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:

A sample size of 551 is required.

Step-by-step explanation:

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

[tex]\alpha = \frac{1 - 0.95}{2} = 0.025[/tex]

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

That is z with a pvalue of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.

Now, find the margin of error M as such

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

Population standard deviation of 7 hours.

This means that [tex]\sigma = 7[/tex]

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

This is n for which M = 0.585. So

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

[tex]0.585 = 1.96\frac{7}{\sqrt{n}}[/tex]

[tex]0.585\sqrt{n} = 1.96*7[/tex]

[tex]\sqrt{n} = \frac{1.96*7}{0.585}[/tex]

[tex](\sqrt{n})^2 = (\frac{1.96*7}{0.585})^2[/tex]

[tex]n = 550.04[/tex]

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.

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