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

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

b. n = 166

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".

represent the sample mean for the sample  

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:

   (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:

  (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 , replacing into formula (b) we got:

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

b. n = 166

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