Question

Hypothesis Testing_02 (when o is manager of a fleet of automobiles is testine two brands of tires and assigns one time of each brand at rande new rear wheels of eleht cars and runs the cars until the tires wear out. The data in kilometers follow Brand 01 36,925 45,300 36,240 32.100 37,210 48,360 38,200 33,500 Brand 02 34,318 42,280 35,500 31,950 38,015 47,800 37,810 33,215 a. You need to test if the mean wear out distance for Brand 01 exceeds that of Brand 02. Write down the null and alternative hypotheses to be tested. Clearly define all the terms. b. What is the type of statistical test that that should be used? Explain. c. Test the hypotheses at a = 0.05, using the critical value approach. What is your decision on null hypothesis(H)? Interpret the test results. d. What is the P Value for the test? Test the hypotheses using the P Value. Do you get the same answer as part (c)? e. Construct a 95% confidence interval for the mean difference of wear out distances. Test the hypotheses using the confidence interval approach. Do you get the same answer as parts (c) and (d)?

Answer 1

TWO-Sample T-Test and ci: Brand01, Brando2 Two-sample I for Brand01 vs Brando2 N Mean StDev SE Mean Brand01 8 38479 5590 1976 Brando2 8 37611 5244 1854 Difference = mu (Brand01) - mu (Brand02) Estimate for difference: 868 95% lower bound for difference: -3931 T-Test of difference = 0 (vs >): I-Value = 0.32 P-Value = 0.377 A НТВ >

Ans. a. Hypothesis : _ HO: 41 = el2 us 4 : th>u2 where ell is the mean of brandol - 12 is the mean of brandon. and b. Here Two sample t-test is used here. C: 6:0.05 critical value = 1.7613 Test statistic te0.32 si Accept Ho. 1.7613 Pvalue = 0.372 > 0.05 i Accept Ho. i we get same aus as - part o. 95%.c.1 for mean difference (51 - R2) 1 taiem-2, a * /sie st (138479.38-37611) § 1-7613 x 31251217 + 27498572 ( 868-375 1 4773.001) (-3904.63 ; 5641-376) Yes we get saune aus as part cand d.