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Sagot :
The margin of error for 16 eggs with sample means 2 ounce and standard deviation as 42 ounces is 18.407 .
In the question ,
it is given that ,
the number of eggs = 16
the sample means weight is = 2 ounce .
the standard deviation is = 42 ounces .
we have to find the margin of error .
degree of freedom is 16 - 1 = 15 .
and the t score for the 90% confidence interval is
t₁₅,₀.₀₅ = 1.753
So , the margin of error is = t₁₅,₀.₀₅×(standard deviation/√n)
Substituting the values ,
we get ,
= 1.753×(42/√16)
On simplifying further ,
we get ,
= 18.407
Therefore , the margin of error is = 18.407 .
The given question is incomplete , the complete question is
A sample of 16 eggs yields a sample mean weight of 2.00 ounces and a sample standard deviation = 42.0 ounces, find the margin of error in estimating a confidence interval estimate at the 90% level of confidence. Round your answer to three decimal places. You may assume that egg weights are normally distributed .
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