Question

The​ quality-control manager at a compact fluorescent light bulb​ (CFL) factory needs to determine whether the mean life of a large shipment of CFLs is equal to 7,487 hours. The population standard deviation is 1,080 hours. A random sample of 81 light bulbs indicates a sample mean life of 7,187 hours. a. At the 0.05 level of​ significance, is there evidence that the mean life is different from 7,487 hours? b. Compute the​ p-value and interpret its meaning. c. Construct a 95​% confidence interval estimate of the population mean life of the light bulbs. d. Compare the results of​ (a) and​ (c). What conclusions do you​ reach?

Answer 1

a)  

One sample z test:

Claim : The mean life is different from 7487 hours

H₀ : µ = 7487 vs H_a : µ ≠ 7487

Given : = 7187, σ = 1080 , n = 81

Population standard deviation σ is known therefore we use z statistic.

Test statistic:

Z = =

Z = -2.50

Critical value : α = 0.05

As H_a contain ≠ sign , this is two tail test,

Therefore Z_{(α/2 )} = 1.96

Critical region: Reject H₀, if |Z | ≥ 1.96 Or fail to reject H0 , if | Z | < 1.96

Decision: z = -2.50 , | z | = 2.50

As | Z | is greater than 1.96 , we reject H₀

Conclusion: There is significant evidence that the mean life is different from 7487 hours

b) P-value = 2*P( z ≤ -2.50 ) = 2*0.0062 = 0.0124

The probability that sample mean fall outside the 2.5 standard deviation of the population mean is 0.0124

c)

Given confidence level = 0.95

α = 1- 0.95 = 0.05 , α/2 = 0.025 , So z_α/2 = 1.96

Confidence interval =

=

= ( 6951.8 , 7422.2 )

Population mean 7484 do not fall in the 95% confidence interval.

d) As population mean 7484 do not fall in the 95% confidence interval, we conclude the same conclusion that the mean life is different from 7487 hours.

So both test and confidence interval gives the same result.