Question

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, he calculated a mean a 396 and a standard deviation of 100. The 95% confidence interval results in an interval of 396 +/- 24.9793. Identify the sampling error or margin of error.

a) 396

b) +/- 24.9793

c) 100/√64

d) 95%

Answer 1

Solution:

Degrees of freedom = df = n - 1 = 63

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,63 = 1.998

Margin of error = E = t_/2,df * (s /n)

= 1.998 * ( 100/ 64)

= 24.9793

Option b is correct.