Answer : (2.67, 3.53)
Step to step explanation:
Confidence interval for mean, when population standard deviation is unknown:
[tex]\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}[/tex]
, where [tex]\overline{x}[/tex] = sample mean
n= sample size
s= sample standard deviation
[tex]t_{\alpha/2}[/tex] = Critical t-value for n-1 degrees of freedom
We assume the family size is normal distributed.
Given, n= 31 , [tex]\overline{x}=3.1[/tex], s= 1.42 ,
[tex]\alpha=1-0.9=0.10[/tex]
Critical t value for [tex]\alpha/2=0.05[/tex] and degree of 30 freedom
[tex]t_{\alpha/2}[/tex] = 1.697 [By t-table]
The required confidence interval:
[tex]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)[/tex]
Hence, the 90% confidence interval for the estimate = (2.67, 3.53)
