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

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The formula for the 95% prediction interval for a new observation is: Ŷ ± t(0.025, n-2) * s(y-ŷ) where Ŷ is the predicted value of the response variable, t(0.025, n-2) is the 2.5th percentile of the t-distribution with n-2 degrees of freedom, s(y-ŷ) is the standard error of the estimate, and n is the sample size. We are given: - n = 10 - ΣX = 32 - ΣX^2 = 134 - ΣXY = 1.66 Let's calculate the slope and intercept of the regression line: slope, b = ΣXY / ΣX^2 = 1.66 / 134 = 0.0124 intercept, a = ȳ - b x̄ = (ΣY / n) - b (ΣX / n) = (s_y∙x / n) - b (ΣX / n) = (1.66 / 10) - (0.0124 * 32 / 10) = 0.1308 So, the regression equation is: y = 0.1308 + 0.0124x When x = 4, the predicted value of y is: ŷ = 0.1308 + 0.0124(4) = 0.1836 Now, let's calculate the standard error of the estimate: s(y-ŷ) = sqrt[ Σ(y - ŷ)^2 / (n-2) ] s(y-ŷ) = sqrt[ Σ(y - a - b(x - x̄))^2 / (n-2) ] s(y-ŷ) = sqrt[ Σ(y - a - bx)^2 / (n-2) ] s(y-ŷ) = sqrt[ Σ(y^2 - 2ay - 2bxy + 2abx + b^2x^2 - 2abx̄ + a^2) / (n-2) ] s(y-ŷ) = sqrt[ Σy^2 - 2aΣy - 2bΣ(xy) + 2abΣ(x) + b^2Σ(x^2) - 2abΣ(x) + a^2(n-2) ] / (n-2) s(y-ŷ) = sqrt[ Σy^2 - 2aΣy - 2bΣ(xy) + b^2Σ(x^2) + a^2(n-2) ] / (n-2) s(y-ŷ) = sqrt[ Σy^2 - 2(0.1308)(Σy) - 2(0.0124)(1.66) + (0.0124)^2(134) + (0.1308)^2(10-2)] / (10-2) s(y-ŷ) = 0.4603 Finally, using a t-distribution table or calculator, we find t(0.025, 8) = 2.306. Therefore, the 95% prediction interval for a new observation with x = 4 is: 0.1836 ± 2.306(0.4603) 0.1836 ± 1.0597 The interval is (0.1239, 0.2433) when rounded to four decimal places. Note: Choice A is not possible as the lower limit of a prediction interval is never zero. Choice B is not correct as it is a confidence interval for the mean and not a prediction interval for a single new observation. Choice D is also incorrect as it doesn't include the actual predicted value.

A Regression Analysis Yields the Following Information: ; N