Question

One person wants to get a 95% z-confidence interval with a margin of error of at most 15 based on a population standard deviation of 60. What is the minimum sample size needed?

Answer 1

`n=((1.96(60))/(15))^2 =61.46 \approx 62`

So the answer for this case would be n=62 rounded up to the nearest integer

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`\bar X` represent the sample mean for the sample

`\mu` population mean

`\sigma=60` represent the population standard deviation

n represent the sample size (variable of interest)

Confidence =95% or 0.95

The margin of error is given by this formula:

`ME=z_(\alpha/2)(\sigma)/(√(n))` (1)

And on this case we have that ME =15, and we are interested in order to find the value of n, if we solve n from equation (1) we got:

`n=((z_(\alpha/2) \sigma)/(ME))^2` (2)

The critical value for 95% of confidence interval is provided,
`z_(\alpha/2)=1.96`, replacing into formula (2) we got:

`n=((1.96(60))/(15))^2 =61.46 \approx 62`

So the answer for this case would be n=62 rounded up to the nearest integer