Question

The tensile strength of a fiber used in manufactuing cloth is often interest to the purchaser. Previous experience indicates that the standard deviation of tensile strength is 5psi. A random sample of fifteen fiber specimens is selected and the average tesile strength is found to be 122psi. Assuming a normal distribution, find 99% confidence interval for the mean tensile strength.

Answer 1

Solution :

Given that,

= 122

s = 5

n = 15

Degrees of freedom = df = n - 1 = 15 - 1 = 14

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,14 = 2.977

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

= 2.977 * (5 / 15)

= 3.8

The 99% confidence interval estimate of the population mean is,

- E < < + E

122 - 3.8 < < 122 + 3.8

118.2 < < 125.8

(118.2 , 125.8)