A random sample of 79 companies from the Forbes 500 list (which actually consists of nearly 800 companies)was selected,and the relationship between sales (in hundreds of thousands of dollars)and profits (in hundreds of thousands of dollars)was investigated by regression.The following results were obtained from statistical software. Suppose the researchers conducting this study wish to estimate the profits (in hundreds of thousands of dollars)for companies that had sales (in hundreds of thousands of dollars)of 500.The following results were obtained from statistical software. If the researchers wish to estimate the profits for a particular company that had sales of 500,what would be a 95% prediction interval for the profits?
A)(-1066.4,805.6)
B)(-248.5,-12.3)
C)-130.4 ± 59.3
D)500 ± 59.3

Question

A random sample of 79 companies from the Forbes 500 list (which actually consists of nearly 800 companies)was selected,and the relationship between sales (in hundreds of thousands of dollars)and profits (in hundreds of thousands of dollars)was investigated by regression.The following results were obtained from statistical software.   Suppose the researchers conducting this study wish to estimate the profits (in hundreds of thousands of dollars)for companies that had sales (in hundreds of thousands of dollars)of 500.The following results were obtained from statistical software.  If the researchers wish to estimate the profits for a particular company that had sales of 500,what would be a 95% prediction interval for the profits?
A)(-1066.4,805.6) 
B)(-248.5,-12.3) 
C)-130.4 ± 59.3 
D)500 ± 59.3

Quizplus · Accepted Answer

To calculate the prediction interval, we use the formula:

y = b0 + b1x ± t*SE(y)

where
y = predicted value of Y for given X
b0 and b1 = intercept and slope from regression results
t = t-value from t-distribution based on desired degree of confidence
SE(y) = standard error of the estimate, calculated as sqrt[MSE(1 + 1/n + (x - xbar)^2/SSx)]

Plugging in the values:
x = 500 (sales)
b0 = -1066.4
b1 = 2.157
MSE = 47076.17
n = 79
xbar = 402.302
SSx = 201302.098
SE(y) = sqrt[47076.17(1 + 1/79 + (500 - 402.302)^2/201302.098)] = 59.3

For a 95% prediction interval, t-value with 77 degrees of freedom (n-2) is 1.990.
Thus, y = -1066.4 + 2.157(500) ± 1.990(59.3) = (-1066.4, 805.6)

Therefore, the best choice is (A) with the prediction interval (-1066.4, 805.6).

A Random Sample of 79 Companies from the Forbes 500