Question

A simple random sample of electronic components wil be selected to test for the mean lifetime in hours. Assume that component lifetimes are normally distributed with population standard deviation of 16 hours. How many components must be sampled so that a 99% confidence interval will have margin of error of 3 hours

Answer 1

Answer:

189 components must be sampled.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$.

That is z with a pvalue of $1 - 0.005 = 0.995$, so Z = 2.575.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

Assume that component lifetimes are normally distributed with population standard deviation of 16 hours.

This means that $\sigma = 16$

How many components must be sampled so that a 99% confidence interval will have margin of error of 3 hours?

n components must be sampled.

n is found when M = 3. So

$M = z\frac{\sigma}{\sqrt{n}}$

$3 = 2.575\frac{16}{\sqrt{n}}$

$3\sqrt{n} = 2.575*16$

$\sqrt{n} = \frac{2.575*16}{3}$

$(\sqrt{n})^2 = (\frac{2.575*16}{3})^2$

$n = 188.6$

Rounding up:

189 components must be sampled.