Question

the quality control manager at a compact fluorecent light bulb factory needs to determine whether the mean life of a large shipment of CFL is equal to 7500 hours .the population sdtandard deviation is 1000 hours . A random sample of 64 CFL indicate a sample mean life of 7250 hours

Answer 1

Solution:

    We are given that: The population standard deviation is 1000 hours.

That is:

A random sample of 64 CFL indicate a sample mean life of 7250 hours

Sample size = n = 64

Sample mean =

The quality control manager at a compact fluorescent light bulb factory needs to determine whether the mean life of a large shipment of CFL is equal to 7500 hours.

Thus we use the following steps:

Step 1) State null and alternative hypothesis:

since we have to test mean is equal to 7500 or not, this is a nondirectional hypothesis, thus we use the two-tailed test.

Step 2) Find the test statistic:

Step 3) Find z critical value:

Since the level of significance is not given, we use level of significance =

Since this is two tailed, we find: Area =

Look in z table for area = 0.0250 or its closest area and find z value.

Area 0.0250 corresponds to -1.9 and 0.06, that is : z = -1.96

This is two-tailed test thus z critical values are: ( -1.96 , 1.96)

Step 4) Rejection region/ Decision rule:
Reject H0 if z test statistic value < -1.96   or z test statistic value > 1.96, otherwise we fail to reject H0..

Since z test statistic value = -2.00 < z critical value = -1.96 , we reject null hypothesis H0.

Step 5) Conclusion: Since we rejected null hypothesis H0 , we conclude that: the mean life of a large shipment of CFL is different from 7500 hours.