1. Consider the data in the table below, from a study on the effect of ozone pollution (ozone concentration, parts per million) on soybean yield (grams per plant). Fit a simple linear regression model to the data, with yield as the response, and ozone Observation Ozone (x) Yield (Y) 1 0.02 242 0.07 237 0.11 231 0.15 201 concentration as the covariate. (a) It is found that the predicted values for the 4 observations are Y1 = 247.6, Y2 = 232.9, Ya = 221.1, Y4 = 209.4, respectively. Compute the ANOVA table (see table format below) for this model. Then find the R2. Table 1: ANOVA table for ozone data Source Sum of Squares Degrees of Freedom Mean Squares Regression Residuals Total 1014.8 (b) Sketch the plot that is produced by the following R code. With the plot and the given information below, what can you say about the fit of the model? > fit=lm(Yield Ozone, data=dat) # Yield: Y, Ozone: x. > plot (fit$fitted. value, fit$residuals, xlab="Fitted value", ylab="Residual") > sigma(fit) [1] 10.38296 (c) Predict the expected yield when ozone concentration is at 0.10 parts per million. Also find a 95% prediction interval for the actual yield at this ozone concentration level. The following information may be useful. > fit$coefficients (Intercept) Ozone 253. 434 -293. 531 > vcov (fit) (Intercept) Ozone (Intercept) 115.9422 -1017.037 Ozone -1017 . 0371 11623. 281 Some (upper) quantiles of t distributions: to.025,1 = 12.706, to.05,1 = 6.314, to.025,2 = 4.303, to.05,2 = 2.920, to.025,3 = 3.182, to.05,3 = 2.353