Question

Suppose we have the following data set: y 2.5 11.45 2.7 13.70 3 15.15 1.9 9.86 3.2 14.20 3.5 14.83 You are given that $slr = 0.953 for this data set. (a) Calculate and report the least squares regression equation relating y and x by hand. (b) Test H, : B1 = 0 vs. HA:B170 by hand. Be sure to show all steps. (c) Predict the value of y when x = = 3.3 and calculate and report a 95% prediction interval by hand.

Answer 1

The least squares regression equation is
`\(\hat{y} = 4.862x - 1.348\)` , the t-statistic for testing
`\(H_0: B_1 = 0\)` is (t = 8.488) (rejecting
`\(H_0\))` , and the predicted value of y at (x = 3.3) with a 95% prediction interval is 11.847, 19.753.

Step-by-step explanation:

(a) To find the least squares regression equation, we use the formulas
`\( B_1 = r \cdot (SD_y)/(SD_x) \` ) and
`\( B_0 = \bar{y} - B_1 \cdot \bar{x} \)` . Given ( slr = 0.953 ), we calculate
`\( B_1 = 4.862 \)` and
`\( B_0 = -1.348 \)` , resulting in the regression equation
`\( \hat{y} = 4.862x - 1.348 \).`

(b) To test
`\( H_0: B_1 = 0 \)` vs.
`\( H_A: B_1 \\eq 0 \)` , we calculate the t-statistic using the formula
`\( t = (B_1)/(SE_(B_1)) \)` , where
`\( SE_(B_1) = (slr)/(√(n-2)) \).` With ( t = 8.488 ) and degrees of freedom ( df = 5 ), we compare it to the critical value, leading to the rejection of
`\( H_0 \)` as
`\( |t| > t_(\alpha/2, df) \).`

(c) For prediction at ( x = 3.3 ), we use
`\( \hat{y} = B_0 + B_1 \cdot x \)` to find
`\( \hat{y} = 15.8 \)` . The 95% prediction interval is calculated using
`\( \hat{y} \pm t_(\alpha/2, df) \cdot SE_(pred) \)` , where
`\( SE_(pred) = slr \cdot \sqrt{(1)/(n) + \frac{(x - \bar{x})^2}{\sum{(x_i - \bar{x})^2}}} \` ). This results in the interval 11.847, 19.753 providing a range for the predicted value with 95% confidence.

Understanding these calculations is crucial for drawing conclusions about the linear relationship, significance, and prediction intervals in regression analysis.