Question

Find 90% confidence interval for the mean of normal distribution if a sample of size 7 gave the values 9,16,10,14,8,13,14

Answer 1

$\mu=(9+16+10+14+8+13+14)/7=12$

Z(90%)=1.645

$\sigma=\sqrt{\frac{\sum(x-\mu)^2}{n-1}}=\sqrt{\frac{9+16+4+4+16+1+4}{6}}=3$

$CI=\mu \plusmn z\frac{\sigma}{\sqrt{n}}=12 \plusmn 1.645 \frac{3}{\sqrt{7}}=12 \plusmn 1.87$