The summary statistics for a certain set of points are: $n = 17 , s _ { \mathrm { e } } = 2.880 , \sum ( x - \bar { x } ) ^ { 2 } = 19.241$, and $b _ { 1 } = 1.839$. Assume the conditions of the linear model hold. A $95 \%$ confidence interval for $\beta _ { 1 }$ will be constructed. What is the margin of error?
A) 41.002571
B) 1.391921
C) 1.146365
D) 1.399143

Question

Quizplus · Accepted Answer

The margin of error is $t^* s_e \sqrt{1/n + x^2/(n\sum(x-\bar{x})^2)}$. Plugging in the given values, we get $t^* s_e \sqrt{1/n + x^2/(n\sum(x-\bar{x})^2)} = 2.131884 \times 2.880 \times \sqrt{1/17 + 1.839^2/(17\times19.241)} = 1.399143$.

The Summary Statistics for a Certain Set of Points Are n=17,se=2.880,∑(x−xˉ)2=19.241n = 17 , s _ { \mathrm { e } } = 2.880 , \sum ( x - \bar { x } ) ^ { 2 } = 19.241n=17,se​=2.880,∑(x−xˉ)2=19.241

The Summary Statistics for a Certain Set of Points Are $n = 17 , s _ { \mathrm { e } } = 2.880 , \sum ( x - \bar { x } ) ^ { 2 } = 19.241$