Question

You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately σ = 58.2 σ=58.2. You would like to be 99% confident that your estimate is within 1 of the true population mean. How large of a sample size is required? Do not round mid-calculation.

Answer 1

`n=((2.58(58.2))/(1))^2 =22546.82 \approx 22547`

So the answer for this case would be n=22547 rounded up to the nearest integer

Explanation:

Let's define some notation

`\bar X` represent the sample mean

`\mu` population mean (variable of interest)

`\sigma=58.2` represent the population standard deviation

n represent the sample size

`ME =1` represent the margin of error desire

The margin of error is given by this formula:

`ME=z_(\alpha/2)(\sigma)/(√(n))` (a)

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

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

The critical value for 99% of confidence interval now can be founded using the normal distribution. The significance would be
`\alpha=0.01` and the critical value
`z_(\alpha/2)=2.58`, replacing into formula (b) we got:

`n=((2.58(58.2))/(1))^2 =22546.82 \approx 22547`

So the answer for this case would be n=22547 rounded up to the nearest integer