Question

An achievement test given to high school seniors is used to predict college GPA. The following relationship between the test scores and college GPA exists within the entire population of students: 1. Y 0.10+3.94 X+ E In the above, Y (response) is the college GPA of a given student, X is the student's score on the achievement test (predictor, scored as proportion correct, between 0 and 1), and e is random error, with E~N(0,.005), where the value.005 is the variance of e. Note: The equation above, as indicated, is a population regression equation. Suppose a sample of 20 students is obtained, resulting in a sample mean achievement test score of.791 and a sample variance achievement test score of.00242. a. Use the information above to completely state the distribution of the statistic, b, the estimator b. Use the information above to completely state the distribution of the statistic, bo, the estimator c. Use the above to calculate the probability that the sample of 20 students indicated above results for B1, for a sample of size 20 with these predictor values. [Hint: See Lecture 4 notes.] for Bo, for a sample of size 20 with these predictor values. in a value of b, as small or smaller than 3.76.

Answer 1

$(a)~S_{xx}=(20-1)*sample~variance~achievement~test~score\\ =19*0.00242=0.04598\\ b_1=\frac{\sum_{i=1}^n(x_i-\bar{x})y_i}{S_{xx}}\\ E(b_1)=\frac{\sum_{i=1}^n(x_i-\bar{x})E(y_i)}{S_{xx}}=\frac{\beta_0\sum_{i=1}^n(x_i-\bar{x})+\beta_1\sum_{i=1}^n(x_i-\bar{x})x_i}{S_{xx}}=\beta_1,\\~since~\sum_{i=1}^n(x_i-\bar{x})=0\\ Var(b_1)=\frac{\sum_{i=1}^n(x_i-\bar{x})^2Var(y_i)}{S_{xx}^2}=\frac{\sigma^2}{S_{xx}}=0.005/0.04598=0.1087\\ Since~b_1~is~a~linear~function~of~independent~y_i's~and~y_i's~follow\\ independently~normal~distribution~then\\ b_1\sim N\left(\beta_1,0.1087\right ).\\ Since~\beta_1=3.94~so~b_1\sim N\left(3.94,0.1087\right )\\$

$(b)~b_0=\bar{y}-b_1\bar{x}=\frac{1}{n}\sum_{i=1}^ny_i-\bar{x}\frac{\sum_{i=1}^n(x_i-\bar{x})y_i}{S_{xx}}\\ =\sum_{i=1}^n\left(\frac{1}{n}-\frac{\bar{x}(x_i-\bar{x})}{S_{xx}} \right )y_i\\ E(b_0)=\sum_{i=1}^n\left(\frac{1}{n}-\frac{\bar{x}(x_i-\bar{x})}{S_{xx}} \right )E(y_i)=\sum_{i=1}^n\left(\frac{1}{n}-\frac{\bar{x}(x_i-\bar{x})}{S_{xx}} \right )(\beta_0&plus;\beta_1x_i)\\ =\beta_0&plus;\beta_1\bar{x}-\beta_1\bar{x},~since~\sum_{i=1}^n(x_i-\bar{x})=0~and~\sum_{i=1}^n(x_i-\bar{x})x_i=S_{xx}\\ =\beta_0=0.10\\ Var(b_0)=\sum_{i=1}^n\left(\frac{1}{n}-\frac{\bar{x}(x_i-\bar{x})}{S_{xx}} \right )^2Var(y_i)=\sigma^2\left(\frac{1}{n}&plus;\frac{\bar{x}^2}{S_{xx}} \right )\\= 0.005*(1/20&plus;(0.791)^2/0.04598) =0.0683.\\ Since~b_0~is~a~linear~fuction~of~y_i's~and~Y_i's~are~independently~normal~variables\\ so~b_0\sim N(0.10,~0.0683).\\ c.~P(b_1<3.76)=\Phi((3.76-3.94)/0.1087)=\Phi(-1.6559)\\=1-\Phi(1.6559)=0.0489$