Question

A sample of 81 60-watt lightbulbs produced by Dynamics, Inc. obtained a mean lifetime of 1347 hours with a variance of 729 hours. A sample of 50 60-watt lightbulbs produced by National Electronics Corporation obtained a mean lifetime of 1282 hours with a variance of 800 hours.

a) Calculate a 99% confidence interval for μ1- μ2, the difference between the true mean lifetimes of the two type of lightbulbs. Answer to one decimal place, the lower bound and the upper bound of the interval should be correct to one decimal place.

b) Test H₀: μ₁- μ₂ = 50 versus H₁ : μ₁- μ₂ > 50 by setting α = .01 . We can reject the Null Hypothesis.

True or False?

Answer 1

a) We need to construct the 99% confidence interval for the difference between the population means μ_1​−μ_2​, for the case that the population standard deviations are not known. The following information has been provided about each of the samples:

Sample Mean 1 ( )	1347
Sample Standard Deviation 1 (s1​)	729
Sample Size 1 (n1​)	81
Sample Mean 2 ()	1282
Sample Standard Deviation 2 (s2​)	800
Sample Size 2 (n2)	50

Based on the information provided, we assume that the population variances are equal, so then the number of degrees of freedom are df=n₁​+n_{2 ​}− 2= 81+50−2 = 129.

The critical value for α=0.01 and df=129 degrees of freedom is t₁₂₉ = 2.614.

The pooled standard deviation

The standard error is computed as follows:

Now, we finally compute the confidence interval:

Therefore, the 99% confidence interval for the difference between the population means​ is −290.8 < μ_1​−μ₂​ < 420.8

b) Now ,we want to test

H₀: μ₁ - µ₂ = 50

H₁: μ₁ - µ₂ > 50

This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

The critical value for this right-tailed test is t_c = 2.356, for α=0.01 and df = 129.

Test Statistics

= 0.1102

Since it is observed that t= 0.1102 ≤ t_c​=2.356 , it is then concluded that the null hypothesis is not rejected.