Worked solution P01-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 P01-Easy-2. The row-A conditional proportion is 18/(18+12)=0.600. Compare it with row B's 0.410; unequal conditional proportions indicate association in the table. $$P(\text{Yes}\mid A)=\frac{18}{30}=0.600.$$ The grand total 69 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 P01-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 P01-Easy-4. $P(A\cup B)=0.51$ and $P(A\mid B)=0.634$. $$P(A\cup B)=0.36+0.41-0.26=0.51,P(A\mid B)=\frac{0.26}{0.41}=0.634.$$ 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 P01-Easy-5. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{83.0-71}{6}=2.00,P(X\le 83.0)=\Phi (2.00)\approx 0.9772.$$ 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 P01-Easy-6. The exact probability is 0.22411; the model mean is 3.840. $$P(X=3)=\left(\genfrac{}{}{0ex}{}{12}{3}\right)(0.32{)}^3(0.68{)}^9=0.22411.$$ Interpretation: Across many repetitions of 12 trials, the average number of successes approaches $np=3.840$. 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 P01-Easy-7. The sampling distribution has mean 75.000, standard error 1.4832, and the probability is approximately 0.1587 under a justified normal approximation. $${\mu }_{\overline{X}}=75,{\sigma }_{\overline{X}}=\frac{11}{\sqrt{55}}=1.4832,z=1.$$ 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 P01-Easy-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 P01-Easy-9. The t interval is (70.225, 76.575). $$\overline{x}\pm t^{*}\frac{s}{\sqrt{n}}=73.40\pm 3.175.$$ 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 P01-Easy-10. The statistic is z=2.592; the two-sided p-value is 0.0096. $$z=\frac{0.6119-0.50}{\sqrt{0.50(0.50)/134}}=2.592.$$ 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 P01-Easy-11. The statistic is t=1.740 with df=34; p=0.0909. $$t=\frac{52.50-50}{8.50/\sqrt{35}}=1.740.$$ 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 P01-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 P01-Easy-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 P01-Easy-14. The mean is 18.29, the median is 18.00, and the sample standard deviation is 4.15. $$\overline{x}=\frac{\sum x_i}{7}=18.29,s=\sqrt{\frac{\sum (x_i-\overline{x}{)}^2}{6}}=4.15.$$ 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 P01-Easy-15. The row-A conditional proportion is 24/(24+18)=0.571. Compare it with row B's 0.487; unequal conditional proportions indicate association in the table. $$P(\text{Yes}\mid A)=\frac{24}{42}=0.571.$$ 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 P01-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 P01-Easy-17. $P(A\cup B)=0.51$ and $P(A\mid B)=0.585$. $$P(A\cup B)=0.34+0.41-0.24=0.51,P(A\mid B)=\frac{0.24}{0.41}=0.585.$$ 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 P01-Tough-1. The mean is 18.29, the median is 18.00, and the sample standard deviation is 4.15. $$\overline{x}=\frac{\sum x_i}{7}=18.29,s=\sqrt{\frac{\sum (x_i-\overline{x}{)}^2}{6}}=4.15.$$ 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 P01-Tough-2. The row-A conditional proportion is 29/(29+17)=0.630. Compare it with row B's 0.432; unequal conditional proportions indicate association in the table. $$P(\text{Yes}\mid A)=\frac{29}{46}=0.630.$$ 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 P01-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 P01-Tough-4. $P(A\cup B)=0.46$ 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 P01-Tough-5. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{75.0-63}{6}=2.00,P(X\le 75.0)=\Phi (2.00)\approx 0.9772.$$ 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 P01-Tough-6. The exact probability is 0.16698; the model mean is 4.510. $$P(X=3)=\left(\genfrac{}{}{0ex}{}{11}{3}\right)(0.41{)}^3(0.59{)}^8=0.16698.$$ Interpretation: Across many repetitions of 11 trials, the average number of successes approaches $np=4.510$. 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 P01-Tough-7. The sampling distribution has mean 59.000, standard error 1.0787, and the probability is approximately 0.1587 under a justified normal approximation. $${\mu }_{\overline{X}}=59,{\sigma }_{\overline{X}}=\frac{8}{\sqrt{55}}=1.0787,z=1.$$ 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 P01-Tough-8. The one-proportion z interval is (0.4567, 0.6299). $$\widehat{p}\pm 1.96\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}=0.5433\pm 0.0866.$$ 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=69, failures=58. Error to reject: Do not say there is a 95 percent probability this fixed population proportion lies in the computed interval.

Worked solution P01-Tough-9. The t interval is (70.350, 76.850). $$\overline{x}\pm t^{*}\frac{s}{\sqrt{n}}=73.60\pm 3.250.$$ 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 P01-Tough-10. The statistic is z=2.073; the two-sided p-value is 0.0381. $$z=\frac{0.5896-0.50}{\sqrt{0.50(0.50)/134}}=2.073.$$ 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 P01-Tough-11. The statistic is t=3.000 with df=34; p=0.0050. $$t=\frac{53.60-50}{7.10/\sqrt{35}}=3.000.$$ 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 P01-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 P01-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 P01-Tough-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 P01-Tough-15. The row-A conditional proportion is 23/(23+14)=0.622. Compare it with row B's 0.333; unequal conditional proportions indicate association in the table. $$P(\text{Yes}\mid A)=\frac{23}{37}=0.622.$$ 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 P01-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 P01-Tough-17. $P(A\cup B)=0.43$ 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 P01-Toughest-1. The mean is 18.29, the median is 18.00, and the sample standard deviation is 4.15. $$\overline{x}=\frac{\sum x_i}{7}=18.29,s=\sqrt{\frac{\sum (x_i-\overline{x}{)}^2}{6}}=4.15.$$ 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 P01-Toughest-2. The row-A conditional proportion is 18/(18+18)=0.500. Compare it with row B's 0.409; unequal conditional proportions indicate association in the table. $$P(\text{Yes}\mid A)=\frac{18}{36}=0.500.$$ 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 P01-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 P01-Toughest-4. $P(A\cup B)=0.54$ and $P(A\mid B)=0.756$. $$P(A\cup B)=0.44+0.41-0.31=0.54,P(A\mid B)=\frac{0.31}{0.41}=0.756.$$ 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 P01-Toughest-5. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{86.0-62}{12}=2.00,P(X\le 86.0)=\Phi (2.00)\approx 0.9772.$$ 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 P01-Toughest-6. The exact probability is 0.13657; the model mean is 4.900. $$P(X=3)=\left(\genfrac{}{}{0ex}{}{14}{3}\right)(0.35{)}^3(0.65{)}^{11}=0.13657.$$ Interpretation: Across many repetitions of 14 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 P01-Toughest-7. The sampling distribution has mean 51.000, standard error 1.3484, and the probability is approximately 0.1587 under a justified normal approximation. $${\mu }_{\overline{X}}=51,{\sigma }_{\overline{X}}=\frac{10}{\sqrt{55}}=1.3484,z=1.$$ 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 P01-Toughest-8. The one-proportion z interval is (0.3623, 0.5353). $$\widehat{p}\pm 1.96\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}=0.4488\pm 0.0865.$$ 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=57, failures=70. Error to reject: Do not say there is a 95 percent probability this fixed population proportion lies in the computed interval.

Worked solution P01-Toughest-9. The t interval is (69.723, 76.677). $$\overline{x}\pm t^{*}\frac{s}{\sqrt{n}}=73.20\pm 3.477.$$ 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 P01-Toughest-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 P01-Toughest-11. The statistic is t=2.219 with df=34; p=0.0333. $$t=\frac{52.70-50}{7.20/\sqrt{35}}=2.219.$$ 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 P01-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 P01-Toughest-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 P01-Toughest-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 P01-Toughest-15. The row-A conditional proportion is 21/(21+19)=0.525. Compare it with row B's 0.385; unequal conditional proportions indicate association in the table. $$P(\text{Yes}\mid A)=\frac{21}{40}=0.525.$$ 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 P01-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 P01-Toughest-17. $P(A\cup B)=0.58$ and $P(A\mid B)=0.730$. $$P(A\cup B)=0.48+0.37-0.27=0.58,P(A\mid B)=\frac{0.27}{0.37}=0.730.$$ 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.