The population estimated minimum sample size will be 11 , and the critical value will be 1.28.
The average gorilla weight is being calculated as a range estimate by a researcher (in pounds).
She wants to be within 7.1 pounds of the actual average with 80% accuracy.
Previous research has shown that the standard deviation of the weights of gorillas is 18.2 pounds.
a) Since the confidence is 0.80 or 80% then the value of α is 0.20 and and we can calculate the critical value. Then the value of the z-score will be z = 1.28.
b) The margin of error is given as
M.E.= z*(standard deviation/√n)
We have,
Margin of error = 7.1
From the above formula we get;
n = ((z* standard deviation )/marginal error)²
n = ((1.28*18.2)/7.1)²
= 10.76 ≈ 11
Therefore, the researcher should take 11 samples.
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