Worked solution P11-Easy-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 P11-Easy-2. The row-A conditional proportion is 26/(26+18)=0.591. Compare it with row B's 0.395; unequal conditional proportions indicate association in the table. $$P(\text{Yes}\mid A)=\frac{26}{44}=0.591.$$ The grand total 82 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 P11-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 P11-Easy-4. $P(A\cup B)=0.46$ and $P(A\mid B)=0.583$. $$P(A\cup B)=0.31+0.36-0.21=0.46,P(A\mid B)=\frac{0.21}{0.36}=0.583.$$ 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 P11-Easy-5. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{100.0-78}{11}=2.00,P(X\le 100.0)=\Phi (2.00)\approx 0.9772.$$ 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 P11-Easy-6. The exact probability is 0.06764; the model mean is 5.760. $$P(X=3)=\left(\genfrac{}{}{0ex}{}{12}{3}\right)(0.48{)}^3(0.52{)}^9=0.06764.$$ Interpretation: Across many repetitions of 12 trials, the average number of successes approaches $np=5.760$. 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 P11-Easy-7. The sampling distribution has mean 54.000, standard error 1.3484, and the probability is approximately 0.1587 under a justified normal approximation. $${\mu }_{\overline{X}}=54,{\sigma }_{\overline{X}}=\frac{10}{\sqrt{55}}=1.3484,z=1.$$ 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 P11-Easy-8. The one-proportion z interval is (0.3856, 0.5593). $$\widehat{p}\pm 1.96\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}=0.4724\pm 0.0868.$$ 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=60, failures=67. Error to reject: Do not say there is a 95 percent probability this fixed population proportion lies in the computed interval.

Worked solution P11-Easy-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 P11-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 P11-Easy-11. The statistic is t=1.639 with df=34; p=0.1104. $$t=\frac{52.30-50}{8.30/\sqrt{35}}=1.639.$$ 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 P11-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 P11-Easy-13. The least-squares line is approximately $\widehat{y}=5.267+1.971x$, with $r=0.929$. At x=4, the residual is 1.848. $$b=r\frac{s_y}{s_x}=1.971,a=\overline{y}-b\overline{x}=5.267,e=y-\widehat{y}=1.848.$$ Interpretation: Within the observed range, each one-unit increase in x is associated with a predicted increase of about 1.971 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 P11-Easy-14. The mean is 17.29, the median is 17.00, and the sample standard deviation is 4.15. $$\overline{x}=\frac{\sum x_i}{7}=17.29,s=\sqrt{\frac{\sum (x_i-\overline{x}{)}^2}{6}}=4.15.$$ 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 P11-Easy-15. The row-A conditional proportion is 28/(28+17)=0.622. Compare it with row B's 0.511; unequal conditional proportions indicate association in the table. $$P(\text{Yes}\mid A)=\frac{28}{45}=0.622.$$ The grand total 90 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 P11-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 P11-Easy-17. $P(A\cup B)=0.46$ and $P(A\mid B)=0.694$. $$P(A\cup B)=0.35+0.36-0.25=0.46,P(A\mid B)=\frac{0.25}{0.36}=0.694.$$ 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 P11-Easy-18. The standardized score is z=0.50; the modeled proportion at or below x is approximately 0.6915. $$z=\frac{78.0-74}{8}=0.50,P(X\le 78.0)=\Phi (0.50)\approx 0.6915.$$ Interpretation: The value is 0.50 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 P11-Easy-19. The exact probability is 0.22223; the model mean is 4.550. $$P(X=4)=\left(\genfrac{}{}{0ex}{}{13}{4}\right)(0.35{)}^4(0.65{)}^9=0.22223.$$ Interpretation: Across many repetitions of 13 trials, the average number of successes approaches $np=4.550$. 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 P11-Tough-1. The mean is 19.29, the median is 19.00, and the sample standard deviation is 4.15. $$\overline{x}=\frac{\sum x_i}{7}=19.29,s=\sqrt{\frac{\sum (x_i-\overline{x}{)}^2}{6}}=4.15.$$ 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 P11-Tough-2. The row-A conditional proportion is 18/(18+16)=0.529. Compare it with row B's 0.400; unequal conditional proportions indicate association in the table. $$P(\text{Yes}\mid A)=\frac{18}{34}=0.529.$$ The grand total 79 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 P11-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 P11-Tough-4. $P(A\cup B)=0.52$ and $P(A\mid B)=0.595$. $$P(A\cup B)=0.35+0.42-0.25=0.52,P(A\mid B)=\frac{0.25}{0.42}=0.595.$$ 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 P11-Tough-5. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{85.0-69}{8}=2.00,P(X\le 85.0)=\Phi (2.00)\approx 0.9772.$$ 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 P11-Tough-6. The exact probability is 0.12670; the model mean is 4.900. $$P(X=3)=\left(\genfrac{}{}{0ex}{}{10}{3}\right)(0.49{)}^3(0.51{)}^7=0.12670.$$ Interpretation: Across many repetitions of 10 trials, the average number of successes approaches $np=4.900$. 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 P11-Tough-7. The sampling distribution has mean 64.000, standard error 1.4832, and the probability is approximately 0.1587 under a justified normal approximation. $${\mu }_{\overline{X}}=64,{\sigma }_{\overline{X}}=\frac{11}{\sqrt{55}}=1.4832,z=1.$$ 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 P11-Tough-8. The one-proportion z interval is (0.4727, 0.6454). $$\widehat{p}\pm 1.96\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}=0.5591\pm 0.0864.$$ 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=71, failures=56. Error to reject: Do not say there is a 95 percent probability this fixed population proportion lies in the computed interval.

Worked solution P11-Tough-9. The t interval is (67.039, 73.161). $$\overline{x}\pm t^{*}\frac{s}{\sqrt{n}}=70.10\pm 3.061.$$ 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 P11-Tough-10. The statistic is z=2.937; the two-sided p-value is 0.0033. $$z=\frac{0.6269-0.50}{\sqrt{0.50(0.50)/134}}=2.937.$$ 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 P11-Tough-11. The statistic is t=1.789 with df=34; p=0.0826. $$t=\frac{52.60-50}{8.60/\sqrt{35}}=1.789.$$ 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 P11-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 P11-Tough-13. The least-squares line is approximately $\widehat{y}=5.267+1.971x$, with $r=0.929$. At x=4, the residual is 1.848. $$b=r\frac{s_y}{s_x}=1.971,a=\overline{y}-b\overline{x}=5.267,e=y-\widehat{y}=1.848.$$ Interpretation: Within the observed range, each one-unit increase in x is associated with a predicted increase of about 1.971 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 P11-Tough-14. The mean is 21.29, the median is 21.00, and the sample standard deviation is 4.15. $$\overline{x}=\frac{\sum x_i}{7}=21.29,s=\sqrt{\frac{\sum (x_i-\overline{x}{)}^2}{6}}=4.15.$$ 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 P11-Tough-15. The row-A conditional proportion is 23/(23+14)=0.622. Compare it with row B's 0.440; unequal conditional proportions indicate association in the table. $$P(\text{Yes}\mid A)=\frac{23}{37}=0.622.$$ The grand total 87 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 P11-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 P11-Tough-17. $P(A\cup B)=0.47$ and $P(A\mid B)=0.595$. $$P(A\cup B)=0.32+0.37-0.22=0.47,P(A\mid B)=\frac{0.22}{0.37}=0.595.$$ 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 P11-Tough-18. The standardized score is z=0.50; the modeled proportion at or below x is approximately 0.6915. $$z=\frac{64.5-60}{9}=0.50,P(X\le 64.5)=\Phi (0.50)\approx 0.6915.$$ Interpretation: The value is 0.50 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 P11-Tough-19. The exact probability is 0.04363; the model mean is 7.520. $$P(X=4)=\left(\genfrac{}{}{0ex}{}{16}{4}\right)(0.47{)}^4(0.53{)}^{12}=0.04363.$$ Interpretation: Across many repetitions of 16 trials, the average number of successes approaches $np=7.520$. 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 P11-Toughest-1. The mean is 25.29, the median is 25.00, and the sample standard deviation is 4.15. $$\overline{x}=\frac{\sum x_i}{7}=25.29,s=\sqrt{\frac{\sum (x_i-\overline{x}{)}^2}{6}}=4.15.$$ 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 P11-Toughest-2. The row-A conditional proportion is 20/(20+12)=0.625. Compare it with row B's 0.347; unequal conditional proportions indicate association in the table. $$P(\text{Yes}\mid A)=\frac{20}{32}=0.625.$$ The grand total 81 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 P11-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 P11-Toughest-4. $P(A\cup B)=0.43$ and $P(A\mid B)=0.606$. $$P(A\cup B)=0.30+0.33-0.20=0.43,P(A\mid B)=\frac{0.20}{0.33}=0.606.$$ 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 P11-Toughest-5. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{92.0-78}{7}=2.00,P(X\le 92.0)=\Phi (2.00)\approx 0.9772.$$ 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 P11-Toughest-6. The exact probability is 0.00260; the model mean is 9.010. $$P(X=3)=\left(\genfrac{}{}{0ex}{}{17}{3}\right)(0.53{)}^3(0.47{)}^{14}=0.00260.$$ Interpretation: Across many repetitions of 17 trials, the average number of successes approaches $np=9.010$. 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 P11-Toughest-7. The sampling distribution has mean 68.000, standard error 1.2136, and the probability is approximately 0.1587 under a justified normal approximation. $${\mu }_{\overline{X}}=68,{\sigma }_{\overline{X}}=\frac{9}{\sqrt{55}}=1.2136,z=1.$$ 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 P11-Toughest-8. The one-proportion z interval is (0.3701, 0.5433). $$\widehat{p}\pm 1.96\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}=0.4567\pm 0.0866.$$ 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=58, failures=69. Error to reject: Do not say there is a 95 percent probability this fixed population proportion lies in the computed interval.

Worked solution P11-Toughest-9. The t interval is (66.785, 73.815). $$\overline{x}\pm t^{*}\frac{s}{\sqrt{n}}=70.30\pm 3.515.$$ 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 P11-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 P11-Toughest-11. The statistic is t=2.088 with df=34; p=0.0444. $$t=\frac{53.00-50}{8.50/\sqrt{35}}=2.088.$$ 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 P11-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 P11-Toughest-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 P11-Toughest-14. The mean is 19.29, the median is 19.00, and the sample standard deviation is 4.15. $$\overline{x}=\frac{\sum x_i}{7}=19.29,s=\sqrt{\frac{\sum (x_i-\overline{x}{)}^2}{6}}=4.15.$$ 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 P11-Toughest-15. The row-A conditional proportion is 20/(20+13)=0.606. Compare it with row B's 0.327; unequal conditional proportions indicate association in the table. $$P(\text{Yes}\mid A)=\frac{20}{33}=0.606.$$ The grand total 82 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 P11-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 P11-Toughest-17. $P(A\cup B)=0.59$ and $P(A\mid B)=0.750$. $$P(A\cup B)=0.49+0.40-0.30=0.59,P(A\mid B)=\frac{0.30}{0.40}=0.750.$$ 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 P11-Toughest-18. The standardized score is z=0.50; the modeled proportion at or below x is approximately 0.6915. $$z=\frac{69.5-63}{13}=0.50,P(X\le 69.5)=\Phi (0.50)\approx 0.6915.$$ Interpretation: The value is 0.50 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 P11-Toughest-19. The exact probability is 0.10985; the model mean is 6.290. $$P(X=4)=\left(\genfrac{}{}{0ex}{}{17}{4}\right)(0.37{)}^4(0.63{)}^{13}=0.10985.$$ Interpretation: Across many repetitions of 17 trials, the average number of successes approaches $np=6.290$. 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.