Question

9.14 The quality-control manager at a compact flu-orescent light bulb (CFL) factory needs to determine whether the mean life of a large shipment of CFLs is equal to 7,500 hours. The population standard deviation is 1,000 hours. A random sample of 64 CFLs indicates a sample mean life of 7,250 hours.

a. At the 0.05 level of significance, is there evidence that the mean life is different from 7,500 hours?

b. Compute the p-value and interpret its meaning.

c. Construct a 95% confidence interval estimate of the population mean life of the CFLs.

d. Compare the results of (a) and (c). What conclusions do you reach?

Need in EXCEL, please.

Thank you in advance.

Answer 1

To Test :-

H₀ :-  

H₁ :-  

Test Statistic :-

Z = -2

Test Criteria :-

Reject null hypothesis if

Result :- Reject null hypothesis

P value = P ( Z < - 2 ) = 0.02275

SInce the alternative hypothesis is two sided, we need to multiply the probability by 2

P value = 2 * 0.02275 = 0.04550

Confidence Interval :-

Lower Limit =

Lower Limit = 7005.0045

Upper Limit =

Upper Limit = 7494.9955

95% Confidence interval is ( 7005.0045 , 7494.9955 )

Since   does not lies in the interval  ( 7005.0045 , 7494.9955 ) hence we reject null hypothesis.

To calculate P value in excel

=2*(1-NORMSDIST(2) )