Summary output of excel linear regression tool
SUMMARY OUTPUT | ||||||||
Regression Statistics | ||||||||
Multiple R | 0.972564957 | |||||||
R Square | 0.945882596 | |||||||
Adjusted R Square | 0.940962832 | |||||||
Standard Error | 116.9172647 | |||||||
Observations | 13 | |||||||
ANOVA | ||||||||
df | SS | MS | F | Significance F | ||||
Regression | 1 | 2628150.655 | 2628150.655 | 192.2617825 | 2.59973E-08 | |||
Residual | 11 | 150366.1145 | 13669.64677 | |||||
Total | 12 | 2778516.769 | ||||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | Lower 95.0% | Upper 95.0% | |
Intercept | -56.91578141 | 95.34450385 | -0.59694874 | 0.562626736 | -266.7676195 | 152.9360567 | -266.7676195 | 152.9360567 |
Xi | 1.193024667 | 0.086040503 | 13.8658495 | 2.59973E-08 | 1.003650796 | 1.382398537 | 1.003650796 | 1.382398537 |
A ] Estimator of X that model has suggested is 1.19
it suggests that if total earnings will increase by approx 19%
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B] there are 123 offices
as suggested in hint X =128200, X_mean = 128200/123 = 1042.27
Mean Y :
Mean Y_predicted = X * 1.19302 - 56.9158
= 1042.27 * 1.19302 - 56.9158
= 1186.53
for lower 95% confidence interval :
Mean Y_predicted = X * 1.0036 - 266.76761947246
= 1042.27 * 1.0036 -266.76761947246
= 779.25
for Upper 95% confidence interval :
Mean Y_predicted = X * 1.3823 + 152.936
= 1042.27 * 1.3823 + 152.936
= 1593.66
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C] There are 123 offices
as suggested in hint X =128200
Mean Y :
Y_predicted = X * 1.19302 - 56.9158
= 128200 * 1.19302 - 56.9158
= 152888.24
for lower 95% confidence interval :
Y_predicted = X * 1.0036 - 266.76761947246
= 128200 * 1.0036 -266.76761947246
= 128394.7524
for Upper 95% confidence interval :
Y_predicted = X * 1.3823 + 152.936
= 128200 * 1.3823 + 152.936
= 177363.796
Corresponding 3-month period of the previous year. A simple random sample of 13 district offices ...
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