Question

1. A type of roller bearing has an established mean time to failure of 6000 hours. A sample of 45, using a changed lubricant, gave a mean life of 6400 hours and a standard deviation of 450 hours. Using a 1% significance level, test the claim that the new lubricant has resulted in a significant change in mean life.

Answer 1

Given : Sample size=n=45

Sample mean=

X = 6400

Sample standard deviation=s=450

Population mean=

μα = 6000

Significance level=

a = 0.01

1) Hypothesis :
Ho :μ = 6000
Vs  
00097 11: "H

Here , the population standard deviation is unknown but the sample size is greater than 30.

Therefore , use Z-test

2) The test statistic is ,

(1‘0)N+ 017 - x = 7

6400 – 6000 - 450/45 = 5.96

3) Critical value :

Za/2 = 20.01/2 = +2.58

Rejection rule : If , then reject Ho , accept otherwise

ZSTAT > Za/2

Reject Ho A at Ho à Reject Ho y roul to 11 -Za 2-0 2

4) Decision : Here ,the value of the test statistic is in the rejection region. i.e.

ZSTAT = 5.96 > Za/2 = 2.58

Therefore , reject Ho

5) Conclusion : Hence , there is sufficient evidence to support the claim that the new lubricant has resulted in a significant change in mean life.