Question

Assume the population is normally distributed with: 

Xbar = 96.44           S =10.2        n=15                 t=1.76

Construct a​ 90% confidence interval estimate for the population​ mean, μ

Answer 1

$c = 0.90 \\ n = 15 \\ \bar{x} = 96.44 \\ t_{α/2}= 1.76$

The margin of error is:

$E = t_{α/2} \times \frac{S}{\sqrt{n}} = 1.76 \times \frac{10.2}{\sqrt{15}} = 4.63$

The confidence interval:

$\bar{x} -E < μ < \bar{x} +E \\ 96.44 -4.63 < μ < 96.44 +4.63 \\ 91.81 < μ < 101.07$