Question

The following data have been collected for a sample from a normal population: 5, 10, 8, 11, 

17, 6, 15, 23, (a) What is the point estimate of population mean and standard deviation? (b) What 

is the confidence interval for population mean at 99 per cent confidence interval?

Answer 1

We have given the population : 5, 10, 8, 11, 17, 6, 15, 23

 a.) The point estimate of population mean $= \dfrac{5+10+8+11+17+6+15+23}{8} = 11.875$

The point estimate of population standard deviation

$= \sqrt{\dfrac{(5-11.875)^2+(10-11.875)^+(8-11.875)^2+(11-11.875)^2+(17-11.875)^2+(6-11.875)^2+(15-11.875)^2+(23-11.875)^2}{8-1}} \\= \sqrt{\dfrac{47.26+3.51+15.01+0.76+26.26+34.51+9.76+123.76}{7}}\\ = 6.10$

b.) The margin of error $= t_{\alpha/2}\times \dfrac{s}{\sqrt{n}}$

$= 3.45 \times \dfrac{6.10}{\sqrt{8}} = 7.46$

Then confidence interval becomes

$\bar x-7.46<\mu<\bar x +7.4\\\implies 11.875-7.46< \mu<11.875+7.46\\ \implies 4.41<\mu<19.33$