How large of a sample size do I need to be 95% confident that the half-length is no more than 2 tons for a sample that has an average of 865 tons and an estimated standard deviation of 78?
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Solution
standard deviation =s = =78
Margin of error = E = 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 ( Using z table ( see the 0.025 value in standard normal (z) table corresponding z value is 1.96 )
sample size = n = [Z/2* / E] 2
n = ( 1.96*78 / 2)2
n =5843
Sample size = n =5843
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