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 7,500 hours. The population standard deviation is 1,200 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?

Answer 1

a. The hypotheses are

Rejection region:

Reject the null hypothesis if |Z_0bs|>Z_0.025=1.96

Test statistic calculated as

hence calculated as -1.67

P: value

P value calculated from the t stststics table as

0.095

Conclusion:

Since |Z_obs|<Z_0.05 and P value is >level of significace hence we fail to reject the null hypothesis, hence we can say that we do not have enough evidence to support the claim that mean life is different from 7500.

p value computed as 0.095

Interpretation :

If the p-value is less than 0.05, we reject the null hypothesis that there's no difference between the means and conclude that a significant difference does exist. But in here we have P value greatr than the level of significnce hence we fail to reject our null hyothesis and have no evidence to support our claim.

c.

The formula for estimation is:

μ = ± Z(s)

where:

= sample mean
Z = Z statistic determined by confidence level
s = standard error = √(s2/n)

Calculation

M = 7250
Z = 1.96
s = √(12002/64) = 150

μ = M ± Z(s)
μ = 7250 ± 1.96*150
μ = 7250 ± 293.99

95% CI [6956.01, 7543.99].

We can be 95% confident that the population mean (μ) falls between 6956.01 and 7543.99.

d.)Hence from above calcuation in both parts it is clear that $7500 Lies in the confidence interval and also we failed to reject null hypothesis which was hence we conclude that we cnnot support that the mean is different from $7500.