▶️ Answer/Explanation
(a)
The scatterplot reveals a strong, positive, roughly linear association between the mass and length of bullfrogs. There are no points that seriously deviate from the straight-line pattern exhibited by most of the points in the plot.
(b)
The slope of the least-squares regression line is \( 6.086 \). This indicates that for each additional millimeter of length, the predicted mass of a bullfrog increases by 6.086 grams.
(c)
The coefficient of determination \( r^2 \approx 0.819 \) indicates that 81.9% of the variation in bullfrog mass can be explained by the linear relationship with bullfrog length as described by the least-squares regression line.
(d)
(i) The bullfrog with the largest absolute value residual has a length of approximately 162 mm and a mass of approximately 356 g.
(ii) The least-squares regression line overestimates the mass of this bullfrog. This is evident because the point for this bullfrog is below the regression line in Plot 2, meaning the actual mass is less than the predicted mass.
▶️ Answer/Explanation
(a)
- Treatments: The new drug and the placebo.
- Experimental units: The \(72\) people participating in the experiment.
- Response variable: The improvement in acne severity, measured on a scale from \(0\) to \(100\).
(b)
The severity of acne differs from one twin pair to another. This initial acne severity is a potential confounding variable. In a completely randomized design, this variability between pairs could obscure the true effect of the treatment.
A statistical advantage of using a matched-pairs design is that it controls for the variability in initial acne severity between the pairs of twins. By pairing individuals who are similar (in this case, identical twins), the design isolates the effect of the treatment from the effect of initial severity. This reduction in variability provides a more precise estimate of the treatment effect (the difference in improvement between the drug and the placebo) and increases the statistical power of the experiment, making it easier to detect a significant difference between the drug and the placebo if one truly exists.
(c)
The random assignment should be performed separately for each of the \(36\) pairs of twins. One valid method is as follows:
For each pair of twins, flip a fair coin.
- If the coin lands on heads, one twin (e.g., the older twin) receives the new drug and the other twin receives the placebo.
- If the coin lands on tails, the assignment is reversed: the same twin receives the placebo and the other receives the new drug.
This process is repeated for all \(36\) pairs, ensuring that within each pair, each twin has a \(50\%\) chance of receiving either treatment, and the assignment for one pair is independent of all other pairs.
▶️ Answer/Explanation
(a)
The amount of shampoo, \( A \), is normally distributed with \( \mu = 0.60 \) and \( \sigma = 0.04 \).
Calculate the z-score:
\( z = \frac{0.50 – 0.60}{0.04} = -2.5 \)
Find the probability:
\( P(A < 0.50) = P(Z < -2.5) \approx 0.0062 \)
So, \(\boxed{P \approx 0.0062}\).
(b)
(i) Let \( X \) be the number of underfilled bottles in a box of \( 10 \). Then \( X \sim \mathrm{Binomial}(n = 10, p = 0.0062) \).
(ii) A crate is rejected if \( X \ge 2 \). Using the complement rule:
\( P(X \ge 2) = 1 – P(X \le 1) = 1 – [P(X=0) + P(X=1)] \)
\( P(X=0) = \binom{10}{0}(0.0062)^0(0.9938)^{10} \approx 0.9396 \)
\( P(X=1) = \binom{10}{1}(0.0062)^1(0.9938)^9 \approx 0.0584 \)
\( P(X \ge 2) = 1 – (0.9396 + 0.0584) = 0.002 \) (more precisely \( 0.0017 \))
So, \(\boxed{P \approx 0.0017}\).
(c)
For the original programming:
\( P(A < 0.5) = P\left(Z < \frac{0.5 – 0.60}{0.04}\right) = P(Z < -2.5) \approx 0.0062 \).
For the adjusted programming:
\( P(A < 0.5) = P\left(Z < \frac{0.5 – 0.56}{0.03}\right) = P(Z < -2.0) \approx 0.0228 \).
Because the probability of an underfilled bottle is greater for the adjusted programming (\( 0.0228 \) vs \( 0.0062 \)), this would result in more rejected shipments. ✅ Recommendation: The company should continue with the original machine programming because it yields a lower probability of underfilling and therefore fewer rejected shipments.
▶️ Answer/Explanation
(a)
State: We will construct a one-sample z-interval for a population proportion.
Let \( p \) = the true proportion of all teenagers in the United States who would respond that they use a video streaming service every day.
Plan: We verify the conditions:
• Random: The sample was randomly selected.
• 10% Condition: \( 920 \) is less than 10% of all teenagers in the United States.
• Large Counts: \( n\hat{p} = 920 \times 0.59 = 542.8 \geq 10 \) and \( n(1-\hat{p}) = 920 \times 0.41 = 377.2 \geq 10 \)
Do: The \( 95\% \) confidence interval is:
\( \hat{p} \pm z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = 0.59 \pm 1.96\sqrt{\frac{0.59 \times 0.41}{920}} \)
\( = 0.59 \pm 1.96\sqrt{0.000263} \approx 0.59 \pm 1.96 \times 0.0162 \approx 0.59 \pm 0.0318 \)
The interval is \( (0.5582, 0.6218) \)
Conclude: We are \( 95\% \) confident that the true proportion of all teenagers in the United States who would respond that they use a video streaming service every day is between \( 0.558 \) and \( 0.622 \).
(b)
Yes, the sample data provide convincing statistical evidence that the proportion is not \( 0.5 \).
Justification: The \( 95\% \) confidence interval \( (0.558, 0.622) \) does not contain \( 0.5 \). Since \( 0.5 \) is not a plausible value for the population proportion based on this interval, we have convincing evidence that the true proportion is different from \( 0.5 \).
▶️ Answer/Explanation
(a)
To find the medians, we locate the middle value for each group.
Dark Chocolate Group: There are \(13\) participants. The median is the value of the \(\frac{13+1}{2} = 7^{\text{th}}\) observation. Counting from the left on the dotplot, the \(7^{\text{th}}\) value is \(7\) mmHg.
White Chocolate Group: There are \(25 – 13 = 12\) participants. The median is the average of the \(\frac{12}{2} = 6^{\text{th}}\) and \(7^{\text{th}}\) observations. Counting from the left, the \(6^{\text{th}}\) value is \(-1\) mmHg and the \(7^{\text{th}}\) value is \(1\) mmHg. The median is \(\frac{-1 + 1}{2} = 0\) mmHg.
Comparison: The median reduction in blood pressure for the dark chocolate group (\(7\) mmHg) is greater than the median reduction for the white chocolate group (\(0\) mmHg).
(b)
The researcher’s conclusion is not necessarily true because the observed difference in sample means (\(5.66\) mmHg) could be due to sampling variability. The random assignment of participants into two groups can result in a difference between the groups by chance alone, even if there is no true difference in the effects of the two types of chocolate. A single sample difference is not sufficient evidence. A formal statistical inference procedure is needed to determine the probability of observing a difference this large or larger purely by chance, assuming no real effect exists.
(c)
The observed difference in the mean reduction in blood pressure between the two groups is \(\bar{x}_{\text{dark}} – \bar{x}_{\text{white}} = 6.08 – 0.42 = 5.66\) mmHg.
The simulation was conducted under the assumption of no difference. We need to find the probability of observing a difference of \(5.66\) mmHg or greater in this simulation. Looking at the dotplot of simulation results, we count the number of trials where the simulated difference in means was \(5.66\) or more. There is one dot at \(6\), one at \(7\), and one at \(8\). Thus, \(3\) out of the \(120\) trials resulted in a difference of \(5.66\) mmHg or greater.
The estimated p-value is \(P(\text{difference} \geq 5.66) = \frac{3}{120} = 0.025\).
Since the p-value of \(0.025\) is less than the significance level of \(\alpha = 0.05\), we reject the null hypothesis. There is convincing statistical evidence that adding dark chocolate to a daily diet will result in a greater mean reduction in blood pressure than adding white chocolate to a daily diet for people similar to those in this study.
▶️ Answer/Explanation
(a)
(i) Relative frequency table:
| Clinic A | Clinic B | |
|---|---|---|
| Unsuccessful treatment | \( \frac{51}{139} \approx 0.367 \) | \( \frac{33}{68} \approx 0.485 \) |
| Successful treatment | \( \frac{88}{139} \approx 0.633 \) | \( \frac{35}{68} \approx 0.515 \) |
(ii) Clinic A is more successful. Clinic A has a higher success rate (\( 63.3\% \)) compared to Clinic B (\( 51.5\% \)).
(b)
No, a statistically significant result would not allow researchers to conclude causation. This is an observational study (not a randomized experiment) where patients were not randomly assigned to clinics. There may be confounding variables, such as allergy severity, that affect both the clinic a patient attends and the success of treatment.
(c)
(i)
• Clinic A: More successful in treating mild allergies (\( \frac{78}{104} = 0.75 \)) than severe (\( \frac{10}{35} \approx 0.286 \))
• Clinic B: More successful in treating mild allergies (\( \frac{11}{12} \approx 0.917 \)) than severe (\( \frac{24}{56} \approx 0.429 \))
(ii)
• Clinic A: More likely to treat mild allergy sufferers (\( \frac{104}{139} \approx 0.748 \)) than severe (\( \frac{35}{139} \approx 0.252 \))
• Clinic B: More likely to treat severe allergy sufferers (\( \frac{56}{68} \approx 0.824 \)) than mild (\( \frac{12}{68} \approx 0.176 \))
(d)
The overall success rate favored Clinic A in part (a-ii), which is different from the physician’s conclusion. This is an example of Simpson’s Paradox. Clinic A treats mostly mild cases (\( 74.8\% \)) which have higher success rates, while Clinic B treats mostly severe cases (\( 82.4\% \)) which have lower success rates. When we look within each severity level, Clinic B actually has higher success rates for both mild (\( 91.7\% \) vs \( 75.0\% \)) and severe (\( 42.9\% \) vs \( 28.6\% \)) allergies. The overall success rates are misleading because the clinics treat different types of patients.