Question

A travel agent wants to determine how much the average client is willing to pay for a weekend at an all-expense paid resort. The agent surveys 30 clients and finds that the average willingness to pay is $2,500 with a standard deviation of $840. However, the travel agent is not satisfied and wants to be 95% confident that the sample mean falls within $150 of the true average. What is the minimum number of clients the travel agent should survey

Answer 1

Answer:

$n=(\frac{1.960(840)}{150})^2 =120.47 \approx 121$

So the answer for this case would be n=12 rounded up to the next integer

Explanation:

$\bar X=2500$ represent the sample mean

$\mu$ population mean (variable of interest)

s=840 represent the sample standard deviation

n represent the sample size  

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 =150 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 95% of confidence interval, the significance level if 5% and the critical value would be $z_{\alpha/2}=1.960$, replacing into formula (b) we got:

$n=(\frac{1.960(840)}{150})^2 =120.47 \approx 121$

So the answer for this case would be n=12 rounded up to the next integer