Solution:
In excel go to
Data>Data analysis>regression
you will get
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SUMMARY OUTPUT |
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Regression Statistics |
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| Multiple
R |
0.804387228 |
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| R
Square |
0.647038813 |
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| Adjusted
R Square |
0.529385084 |
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| Standard
Error |
3167.333553 |
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Observations |
9 |
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ANOVA |
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df |
SS |
MS |
F |
Significance F |
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Regression |
2 |
1.1E+08 |
55171176 |
5.4995181 |
0.043972469 |
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Residual |
6 |
60192011 |
10032002 |
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Total |
8 |
1.71E+08 |
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Coefficients |
Standard Error |
t Stat |
P-value |
Lower 95% |
Upper 95% |
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Intercept |
54187.6859 |
3044.864 |
17.79642 |
2.023E-06 |
46737.17125 |
61638.2 |
| Years of
experience |
1206.385977 |
384.0964 |
3.140842 |
0.0200465 |
266.5359348 |
2146.236 |
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Average evaluation |
-624.0392499 |
589.5046 |
-1.05858 |
0.3305343 |
-2066.505 |
818.4265 |
For avergae evaluation
t=-1.05858
p=0.3305
p>0.05
Not significant predictor at 5% level of signifcance.
he eval score is NOT a significant predictor of salary at the 5%
level.
FALSE
