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 sa of n -22 indicates a sample mean tread wear index of 242.2 and a sample standard deviation of 21.7 Complete parts (o) through (c) a. Assuming that the population of tread wear indexes is normally distributed, construct a 90% confidence interval estimate mple of the population mean tread wear in ex for tires produced by this manufacturer under this brand name. sus (Round to two decimal places as needed)

Answer 1

n = Sample Size = 22

$\bar{x}$ = Sample Mean = 242.2

s = Sample Sd = 21.7

SE = s/ $\sqrt{n}$

= 21.7/ $\sqrt{22}$ = 4.6265

$\alpha$ = 0.10

ndf = 22 - 1 = 21

From Table, critical values of t = $\pm$ 1.7207

Confidence interval:

242.2 $\pm$ (1.7207 X 4.6265)

= 242.2 $\pm$ 7.9608

=( 234.24, 250.16)

So,

Confidence Interval:

234.24 $\leq$ $\mu$ $\leq$ 250.16