Using Excel, go to Data, select Data Analysis, choose Regression.
SUMMARY OUTPUT | ||||||
Regression Statistics | ||||||
Multiple R | 0.945 | |||||
R Square | 0.893 | |||||
Adjusted R Square | 0.875 | |||||
Standard Error | 0.611 | |||||
Observations | 8 | |||||
ANOVA | ||||||
df | SS | MS | F | Significance F | ||
Regression | 1 | 18.633 | 18.633 | 49.863 | 0.000 | |
Residual | 6 | 2.242 | 0.374 | |||
Total | 7 | 20.875 | ||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | |
Intercept | 0.526 | 0.397 | 1.327 | 0.233 | -0.444 | 1.497 |
x | 0.626 | 0.089 | 7.061 | 0.000 | 0.409 | 0.843 |
a) y = 0.526 + 0.626x
b) H0: β1 = 0, There is not sufficient evidence that x contributes information for linear prediction of y
H1: β1 ≠ 0, There is sufficient evidence that x contributes information for linear prediction of y
c) Test statistic: t-test
Degrees of freedom = n-2 = 8-2 = 6
Level of significance = 0.05
Test statistic (t Stat) = 7.061
Critical value (Using Excel function T.INV.2T(probability,df)) = T.INV.2T(0.05,6) = 2.45
Since test statistic is more than 0.05, we reject the null hypothesis.
There is sufficient evidence that x contributes information for linear prediction of y.
d) 95% confidence interval for β1: (0.409, 0.843)
2. [40 points) Consider the following pairs of observations: X 5 3 2 6 6 0...
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