Worked solution P09-Easy-1. The mean is 29.29, the median is 29.00, and the sample standard deviation is 4.15. $$\overline{x}=\frac{\sum x_i}{7}=29.29,s=\sqrt{\frac{\sum (x_i-\overline{x}{)}^2}{6}}=4.15.$$ 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 P09-Easy-2. The row-A conditional proportion is 29/(29+13)=0.690. Compare it with row B's 0.471; unequal conditional proportions indicate association in the table. $$P(\text{Yes}\mid A)=\frac{29}{42}=0.690.$$ The grand total 93 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 P09-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 P09-Easy-4. $P(A\cup B)=0.50$ 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 P09-Easy-5. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{79.0-63}{8}=2.00,P(X\le 79.0)=\Phi (2.00)\approx 0.9772.$$ 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 P09-Easy-6. The exact probability is 0.00330; the model mean is 8.840. $$P(X=3)=\left(\genfrac{}{}{0ex}{}{17}{3}\right)(0.52{)}^3(0.48{)}^{14}=0.00330.$$ Interpretation: Across many repetitions of 17 trials, the average number of successes approaches $np=8.840$. 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 P09-Easy-7. The sampling distribution has mean 67.000, standard error 1.2136, and the probability is approximately 0.1587 under a justified normal approximation. $${\mu }_{\overline{X}}=67,{\sigma }_{\overline{X}}=\frac{9}{\sqrt{55}}=1.2136,z=1.$$ 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 P09-Easy-8. The one-proportion z interval is (0.4170, 0.5909). $$\widehat{p}\pm 1.96\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}=0.5039\pm 0.0870.$$ 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=64, failures=63. Error to reject: Do not say there is a 95 percent probability this fixed population proportion lies in the computed interval.

Worked solution P09-Easy-9. The t interval is (68.039, 74.161). $$\overline{x}\pm t^{*}\frac{s}{\sqrt{n}}=71.10\pm 3.061.$$ 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 P09-Easy-10. The statistic is z=1.901; the two-sided p-value is 0.0574. $$z=\frac{0.5821-0.50}{\sqrt{0.50(0.50)/134}}=1.901.$$ 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 P09-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 P09-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 P09-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 P09-Easy-14. 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 P09-Tough-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 P09-Tough-2. The row-A conditional proportion is 24/(24+18)=0.571. Compare it with row B's 0.429; unequal conditional proportions indicate association in the table. $$P(\text{Yes}\mid A)=\frac{24}{42}=0.571.$$ 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 P09-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 P09-Tough-4. $P(A\cup B)=0.49$ and $P(A\mid B)=0.513$. $$P(A\cup B)=0.30+0.39-0.20=0.49,P(A\mid B)=\frac{0.20}{0.39}=0.513.$$ 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 P09-Tough-5. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{86.0-68}{9}=2.00,P(X\le 86.0)=\Phi (2.00)\approx 0.9772.$$ 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 P09-Tough-6. The exact probability is 0.15261; the model mean is 4.680. $$P(X=3)=\left(\genfrac{}{}{0ex}{}{12}{3}\right)(0.39{)}^3(0.61{)}^9=0.15261.$$ Interpretation: Across many repetitions of 12 trials, the average number of successes approaches $np=4.680$. 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 P09-Tough-7. The sampling distribution has mean 70.000, standard error 1.0787, and the probability is approximately 0.1587 under a justified normal approximation. $${\mu }_{\overline{X}}=70,{\sigma }_{\overline{X}}=\frac{8}{\sqrt{55}}=1.0787,z=1.$$ 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 P09-Tough-8. The one-proportion z interval is (0.4969, 0.6684). $$\widehat{p}\pm 1.96\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}=0.5827\pm 0.0858.$$ 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=74, failures=53. Error to reject: Do not say there is a 95 percent probability this fixed population proportion lies in the computed interval.

Worked solution P09-Tough-9. The t interval is (67.847, 74.953). $$\overline{x}\pm t^{*}\frac{s}{\sqrt{n}}=71.40\pm 3.553.$$ 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 P09-Tough-10. The statistic is z=1.901; the two-sided p-value is 0.0574. $$z=\frac{0.5821-0.50}{\sqrt{0.50(0.50)/134}}=1.901.$$ 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 P09-Tough-11. The statistic is t=2.133 with df=34; p=0.0403. $$t=\frac{53.10-50}{8.60/\sqrt{35}}=2.133.$$ 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 P09-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 P09-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 P09-Tough-14. 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 P09-Toughest-1. The mean is 15.29, the median is 15.00, and the sample standard deviation is 4.15. $$\overline{x}=\frac{\sum x_i}{7}=15.29,s=\sqrt{\frac{\sum (x_i-\overline{x}{)}^2}{6}}=4.15.$$ 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 P09-Toughest-2. The row-A conditional proportion is 27/(27+12)=0.692. Compare it with row B's 0.422; unequal conditional proportions indicate association in the table. $$P(\text{Yes}\mid A)=\frac{27}{39}=0.692.$$ The grand total 84 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 P09-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 P09-Toughest-4. $P(A\cup B)=0.48$ and $P(A\mid B)=0.600$. $$P(A\cup B)=0.38+0.25-0.15=0.48,P(A\mid B)=\frac{0.15}{0.25}=0.600.$$ 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 P09-Toughest-5. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{93.0-79}{7}=2.00,P(X\le 93.0)=\Phi (2.00)\approx 0.9772.$$ 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 P09-Toughest-6. The exact probability is 0.18651; the model mean is 4.300. $$P(X=3)=\left(\genfrac{}{}{0ex}{}{10}{3}\right)(0.43{)}^3(0.57{)}^7=0.18651.$$ Interpretation: Across many repetitions of 10 trials, the average number of successes approaches $np=4.300$. 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 P09-Toughest-7. The sampling distribution has mean 77.000, standard error 1.4832, and the probability is approximately 0.1587 under a justified normal approximation. $${\mu }_{\overline{X}}=77,{\sigma }_{\overline{X}}=\frac{11}{\sqrt{55}}=1.4832,z=1.$$ 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 P09-Toughest-8. The one-proportion z interval is (0.5132, 0.6837). $$\widehat{p}\pm 1.96\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}=0.5984\pm 0.0853.$$ 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=76, failures=51. Error to reject: Do not say there is a 95 percent probability this fixed population proportion lies in the computed interval.

Worked solution P09-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 P09-Toughest-10. The statistic is z=1.901; the two-sided p-value is 0.0574. $$z=\frac{0.5821-0.50}{\sqrt{0.50(0.50)/134}}=1.901.$$ 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 P09-Toughest-11. The statistic is t=2.042 with df=34; p=0.0489. $$t=\frac{52.90-50}{8.40/\sqrt{35}}=2.042.$$ 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 P09-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 P09-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 P09-Toughest-14. The mean is 27.29, the median is 27.00, and the sample standard deviation is 4.15. $$\overline{x}=\frac{\sum x_i}{7}=27.29,s=\sqrt{\frac{\sum (x_i-\overline{x}{)}^2}{6}}=4.15.$$ 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.