Question

A survey is to be conducted to determine the average driving in miles by Minnesota State University, Mankato students. The investigator wants to know how large the sample should be so that he/she can be 92% confident on the estimate and the estimate is within 1.5 miles of the true average. A similar study conducted in past and it was found that the standard deviation of the students’ driving distances was 8.2 miles.

Answer 1

Answer:

$n=(\frac{1.75(8.2)}{1.5})^2 =91.52 \approx 92$

So the answer for this case would be n=92 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 (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

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 =1.5 and we are interested in order to find the value of n, if we solve n from equation (b) we got:

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

The critical value for 92% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.04;0;1)", and we got $z_{\alpha/2}=1.75$, replacing into formula (b) we got:

$n=(\frac{1.75(8.2)}{1.5})^2 =91.52 \approx 92$

So the answer for this case would be n=92 rounded up to the nearest integer