AP Statistics 2.4 Representing the Relationship Between Two Quantitative Variables- FRQs - Exam Style Questions
▶️ Answer/Explanation
(a)
There is a strong, positive, and roughly linear relationship between the chest circumference and weight of male tule elk. There are no obvious outliers or influential points that deviate from the linear pattern.
(b)
(i) Predicted weight \( = -350.3 + 3.7455(145.9) \approx -350.3 + 546.47 \approx 196.17\) kg.
\(\boxed{\text{Predicted weight} \approx 196.17 \text{ kg}}\)
(ii) Residual = Actual – Predicted
Residual \( = 204.3 – 196.17 = 8.13\) kg.
\(\boxed{\text{Residual} \approx 8.13 \text{ kg}}\)
(c)
For each additional centimeter of chest circumference, the predicted weight of a male tule elk increases by approximately \(3.7455\) kilograms.
(d)
(i) We need to find the p-value for a t-test statistic of \(3.408\) with degrees of freedom \(df = n-2 = 30-2 = 28\). Since the alternative hypothesis is two-sided (\(H_a: \beta \ne 4.5\)), the p-value is \(2 \times P(t_{28} > 3.408)\).
Using a t-table or calculator, this probability is approximately \(0.002\).
\(\boxed{\text{p-value} \approx 0.002}\)
(ii) Because the p-value (\(\approx 0.002\)) is less than the significance level (\(\alpha=0.05\)), the wildlife biologist should reject the null hypothesis. There is convincing statistical evidence that the slope of the population regression line for male tule elk is different from \(4.5\) kg/cm.