Question

1. 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 at least 7,500 hours. A random sample of 64 CFLs indicates a sample mean life of 7,250 hours. At the 0.05 significance level, is there evidence that the mean life is less than 7,500 hours?

1. Assume the population standard deviation is 1,000 hours.
2. Assume the sample standard deviation is 1,000 hours.

Answer 1

a)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 7500
Alternative Hypothesis, Ha: μ < 7500

Rejection Region
This is left tailed test, for α = 0.05
Critical value of z is -1.645.
Hence reject H0 if z < -1.645

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (7250 - 7500)/(1000/sqrt(64))
z = -2

P-value Approach
P-value = 0.0228
As P-value < 0.05, reject the null hypothesis.

yes, there is sufficient evidence that the mean life is less than 7,500 hour

b)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 7500
Alternative Hypothesis, Ha: μ < 7500

Rejection Region
This is left tailed test, for α = 0.05 and df = 63
Critical value of t is -1.669.
Hence reject H0 if t < -1.669

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (7250 - 7500)/(1000/sqrt(64))
t = -2

P-value Approach
P-value = 0.0249
As P-value < 0.05, reject the null hypothesis.

yes, there is sufficient evidence that the mean life is less than 7,500 hour