In: Statistics and Probability
The data in the table represent the weights of various domestic cars and their miles per gallon in the city for the 2008 model year. For the data from the first 11 cars, the least-squares regression line is y=−0.0062x+42.4755. A twelfth car weighs 2,705 pounds and gets 14 miles per gallon. Compute the coefficient of determination of the expanded data set (including the twelfth car). What effect does the addition of the twelfth car to the data set have on Rsquared2?
Car Weight_(pounds)_x
Miles_per_Gallon_y
1 3765 21
2 3980 19
3 3532 22
4 3174 21
5 2582 27
6 3729 18
7 2601 26
8 3775 18
9 3313 19
10 2991 26
11 2753 26
12 2705 14
1) The coefficient of determination of the expanded data set is
2)How does the addition of the twelfth car to the data set affect Rsquared2? Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a) It increases by __
b) It decreases by __
c) it does not affect it
Solution: We can use the excel regression data analysis tool to find the answer to the given questions. The excel output is given below:
SUMMARY OUTPUT | ||||||
Regression Statistics | ||||||
Multiple R | 0.4988 | |||||
R Square | 0.2488 | |||||
Adjusted R Square | 0.1737 | |||||
Standard Error | 3.7269 | |||||
Observations | 12 | |||||
ANOVA | ||||||
df | SS | MS | F | Significance F | ||
Regression | 1 | 46.01604742 | 46.01604742 | 3.312875614 | 0.098755827 | |
Residual | 10 | 138.9006192 | 13.89006192 | |||
Total | 11 | 184.9166667 | ||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | |
Intercept | 34.3965 | 7.2120 | 4.7694 | 0.0008 | 18.3272 | 50.4658 |
Weight_(pounds)_x | -0.0040 | 0.0022 | -1.8201 | 0.0988 | -0.0089 | 0.0009 |
1) The coefficient of determination of the expanded data set is
Answer:
2)How does the addition of the twelfth car to the data set affect Rsquared2? Select the correct choice below and, ifnecessary, fill in the answer box to complete your choice.
b) It decreases by 0.5365
Explanation:
R-square for 11 pairs of data set is 0.7853 and the R-square for the 12 pairs of data set is 0.2488. Therefore, the R-square is decreased by 0.7853-0.2488=0.5365