A consumer organization wants to estimate the actual tread wear index of a brand

Question

A consumer organization wants to estimate the actual tread wear index of a brand name of tires that claims "graded 250" on the sidewall of the tire. A random sample of n = 17 indicates a sample mean tread wear index of 243.8 and a sample standard deviation of 24.9. Complete parts (a) through (b) a. Assuming that the population of tread wear indexes is normally distributed, construct a 90% confidence interval estimate of the population mean tread wear index for tires produced by this manufacturer under this brand name. «D (Round to two decimal places as needed.) b. Do you think that the consumer organization should accuse the manufacturer of producing tires that do not meet the perfomance information on the sidewall of the tire? Explain. No, because a grade of 250 is not in the interval. B. A. Yes, because a grade of 250 is in the interval. ° C. Yes, because a grade of 250 is not in the interval. 0 D. No, because a grade of 250 is in the interval

Explanation / Answer

Mean is 243.8 and s is 24.9, standard error SE is s/sqrt(N)=24.9/sqrt(17)=6.039

for 90% confidence, the z value is 1.65

thus lower limit is mean-z*SE=243.8-1.65*6.039=233.84

upper limit is mean+z*SE=243.8+1.65*6.039=253.76

thus, the value of 250 is in the interval so option D

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