Question

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?

Answer #1

:

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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