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) $2722.34 \pm 2.00(259.92)$
() $2722.34 \pm 1.67\left(\frac{259.92}{\sqrt{64}} ight)$
(ㄷ) $2722.34 \pm 2.00\left(\frac{259.92}{\sqrt{64}} ight)$

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) $2722.34 \pm 2.00(259.92)$
() $2722.34 \pm 1.67\left(\frac{259.92}{\sqrt{64}}ight)$
(ㄷ) $2722.34 \pm 2.00\left(\frac{259.92}{\sqrt{64}}ight)$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the 95% confidence interval for the slope of the least squares regression line, we need to use the t-distribution critical value for 95% confidence and 63 degrees of freedom. ‖ ‖ step 2 ⋮ The t-distribution critical value for 95% confidence and 63 degrees of freedom is approximately 2.00 (from asksia-ll calculation list). ‖ ‖ step 3 ⋮ The standard error of the slope (SE Coef) is given as 259.92. Since the sample size is 64, we divide the SE Coef by the square root of the sample size to get the standard error of the estimate of the slope. ‖ ‖ step 4 ⋮ Calculating the standard error of the estimate: $SE_{estimate} = \frac{259.92}{\sqrt{64}} = \frac{259.92}{8} = 32.49$. ‖ ‖ step 5 ⋮ Multiplying the t-distribution critical value by the standard error of the estimate gives us the margin of error for the 95% confidence interval. ‖ ‖ step 6 ⋮ Calculating the margin of error: $ME = 2.00 imes 32.49 = 64.98$. ‖ ‖ step 7 ⋮ The 95% confidence interval for the slope is then $2722.34 \pm ME$. ‖ ‖ step 8 ⋮ Substituting the margin of error into the confidence interval formula: $2722.34 \pm 64.98$. ‖ ***ㄷ*** ∻Key Concept∻ ⚹Confidence Interval for Slope⚹ ∻Explanation∻ ⚹The 95% confidence interval for the slope of the least squares regression line is calculated using the t-distribution critical value, the standard error of the slope, and the sample size.⚹◊What is the formula to calculate the 95% confidence interval for the slope of the least squares regression line?⍭ Generate me a similar question◊

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