Janine is studying the relationship between the size of a diamond (in carats) and its price. She obtains size and price data for a random sample of 64 diamonds. Here is computer output from a least-squares regression analysis on her sample:
\begin{tabular}{lrrrr}
Predictor & Coef & SE Coef & T & P \
\hline Constant & 8.00 & 60.76 & 0.13 & 0.90 \
Carat & 2722.34 & 259.92 & 10.47 & 0.00 \
S $=132.30$ & R-sq $=63.9 \%$ & & &
\end{tabular}

Assume that all conditions for inference have been met.
Which of these is a $95 \%$ confidence interval for the slope of the least squares regression line?

Choose 1 answer:
(A) $2722.34 \pm 1.67(259.92)$
(B) $2722.34 \pm 1.96(259.92)$
(c) $2722.34 \pm 2.00(259.92)$
() $2722.34 \pm 1.67\left(\frac{259.92}{\sqrt{64}} ight)$
() $2722.34 \pm 2.00\left(\frac{259.92}{\sqrt{64}} ight)$

Question

Janine is studying the relationship between the size of a diamond (in carats) and its price. She obtains size and price data for a random sample of 64 diamonds. Here is computer output from a least-squares regression analysis on her sample:
\begin{tabular}{lrrrr} 
Predictor & Coef & SE Coef & T & P \
\hline Constant & 8.00 & 60.76 & 0.13 & 0.90 \
Carat & 2722.34 & 259.92 & 10.47 & 0.00 \
S $=132.30$ & R-sq $=63.9 \%$ & & &
\end{tabular}

Assume that all conditions for inference have been met.
Which of these is a $95 \%$ confidence interval for the slope of the least squares regression line?

Choose 1 answer:
(A) $2722.34 \pm 1.67(259.92)$
(B) $2722.34 \pm 1.96(259.92)$
(c) $2722.34 \pm 2.00(259.92)$
() $2722.34 \pm 1.67\left(\frac{259.92}{\sqrt{64}}ight)$
() $2722.34 \pm 2.00\left(\frac{259.92}{\sqrt{64}}ight)$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To construct a 95% confidence interval for the slope of the least squares regression line, we need to use the standard error of the slope and the appropriate t* value for the given confidence level and degrees of freedom. ‖ ‖ step 2 ⋮ The degrees of freedom (df) for the t-distribution in a regression analysis is n - 2, where n is the number of observations. Here, n = 64, so df = 64 - 2 = 62. ‖ ‖ step 3 ⋮ For a 95% confidence interval and 62 degrees of freedom, the t* value is approximately 2.00 (this can be found in t-distribution tables or using statistical software). ‖ ‖ step 4 ⋮ The standard error of the slope (SE Coef for Carat) is given as 259.92. To find the standard error of the estimate for the slope, we divide this by the square root of the number of observations, which is √64 = 8. ‖ ‖ step 5 ⋮ The 95% confidence interval for the slope is calculated as: slope estimate ± t* × (SE of slope estimate). Plugging in the values, we get: 2722.34 ± 2.00 × (259.92 / 8). ‖ ‖ step 6 ⋮ Simplifying the expression in step 5, we calculate the margin of error: 2.00 × (259.92 / 8) = 2.00 × 32.49 = 64.98. ‖ ‖ step 7 ⋮ The 95% confidence interval is therefore: 2722.34 ± 64.98. ‖ ∻1 Answer∻ ***E*** ∻Key Concept∻ ⚹Confidence Interval for Slope in Regression⚹ ∻Explanation∻ ⚹The confidence interval for the slope of a regression line is constructed using the estimated slope, the standard error of the slope, and the t* value corresponding to the desired confidence level and degrees of freedom.⚹◊What formula is used to calculate the confidence interval for the slope of the least squares regression line?⍭ Generate me a similar question◊

Sia