COMM 291 ASSIGNMENT 6

Gir and vs putts show that the constant variance

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Unformatted text preview: equation: y = 0.828426 + 0.004968GIR – 0.514619Putts Note that the R-square drops only a little bit, from 11% to 9.5%, not a significant drop. e) Residual plots vs. GIR and vs. Putts show that the constant variance assumption appears to be satisfied. However, normality does not; note the greater clustering below the 0 line and sparser clustering above, indicating skewness. The normal probability plot is not a straight line, which also indicates nonnormality. A histogram (not shown here) confirms the non-normality. Independence cannot be judged easily by the residual plots, but is a reasonable assumption here. Question 3: Simple Linear Regression Excel Template SUMMARY OUTPUT Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations ANOVA (1) (2) (3) (4) (5) Df (6) (7) (8) Regression Residual Total Intercept X Variable (1) (2) (3) (4) (5) (6 ) (7) (8) (9) (10) (11) (12) (13) (14) (15) (16) (17) Coefficients (16) (17) MS (12) (13) SS (9) (10) (11) Standard Error (18) (19) t Stat (20) (21) F (14) P- value (22) (23) = Positive square root of (2) = SSM/SST = r2, where r = correlation coefficient = 1-(1-R2)[(n-1)/(n-k-1)] = se = Root MSE = √(13) =n =1 (In multiple regression (6) = k) = n -2 (In multiple regression (7) = n–k–1) = n -1 = SSM = SST...
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