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

A sample size of 551 is required.

Explanation:

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

`\alpha = (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(\sigma)/(√(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(\sigma)/(√(n))`

`0.585 = 1.96(7)/(√(n))`

`0.585√(n) = 1.96*7`

`√(n) = (1.96*7)/(0.585)`

`(√(n))^2 = ((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.