How large a sample is required if we want to have 90% confidence that the averag

Question

How large a sample is required if we want to have 90% confidence that the average tensile strength of a wood panel is within 0.4 (margin of error) of the expected value if the variance is known to be 2.0? If the observed concentration readings are 2.56, 2.87, 2.06, 2.80, and 2.43, determine 95% confidence interval for the mean concentration assuming variance (sigma^2) = 0.005. If the average weight for a male population is approximately normally distributed with a mean (mu) of 196 pounds and a standard deviation (sigma) of 35 pounds. Ask: What is the lower 10^th percentile of the weight for the male population?

Explanation / Answer

Q3.

Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.1% LOS is = 1.64 ( From Standard Normal Table )
Standard Deviation ( S.D) = 2
ME =0.4
n = ( 1.64*2/0.4) ^2
= (3.28/0.4 ) ^2
= 67.24 ~ 68      

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