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 7,531 hours. The population standard deviation is 1,080 hours. A random sample of 8181 light bulbs indicates a sample mean life of 7,231 hours.

a. At the 0.05 level of​ significance, is there evidence that the mean life is different from 7,531 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?

A) Let μ be the population mean. Determine the null​ hypothesis, H0​, and the alternative​ hypothesis, H1.

H0​: μ=

H1​: μ≠

What is the test​ statistic?

ZSTAT=       ​(Round to two decimal places as​ needed.)

What​ is/are the critical​ value(s)?

​(Round to two decimal places as needed. Use a comma to separate answers as​ needed.)

What is the final​ conclusion?

A. Fail to reject H0. There is not sufficient evidence to prove that the mean life is different from 7,531 hours.

B. Reject H0. There is sufficient evidence to prove that the mean life is different from 7,531 hours.

C. Fail to reject H0. There is sufficient evidence to prove that the mean life is different from 7,531 hours.

D. Reject H0. There is not sufficient evidence to prove that the mean life is different from 7,531 hours.

B) What is the​ p-value?

​(Round to three decimal places as​ needed.)

Interpret the meaning of the​ p-value. Choose the correct answer below.

A. Reject H0. There is sufficient evidence to prove that the mean life is different from 7,531 hours.

B. Fail to reject H0. There is sufficient evidence to prove that the mean life is different from 7,531 hours.

C. Reject H0. There is not sufficient evidence to prove that the mean life is different from 7,531 hours.

D. Fail to reject H0. There is not sufficient evidence to prove that the mean life is different from 7,531 hours.

Answer 1

a)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 7531
Alternative Hypothesis, Ha: μ ≠ 7531

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (7231 - 7531)/(1080/sqrt(81))
z = -2.50

Rejection Region
This is two tailed test, for α = 0.05
Critical value of z are -1.96 and 1.96.
Hence reject H0 if z < -1.96 or z > 1.96

B. Reject H0. There is sufficient evidence to prove that the mean life is different from 7,531 hours.

b)

P-value Approach
P-value = 0.012
As P-value < 0.05, reject the null hypothesis.

B. Reject H0. There is sufficient evidence to prove that the mean life is different from 7,531 hours.

c)

sample mean, xbar = 7231
sample standard deviation, σ = 1080
sample size, n = 81

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

ME = zc * σ/sqrt(n)
ME = 1.96 * 1080/sqrt(81)
ME = 235.2

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (7231 - 1.96 * 1080/sqrt(81) , 7231 + 1.96 * 1080/sqrt(81))
CI = (6996 , 7466)

d

Th econclusions are same in a) and c)