Question

A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. Of the 32 tires surveyed, the mean lifespan was 46,600 miles with a standard deviation of 9,800 miles. Using alpha = 0.05, is the data highly consistent with the claim?

Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

O Part (d) State the distribution to use for the test. (Round your answers to two decimal places.) Х 50000 1511.86 n 1x ) 1 Part (e) What is the test statistic? (If using the z distribution round your answers to two decimal places, and if using the t distribution round y z Part (f) What is the p-value? (Round your answer to four decimal places.)

Answer 1

Solution : given, population mean M= 50000 population Standard deviation, 6 = 8000 Samble Size ne 32 Sample mean, 4 6600 Sample standard deviation, s = 9800 Significance level, alpha = 0.05 20 = Part (d) ď follows normal distribution with mean al and standard deviation o vh Me 50000 67 8000 √32. +474.21 Ñ ~ N( 50000, 1414.21) Part(e) test statistic when population Standard deviation is known Z-test We use Z- X-M o un

Hot M > 5000 o (claim) Hi : M < 50000 (left tailed A les native) This is one failed test. Under Ho, test statistic Z- 46600 50000 80007132 Z -2.40 - 3400 1414.21 Z=-2.40 Part (f) P-value = PLZE - 2.40) 0.0082 [using normal tables ] Part (h) We have L=0005 P-value = 0.0082 Since , IP- Value < & 0.0082 <0.05 So, this is significant & we reject the null hypothesis (Ho), and conclude that Data is not highly consistent with the claim

Past i Confidence level = 95% = 0.95 Significance level = 1-0.95 = 0.05 95% confidence interval for true mean (M) M х + Zo Nъ where Za is the critical value of z at 0.05 level Value of Zoos 1.96 [from normal tables

u 44 600 + 1.46 xBooo 22 - - u M ч 66 oo + 1.46 x 1ч1 ч. 21) (46600 211). 85 ) 46 600 — 21+. 85, 966 oot 217 • 85 цE (458 28. 15 , ца321.B5) M Ч382 8 Ч9 342 С