For a sample of size 16, the following values were obtained: $b _ { 0 } = 3.76 , b _ { 1 } = 4.21$ $s _ { \mathrm { e } } = 3.72 , \sum ( x - \bar { x } ) ^ { 2 } = 199.02 , \text { and } \bar { x } = 3.54$
Construct a 99% confidence interval for the mean response when x=1 .
A) (4.56,11.38)
B) (0,10.98)
C) (4.40,11.54)
D) (4.96,10.98)

Question

For a sample of size 16, the following values were obtained: $b _ { 0 } = 3.76 , b _ { 1 } = 4.21$ $s _ { \mathrm { e } } = 3.72 , \sum ( x - \bar { x } ) ^ { 2 } = 199.02 , \text { and } \bar { x } = 3.54$
Construct a  99%  confidence interval for the mean response when  x=1 .
A)  (4.56,11.38) 
B)  (0,10.98) 
C)  (4.40,11.54) 
D)  (4.96,10.98)

Quizplus · Accepted Answer

The mean response when $x = 1$ is given by $\hat{y} = b_0 + b_1x = 3.76 + 4.21(1) = 7.97$. To construct a 99% confidence interval for the mean response, we use the formula $\hat{y} \pm t_{\alpha/2} \cdot s_{\hat{y}}$, where $s_{\hat{y}} = s_e \sqrt{\frac{1}{n} + \frac{(x - \bar{x})^2}{\sum (x - \bar{x})^2}}$. Given $s_e = 3.72$, $n = 16$, $\bar{x} = 3.54$, and $\sum (x - \bar{x})^2 = 199.02$, for $x = 1$, $s_{\hat{y}} = 3.72 \sqrt{\frac{1}{16} + \frac{(1 - 3.54)^2}{199.02}}$. The t-value for a 99% confidence interval with $df = n - 2 = 14$ is approximately 2.977 (from t-distribution table). Plugging in the values and calculating gives the interval (4.56, 11.38).

For a Sample of Size 16, the Following Values Were b0=3.76,b1=4.21b _ { 0 } = 3.76 , b _ { 1 } = 4.21b0​=3.76,b1​=4.21

For a Sample of Size 16, the Following Values Were $b _ { 0 } = 3.76 , b _ { 1 } = 4.21$