1 Check the normality assumption 2 Interpret the coefficient

1. Check the normality assumption.

2. Interpret the coefficient estimate in the context of the problem.

3. Interpret the R-square value in the context of the problem.

4. Predict with 99% confidence the average evaluation score of all the employees who have worked in the company for seven and a half years.

Minitab output:

Analysis of Variance

Source DF Adj SS Adj MS F-Value P-Value

Regression 1 4824 4824.5 8.42 0.004

EMP@EOY 1 4824 4824.5 8.42 0.004

Error 142 81348 572.9

Lack-of-Fit 114 64866 569.0 0.97 0.569

Pure Error 28 16482 588.6

Total 143 86172

Model Summary

S R-sq R-sq(adj) R-sq(pred)

23.9348 5.60% 4.93% 2.61%

Coefficients

Term Coef SE Coef T-Value P-Value VIF

Constant 127.86 2.85 44.81 0.000

EMP@EOY 1.494 0.515 2.90 0.004 1.00

Regression Equation

SCORE = 127.86 + 1.494 EMP@EOY

Fits and Diagnostics for Unusual Observations

Obs SCORE Fit Resid Std Resid

5 102.00 148.46 -46.46 -1.99 X

9 99.00 146.93 -47.93 -2.05 R X

52 176.00 153.63 22.37 0.98 X

56 61.00 134.92 -73.92 -3.10 R

83 75.00 144.15 -69.15 -2.93 R

84 157.00 146.93 10.07 0.43 X

110 143.00 149.19 -6.19 -0.27 X

118 80.00 128.58 -48.58 -2.04 R

122 137.00 146.84 -9.84 -0.42 X

124 154.00 153.05 0.95 0.04 X

129 171.00 154.99 16.01 0.71 X

133 40.00 129.30 -89.30 -3.75 R

SCORE = 127.86 + 1.494 EMP@EOY

for each value of EMP EOY you will have an increase of 1.494 in SCORE

R square = 5.60%

that means that only 5.60% of the data can be represent by the model