SUMMARY OUTPUT | ||||||||
Regression Statistics | ||||||||
Multiple R | 0.961422643 | |||||||
R Square | 0.924333499 | |||||||
Adjusted R Square | 0.902714499 | |||||||
Standard Error | 12.83060262 | |||||||
Observations | 10 | |||||||
ANOVA | ||||||||
df | SS | MS | F | Significance F | ||||
Regression | 2 | 14077.22945 | 7038.614727 | 42.75560781 | 0.000119169 | |||
Residual | 7 | 1152.370546 | 164.6243636 | |||||
Total | 9 | 15229.6 | ||||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | Lower 95.0% | Upper 95.0% | |
Intercept | -18.73506579 | 18.14254204 | -1.032659356 | 0.336115885 | -61.63536067 | 24.1652291 | -61.63536067 | 24.1652291 |
x1 | 2.014081562 | 0.249470879 | 8.073413489 | 8.59808E-05 | 1.424176671 | 2.603986454 | 1.424176671 | 2.603986454 |
x2 | 4.728069085 | 0.957466177 | 4.938105594 | 0.001678746 | 2.464021343 | 6.992116827 | 2.464021343 | 6.992116827 |
a)
Hypothesis:
Test statistic = F = 42.76
P value = 0.000 < 0.05 (significance level)
b)
Hypothesis :
Test stat : t = 8.07
P value = 0.000 < 0.05 (significance level)
c)
Hypothesis :
Test stat : t = 4.94
P value = 0.002 < 0.05 (significance level)
You may need to use the appropriate technology to answer this question. Consider the following data...
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