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

Answer:

$n=(\frac{1.96(60)}{15})^2 =61.46 \approx 62$  

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

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

$\sigma=60$ 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:  

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

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

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

$n=(\frac{1.96(60)}{15})^2 =61.46 \approx 62$  

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