Question

A consumer organization wants to estimate the actual tread wear index of a brand name of tires that claims​ "graded

200 on the sidewall of the tire. A random sample ofn equals 18indicates a sample mean tread wear index of192.5

and a sample standard deviation of22.7

Complete parts​ (a) through​ (c).

a.a.	Assuming that the population of tread wear indexes is normally​ distributed, construct a90 % confidence interval estimate of the population mean tread wear index for tires produced by this manufacturer under this brand name.

Answer 1

Solution :

degrees of freedom = n - 1 = 18 - 1 = 17

t/2,df = t_0.05,17 = 1.740

Margin of error = E = t_/2,df * (s /n)

= 1.740 * ( 22.7/ 18)

Margin of error = E = 9.31

The 90% confidence interval estimate of the population mean is,

- E < < + E

192.5 - 9.31 < < 192.5 + 9.31

( 183.19 < < 201.81 )