(p=0.20\) or \(p=0.80\). 4. Complementary proportions have the same product \(p(1-p)\), so they produce the same standard deviation. 5. The sampling-distribution means are \(0.20\) and \(0.80\), respectively.

a) \(p=0.20\) or \(p=0.80\). b) Complementary proportions have equal sampling variability. c) The corresponding means are \(0.20\) and \(0.80\).

- Convert the sample-proportion event into a whole-number count event. - For the approximation, adjust the count boundary by half a unit before standardizing. - Compare the exact and approximate probabilities numerically. - Relate the discrepancy to how close the expected rare-outcome count is to the usual large-count threshold.

1. The event \(\hat p\le0.010\) is equivalent to \(X\le5\). 2. Since \(X\sim\operatorname{Bin}(500,0.02)\), the exact binomial probability is \(\sum_{k=0}^{5}\binom{500}{k}(0.02)^k(0.98)^{500-k}\approx0.06519\). 3. The count distribution has mean \(\mu_X=500(0.02)=10\) and standard deviation \(\sigma_X=\sqrt{500(0.02)(0.98)}\approx3.1305\). 4. With continuity correction, \(P(X\le5)\) is approximated by \(P(Y\le5.5)\) for a normal variable \(Y\) with that mean and standard deviation. 5. The standardized boundary is \(z=\frac{5.5-10}{3.1305\ldots}\approx-1.437\), giving \(P(X\le5)\approx0.07529\). 6. The expected success count is exactly \(10\), so the usual large-count condition is only just met; the noticeable discrepancy between the exact and approximate probabilities is therefore unsurprising.

a) \(\sum_{k=0}^{5}\binom{500}{k}(0.02)^k(0.98)^{500-k}\approx0.06519\). b) \(0.07529\). c) The normal approximation is reasonably close but not highly accurate; the expected success count is exactly at the usual minimum large-count threshold.

- Determine which whole-number success counts round to the displayed percentage. - Work with the success-count distribution so that the continuity correction can be applied naturally. - Convert the corrected count boundaries to standardized values before finding the middle area.

1. Reporting \(40\%\) requires the success count to be \(99\), \(100\), or \(101\), because these correspond to sample percentages \(39.6\%\), \(40.0\%\), and \(40.4\%\). 2. For \(X\sim\operatorname{Bin}(250,0.37)\), the mean is \(92.5\) and the standard deviation is \(\sqrt{250(0.37)(0.63)}\approx7.6338\). 3. With a continuity correction, approximate \(P(99\le X\le101)\) by \(P(98.5<Y<101.5)\) for a normal variable \(Y\) with that mean and standard deviation. 4. The standardized bounds are approximately \(0.786\) and \(1.179\). 5. The normal approximation is approximately \(0.0967\).

Approximately \(0.0967\).