Question

1.    Find 95% confidence limits for the mean of a normally distributed population from which the following samples was taken 15,17,10,18,16,9,7,11,13,14.

Answer 1

$c = 0.95 \\ n = 10 \\ \bar{x} = \frac{15+17+10+18+16+9+7+11+13+14}{10} = 13 \\ s = \sqrt{ \frac{(15-13)^2 + (17-13)^2 + (10-13)^2 + (18-13)^2 + (16-13)^2 + (9-13)^2 + (7-13)^2 + (11-13)^2 + (13-13)^2 + (14-13)^2}{10-1} } \\ = \sqrt{ \frac{4 + 16 + 9 + 25 + 9 + 16 + 36 + 4 + 0 + 1}{9} } \\ = 3.651 \\ df = n-1 = 10 -1 = 9 \\ α = \frac{1-c}{2} = 0.025 \\ t_{α/2}= 2.26216$

The margin of error is:

$E = t_{α/2} \times \frac{s}{\sqrt{n}} = 2.262 \times \frac{3.651}{\sqrt{10}} = 2.6118$

The confidence interval:

$\bar{x} -E < μ < \bar{x} +E \\ 10 -2.61 < μ < 10 +2.61 \\ 7.39 < μ < 12.61$