Question

SUMMARY OUTPUT Regression Statistics Multiple R 0.909785963 R Square 0.827710499 Adjusted R Square 0.826591736 Standard Error...

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.
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Homework Answers

Answer #1

(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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