Question

Battery life for a hand-held computer is normally distrituted and has a population standard deviation of 7 hours. Suppose you need to estimate a confidence interval estimate at the 95% level of confidence for the mean life of these batteries. Determine the sample size required to have a margin of error of 0.585 hours. Round up to the nearest whole number.

Answer 1

Answer:

A sample size of 551 is required.

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:

$\alpha = \frac{1 - 0.95}{2} = 0.025$

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 $1 - 0.025 = 0.975$, so Z = 1.96.

Now, find the margin of error M as such

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

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

Population standard deviation of 7 hours.

This means that $\sigma = 7$

Determine the sample size required to have a margin of error of 0.585 hours.

This is n for which M = 0.585. So

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

$0.585 = 1.96\frac{7}{\sqrt{n}}$

$0.585\sqrt{n} = 1.96*7$

$\sqrt{n} = \frac{1.96*7}{0.585}$

$(\sqrt{n})^2 = (\frac{1.96*7}{0.585})^2$

$n = 550.04$

Rounding up(as for a sample of 550 the margir of error will be a bit above the desired target):

A sample size of 551 is required.