Question

SUMMARY OUTPUT Regression Statistics Multiple R 0.9655 R Square 0.9321 Adjusted R Square 0.9307 Standard Error 0.5383 Observations 50 ANOVA df F 659.4383 Significance F 1.07386E-29 Regression Residual Total 1 48 49 SSM S 191.0842089 191.084209 13.90887066 0.28976814 204.9930796 Intercept Increase in profits (%) Coefficients Standard Error 2.28990 .0910 0.9513 0.0370 Stat 25.17540 25.6795 P-value .0000 0.0000 Lower 95% 2 .1070 0.8768 Upper 95% Lower 95.0%Jpper 95.0% 2.4728 2.1070 2.4728 1.0258 0.8768 10258 Increase in Manager's Salary (%) 4,00 2.00 1.00 -4.000 3.000 2.000 1.000 0.000 -1.00 1.000 2.000 3.000 4.000 5.000 6.000

Answer 1

Regression equation : y = 2.2899 +0.9513.r; where, y = Increase in Manager's salary (%), I = Increase in Profits (%) Since p-value of F test is very low < 0.01, so overall regression equation is significant. Moreover Rº = 0.9321 i.e. 93.21% of Increase in Manager's salary is explained by this regression equation and standard error of regression equation = 0.5383 is very low and from scatter plot we observe that there is a strong positive correlation between x andy hence this model is useful for prediction. From residual plot it is observed that there is no pattern and the points are randomly distributed on the both sides of r - aris hence (i) assumption of linearity holds (ii) E(ei) = 0 where, e;= random error corresponding to ith observation (ii) Var(ei) = constant hence equal variance assumption holds.

From normal probability plot it is observed that assumption of normality holds since points are very closed to the reference line passing through origin. Yes, since p-value corresponding "Increase in profits" = 0,0000 < 0.01 so "change in division profit" appear to be a good predictor for the manager's pay raise. For increase in profits = r = -0.51, then 9 = 2.2899 +0.9513 * (-0.51) = 1.80 But r = 20 lies outside range of r so we do not know the relationship between (2, y) outside the range of rhence we can't predict y when I = 20.