Worked solution P16-Easy-1. The mean is 28.29, the median is 28.00, and the sample standard deviation is 4.15. $$\overline{x}=\frac{\sum x_i}{7}=28.29,s=\sqrt{\frac{\sum (x_i-\overline{x}{)}^2}{6}}=4.15.$$ Interpretation: The mean is the balance point; s describes a typical distance from the mean in the variable's units. Validity: Use resistant summaries such as median and IQR when skewness or outliers make mean and standard deviation misleading. Error to reject: Do not divide the sample variance by n when the requested quantity is the conventional sample standard deviation s.

Worked solution P16-Easy-2. The row-A conditional proportion is 23/(23+12)=0.657. Compare it with row B's 0.356; unequal conditional proportions indicate association in the table. $$P(\text{Yes}\mid A)=\frac{23}{35}=0.657.$$ The grand total 80 is not the conditional denominator. Interpretation: The comparison concerns distributions within row groups and does not by itself establish causation. Validity: Use the group named after the conditioning bar as the denominator and report which variable defines each row and column. Error to reject: Dividing by the grand total produces a joint relative frequency, not a conditional relative frequency.

Worked solution P16-Easy-3. State experimental units, treatments, response, assignment mechanism, comparison, replication, controls, and the intended scope of inference. With 120 units and four treatments, equal random assignment gives $120/4=30$ units per treatment before attrition. Interpretation: Random assignment balances lurking variables in expectation and supports a causal comparison; random sampling is what supports generalization. Validity: Interference should be limited, treatment implementation consistent, and response measurement blind where feasible. Error to reject: A large observational comparison is not automatically an experiment, because sample size does not replace random assignment.

Worked solution P16-Easy-4. $P(A\cup B)=0.42$ and $P(A\mid B)=0.630$. $$P(A\cup B)=0.32+0.27-0.17=0.42,P(A\mid B)=\frac{0.17}{0.27}=0.630.$$ Interpretation: The conditional probability restricts the reference group to outcomes where B occurs. Validity: Check that probabilities are coherent and that the denominator event has positive probability. Error to reject: Adding P(A) and P(B) without subtracting the intersection double-counts outcomes in both events.

Worked solution P16-Easy-5. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{77.0-65}{6}=2.00,P(X\le 77.0)=\Phi (2.00)\approx 0.9772.$$ Interpretation: The value is 2.00 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P16-Easy-6. The exact probability is 0.11096; the model mean is 5.250. $$P(X=3)=\left(\genfrac{}{}{0ex}{}{15}{3}\right)(0.35{)}^3(0.65{)}^{12}=0.11096.$$ Interpretation: Across many repetitions of 15 trials, the average number of successes approaches $np=5.250$. Validity: Verify binary outcomes, independent trials, fixed n, and constant success probability before using the binomial model. Error to reject: Do not omit the combination factor; it counts the different orders producing exactly k successes.

Worked solution P16-Easy-7. The sampling distribution has mean 80.000, standard error 1.4832, and the probability is approximately 0.1587 under a justified normal approximation. $${\mu }_{\overline{X}}=80,{\sigma }_{\overline{X}}=\frac{11}{\sqrt{55}}=1.4832,z=1.$$ Interpretation: The sampling distribution describes variation among statistics from repeated samples, not variation among individual observations. Validity: Use an independent random sample; if sampling without replacement, the population should be at least $10n=550$. Normality is exact for a normal population and approximate through CLT otherwise. Error to reject: Increasing n reduces the standard error but does not change the population standard deviation or make the raw observations normal.

Worked solution P16-Easy-8. The one-proportion z interval is (0.3239, 0.4950). $$\widehat{p}\pm 1.96\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}=0.4094\pm 0.0855.$$ Interpretation: Confidence describes the long-run success rate of the interval-producing method; the contextual interval estimates the named population parameter. Validity: Check randomization, the 10 percent condition if sampling without replacement, and large counts: successes=52, failures=75. Error to reject: Do not say there is a 95 percent probability this fixed population proportion lies in the computed interval.

Worked solution P16-Easy-9. The t interval is (69.977, 76.023). $$\overline{x}\pm t^{*}\frac{s}{\sqrt{n}}=73.00\pm 3.023.$$ Interpretation: The interval estimates a population mean in the original measurement units and must name the represented population. Validity: Use a random sample or randomized design, independence, and a population shape or sample size that makes the t procedure reliable. Error to reject: Use t rather than z when sigma is unknown and sample standard deviation s estimates it.

Worked solution P16-Easy-10. The statistic is z=3.455; the two-sided p-value is 0.0005. $$z=\frac{0.6493-0.50}{\sqrt{0.50(0.50)/134}}=3.455.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P16-Easy-11. The statistic is t=1.479 with df=34; p=0.1483. $$t=\frac{52.00-50}{8.00/\sqrt{35}}=1.479.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P16-Easy-12. The statistic is 6.418, df=2, and p=0.0404; expected counts are computed from row and column totals. $$E_{ij}=\frac{(\text{row total})(\text{column total})}{124},{\chi }^2=6.418,df=(2-1)(3-1)=2.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Worked solution P16-Easy-13. The least-squares line is approximately $\widehat{y}=5.267+2.971x$, with $r=0.967$. At x=4, the residual is 1.848. $$b=r\frac{s_y}{s_x}=2.971,a=\overline{y}-b\overline{x}=5.267,e=y-\widehat{y}=1.848.$$ Interpretation: Within the observed range, each one-unit increase in x is associated with a predicted increase of about 2.971 y-units. Validity: Inspect form, unusual points, and residual pattern before relying on a linear model or extrapolating. Error to reject: Correlation and slope describe association; neither proves that changing x causes y to change.

Worked solution P16-Easy-14. The mean is 20.29, the median is 20.00, and the sample standard deviation is 4.15. $$\overline{x}=\frac{\sum x_i}{7}=20.29,s=\sqrt{\frac{\sum (x_i-\overline{x}{)}^2}{6}}=4.15.$$ Interpretation: The mean is the balance point; s describes a typical distance from the mean in the variable's units. Validity: Use resistant summaries such as median and IQR when skewness or outliers make mean and standard deviation misleading. Error to reject: Do not divide the sample variance by n when the requested quantity is the conventional sample standard deviation s.

Worked solution P16-Easy-15. The row-A conditional proportion is 26/(26+17)=0.605. Compare it with row B's 0.426; unequal conditional proportions indicate association in the table. $$P(\text{Yes}\mid A)=\frac{26}{43}=0.605.$$ The grand total 97 is not the conditional denominator. Interpretation: The comparison concerns distributions within row groups and does not by itself establish causation. Validity: Use the group named after the conditioning bar as the denominator and report which variable defines each row and column. Error to reject: Dividing by the grand total produces a joint relative frequency, not a conditional relative frequency.

Worked solution P16-Easy-16. State experimental units, treatments, response, assignment mechanism, comparison, replication, controls, and the intended scope of inference. With 120 units and four treatments, equal random assignment gives $120/4=30$ units per treatment before attrition. Interpretation: Random assignment balances lurking variables in expectation and supports a causal comparison; random sampling is what supports generalization. Validity: Interference should be limited, treatment implementation consistent, and response measurement blind where feasible. Error to reject: A large observational comparison is not automatically an experiment, because sample size does not replace random assignment.

Worked solution P16-Easy-17. $P(A\cup B)=0.50$ and $P(A\mid B)=0.677$. $$P(A\cup B)=0.40+0.31-0.21=0.50,P(A\mid B)=\frac{0.21}{0.31}=0.677.$$ Interpretation: The conditional probability restricts the reference group to outcomes where B occurs. Validity: Check that probabilities are coherent and that the denominator event has positive probability. Error to reject: Adding P(A) and P(B) without subtracting the intersection double-counts outcomes in both events.

Worked solution P16-Tough-1. The mean is 16.29, the median is 16.00, and the sample standard deviation is 4.15. $$\overline{x}=\frac{\sum x_i}{7}=16.29,s=\sqrt{\frac{\sum (x_i-\overline{x}{)}^2}{6}}=4.15.$$ Interpretation: The mean is the balance point; s describes a typical distance from the mean in the variable's units. Validity: Use resistant summaries such as median and IQR when skewness or outliers make mean and standard deviation misleading. Error to reject: Do not divide the sample variance by n when the requested quantity is the conventional sample standard deviation s.

Worked solution P16-Tough-2. The row-A conditional proportion is 22/(22+16)=0.579. Compare it with row B's 0.385; unequal conditional proportions indicate association in the table. $$P(\text{Yes}\mid A)=\frac{22}{38}=0.579.$$ The grand total 77 is not the conditional denominator. Interpretation: The comparison concerns distributions within row groups and does not by itself establish causation. Validity: Use the group named after the conditioning bar as the denominator and report which variable defines each row and column. Error to reject: Dividing by the grand total produces a joint relative frequency, not a conditional relative frequency.

Worked solution P16-Tough-3. State experimental units, treatments, response, assignment mechanism, comparison, replication, controls, and the intended scope of inference. With 120 units and four treatments, equal random assignment gives $120/4=30$ units per treatment before attrition. Interpretation: Random assignment balances lurking variables in expectation and supports a causal comparison; random sampling is what supports generalization. Validity: Interference should be limited, treatment implementation consistent, and response measurement blind where feasible. Error to reject: A large observational comparison is not automatically an experiment, because sample size does not replace random assignment.

Worked solution P16-Tough-4. $P(A\cup B)=0.55$ and $P(A\mid B)=0.737$. $$P(A\cup B)=0.45+0.38-0.28=0.55,P(A\mid B)=\frac{0.28}{0.38}=0.737.$$ Interpretation: The conditional probability restricts the reference group to outcomes where B occurs. Validity: Check that probabilities are coherent and that the denominator event has positive probability. Error to reject: Adding P(A) and P(B) without subtracting the intersection double-counts outcomes in both events.

Worked solution P16-Tough-5. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{81.0-69}{6}=2.00,P(X\le 81.0)=\Phi (2.00)\approx 0.9772.$$ Interpretation: The value is 2.00 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P16-Tough-6. The exact probability is 0.07954; the model mean is 5.780. $$P(X=3)=\left(\genfrac{}{}{0ex}{}{17}{3}\right)(0.34{)}^3(0.66{)}^{14}=0.07954.$$ Interpretation: Across many repetitions of 17 trials, the average number of successes approaches $np=5.780$. Validity: Verify binary outcomes, independent trials, fixed n, and constant success probability before using the binomial model. Error to reject: Do not omit the combination factor; it counts the different orders producing exactly k successes.

Worked solution P16-Tough-7. The sampling distribution has mean 54.000, standard error 1.2136, and the probability is approximately 0.1587 under a justified normal approximation. $${\mu }_{\overline{X}}=54,{\sigma }_{\overline{X}}=\frac{9}{\sqrt{55}}=1.2136,z=1.$$ Interpretation: The sampling distribution describes variation among statistics from repeated samples, not variation among individual observations. Validity: Use an independent random sample; if sampling without replacement, the population should be at least $10n=550$. Normality is exact for a normal population and approximate through CLT otherwise. Error to reject: Increasing n reduces the standard error but does not change the population standard deviation or make the raw observations normal.

Worked solution P16-Tough-8. The one-proportion z interval is (0.4013, 0.5751). $$\widehat{p}\pm 1.96\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}=0.4882\pm 0.0869.$$ Interpretation: Confidence describes the long-run success rate of the interval-producing method; the contextual interval estimates the named population parameter. Validity: Check randomization, the 10 percent condition if sampling without replacement, and large counts: successes=62, failures=65. Error to reject: Do not say there is a 95 percent probability this fixed population proportion lies in the computed interval.

Worked solution P16-Tough-9. The t interval is (69.474, 76.126). $$\overline{x}\pm t^{*}\frac{s}{\sqrt{n}}=72.80\pm 3.326.$$ Interpretation: The interval estimates a population mean in the original measurement units and must name the represented population. Validity: Use a random sample or randomized design, independence, and a population shape or sample size that makes the t procedure reliable. Error to reject: Use t rather than z when sigma is unknown and sample standard deviation s estimates it.

Worked solution P16-Tough-10. The statistic is z=2.246; the two-sided p-value is 0.0247. $$z=\frac{0.5970-0.50}{\sqrt{0.50(0.50)/134}}=2.246.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P16-Tough-11. The statistic is t=2.784 with df=34; p=0.0087. $$t=\frac{54.00-50}{8.50/\sqrt{35}}=2.784.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P16-Tough-12. The statistic is 6.418, df=2, and p=0.0404; expected counts are computed from row and column totals. $$E_{ij}=\frac{(\text{row total})(\text{column total})}{124},{\chi }^2=6.418,df=(2-1)(3-1)=2.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Worked solution P16-Tough-13. The least-squares line is approximately $\widehat{y}=5.267+3.971x$, with $r=0.981$. At x=4, the residual is 1.848. $$b=r\frac{s_y}{s_x}=3.971,a=\overline{y}-b\overline{x}=5.267,e=y-\widehat{y}=1.848.$$ Interpretation: Within the observed range, each one-unit increase in x is associated with a predicted increase of about 3.971 y-units. Validity: Inspect form, unusual points, and residual pattern before relying on a linear model or extrapolating. Error to reject: Correlation and slope describe association; neither proves that changing x causes y to change.

Worked solution P16-Tough-14. The mean is 16.29, the median is 16.00, and the sample standard deviation is 4.15. $$\overline{x}=\frac{\sum x_i}{7}=16.29,s=\sqrt{\frac{\sum (x_i-\overline{x}{)}^2}{6}}=4.15.$$ Interpretation: The mean is the balance point; s describes a typical distance from the mean in the variable's units. Validity: Use resistant summaries such as median and IQR when skewness or outliers make mean and standard deviation misleading. Error to reject: Do not divide the sample variance by n when the requested quantity is the conventional sample standard deviation s.

Worked solution P16-Tough-15. The row-A conditional proportion is 28/(28+18)=0.609. Compare it with row B's 0.396; unequal conditional proportions indicate association in the table. $$P(\text{Yes}\mid A)=\frac{28}{46}=0.609.$$ The grand total 99 is not the conditional denominator. Interpretation: The comparison concerns distributions within row groups and does not by itself establish causation. Validity: Use the group named after the conditioning bar as the denominator and report which variable defines each row and column. Error to reject: Dividing by the grand total produces a joint relative frequency, not a conditional relative frequency.

Worked solution P16-Tough-16. State experimental units, treatments, response, assignment mechanism, comparison, replication, controls, and the intended scope of inference. With 120 units and four treatments, equal random assignment gives $120/4=30$ units per treatment before attrition. Interpretation: Random assignment balances lurking variables in expectation and supports a causal comparison; random sampling is what supports generalization. Validity: Interference should be limited, treatment implementation consistent, and response measurement blind where feasible. Error to reject: A large observational comparison is not automatically an experiment, because sample size does not replace random assignment.

Worked solution P16-Tough-17. $P(A\cup B)=0.59$ and $P(A\mid B)=0.737$. $$P(A\cup B)=0.49+0.38-0.28=0.59,P(A\mid B)=\frac{0.28}{0.38}=0.737.$$ Interpretation: The conditional probability restricts the reference group to outcomes where B occurs. Validity: Check that probabilities are coherent and that the denominator event has positive probability. Error to reject: Adding P(A) and P(B) without subtracting the intersection double-counts outcomes in both events.

Worked solution P16-Toughest-1. The mean is 24.29, the median is 24.00, and the sample standard deviation is 4.15. $$\overline{x}=\frac{\sum x_i}{7}=24.29,s=\sqrt{\frac{\sum (x_i-\overline{x}{)}^2}{6}}=4.15.$$ Interpretation: The mean is the balance point; s describes a typical distance from the mean in the variable's units. Validity: Use resistant summaries such as median and IQR when skewness or outliers make mean and standard deviation misleading. Error to reject: Do not divide the sample variance by n when the requested quantity is the conventional sample standard deviation s.

Worked solution P16-Toughest-2. The row-A conditional proportion is 27/(27+13)=0.675. Compare it with row B's 0.368; unequal conditional proportions indicate association in the table. $$P(\text{Yes}\mid A)=\frac{27}{40}=0.675.$$ The grand total 78 is not the conditional denominator. Interpretation: The comparison concerns distributions within row groups and does not by itself establish causation. Validity: Use the group named after the conditioning bar as the denominator and report which variable defines each row and column. Error to reject: Dividing by the grand total produces a joint relative frequency, not a conditional relative frequency.

Worked solution P16-Toughest-3. State experimental units, treatments, response, assignment mechanism, comparison, replication, controls, and the intended scope of inference. With 120 units and four treatments, equal random assignment gives $120/4=30$ units per treatment before attrition. Interpretation: Random assignment balances lurking variables in expectation and supports a causal comparison; random sampling is what supports generalization. Validity: Interference should be limited, treatment implementation consistent, and response measurement blind where feasible. Error to reject: A large observational comparison is not automatically an experiment, because sample size does not replace random assignment.

Worked solution P16-Toughest-4. $P(A\cup B)=0.57$ and $P(A\mid B)=0.750$. $$P(A\cup B)=0.47+0.40-0.30=0.57,P(A\mid B)=\frac{0.30}{0.40}=0.750.$$ Interpretation: The conditional probability restricts the reference group to outcomes where B occurs. Validity: Check that probabilities are coherent and that the denominator event has positive probability. Error to reject: Adding P(A) and P(B) without subtracting the intersection double-counts outcomes in both events.

Worked solution P16-Toughest-5. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{93.0-73}{10}=2.00,P(X\le 93.0)=\Phi (2.00)\approx 0.9772.$$ Interpretation: The value is 2.00 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P16-Toughest-6. The exact probability is 0.03177; the model mean is 6.750. $$P(X=3)=\left(\genfrac{}{}{0ex}{}{15}{3}\right)(0.45{)}^3(0.55{)}^{12}=0.03177.$$ Interpretation: Across many repetitions of 15 trials, the average number of successes approaches $np=6.750$. Validity: Verify binary outcomes, independent trials, fixed n, and constant success probability before using the binomial model. Error to reject: Do not omit the combination factor; it counts the different orders producing exactly k successes.

Worked solution P16-Toughest-7. The sampling distribution has mean 72.000, standard error 1.4832, and the probability is approximately 0.1587 under a justified normal approximation. $${\mu }_{\overline{X}}=72,{\sigma }_{\overline{X}}=\frac{11}{\sqrt{55}}=1.4832,z=1.$$ Interpretation: The sampling distribution describes variation among statistics from repeated samples, not variation among individual observations. Validity: Use an independent random sample; if sampling without replacement, the population should be at least $10n=550$. Normality is exact for a normal population and approximate through CLT otherwise. Error to reject: Increasing n reduces the standard error but does not change the population standard deviation or make the raw observations normal.

Worked solution P16-Toughest-8. The one-proportion z interval is (0.3163, 0.4868). $$\widehat{p}\pm 1.96\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}=0.4016\pm 0.0853.$$ Interpretation: Confidence describes the long-run success rate of the interval-producing method; the contextual interval estimates the named population parameter. Validity: Check randomization, the 10 percent condition if sampling without replacement, and large counts: successes=51, failures=76. Error to reject: Do not say there is a 95 percent probability this fixed population proportion lies in the computed interval.

Worked solution P16-Toughest-9. The t interval is (69.096, 76.504). $$\overline{x}\pm t^{*}\frac{s}{\sqrt{n}}=72.80\pm 3.704.$$ Interpretation: The interval estimates a population mean in the original measurement units and must name the represented population. Validity: Use a random sample or randomized design, independence, and a population shape or sample size that makes the t procedure reliable. Error to reject: Use t rather than z when sigma is unknown and sample standard deviation s estimates it.

Worked solution P16-Toughest-10. The statistic is z=1.382; the two-sided p-value is 0.1669. $$z=\frac{0.5597-0.50}{\sqrt{0.50(0.50)/134}}=1.382.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P16-Toughest-11. The statistic is t=2.891 with df=34; p=0.0067. $$t=\frac{54.30-50}{8.80/\sqrt{35}}=2.891.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P16-Toughest-12. The statistic is 6.418, df=2, and p=0.0404; expected counts are computed from row and column totals. $$E_{ij}=\frac{(\text{row total})(\text{column total})}{124},{\chi }^2=6.418,df=(2-1)(3-1)=2.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use random sampling or random assignment as appropriate, independent observations, and sufficiently large expected counts. Error to reject: Check expected counts, not observed counts; a large cell contribution identifies where observed and expected differ most.

Worked solution P16-Toughest-13. The least-squares line is approximately $\widehat{y}=5.267+3.971x$, with $r=0.981$. At x=4, the residual is 1.848. $$b=r\frac{s_y}{s_x}=3.971,a=\overline{y}-b\overline{x}=5.267,e=y-\widehat{y}=1.848.$$ Interpretation: Within the observed range, each one-unit increase in x is associated with a predicted increase of about 3.971 y-units. Validity: Inspect form, unusual points, and residual pattern before relying on a linear model or extrapolating. Error to reject: Correlation and slope describe association; neither proves that changing x causes y to change.

Worked solution P16-Toughest-14. The mean is 16.29, the median is 16.00, and the sample standard deviation is 4.15. $$\overline{x}=\frac{\sum x_i}{7}=16.29,s=\sqrt{\frac{\sum (x_i-\overline{x}{)}^2}{6}}=4.15.$$ Interpretation: The mean is the balance point; s describes a typical distance from the mean in the variable's units. Validity: Use resistant summaries such as median and IQR when skewness or outliers make mean and standard deviation misleading. Error to reject: Do not divide the sample variance by n when the requested quantity is the conventional sample standard deviation s.

Worked solution P16-Toughest-15. The row-A conditional proportion is 30/(30+13)=0.698. Compare it with row B's 0.400; unequal conditional proportions indicate association in the table. $$P(\text{Yes}\mid A)=\frac{30}{43}=0.698.$$ The grand total 88 is not the conditional denominator. Interpretation: The comparison concerns distributions within row groups and does not by itself establish causation. Validity: Use the group named after the conditioning bar as the denominator and report which variable defines each row and column. Error to reject: Dividing by the grand total produces a joint relative frequency, not a conditional relative frequency.

Worked solution P16-Toughest-16. State experimental units, treatments, response, assignment mechanism, comparison, replication, controls, and the intended scope of inference. With 120 units and four treatments, equal random assignment gives $120/4=30$ units per treatment before attrition. Interpretation: Random assignment balances lurking variables in expectation and supports a causal comparison; random sampling is what supports generalization. Validity: Interference should be limited, treatment implementation consistent, and response measurement blind where feasible. Error to reject: A large observational comparison is not automatically an experiment, because sample size does not replace random assignment.

Worked solution P16-Toughest-17. $P(A\cup B)=0.41$ and $P(A\mid B)=0.645$. $$P(A\cup B)=0.30+0.31-0.20=0.41,P(A\mid B)=\frac{0.20}{0.31}=0.645.$$ Interpretation: The conditional probability restricts the reference group to outcomes where B occurs. Validity: Check that probabilities are coherent and that the denominator event has positive probability. Error to reject: Adding P(A) and P(B) without subtracting the intersection double-counts outcomes in both events.