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

Solution:-

a) 99% confidence interval for μ₁- μ₂ is C.I = (51.9, 78.1).

C.I = (1347 - 1282) + 2.615*5

C.I = 65 + 13.075

C.I = (51.925, 78.08)

Lower bound = 51.9

Upper bound = 78.1

b) True, We can reject the Null Hypothesis.

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u₁ - u₂ = 50
Alternative hypothesis: u₁ - u₂ > 50

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the mean difference between sample means is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.01. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 2.615
DF = 129

t = [ (x₁ - x₂) - d ] / SE

t = 5.74

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is thesize of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.

The observed difference in sample means produced a t statistic of 5.74.

Therefore, the P-value in this analysis is 0.00

Interpret results. Since the P-value (0.00) is less than the significance level (0.01), hence we have to reject the null hypothesis.

We can reject the Null Hypothesis.