Checking conditions for inference practice

Random. The 240 riders are stated to be a random sample of the agency's riders, so the randomization condition is met. Say what makes it random rather than writing the word 'random' by itself.

10%. Sampling is without replacement, so the sample size $n$ must be at most 10% of the population size $N$: $n \le 0.10N$. Here $0.10(24000) = 2400$, and $240 \le 2400$, so the condition is met.

Large Counts. For a test, the counts are checked at the hypothesized value $p_0 = 0.35$, not at the sample proportion. $np_0 = 240(0.35) = 84$ and $n(1 - p_0) = 240(0.65) = 156$.

Confirm both counts clear 10. $84 \ge 10$ and $156 \ge 10$, so the condition is met. As an arithmetic check, $84 + 156 = 240$, which is $n$.

Note what plays no role here. The sample proportion $\hat{p} = \frac{96}{240} = 0.40$ is the statistic you will standardize next, but it does not enter the Large Counts check for a test.

Conclude. All three conditions are met, so a one-sample z-test for a population proportion is justified for these data.

(a) Random and 10%. The 60 members are a random sample, and $0.10(900) = 90$ with $60 \le 90$, so both conditions are met.

(a) Large Counts at $p_0$. A test has a hypothesized value, so use $p_0 = 0.20$: $np_0 = 60(0.20) = 12$ and $n(1 - p_0) = 60(0.80) = 48$. Both are at least 10, so the condition is met and the test is justified.

(b) Random and 10% again. These do not depend on the procedure, so they are still met.

(b) Normality at $\hat{p}$. An interval has no hypothesized value, so the condition uses the observed counts: successes $n\hat{p} = 8$ and failures $n(1 - \hat{p}) = 60 - 8 = 52$.

(b) Compare to 10. The 52 failures are fine, but the 8 successes fall below 10, so the condition fails and a one-sample z-interval is not justified for these data.

(c) Explain the gap. A test asks whether this sample would be surprising if $p$ were $0.20$, so its sampling distribution is centered at $0.20$ and the counts are checked there. An interval has no such anchor and must use $\hat{p} = \frac{8}{60} \approx 0.133$, and at that rate 60 members produce too few successes for the normal approximation.

(c) Take the general lesson. The same dataset can justify one procedure and rule out another, so check the counts for the procedure you are actually running.

Random. The 46 employees are a random sample of the hospital's employees, so the condition is met.

10%. Sampling is without replacement, and $0.10(3200) = 320$ with $46 \le 320$, so the condition is met.

Sample data condition. The population shape is unknown, but $n = 46 \ge 30$, which satisfies the condition on its own. You do not need a plot and you do not need to assume the commute times are normal.

Say why $n \ge 30$ is enough. By the central limit theorem, the sampling distribution of $\bar{x}$ is approximately normal for a sample this size even when the population of individual commute times is skewed.

Pick the right family of curves. The population standard deviation $\sigma$ (sigma) is unknown, so you use $s$ and a t-distribution with degrees of freedom $df = n - 1 = 46 - 1 = 45$, not a z-interval.

Find the standard error. $SE = \frac{s}{\sqrt{n}} = \frac{9.6}{\sqrt{46}} = \frac{9.6}{6.7823} \approx 1.415$ minutes.

(a) Random and 10%. The 14 customers are randomly selected, and $0.10(300) = 30$ with $14 \le 30$, so both are met.

(a) Sample data condition. Here $n = 14 < 30$, so the sample size alone does not settle it and you fall back on the shape of the data. Roughly symmetric with no outliers means free from strong skewness and outliers, so the condition is met.

(a) Conclude. All three conditions are met, so a one-sample t-interval with $df = 14 - 1 = 13$ is justified.

(b) Recheck each condition separately. Random and 10% are unchanged and still met, because changing the shape of the data does not change how the sample was chosen.

(b) Apply the sample data condition. With $n = 14 < 30$ and a plot showing strong right skew plus a clear outlier at 95 minutes, the condition fails. The one-sample t-interval is not justified, so describe the data instead of reporting an interval.

(c) Name what would fix it. Either a larger sample, since $n \ge 30$ removes the need to inspect the plot, or credible information that the population of visit times is approximately normal.

(c) Name what would not fix it. Lowering the confidence level, using a different formula for the standard error, or dropping the outlier without a documented reason. The condition is about the shape of the data relative to $n$.

Confirm the table. The 'not sorted correctly' counts are $60 - 39 = 21$, $60 - 45 = 15$, and $60 - 30 = 30$. Column totals are $39 + 45 + 30 = 114$ and $21 + 15 + 30 = 66$, and $114 + 66 = 180$.

Random. Volunteers were randomly assigned to the three labels, which meets the randomized-experiment version of the randomization condition for a test for homogeneity.

10%. This condition is not needed here, because the data come from a randomized experiment rather than from sampling without replacement out of a population.

Set up the expected counts. $E = \frac{(\text{row total})(\text{column total})}{\text{grand total}}$, computed under the null hypothesis that the three labels give the same distribution of outcomes.

Compute the 'sorted correctly' column. Every row total is 60, so each expected count is $\frac{60(114)}{180} = \frac{6840}{180} = 38$.

Compute the 'not sorted correctly' column. Each expected count is $\frac{60(66)}{180} = \frac{3960}{180} = 22$.

Check the arithmetic. Each row gives $38 + 22 = 60$, and the columns give $3(38) = 114$ and $3(22) = 66$, matching the observed totals.

Apply the rule. All six expected counts, $38, 38, 38, 22, 22, 22$, are greater than 5, so the expected counts condition is met and the $\chi^2$ test for homogeneity is justified. Note that the small observed count of 15 is irrelevant, since the condition is about expected counts.

Find the sample proportions. $\hat{p}_1 = \frac{117}{180} = 0.65$ for the paper prompt and $\hat{p}_2 = \frac{135}{180} = 0.75$ for the text prompt.

(a) Random and 10%. Drivers were randomly assigned to the two prompts, which meets the randomized-experiment version of the condition, and the 10% condition is unnecessary for a randomized experiment.

(a) Pooled proportion. A test assumes $p_1 = p_2$, so the counts are checked at the combined proportion $\hat{p}_c = \frac{117 + 135}{180 + 180} = \frac{252}{360} = 0.70$.

(a) Expected counts for the test. For each group, $180(0.70) = 126$ and $180(0.30) = 54$. All four values $126, 54, 126, 54$ are at least 10, so the two-sample z-test is justified.

(b) Observed counts for the interval. An interval estimates a difference and never assumes the proportions are equal, so it uses each group's own counts: 117 and $180 - 117 = 63$ for the paper prompt, 135 and $180 - 135 = 45$ for the text prompt.

(b) Compare to 10. All four observed counts are at least 10, so the two-sample z-interval is also justified.

(c) Explain the difference. The test pools because its null hypothesis says the two proportions share one common value, and the best estimate of that value uses both samples. The interval has no null value to borrow, so it keeps the groups separate.

(c) Note the shared discipline. Both procedures happened to pass here, but each required four specific counts written down, not the phrase 'large counts is met'.

Find the sample proportion. $\hat{p} = \frac{1080}{1500} = 0.72$.

(a) 10%. The show has several million listeners, so 1,500 is far below 10% of the population. This condition is met with room to spare.

(a) Large Counts. At $p_0 = 0.65$, $np_0 = 1500(0.65) = 975$ and $n(1 - p_0) = 1500(0.35) = 525$. Both are far above 10, so this condition is met.

(a) Random. This one fails. Respondents selected themselves by choosing to visit the page, which makes this a voluntary response sample, not a random sample of the show's listeners.

(a) Say why the failure matters. Listeners with strong feelings about ads are the ones most likely to answer, so $\hat{p} = 0.72$ estimates the skip rate among people motivated to respond, not among all listeners.

(b) Conclude. The test is not justified. Two of the three conditions pass easily but that does not rescue the procedure, because randomization is the condition that lets you treat $\hat{p}$ as an unbiased estimate of $p$ and lets the standard error formula describe the real sampling variability.

(c) Answer the sample size question. No. Bias does not shrink as $n$ grows; only the standard error does, since $SE$ has $\sqrt{n}$ in the denominator.

(c) Describe the result. With 15,000 voluntary responses you would get a narrower interval centered on the same wrong value, which is a more confident wrong answer rather than a better one.

(a) Random and 10%. The 150 attendees are a random sample, and $0.10(18000) = 1800$ with $150 \le 1800$, so both are met.

(a) Normality at $\hat{p}$. An interval uses observed counts: successes $= 27$ and failures $= 150 - 27 = 123$. Both are at least 10, so the condition is met and the one-sample z-interval is justified, with $\hat{p} = \frac{27}{150} = 0.18$.

(b) Random and 10%. A test for independence needs one random sample from a single population, which is what the organizers have, and $150 \le 1800$ again, so both are met.

(b) Set up the expected counts. $E = \frac{(\text{row total})(\text{column total})}{150}$, where the column totals are 27 for the shuttle and 123 for no shuttle, and the row totals are 30, 96, and 24.

(b) Compute the shuttle column. Under 18: $\frac{30(27)}{150} = \frac{810}{150} = 5.4$. Ages 18 to 59: $\frac{96(27)}{150} = \frac{2592}{150} = 17.28$. Ages 60 and over: $\frac{24(27)}{150} = \frac{648}{150} = 4.32$.

(b) Compute the no-shuttle column. Under 18: $\frac{30(123)}{150} = \frac{3690}{150} = 24.6$. Ages 18 to 59: $\frac{96(123)}{150} = \frac{11808}{150} = 78.72$. Ages 60 and over: $\frac{24(123)}{150} = \frac{2952}{150} = 19.68$.

(b) Check the arithmetic. The shuttle column gives $5.4 + 17.28 + 4.32 = 27$ and the other column gives $24.6 + 78.72 + 19.68 = 123$, matching the observed totals.

(b) Apply the rule. The expected count $4.32$ for '60 and over, shuttle' is not greater than 5, so the expected counts condition fails and the $\chi^2$ test for independence is not justified. One failing cell is enough to stop the procedure.

(c) Explain the split. Conditions attach to procedures, not to datasets. The interval only needs enough shuttle riders in the sample overall, and 27 clears the bar, while the chi-square test needs enough expected in every one of the six cells, and the 60-and-over group has only 24 people crossed with a column that is just 18% of the sample.

(c) Say what would fix it. A larger overall sample, or a design that surveys more attendees aged 60 and over, would raise that expected count above 5. Choosing a larger $\alpha$ (alpha, the significance level) would not, and neither would pointing at the observed counts, since the rule is about expected counts.