Mixed inference practice: choosing the procedure

(a) Read the structure first. There is one sample, the response is categorical (bus or not bus), and the office wants an estimate with a margin of error rather than a decision about a claimed value. One sample, a categorical response, and an estimation goal give a one-proportion z-interval.

(b) Conditions. Random: a simple random sample was taken. 10%: $250 \le 0.10 \times 9400 = 940$, so sampling without replacement is fine. Large counts: 90 successes and $250 - 90 = 160$ failures, both at least 10.

(c) Point estimate. $\hat{p} = \dfrac{90}{250} = 0.36$.

(c) Standard error. $SE = \sqrt{\dfrac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\dfrac{0.36 \times 0.64}{250}} = \sqrt{\dfrac{0.2304}{250}} = \sqrt{0.0009216} = 0.03036$.

(c) Margin of error. For 95% confidence $z^* = 1.96$, so $ME = 1.96 \times 0.03036 = 0.0595$.

(c) Interval and interpretation. $0.36 \pm 0.0595$ gives $(0.3005, 0.4195)$. You are 95% confident that the interval from 0.300 to 0.420 captures the true proportion of undergraduates at this campus who commute by bus.

(a) Structure. One sample, a quantitative response (weight in grams), and a claimed value to test against, with the population standard deviation $\sigma$ unknown. One sample plus a quantitative response plus unknown $\sigma$ gives a one-sample t-test.

(b) Hypotheses. $H_0$: $\mu = 340$ grams against $H_a$: $\mu \ne 340$ grams, where $\mu$ is the mean fill weight of all bags in that week's production. The alternative is two-sided because underfilling and overfilling are both departures from the advertised claim.

(b) Conditions. Random sample: stated. 10%: 25 bags is well under 10% of a week's production. Sample data: $n = 25$ is below 30, but the dotplot is roughly symmetric with no outliers, so the t procedure is safe.

(c) Standard error. $SE = \dfrac{s}{\sqrt{n}} = \dfrac{8.0}{\sqrt{25}} = \dfrac{8.0}{5} = 1.6$ grams.

(c) Test statistic. $t = \dfrac{\bar{x} - \mu_0}{SE} = \dfrac{336.4 - 340}{1.6} = \dfrac{-3.6}{1.6} = -2.25$, with $df = 25 - 1 = 24$.

(c) p-value. In the $df = 24$ row, $|t| = 2.25$ falls between 2.064 (upper tail 0.025) and 2.492 (upper tail 0.01), so one tail holds between 0.01 and 0.025 and the two-sided p-value is between 0.02 and 0.05. Technology gives 0.034.

(d) Conclusion. The p-value of about 0.034 is less than $\alpha = 0.05$, so reject $H_0$. There is convincing evidence that the mean fill weight for that week differs from the advertised 340 grams, and the sample mean points to underfilling.

(a) Structure. Two independent groups created by random assignment, and the response is categorical (purchased or did not). Two groups plus a categorical response gives a two-proportion z-test.

(b) Hypotheses. $H_0$: $p_1 = p_2$ against $H_a$: $p_1 \ne p_2$, where $p_1$ is the purchase rate for the redesigned page and $p_2$ is the purchase rate for the current page among visitors like these.

(c) Conditions. Random assignment is stated. Large counts: 145 purchases and $500 - 145 = 355$ non-purchases in the first group, 115 and $500 - 115 = 385$ in the second, all at least 10. The 10% condition is not needed for a randomized experiment.

(c) Sample proportions and the pooled proportion. $\hat{p}_1 = \dfrac{145}{500} = 0.29$ and $\hat{p}_2 = \dfrac{115}{500} = 0.23$. Under $H_0$ the two rates are equal, so pool them: $\hat{p}_c = \dfrac{145 + 115}{500 + 500} = \dfrac{260}{1000} = 0.26$.

(c) Standard error. $SE = \sqrt{\hat{p}_c(1 - \hat{p}_c)\left(\dfrac{1}{n_1} + \dfrac{1}{n_2}\right)} = \sqrt{0.26 \times 0.74 \times \left(\dfrac{1}{500} + \dfrac{1}{500}\right)} = \sqrt{0.1924 \times 0.004} = \sqrt{0.0007696} = 0.027742$.

(c) Test statistic. $z = \dfrac{\hat{p}_1 - \hat{p}_2}{SE} = \dfrac{0.29 - 0.23}{0.027742} = \dfrac{0.06}{0.027742} = 2.163$.

(d) Conclusion. The two-sided p-value is $2 \times P(Z > 2.163) = 2 \times 0.0153 = 0.0306$. Since $0.0306 < 0.05$, reject $H_0$: there is convincing evidence that the purchase rates differ between the two pages. Because visitors were randomly assigned, the design supports a causal conclusion about the page, though only for visitors to this site.

(a) Structure. Two groups, a quantitative response (height), and no natural partner linking a Schedule A seedling to a Schedule B seedling; they are different plants assigned at random. Independent groups plus a quantitative response gives a two-sample t-test. Pairing would require each measurement in one group to be tied to one specific measurement in the other.

(a) Hypotheses. $H_0$: $\mu_1 = \mu_2$ against $H_a$: $\mu_1 \ne \mu_2$, where the two $\mu$ values are the mean six-week heights under the two schedules.

(b) Standard error. Add the variances, never the standard deviations: $SE = \sqrt{\dfrac{s_1^2}{n_1} + \dfrac{s_2^2}{n_2}} = \sqrt{\dfrac{3.2^2}{10} + \dfrac{2.8^2}{10}} = \sqrt{\dfrac{10.24}{10} + \dfrac{7.84}{10}} = \sqrt{1.024 + 0.784} = \sqrt{1.808} = 1.3446$.

(b) Test statistic. $t = \dfrac{\bar{x}_1 - \bar{x}_2}{SE} = \dfrac{24.6 - 21.0}{1.3446} = \dfrac{3.6}{1.3446} = 2.677$.

(c) Conservative degrees of freedom. $df = \min(n_1, n_2) - 1 = 10 - 1 = 9$. In the $df = 9$ row, 2.677 falls between 2.262 (upper tail 0.025) and 2.821 (upper tail 0.01), so the two-sided p-value is between 0.02 and 0.05.

(c) Technology degrees of freedom. A calculator reports $df = 17.69$ and a two-sided p-value of about 0.016. The conservative bound is looser because fewer degrees of freedom put more area in the tails.

(d) Conclusion. Either way the p-value is below 0.05, so reject $H_0$. There is convincing evidence that mean six-week height differs between the two feeding schedules, and because seedlings were randomly assigned, the schedule is a plausible cause.

(a) Structure. Each row is one patient measured twice, so the two columns are not independent; a strong patient reads high in both. Two dependent measurements on the same subjects collapse into one sample of differences, analyzed with a one-sample t procedure, which is the paired t-test.

(b) Differences, after minus before: $31 - 28 = 3$, $33 - 31 = 2$, $29 - 25 = 4$, $36 - 34 = 2$, $30 - 29 = 1$, $31 - 27 = 4$, $34 - 33 = 1$, and $34 - 30 = 4$.

(b) Mean difference. The differences sum to $3 + 2 + 4 + 2 + 1 + 4 + 1 + 4 = 21$, so $\bar{x}_d = \dfrac{21}{8} = 2.625$ kg.

(b) Standard deviation of the differences. Deviations from 2.625 are 0.375, -0.625, 1.375, -0.625, -1.625, 1.375, -1.625, and 1.375. Their squares sum to $0.140625 + 0.390625 + 1.890625 + 0.390625 + 2.640625 + 1.890625 + 2.640625 + 1.890625 = 11.875$, so $s_d = \sqrt{\dfrac{11.875}{7}} = \sqrt{1.6964} = 1.3025$ kg.

(c) Conditions. The patients were randomly selected. With only 8 differences, check them directly: 3, 2, 4, 2, 1, 4, 1, 4 show no strong skewness and no outliers, so the t procedure applies.

(c) Standard error and test statistic. $SE = \dfrac{s_d}{\sqrt{n}} = \dfrac{1.3025}{\sqrt{8}} = \dfrac{1.3025}{2.8284} = 0.4605$, so $t = \dfrac{2.625 - 0}{0.4605} = 5.700$ with $df = 8 - 1 = 7$.

(c) Conclusion. In the $df = 7$ row, 5.700 is past 5.408 (upper tail 0.0005), so the one-sided p-value is under 0.0005; technology gives 0.00037. Reject $H_0$: there is convincing evidence that mean grip strength increased over the four weeks.

(c) What the design does not support. There is no control group, so the increase cannot be credited to the program itself. Ordinary recovery over four weeks would produce the same pattern in these numbers.

(a) Structure. One random sample measured on two categorical variables, which calls for a chi-square test for independence. A two-proportion z-test compares exactly two proportions, and payment method has three categories, so there is no single difference to test. Chi-square handles any number of rows and columns at once.

(a) Hypotheses. $H_0$: payment method and loyalty use are independent among this store's shoppers. $H_a$: payment method and loyalty use are associated.

(b) Expected counts, $\dfrac{\text{row total} \times \text{column total}}{200}$. Card: $\dfrac{100 \times 120}{200} = 60$ and $\dfrac{100 \times 80}{200} = 40$. Cash: $\dfrac{40 \times 120}{200} = 24$ and $\dfrac{40 \times 80}{200} = 16$. Mobile wallet: $\dfrac{60 \times 120}{200} = 36$ and $\dfrac{60 \times 80}{200} = 24$. All six exceed 5.

(b) Terms. Card: $\dfrac{(66 - 60)^2}{60} = \dfrac{36}{60} = 0.6$ and $\dfrac{(34 - 40)^2}{40} = \dfrac{36}{40} = 0.9$. Cash: $\dfrac{(18 - 24)^2}{24} = \dfrac{36}{24} = 1.5$ and $\dfrac{(22 - 16)^2}{16} = \dfrac{36}{16} = 2.25$. Mobile wallet: both observed counts equal their expected counts, so both terms are 0.

(b) Statistic and degrees of freedom. $\chi^2 = 0.6 + 0.9 + 1.5 + 2.25 + 0 + 0 = 5.25$, and $df = (3 - 1)(2 - 1) = 2$.

(c) Conclusion. The critical value for $df = 2$ at $\alpha = 0.05$ is 5.991, and $5.25 < 5.991$, so fail to reject $H_0$; the p-value is about 0.072. There is not convincing evidence that payment method and loyalty use are associated at this store.

(c) How close the call is. A p-value of 0.072 misses the 0.05 line by very little, and most of the statistic (3.75 of 5.25) comes from the cash row, where fewer shoppers used loyalty than independence predicts, and the rest (1.5) from the card row, where more did. That pattern argues for collecting more data rather than declaring the two variables unrelated.

(a) Structure. One sample, a categorical response (fails or passes), and a claimed value of 0.08 to test against, so this is a one-proportion z-test.

(a) Which proportion goes in the standard error. In a test the null value $p_0$ goes in the standard error, $\sqrt{\dfrac{p_0(1 - p_0)}{n}}$, because the whole calculation is carried out assuming $H_0$ is true. Only a confidence interval uses $\hat{p}$ there, which is why an interval and a test on the same data have slightly different standard errors.

(b) Hypotheses. $H_0$: $p = 0.08$ against $H_a$: $p > 0.08$, where $p$ is the true failure rate in the shipment. The alternative is one-sided because the lab is checking only whether the failure rate exceeds the claim.

(b) Conditions. Random sample: stated. 10%: $300 \le 0.10 \times 12000 = 1200$. Large counts under $H_0$: $np_0 = 300 \times 0.08 = 24$ and $n(1 - p_0) = 300 \times 0.92 = 276$, both at least 10.

(c) Sample proportion and standard error. $\hat{p} = \dfrac{33}{300} = 0.11$, and $SE = \sqrt{\dfrac{0.08 \times 0.92}{300}} = \sqrt{\dfrac{0.0736}{300}} = \sqrt{0.00024533} = 0.015663$.

(c) Test statistic and p-value. $z = \dfrac{\hat{p} - p_0}{SE} = \dfrac{0.11 - 0.08}{0.015663} = \dfrac{0.03}{0.015663} = 1.915$. The one-sided p-value is $P(Z > 1.915) = 0.0277$.

(d) Conclusion. Since $0.0277 < 0.05$, reject $H_0$. There is convincing evidence that more than 8% of units in this shipment fail the calibration check.

(a) Why the two-sample test fails. Equal sample sizes and two columns look like two independent groups, but each row is the same rider on the same course, so the two columns move together. A two-sample t procedure requires the two samples to be independent, and pairing breaks that. The correct analysis subtracts within each row and runs a one-sample t on the 8 differences.

(b) Differences, Tire A minus Tire B: $42.1 - 41.3 = 0.8$, $38.6 - 38.0 = 0.6$, $45.0 - 44.6 = 0.4$, $40.2 - 39.4 = 0.8$, $43.5 - 43.1 = 0.4$, $39.8 - 39.6 = 0.2$, $41.7 - 40.5 = 1.2$, and $44.3 - 43.9 = 0.4$.

(b) Mean difference. They sum to $0.8 + 0.6 + 0.4 + 0.8 + 0.4 + 0.2 + 1.2 + 0.4 = 4.8$, so $\bar{x}_d = \dfrac{4.8}{8} = 0.6$ minutes.

(b) Standard deviation of the differences. Deviations from 0.6 are 0.2, 0, -0.2, 0.2, -0.2, -0.4, 0.6, and -0.2, and their squares sum to $0.04 + 0 + 0.04 + 0.04 + 0.04 + 0.16 + 0.36 + 0.04 = 0.72$. So $s_d = \sqrt{\dfrac{0.72}{7}} = \sqrt{0.102857} = 0.3207$ minutes.

(b) Hypotheses and conditions. $H_0$: $\mu_d = 0$ against $H_a$: $\mu_d \ne 0$, where $\mu_d$ is the mean of (Tire A time) minus (Tire B time) for riders like these. The order of the two rides was randomized, and the 8 differences show no strong skewness and no outliers.

(b) Test statistic. $SE = \dfrac{s_d}{\sqrt{n}} = \dfrac{0.3207}{\sqrt{8}} = \dfrac{0.3207}{2.8284} = 0.1134$, so $t = \dfrac{0.6}{0.1134} = 5.291$ with $df = 7$.

(b) p-value and conclusion. In the $df = 7$ row, 5.291 falls between 4.785 (upper tail 0.001) and 5.408 (upper tail 0.0005), so the two-sided p-value is between 0.001 and 0.002; technology gives 0.0011. Reject $H_0$: there is convincing evidence that mean course time differs between the two compounds, with Tire B faster by about 0.6 minutes on average.

(c) Why the answers differ. Rider-to-rider variation is large, with times running from about 38 to 45 minutes, while the tire effect is a steady 0.2 to 1.2 minutes in the same direction for every rider. The two-sample standard error of 1.158 is built out of that rider variation and swamps the effect, while pairing subtracts each rider out and leaves a standard error of 0.1134, about ten times smaller.