Question #319409

A random of 25 sample is drawn from a population with sample mean 105.2 and standard deviation 11.13 Construct a 90% confidence interval. *



Expert's answer

sample mean(x) =105.2

standard deviation (s)=11.13

n=25

Confidence coefficient(t) corresponding to 90% confidence level with a degree of freedom of 24 is 1.711

% C.I =(x- t⋅s√(n)\cdot \dfrac{s}{\surd(n)} ,x+ t⋅s√(n)\cdot \dfrac{s}{\surd(n)})


90 % C.I=(105.2-1.711⋅\cdot 11.13√(25)\dfrac{11.13}{\surd(25 )} ,105.2+1.711⋅\cdot 11.13√(25)\dfrac{11.13}{\surd(25 )} )


90% C.I=(105.2-3.8087,105.2+3.8087)

90% C.I=(101.3913,109.0087)

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