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

Answer:

$n=(\frac{2.58(58.2)}{1})^2 =22546.82 \approx 22547$

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

Step-by-step 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:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$    (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:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$   (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:

$n=(\frac{2.58(58.2)}{1})^2 =22546.82 \approx 22547$

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