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 equals n=25 indicates a sample mean tread wear index of 237.3 and a sample standard deviation of 27.6 Complete parts​ (a) through​ (c).

a.	Assuming that the population of tread wear indexes is normally​ distributed, construct a 99 % confidence interval estimate of the population mean tread wear index for tires produced by this manufacturer under this brand name.

Answer 1

Solution:

Confidence interval for Population mean

Confidence interval = Xbar ± t*S/sqrt(n)

We are given

Xbar = 237.3

S = 27.6

n = 25

df = n – 1 = 25 – 1 = 24

Confidence level = 99%

Critical t value = 2.7969

(by using t-table)

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 237.3 ± 2.7969*27.6/sqrt(25)

Confidence interval = 237.3 ± 2.7969*5.52

Confidence interval = 237.3 ± 15.4391

Lower limit = 237.3 - 15.4391 = 221.86

Upper limit = 237.3 + 15.4391 = 252.74

Required confidence interval = (221.86, 252.74)