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 :-

H_{0} :-

H_{1} :-

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) )

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