The following R output is obtained from a multiple linear regression analysis y
ID: 3258205 • Letter: T
Question
The following R output is obtained from a multiple linear regression analysis y = beta_0 + beta_1x_1 + beta_2 x_2 + elementof, where E(elementof) = 0. Var(elementof) = sigma^2 and observations (x_1i, x_2i, y_i), i = 1, 2, ...., n are independent and identically distributed in which we are interested in the effects of temperature (denoted by x_1) and pressure (denoted by x_2) on the rate of reaction (denoted by y) in a reaction chamber. Call: lm(formula = y ~ x1 + x2) Residuals: Min 1Q Median 3Q Max -0.89961 -0.44920 -0.04601 0.32086 1.44655 Coefficients: Estimate Std. Error t value Pr(> |t|) (Intercept) 2.053 0.268 7.638 0.012 x1 0.910 0.246 3.686 0.037 x2 4.844 0.255 18.989 0.046 Multiple R-squared: 0.9815, Adjusted R-squared: 0.816 1. What is your conclusion for the hypothesis test that there is no effect of the temperature on the rate of reaction? 2. What is the estimated regression line for the effect of the temperature and pressure on the rate of reaction? 3. In the context of this problem, if we fix the pressure value to 0 and we increase the temperature 1 unit, how much change do we expect in the rate of reaction? 4. What percentage of the variability in the data is explained by the estimated regression line? 5. What is the predicted value of the rate of reaction when temperature value is 0, and the pressure value is 0?Explanation / Answer
a) If we assume Significance level 0.05, here p values (0.037) is less than 0.05, we reject the null hypothesis and conclude that there is significant effect of the temperature on the rate of reaction
(b) Estimated Regression line
y = 2.053+0.910x1+4.844x2
(C) If x2 = 0 and x1 = 1 , we expect the change of 0.910 on the rate of reaction
(d) Here multiple R2 Squared is 0.9815 , so 98.15% variability can be explained by the estimated regression line