Hi can you please help me answer these multiple choice questions, no working out is necessary and all relate to the information in the first picture. Thanks :)
Questions 3 - 7 concern the following idea: A trucking company uses the "tyre mania" brand of tyres on all of its trucks, and wonders if it could save money by changing to a brand of tyres that would last longer. They hear of a new brand of tyres called "tyre rific" and wish to investigate whether switching brand of tyres could save them money. They had 20 trucks outfitted with brand new "tyre mania" tyres and 30 trucks fitted with brand new tyres from "tyre rific". For each truck the distance driven was recorded before a tyre change was needed due to worn out tread on one or more of the tyres, at which point all tyres were changed on the truck. The results were analyzed in R via a 2-sample t-test with a 2-sided alternative hypothesis, and some of the output is summarized below (assume a= 0.05 where necessary). mean sd n tyre rific 70.25634 9.699926 30 tyre mania 65.78654 10.857855 20 Two Sample t-test data: distance by tyre t= -1.5219, df = 48, p-value = 0.1346 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -10.375017 1.435424 Levene's Test for Homogeneity of Variance (center = "median") Df F value Pr(>F) group 1 0.7122 0.4029 48 Let uj and u2 denote the mean kms driven before tyre replacement with "tyre mania" and "tyre rific" tyres respectively, and o12 and o2 be the respective variances.Select the relevant null and alternative hypotheses that have been tested in this analysis. Ca. Ho: H2 - H1 = 0 H1 : H2 - H1 0What is the pooled estimate of the standard deviation? Ca. 10.16 Cb. 10.28 Cc. 10.17 C d. 10.30If you were asked to use this analysis to answer the question of whether there is evidence that they should switch to the tyre rific brand of tyres, what is your conclusion? Ca. The mean driving distance for tyre rific tyres is lower than the mean driving distance for tyre mania so they should NOT switch to the tyre rific tyres. Cb. The mean driving distance for tyre rific tyres is higher than the mean driving distance for tyre mania so they should switch to the tyre rific tyres. Cc. 0.1346 > 0.05 so there is no significant evidence that there is any difference in the performance of the two brands of tyre. Cd. 0.4029 > 0.05 so there is no significant evidence that there is any difference in the performance of the two brands of tyre.What did we need to assume about normality in conducting this hypothesis test? Ca. The driving distances for the tyre rific tyres were normally distributed AND the driving distances for the tyre mania tyres were normally distributed b. Nothing thanks to the central limit theorem Cc. All of the driving distances together (ie all 50 data points) come from a Normal Distribution Cd. The test statistic is normally distributedFollowing on from this analysis, if I were to calculate a confidence interval for the mean distance driven by the tyre rific tyres, what degrees of freedom would I use in my test statistic? Ca. 30 + 20 -1 Cb. 30 - 1 Cc. 30 Cd. 30 + 20-2
