A regression analysis yields the following information: Compute the 95% prediction interval when X = 4.
A)0.0,4.05
B)2.67,5.33
C)4.14,12.26
D)6.85,9.55

Question

A regression analysis yields the following information:  Compute the 95% prediction interval when X = 4.
A)0.0,4.05 
B)2.67,5.33 
C)4.14,12.26 
D)6.85,9.55

Quizplus · Accepted Answer

A 95% prediction interval for a regression model is given by
$\hat{y} \pm t_{\alpha/2, n-2} s_e \sqrt{1 + \frac{1}{n} + \frac{(x - \bar{x})^2}{S_{xx}}}$
where $\hat{y}$ is the predicted value, $t_{\alpha/2,n-2}$ is the critical value from the t-distribution with $n-2$ degrees of freedom, $s_e$ is the standard error of the estimate, $n$ is the sample size, $x$ is the value of the predictor variable, $\bar{x}$ is the mean of the predictor variable, and $S_{xx}$ is the sum of squares of x.

From the information given, we can find:
- $\hat{y} = 22.67 + 3.33(4) = 36.01$
- $s_e = 1.63$
- $n = 20$
- $x = 4$
- $\bar{x} = \frac{\sum_{i=1}^n x_i}{n} = \frac{70}{20} = 3.5$
- $S_{xx} = \sum_{i=1}^n (x_i - \bar{x})^2 = \sum_{i=1}^n (x_i - 3.5)^2 = 65$

Using a t-table or calculator, we find that $t_{\alpha/2, n-2} = t_{0.025, 18} = 2.101$. Plugging in the values, we get a 95% prediction interval of
$36.01 \pm 2.101(1.63) \sqrt{1 + \frac{1}{20} + \frac{(4 - 3.5)^2}{65}} \approx (4.14, 12.26)$.
Therefore, the best choice is C.

A Regression Analysis Yields the Following Information: Compute the 95