Question

The lifetime (in hours) of a 60-watt light bulb is a random variable that has a Normal distribution with σ = 30 hours. A random sample of 25 bulbs put on test produced a sample mean lifetime of = 1038. Construct a 92% confidence interval estimate for the mean lifetime μ. If it were desired to cut the confidence interval to half its length while keeping the same 92% level, what size sample would be required to achieve this?

Answer 1

Answer:

a) 92% Confidence interval: (1027.5,1048.5)

b) Sample size = 100

Step-by-step explanation:

We are given the following in the question:

Sample mean, $\bar{x}$ = 1038

Sample size, n = 25

Alpha, α = 0.08

Population standard deviation, σ = 30

a) 92% Confidence interval:

$\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$z_{critical}\text{ at}~\alpha_{0.08} = \pm 1.75$

$1038 \pm 1.75(\frac{30}{\sqrt{25}} ) = 1038 \pm 10.5 = (1027.5,1048.5)$

b) In order to reduce the confidence interval by half, we have to quadruple the sample size.

Thus,

$\text{Sample size} = 25\times 4 = 100$