SUMMARY OUTPUT | ||||||||
Regression Statistics | ||||||||
Multiple R | 0.909785963 | |||||||
R Square | 0.827710499 | |||||||
Adjusted R Square | 0.826591736 | |||||||
Standard Error | 7.177298036 | |||||||
Observations | 156 | |||||||
ANOVA | ||||||||
df | SS | MS | F | Significance F | ||||
Regression | 1 | 38112.05194 | 38112.05194 | 739.8443652 | 1.09619E-60 | |||
Residual | 154 | 7933.095493 | 51.5136071 | |||||
Total | 155 | 46045.14744 | ||||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | Lower 95.0% | Upper 95.0% | |
Intercept | 8.67422449 | 2.447697434 | 3.543830365 | 0.000522385 | 3.838827439 | 13.50962154 | 3.838827439 | 13.50962154 |
X Variable 1 | 0.801382837 | 0.029462517 | 27.20008024 | 1.09619E-60 | 0.743179986 | 0.859585688 | 0.743179986 | 0.859585688 |
(d) | How much of the variation in the sample values of total point earned does the model you estimated in part (b) explain? |
If required, round your answer to two decimal places. | |
% | |
(e) | Mark Sweeney spent 80 hours studying. Use the regression model you estimated in part (b) to predict the total points Mark earned. |
If required, round your answer to the nearest whole number. Do not round intermediate calculations. | |
$ |
(D) Amount of variation is equal to the value of coefficient of determination or R squared
Using the given output data table
R square = 0.8277 or 82.77
So, 82.77% is the variation in the sample values of total point earned does the model you estimated in part (b) explain
(E) Regression equation using the given data output table is
Total points earned = 8.6742 + 0.80138*(number of hours)
we have to find total points earned when number of hours = 80
this gives us
= 8.6742 + 0.80138*(80)
= 72.78
= 73 (rounded to nearest whole number)
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