As an aid to the establishment of personnel requirements, the director of a hospital wishes to estimate the mean number of people who are admitted to the emergency room during a 24-hour period. The director randomly selects 64 different 24-hour periods and determines the number of admissions for each. For this sample, the sample mean= 9.8 and the sample variance=25. If the director wishes to estimate the mean number of admissions per 24-hour period to within 1 admission with 99% reliability, what size sample should she choose?

a. n = 482
b. n = 166
c. n = 154
d. n = 123

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

Solution to the problem

The margin of error is given by this formula:

[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (a)

And on this case we have that ME =1 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

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

The critical value for 99% of confidence interval now can be founded using the normal distribution. And in excel we can use this formula to find it:"=-NORM.INV(0.005;0;1)", and we got [tex]z_{\alpha/2}=2.57583[/tex], replacing into formula (b) we got:

[tex]n=(\frac{2.57583(5)}{1})^2 =165.87 \approx 166[/tex]

So the answer for this case would be n=166 rounded up to the nearest integer

b. n = 166