Question

Question 1 (50 pts): Suppose that a client of yours measure the heights (in inches) of n - 30 wheats grown at locations of various elevations (measured as meters above sea levels). Af- ter some discussion, you decided to fit a linear regression of wheat heights (denoted as yi) on the elevations of the locations (denoted as zi) as follows where ei, E2, . . . , En are i.i.d. errors with Elei] 0 and var(G) σ2. You calculated some descriptive statistics on the data set before transferring it to your colleague. The summary statistics are as follows. 7 TL TI i-1 i-1 Your colleague fitted a simple linear regression using least squares estimation, but somehow lost the original set of data. All that is left is the contaminated R output below. Call: 1m(formula y x) Residuals: Median 0.06319 Mi n 3Q Max 1.62276 Coefficients: Estimate std. Error t value Pr ltl (Intercept) 0.475946 49.01 2e-16 10.46 3.56e-11 0.047799 signif. codes: 0 0.0010.01 0.05 '0.1'1 Residual standard error: 1.045 Multiple R-squared: F-statistic: 109.3 on 1 and 28 DF, p-value: 3.56le- In addition, your colleague vaguely remembered that Σ_1e-30.6, where ei the residual. Your task is to recover as many results as possible before the meeting with your client.

Answer 1

$1.\\ n=error~df+2=28+2=30,~S_{xy}=\sum_{i=1}^n(x_i-\bar{x})(y_i-\bar{y})=2500,\\ ~S_{xx}=\sum_{i=1}^nx_i^2-n\bar{x}^2=325000-30*95.4^2=51965.2\\ S_{yy}=\sum_{i=1}^ny_i^2-n\bar{y}^2=23500-30*27.9^2=147.7\\ \hat{\beta}_1=\frac{S_{xy}}{S_{xx}}=2500/51965.2=0.0481\\ So,~if~we~increase~elevations~of~the~locations~by~1~meter~then~the~estimated\\ height~is~increased~by~0.0481~inches.\\ 2.~\hat{\beta}_0=\bar{y}-\hat{\beta}_1\bar{x}=27.9-0.0481*95.4=23.3113\\ 3.~The~estimated~height~is~23.3112~inches~at~sea~level~i.e.~elevation~of~location~is~zero.\\ 4.~MSE=mean~sum~of~square=\frac{(1-r^2)S_{yy}}{n-2}\\ where,~r^2=coefficient~of~determination=\frac{S_{xy}^2}{S_{xx}S_{yy}}\\ =2500^2/(51965.2*147.7)=0.8143\\ then~MSE=(1-0.8143)*147.7/28=0.9796\\ Estimated~standard~error~of~\hat{\beta}_1=\sqrt{\frac{MSE}{S_{xx}}}=\sqrt{0.9796/51965.2}=0.0043\\=\hat{se}(\hat{\beta}_1)\\$

$95\%~C.I.~for~\beta_1~is\\ \left(\hat{\beta}_1-t_{0.025,28}\times \hat{se}(\hat{\beta}_1) ,~\hat{\beta}_1+t_{0.025,28}\times \hat{se}(\hat{\beta}_1)\right )=(0.0393,~0.0569)\\ where,~t_{0.025,28}=2.0484\\ 5.\\ \widetilde{x}=3.28\times (x-16)~or,~x=16+\widetilde{x}/3.28\\ Now~we~replace~x~by~\widetilde{x}~in~the~model~then~we~get\\ y_i=x_i\beta_1+\beta_0+\epsilon _i\\ =(16+\widetilde{x}_i/3.28)\beta_1+\beta_0+\epsilon _i\\ =(\beta_0+16\beta_1)+\frac{\beta_1}{3.28}\widetilde{x}_i+\epsilon _i\\ =\beta_0^*+\beta_1^*\widetilde{x}_i+\epsilon _i\\\\ where,~\hat{\beta}_0^*=\hat{\beta}_0+16\hat{\beta}_1=23.3113+16*0.0481=24.0809\\ \hat{\beta}_1^*=\frac{\hat{\beta}_1}{3.28}=0.0481/3.28=0.0147\\ So~we~did~not~calculate~the~values~of~least~squares~estimators~for~the~new~regression~model.$