As an aid to the establishment of personnel requirements the

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 mean is 19.8 and sample standard deviation is 5. 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?

Select one:

A. 664

B. 13

C. 166

D. 165

Solution

Margin of Error (half of confidence interval) = 1
The margin of error is defined as the \"radius\" (or half the width) of a confidence interval for a particular statistic.
Level of Confidence = 99
: population standard deviation = 5
(\'z critical value\') from Look-up Table for 99% = 2.576
The Look-up in the Table for the Standard Normal Distribution utilizes the Table\'s cummulative \'area\' feature. The Table shows positve and negative values of (\'z critical\') but since the Standard Normal Distribution is symmetric, only the magnitude of (\'z critical\') is important.


For a Level of Confidence = 99% the corresponding LEFT \'area\' = 0.495. And due to Table\'s symmetric nature, the corresponding RIGHT \'area\' = 0.495 The (\'z critical\') value Look-up is 2.576

Margin of Error = (\'z critical value\') * /SQRT(n)
n = Sample Size

Algebraic solution for n:
n = [(\'z critical value\') * /Margin of Error]²
= [ (2.576 * 5)/1 ]²

Sample Size = 166 for 99% level of confidence