Question

Your firm wants to investigate how much money the typical tourist will spend on their next visit to New York? How many tourists should be included in your sample if you want to be 99% confident that the sample mean is within $20 from the population mean?
A) 56B) 57C) 58D) 59

Answer 1

Answer:

$n=(\frac{2.58(59)}{20})^2 =57.93$

And if we round up the answer we see that the value of n to ensure the margin of error required $\pm=20$ $ is n=58.

C) 58

Step-by-step explanation:

Assuming this :"Your firm wants to investigate how much money the typical tourist will spend on their next visit to New York. How many tourists should be included in your sample if you want to be 99% confident that the sample mean is within $20 from the population mean? From previous studies, we know that the standard deviation is $59."

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".

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma=59)$

We know that the margin of error for a confidence interval is given by:

$Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

The next step would be find the value of $\z_{\alpha/2}$, $\alpha=1-0.99=.01$ and $\alpha/2=0.005$

Using the normal standard table, excel or a calculator we see that:

$z_{\alpha/2}=2.58$

If we solve for n from formula (1) we got:

$\sqrt{n}=\frac{z_{\alpha/2} \sigma}{Me}$

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

And we have everything to replace into the formula:

$n=(\frac{2.58(59)}{20})^2 =57.93$

And if we round up the answer we see that the value of n to ensure the margin of error required $\pm=20$ $ is n=58.

C) 58