Answer to Question #319409 in Statistics and Probability for zee

Question

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. *

Answer 1

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$\cdot \dfrac{s}{\surd(n)}$ ,x+ t$\cdot \dfrac{s}{\surd(n)}$)

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

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

90% C.I=(101.3913,109.0087)