▶️ Answer/Explanation
(a)
On campus proportion: \(\frac{17+7}{33} = \frac{24}{33} \approx 0.727\)
Off campus proportion: \(\frac{25+12}{67} = \frac{37}{67} \approx 0.552\)
(b)
The graph reveals an association between residential status and participation. On-campus students are more likely to participate in extracurricular activities than off-campus students. A larger proportion of off-campus students (\(\approx 45\%\)) participate in no activities compared to on-campus students (\(\approx 27\%\)). Consequently, a larger proportion of on-campus students participate in one activity (\(\approx 52\%\)) compared to off-campus students (\(\approx 37\%\)). The proportions for two or more activities are similar.
(c)
We compare the p-value to a standard significance level, such as \(\alpha = 0.05\). Since the p-value (\(0.23\)) is greater than \(\alpha\) (\(0.05\)), the administrator should **fail to reject the null hypothesis**. There is not convincing statistical evidence to conclude that an association exists between residential status and level of participation in extracurricular activities for all students at the university.
▶️ Answer/Explanation
(a)
\(P(\text{3 women}) = \frac{3}{9} \times \frac{2}{8} \times \frac{1}{7} = \frac{6}{504} = \frac{1}{84} \approx 0.0119\)
Answer: \(\boxed{0.012}\)
(b)
Yes, there is reason to doubt the manager’s claim. The probability of selecting \(3\) women by random chance is only about \(1.2\%\), which is very small. This suggests it’s unlikely this outcome occurred purely by chance.
(c)
No, the proposed process does not correctly simulate the random selection. The dice simulation assumes:
• Independent selections (dice rolls are independent)
• Constant probability of selecting a woman (\( \frac{2}{6} = \frac{1}{3} \))
However, the actual selection is:
• Dependent selections (sampling without replacement)
• Changing probabilities (after each woman is selected, probability decreases)
The dice simulation represents sampling with replacement, not the actual without-replacement scenario.
▶️ Answer/Explanation
(a)
\(z = \frac{140 – 120}{10.5} \approx 1.90\)
\(P(Z > 1.90) = 1 – 0.9713 = 0.0287\)
The probability is approximately \(0.029\).
Answer: \(\boxed{0.029}\)
(b)
High School A would be less likely to lose funding using the suggested plan.
• With \(3\) days: \(\sigma_{\bar{x}} = \frac{10.5}{\sqrt{3}} \approx 6.062\)
• \(z = \frac{140 – 120}{6.062} \approx 3.30\)
• \(P(Z > 3.30) \approx 0.0005\)
The probability decreases from \(0.0287\) to \(0.0005\) because the sampling distribution of the mean has less variability.
(c)
Probability of selecting Monday or Friday in one week: \(\frac{2}{5} = 0.4\)
Since weeks are independent: \((0.4)^3 = 0.064\)
The probability that none of the \(3\) days is Tuesday, Wednesday, or Thursday is \(0.064\).
Answer: \(\boxed{0.064}\)
▶️ Answer/Explanation
(a)
The median is resistant to skewness and outliers, while the mean is not. Income distributions are typically right-skewed, with a small number of very high incomes. These high incomes would pull the mean upward but have little effect on the median. Therefore, the median would provide a better estimate of a typical income than the mean.
\(\boxed{\text{Median is resistant to skewness and outliers}}\)
(b)
Method 2 is better. Method 1 uses voluntary response, which is likely to produce a biased sample. People with higher incomes may be more likely to respond, leading to an overestimate of the mean income. Method 2 uses a random sample with follow-up to ensure responses, which should produce a more representative sample and a better estimate of the population mean income, despite the smaller sample size.
\(\boxed{\text{Method 2}}\)
▶️ Answer/Explanation
State:
We will perform a paired t-test at significance level \(\alpha = 0.05\).
Let \(\mu_{\text{diff}}\) be the population mean difference in purchase price (woman – man).
\(H_0: \mu_{\text{diff}} = 0\)
\(H_a: \mu_{\text{diff}} > 0\)
Plan:
We check the conditions for inference:
1. Random: Random selection of car models and buyers stated.
2. Normality: Sample size (\(n = 8\)) is small, but the dotplot of differences shows no strong skewness or outliers.
Do:
Test statistic: \(t = \frac{\bar{x}_{\text{diff}} – 0}{s_{\text{diff}}/\sqrt{n}} = \frac{585 – 0}{530.71/\sqrt{8}} \approx 3.12\)
Degrees of freedom: \(df = 7\)
p-value: \(P(t > 3.12) \approx 0.008\)
Conclude:
Since p-value (\(0.008\)) < \(\alpha\) (\(0.05\)), we reject \(H_0\).
There is convincing statistical evidence that, on average, women pay more than men in the county for the same car model.
▶️ Answer/Explanation
(a)
Predicted FCR = \(-1.595789 + 0.0372614 \times 175 = 4.925\)
Residual = actual – predicted = \(5.88 – 4.925 = 0.955\)
Interpretation: The car’s actual FCR is \(0.955\) gallons per \(100\) miles greater than predicted for a car of its length.
\(\boxed{0.955}\)
(b i)
The point on graph III with wheel base \(93\) inches and residual approximately \(0.96\) should be circled.
(b ii)
Point B being close to the horizontal line at \(0\) indicates that the car’s actual FCR is very close to the FCR predicted by the regression model using length alone.
(c)
Graph II shows a moderate positive linear association between engine size and residuals from the FCR-length regression. Graph III shows a weak association with no clear pattern between wheel base and these residuals. The association is stronger in graph II than in graph III.
(d)
Jamal should use engine size. The stronger association in graph II indicates that engine size explains more of the variation in FCR that remains after accounting for length, which would improve the prediction model.
\(\boxed{\text{Engine size}}\)