Question

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 24.99 (in 103 10 3 miles ) and the variance was 7.75 7.75 (in 106 10 6 miles2 2 ). For the cars with brand B, the corresponding statistics were 32.92 32.92 (in 103 10 3 miles) and 6.47 6.47 (in 106 10 6 miles2 2 ) 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 s 2 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. ×106 × 10 6 miles 2 2 Part b) Determine a 95% confidence interval for μA−μB μ A − μ B , the difference in the mean 103 10 3 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 H0:μA=μB H 0 : μ A = μ B vs. Ha:μA≠μB H a : μ A ≠ μ B with α=0.05 α = 0.05 . A. Do not reject H0 H 0 since 0 is not in the interval found in part (b). B. Reject H0 H 0 since 0 is not within the interval found in part (b). C. Do not reject H0 H 0 since −7.93 − 7.93 is within the interval found in part (b). D. Do not reject H0 H 0 since 0 is within the interval found in part (b). E. Reject H0 H 0 since 0 is in the interval found in part (b)

Answer 1

Given : = 24.99 , = 32.92 , = 7.75 ,    = 6.47 , n1 = 15 and n2 = 13

a) Pooled variance,

= 7.15923   7.159.

b) Since, the population variances are not known and only sample variances are known, we carry out the study using t-distribution.

In general, (1-)% confidence interval is given as :

where from the t tables,   t0.025, 26 = 2.056

Therefore,( 24.99 - 32.92 - 2.056 * 2.67563 * [ 1/15 + 1/13 ] , 24.99 - 32.92 + 2.056 * 2.67563 * [ 1/15 + 1/13 ] )

( -10.0145431 , -5.845457 )   ( -10.01 , -5.85 )

c) Ho : =   vs Ha :    

Level of significance (l.o.s.) : α = 0.05

95% Confidence interval    ( -10.01 , -5.85 )

Since, our null hypothesis states that   =   i.e.   -   = 0. Our confidence interval show consists of the value 0 in it.

Conclusions : Thus, we reject H0 at 5% l.o.s., since 0 is not within the interval found in part (b).