In: Statistics and Probability
Vehicle Name | SuggestedRetailPrice |
BMW 325Ci 2dr | 30795 |
BMW 325Ci convertible 2dr | 37995 |
BMW 325i 4dr | 28495 |
BMW 325xi 4dr | 30245 |
BMW 330Ci 2dr | 36995 |
BMW 330Ci convertible 2dr | 44295 |
BMW 330i 4dr | 35495 |
BMW 330xi 4dr | 37245 |
BMW 525i 4dr | 39995 |
BMW 530i 4dr | 44995 |
BMW 545iA 4dr | 54995 |
BMW 745i 4dr | 69195 |
BMW 745Li 4dr | 73195 |
Mercedes-Benz C230 Sport 2dr | 26060 |
Mercedes-Benz C240 4dr | 32280 |
Mercedes-Benz C240 4dr | 33480 |
Mercedes-Benz C32 AMG 4dr | 52120 |
Mercedes-Benz C320 4dr | 37630 |
Mercedes-Benz C320 Sport 2dr | 28370 |
Mercedes-Benz C320 Sport 4dr | 35920 |
Mercedes-Benz CL500 2dr | 94820 |
Mercedes-Benz CL600 2dr | 128420 |
Mercedes-Benz CLK320 coupe 2dr (convertible) | 45707 |
Mercedes-Benz CLK500 coupe 2dr (convertible) | 52800 |
Mercedes-Benz E320 4dr | 48170 |
Mercedes-Benz E500 4dr | 57270 |
Mercedes-Benz S430 4dr | 74320 |
Mercedes-Benz S500 4dr | 86970 |
In what follows use any of the following tests/procedures: Regression, multiple regressions, confidence intervals, one-sided t-test or two-sided t-test. All the procedures should be done with 5% P-value or 95% confidence interval
Upload CARS data. SETUP: It is believed that Mercedes’ prices (Suggested price) are different than BMW’s. Given the data your job is to confirm or disprove this belief. (CAREFULL: sort the data in order to extract the needed information).
1. What test/procedure did you perform?
2. What is the P-value/margin of error?
3. Statistical interpretation
4. Conclusion
1. Since Mercedes and BMW are two different brands and prices are collected for these brands, the appropriate test is two sample t test, of course, assuming normality.
Since, we are interested in testing whether Mercedes’ prices (Suggested price) are different than BMW’s or not, the hypothesis is two sided. Then correct option is Two-sided t test. (option b)
2. The boxplot of prices show that the variability of Mercedes is quite higher than that of BMW. Hence, we performed a tw-sided t test assuming different variances and get the p value as .1630
However, if we forcefully assume equal variances, we get the corresponding p value as 0.178.
Thus option e is correct.
3. option b: Since P-value is large we cannot claim that the averages are different.
4. Option b. No. we cannot claim that the above assertion is correct.