Answers to Odd Practice Items

Below are brief final answers to the odd-numbered “Check your understanding” items at the end of each weekly unit, so you can self-check your work. Only the final answer is given — a number, a word, or a one-line conclusion — not the full worked steps. If your answer does not match, redo the matching worked example in that week before moving on. (Weeks 7 and 12 are review-and-exam weeks; their practice lives in the worked exam-style examples inside those units, so they have no items here.)

1Week 1 — Data & Study Design¶

1. width is numerical (continuous); domhand is categorical (nominal).

2. Population = all 3,000 simulated students (parameter: mean $\mu \approx 6.75$ hours); sample = the 40 randomly drawn students (statistic: mean $\bar{x} \approx 6.43$ hours).

3. A simple random sample (SRS).

2Week 2 — Summarizing Numerical Data¶

1. Mean $= 7.125$ hours, median $= 7.0$ hours; the mean is larger, by 0.125 hours.

2. Five-number summary: min $= 5.5$, $Q_1 = 6.375$, median $= 7$, $Q_3 = 7.625$, max $= 9.5$ hours; $IQR = 1.25$ hours.

3. Lower fence $= 4.5$ hours, upper fence $= 9.5$ hours; a 9th value of 13 hours would be flagged as an outlier.

3Week 3 — Summarizing Categorical Data & Tables¶

1. $27.9\%$ of the 240 intakes were Large.

2. About 0.627 ($62.7\%$) of the Medium animals were adopted — a conditional proportion.

3. Sample space $S = \{\text{Drip}, \text{Latte}, \text{Cold Brew}\}$; $P(\text{Drip}\mid\text{Morning}) = 0.450$ ($45.0\%$); the complement is “not Drip (Latte or Cold Brew),” with probability 0.550 ($55.0\%$).

4Week 4 — Probability Foundations¶

1. $P(\text{not Large}) \approx 0.721$ ($72.1\%$), by the complement rule.

2. $P(\text{diamond or ace}) \approx 0.308$ ($30.8\%$); the events are not disjoint (the ace of diamonds is both).

3. $E(X) = \$0.20$.

5Week 5 — Random Variables & Density Curves¶

1. $\mu = E(X) = -\$1.00$ (a $\$1$ average loss per play in the long run).

2. Total expected net winnings across all 200 tickets $= -\$600$ (the club’s expected $\$600$ fundraiser profit).

3. A continuous variable has infinitely many possible values with no finite list to tabulate, and any single exact value has probability 0; a density curve instead gives probability as area under the curve over an interval.

6Week 6 — The Normal Model & z-scores¶

1. The middle 68% of cats weigh between 8.5 and 11.5 lb; about $15.87\%$ weigh less than 8.5 lb.

2. $z_A = 1.5$ and $z_B = 1.8$; the Group-B commuter’s time is more unusual (larger z-score), because each z-score measures distance relative to that group’s own mean and spread.

3. About $2.28\%$ of students sleep less than 5 hours.

7Week 8 — The Normal Distribution: Areas & Cutoffs¶

1. $z \approx -1.94$; $P(X < 6.5) \approx 0.0259$ (about $2.6\%$ of adult cats are below the cutoff).

2. $z = 1.00$; the right tail (above 28) is shaded; $P(X \ge 28) \approx 0.1587$ (about $15.87\%$).

3. A score of about 86.8 or higher places a student in the top 5%.

8Week 9 — Sampling Distributions & Standard Error¶

1. $p = 0.55$ is the parameter (population); $\hat{p} = 0.52$ is the point estimate (sample statistic).

2. $SE_{\hat{p}}(n=50) \approx 0.0704$ and $SE_{\hat{p}}(n=200) \approx 0.0352$; $n$ increased by a factor of 4 and the SE decreased by a factor of 2 ($\sqrt{4} = 2$).

3. The Central Limit Theorem (CLT).

9Week 10 — Confidence Intervals¶

1. $\hat{p} = 0.54$; margin of error $\approx 6.18$ percentage points; 95% CI $\approx (47.8\%, 60.2\%)$.

2. 95% CI $\approx (\$5.32, \$6.88)$.

3. The claim is wrong: “95% confident” describes the long-run success rate of the method, not the probability that one fixed interval contains the true proportion.

10Week 11 — Hypothesis Testing Logic¶

1. $H_0: p = 0.90$ versus $H_A: p < 0.90$; a one-sided (left-tailed) test.

2. The classmate wrongly reads the p-value as $P(H_0 \text{ true})$; correctly, $p \approx 0.0033$ is the chance of a sample this extreme if $H_0$ were true.

3. At $\alpha = 0.05$, reject $H_0$; at $\alpha = 0.01$, fail to reject — the decision depends on the significance level chosen before seeing the data.

11Week 13 — Inference for a Proportion¶

1. 90% confidence interval $= (0.726, 0.874)$, i.e. $(72.6\%, 87.4\%)$.

2. 95% confidence interval $= (0.667, 0.844)$, i.e. $(66.7\%, 84.4\%)$.

3. $z = -1.833$, p-value $\approx 0.0334 < 0.05$, so reject $H_0$: significant evidence that fewer than 60% of subscribers watch primarily on a mobile device.

12Week 14 — Inference for a Mean¶

1. Yes — both conditions (independence; a roughly symmetric small sample with no outliers) hold, so a one-sample t-procedure is appropriate.

2. 90% confidence interval $\approx (7.50, 8.90)$ hours.

3. $t = 2.087$, $df = 19$, p-value $= 0.0253 < 0.05$, so reject $H_0$: significant evidence that subscribers watch more than 90 minutes per day on average.

13Week 15 — Comprehensive Review¶

1. An observational study; a confounding variable (the cat’s own temperament) could explain both the informal meet-and-greet and the faster adoption.

2. Joint.

3. Negative; $z = (9.5 - 11)/1.5 \approx -1$, about 1 standard deviation below the mean.

4. The true rate is the parameter $p$ (unknown); 0.580 is the sample statistic $\hat{p}$.