Question

Please Help!!!

Answer 1

Answer : (2.67, 3.53)

Step to step explanation:

Confidence interval for mean, when population standard deviation is unknown:

$\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}$

, where $\overline{x}$ = sample mean

n= sample size

s= sample standard deviation

$t_{\alpha/2}$ = Critical t-value for n-1 degrees of freedom

We assume the family size is normal distributed.

Given, n= 31 , $\overline{x}=3.1$, s= 1.42 ,

$\alpha=1-0.9=0.10$

Critical t value for $\alpha/2=0.05$ and degree of 30 freedom

$t_{\alpha/2}$ = 1.697  [By t-table]

The required confidence interval:

$3.1\pm ( 1.697)\dfrac{1.42}{\sqrt{31}}\\\\=3.1\pm0.4328\\\\=(3.1-0.4328,\ 3.1+0.4328)=(2.6672,\ 3.5328)\approx(2.67,\ 3.53)$

Hence,  the 90% confidence interval for the estimate = (2.67, 3.53)