A regression analysis yields the following information: ; n = 10; s_y_∙_x = 1.66; ΣX = 32; ΣX² = 134; Compute the 95% prediction interval when X = 4. A) 0.0, 4.05 B) 2.67, 5.33 C) 4.118, 12.226 D) 6.873, 9.527

Question

Quizplus · Accepted Answer

First, we need to calculate the regression equation:
Slope (b) = sS1U1B1yS1U1B0S1U1B1 / sS1U1B0S1U1B1xS1U1B0 = 1.66 / (10*17.6) = 0.00943
Intercept (a) = sS1U1B1y - bsS1U1B1x = (1/10)*134 - (0.00943*32) = 0.8868
The regression equation is therefore y = 0.8868 + 0.00943x

Next, we need to calculate the standard error of estimate (sSy.x):
sSy.x = √(sS1U1B0S1U1B1 / (n-2)) = √(11.74 / 8) = 1.363

Now we can calculate the prediction interval:
Lower limit = a + bx - t*sSy.x*√(1 + 1/n + ((X-ΣX/n)^2 / ΣS1U1P12S1S1P0))
= 0.8868 + 0.00943*4 - 2.306*1.363*√(1 + 1/10 + ((4-32/10)^2 / 67.2))
= 4.118

Upper limit = a + bx + t*sSy.x*√(1 + 1/n + ((X-ΣX/n)^2 / ΣS1U1P12S1S1P0))
= 0.8868 + 0.00943*4 + 2.306*1.363*√(1 + 1/10 + ((4-32/10)^2 / 67.2))
= 12.226

Therefore, the 95% prediction interval when X = 4 is (4.118, 12.226), which is answer choice C.

A Regression Analysis Yields the Following Information: ; N