In: Math
Download the dataset CARS1 from BlackBoard. a. Do not worry about outliers. Assume the data is correct and any outliers will remain in the dataset. b. Do scatterplot and analyze the results. c. Test for correlation (correlation coefficient) d. Regress weight (column 2) against gas mileage in the city (column 1). Make sure you make gas mileage the dependent (Y) variable. e. Determine and fully explain R2 MPG City Weight 19 3545 23 2795 23 2600 19 3515 23 3245 17 3930 20 3115 22 3235 17 3995 22 3115 23 3240 17 4020 18 3220 19 3175 20 3450 19 3225 17 3985 32 2440 29 2500 28 2290
Y( gas mileage) | Weight (X) |
19 | 3545 |
23 | 2795 |
23 | 2600 |
19 | 3515 |
23 | 3245 |
17 | 3930 |
20 | 3115 |
22 | 3235 |
17 | 3995 |
22 | 3115 |
23 | 3240 |
17 | 4020 |
18 | 3220 |
19 | 3175 |
20 | 3450 |
19 | 3225 |
17 | 3985 |
32 | 2440 |
29 | 2500 |
28 | 2290 |
b)
From the scatter plot, we know that when increasing the value of Y decreases the values of weight and vice versa. Hence, the variable Y and weight have a negative association.
c)
Pearson correlation of Y and Weight = -0.871
P-Value = 0.000
Ans: The estimated p-value for correlation test is 0.000 and less than 0.05 level of significance. Hence, we can conclude that the variable Y and weight have significant association at 0.05 level of significance.
d)
The regression equation is
Y = 44.238 - 0.00708 Weight
Predictor | Coef | SE Coef | T | P |
Constant | 44.238 | 3.075 | 14.38 | 0.000 |
Weight | -0.00708 | 0.00094 | -7.53 | 0.000 |
S = 2.13437 R-Sq = 75.9% R-Sq(adj) = 74.6%
The estimated p-value for Weight is 0.000 and less than 0.05 level of significance. Hence, we can conclude that weight is the significant explanatory variable of gas mileage at 0.05 level of significance.
e) The estimated value of R2 is 75.9%. hence, 75.9% variation of the model is explained by the explanatory variable weight.