Question

Consider the following data regarding students' college GPAs and high school GPAs. The estimated regression equation is
Estimated College GPA=2.22+0.3649(High School GPA).Estimated College GPA=2.22+0.3649(High School GPA).

GPAs

College GPA	High School GPA
2.78	3.34
3.70	2.14
2.27	2.09
3.47	2.93
3.14	2.26
3.95	3.66

Step 2 of 3 :  

Compute the mean square error (S²e) for the model. Round your answer to four decimal places.

Answer 1

> College=C(2.78,3.70,2.27,3.47,3.14,3.95) > School=c(3.34,2.14,2.09,2.93, 2.26,3.66) > y=College > X=School > reg=1m(y~) > reg Call: 1m(formula = y - x) Coefficients: (Intercept) 2.2197 0.3649 > summary(reg) Call: Im(formula - y - x) Residuals: -0.6585 0. 6994 -0.7124 0.1811 0.0956 0.3947 Coefficients : Estimate Std. Error t value pr(>t (Intercept) 2. 2197 1.1914 1.863 0.136 0.3649 0.4248 0.859 0.439 Residual standard error: 0.638 on 4 degrees of freedom Adjusted R-squared: -0.05533 Multiple R-squared: 0.1557. p-value: 0.4388 F-statistic: 0.7378 on 1 and 4 DF

Im Turmuid = y X) 6. Residuals: 1 2 3 4 -0.6585 0.6994 -0.7124 0.1811 0.0956 0.3947 coefficients: Estimate std. Error t value pret (Intercept) 2.2197 1.1914 1.863 0.136 0.3649 0.4248 0.859 0.439 Residual standard error: 0.638 on 4 degrees of freedom Multiple R-squared: 0.1557, Adjusted R-squared: F-statistic: 0.7378 on 1 and 4 DF, p-value: 0.4388, > anova (reg) Analysis of variance Table Response: y of Sum Sq Mean Sq F value Pr (>F) 1 0.3003 0.3003 0.7378 0.4388 Residuals 4 1.6280 0.4070
Therefore by using r software

Mean square error = 0.4070