Question

(2 points) Some car tires can develop what is known as "heel and toe" wear if not rotated after a certain mileage. To assess this issue, a consumer group investigated the tire wear on two brands of tire, A and B, say. Fifteen cars were fitted with new brand A tires and thirteen with brand B tires, the cars assigned to brand at random. (Two cars initially assigned to brand B suffered serious tire faults other than heel and toe wear, and were excluded from the study.) The cars were driven in regular driving conditions, and the mileage at which heal and toe wear could be observed was recorded on each car. For the cars with brand A tires, the mean mileage observed was 24.99 in 10 miles) and the variance was 7.75 (in 100 miles). For the cars with brand B, the corresponding statistics were 32.92 (in 10 miles) and 6.47 (in 100 miles ) respectively. The mileage before heal and toe wear is detectable is assumed to be Normally distributed for both brands. Part a) Calculate the pooled variance s2 to 3 decimal places. During intermediate steps to arrive at the answer, make sure you keep as many decimal places as possible so that you can achieve the precision required in this question. X106 miles 2 Part b) Determine a 95% confidence interval for MA - MB, the difference in the mean 109 mileages before heal and toe wear for the two brands of tire. Leave your answer to 2 decimal places.( Part c) Based on the 95% confidence interval constructed in the previous part, which of the following conclusions can be drawn when we test H:HA = MB vs. Ha: NA # HB with a = 0.05. A. Do not reject Ho since 0 is within the interval found in part (b). B. Reject Ho since O is not within the interval found in part (b). C. Do not reject Ho since 0 is not in the interval found in part (b). D. Reject Ho since 0 is in the interval found in part (b). E. Do not reject Ho since -7.93 is within the interval found in part (b).

Answer 1

Sduction : - Given dato 1-15 , 04.99 : S, 547.75 a=!3 F2 = 32.92 f = fuit → pooled vamance so = (n-1) 5,4cm-DS, 5p 3 (15-) (7-45) + (13-(6.ua) nt32 :: ::15+13-2 se : 108:15 + 764 186.15 267 saa = 8:1596 e paired variance so = 21596-xromiler? 7.1596 9:1607 pounded to ideaimas, 951.confidence intewal; for llaisten 05 An fet total of the C5 -(2499-32.43) 1 0.056*4.160265 + +) CiIs -7.93. £ 2056 X .013952: Cil. -93. £ 2.0847 CiI'= (-7793 – 250847 -77193 62.0841 CsI - C -10.0147 –578453) ciI = (-10.01, -5.85 (Rounded to two I decimal places) pot tentreal value at a = .0105 (too failed test of nith 2 = 26 12V

♡ The null and alternative hypothers are to AMB ha Hailla FB The hypothery value o lies outside the Confidence interval E-1000i, 585, so we repuet the null hypothes ts to at &=0005 Reject it since o is not within the intcrral found in poort (6) is option Bis correct