The principal at Riverside High School would like to estimate the mean length of time each day that it takes all the buses to arrive and unload the students. How large a sample is needed if the principal would like to assert with 90% confidence that the sample mean is off by, at most, 7 minutes. Assume that s=14 minutes based on previous studies.
A) 12
B) 13
C) 11
D) 10
E) 35

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".

[tex]\bar X[/tex] represent the sample mean for the sample

[tex]\mu[/tex] population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that [tex]z_{\alpha/2}=1.64[/tex]

The margin of error is given by this formula:

[tex] ME=z_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]

And on this case we have that ME =7 and we are interested in order to find the value of n, if we solve n from the last equation we got:

[tex]n=(\frac{z_{\alpha/2} s}{ME})^2[/tex]

And replacing the values that we got we have:

[tex]n=(\frac{1.64(14)}{7})^2 =10.758 \approx 11[/tex]

C) 11