4. What is the minimum sample size to select to find the mean number of absences per month for school children, within ±0.2 points at a 95 % Confidence Level if it is known that the standard deviation is 1.1 days?
Solution
standard deviation =s = =1.1
Margin of error = E = 0.2
At 95% confidence level the z is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
Z/2 = Z0.025 = 1.96
sample size = n = [Z/2* / E] 2
n = ( 1.96*1.1 / 0.2 )2
n =116.2084
Sample size = n =116
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