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 = 54.5 . You would like to be 99% confident that your estimate is within 5 of the true population mean. How large of a sample size is required

Answer 1

Answer:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$   (3)

$n=(\frac{2.58(54.5)}{5})^2 =790.846 \approx 791$

So the answer for this case would be n=791 rounded up to the nearest integer

Step-by-step 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 (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that $z_{\alpha/2}=2.58$

The margin of error is given by this formula:

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

And on this case we have that ME =+5 and we are interested in order to find the value of n, if we solve n from equation (2) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$   (3)

$n=(\frac{2.58(54.5)}{5})^2 =790.846 \approx 791$

So the answer for this case would be n=791 rounded up to the nearest integer