Question

A marketing researcher wants to find a 96% confidence interval for the mean amount those visitors spend per person per day while visiting a theme park. She knows that the population standard deviation is $12. How large a sample should the researcher select so that the estimate will be within $4 of the population mean?

Answer 1

Answer:

$n=(\frac{2.054(12)}{4})^2 =37.97 \approx 38$

So then the minimum sample to ensure the condition given is n= 38

Step-by-step explanation:

Notation

$\bar X$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

$\sigma=12$ represent the population standard deviation

n represent the sample size  

ME = 4 the margin of error desired

Solution to the problem

When we create a confidence interval for the mean 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 =4 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 96% of confidence interval now can be founded using the normal distribution. The significance is $\alpha=1-0.96 =0.04$. And in excel we can use this formula to find it:"=-NORM.INV(0.02;0;1)", and we got $z_{\alpha/2}=2.054$, replacing into formula (b) we got:

$n=(\frac{2.054(12)}{4})^2 =37.97 \approx 38$

So then the minimum sample to ensure the condition given is n= 38