Patrick B. answered 08/24/19
Math and computer tutor/teacher
99% confidence, alpha = 2.58
1/2 = 16.4/sqrt(N)
Multiplies both sides by 2*sqrt(N):
sqrt(N) = 16.4 *2
N = (32.8)^2 = 1075.84
Ronald L.
asked 08/24/19The ages of a group of 151 randomly selected adult females have a standard deviation of
16.416.4
years. Assume that the ages of female statistics students have less variation than ages of females in the general population, so let
sigmaσequals=16.4
years for the sample size calculation. How many female statistics student ages must be obtained in order to estimate the mean age of all female statistics students? Assume that we want
99%
confidence that the sample mean is within one-half year of the population mean. Does it seem reasonable to assume that the ages of female statistics students have less variation than ages of females in the general population?
The required sample size is:
Patrick B. answered 08/24/19
Math and computer tutor/teacher
99% confidence, alpha = 2.58
1/2 = 16.4/sqrt(N)
Multiplies both sides by 2*sqrt(N):
sqrt(N) = 16.4 *2
N = (32.8)^2 = 1075.84
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