Question

A researcher, interested in estimating a population mean wants to be 95% certain that the length of the confidence interval does not exceed 5. Find the required sample size for his study if the population standard deviation is 20.

Answer 1

Answer:

$n=(\frac{1.96(20)}{5})^2 =61.47 \approx 62$  

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".  

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 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$  

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 =L/2 =5/2 =2.5, since the margin of error is the half of the length for the confidence interval, 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, $z_{\alpha/2}=1.96$ from part a, replacing into formula (2) we got:  

$n=(\frac{1.96(20)}{5})^2 =61.47 \approx 62$  

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