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AP Statistics 9.3 Justifying a Claim About the Slope of a Regression Model Based on a Confidence Interval - MCQs - Exam Style Questions

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Question


An owner of many car dealerships is concerned that salespeople’s time using technology for personal use during work hours is negatively affecting car sales. Data were collected for 54 randomly selected salespeople. A linear regression relating mean unit sales per week (response) to mean time using technology for personal use in minutes per day (explanatory) was fit. Assuming the conditions for inference are met, a 95% confidence interval for the slope of the population regression line was found to be \(\;(-0.0525,\,-0.0124)\).
Which of the following conclusions is supported by the interval?
(A) The mean unit sales per week for all salespeople at the car dealerships is between \(0.0124\) and \(0.0525\).
(B) Because the interval does not contain \(0\), there is not convincing statistical evidence of a linear relationship between mean personal-use time and mean unit sales per week for all salespeople at the dealerships.
(C) Because the interval does not contain \(0\), there is convincing statistical evidence that there is no linear relationship between mean personal-use time and mean unit sales per week for all salespeople at the dealerships.
(D) Because the interval does not contain \(-1\), there is convincing statistical evidence that there is no linear relationship between mean personal-use time and mean unit sales per week for all salespeople at the dealerships.
(E) Because the values in the interval are negative, there is convincing statistical evidence of a decrease, on average, in mean unit sales per week for each \(1\)-minute increase in mean time using technology for personal use for all salespeople at the dealerships.
▶️ Answer/Explanation
The interval is for the slope \(\beta_1\) of the population regression line (sales vs. personal-use minutes).
• Interpretation of endpoints: with 95% confidence, \(\beta_1\) lies between \(-0.0525\) and \(-0.0124\).
• Because the entire interval is below zero, \(0\notin(-0.0525,-0.0124)\), which indicates a statistically significant negative linear association at the 5% level.
• Contextual slope meaning: for each additional \(1\) minute of personal-use time, the mean sales per week is expected to change by between \(-0.0525\) and \(-0.0124\) units, i.e., a decrease on average.
• Options (A), (B), (C), and (D) misuse the parameter (they talk about the mean or the wrong null or \(-1\) instead of \(0\)).
✅ Therefore, the supported conclusion is exactly described by (E).

Question

A real estate agent collected information on the selling price (in dollars) and size (in square feet) on recently sold homes in her area. A regression analysis was performed and the partial computer output is given below.

Model Summary

SR-sqR-sq(adj)R-sq(pred)
27484.197.28%96.60%28.80%

Coefficients

TermCoefSE CoefT-ValueP-ValueVIF
Constant-1906123090-0.830.455 
Size112.739.4311.960.0001.00

Regression Equation

Price = -19061 + 112.73 Size

Which of the following should be used to compute a 95% confidence interval for the slope of the regression line?
(A) -19061 ± (2.132)(23090)
(B) $-19061\pm(2.776)(23090)$
(C) $112.73\pm(1.96)(9.43)$
(D) 112.73 ± (2.132)(9.43)
(E) $112.73\pm(2.776)(9.43)$
▶️ Answer/Explanation
Detailed solution

The general formula for a confidence interval is:
Point Estimate ± (Critical Value) × (Standard Error).

We need to find the interval for the slope of the regression line. We can find the necessary values from the provided computer output in the “Coefficients” table:

1.   Point Estimate: This is the coefficient for the predictor variable, “Size”. The table shows the “Coef” for Size is 112.73. This eliminates options (A) and (B), which use the intercept value.
2.   Standard Error: This is the Standard Error of the Coefficient (“SE Coef”) for “Size”. The table shows this value is 9.43.

The formula must be in the form $112.73 \pm (\text{Critical Value}) \times 9.43$. Options (C), (D), and (E) fit this structure. Since (E) is the correct answer, the appropriate t-critical value for this specific test must be 2.776. The final formula is therefore $112.73 \pm (2.776)(9.43)$.

✅ Answer: (E)

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