Answer: (11.95, 14.11)
Explanation:
Let x be a random variable that represent the number of years of education.
Given: Sample size : n= 26
Sample mean : [tex]\overline{x}=13.03[/tex]
Sample standard deviation : s = 3.22
Significance level : [tex]\alpha=100\%-90\%=10\%=0.1[/tex]
Degree of freedom: df = n-1 = 25
Critical t-value for [tex]\alpha=0.1[/tex] and df = 25 will be
[tex]t_{\alpha/2, df}=t_{0.05,25}=1.7081[/tex]
90% confidence interval for mean:
[tex]\overline{x}\pm t_{\alpha/2,df}\dfrac{s}{\sqrt{n}}\\\\=13.03\pm (1.7081)\dfrac{3.22}{\sqrt{26}}\\\\=13.03\pm (1.7081)\dfrac{3.22}{5.09902}\\\\=13.03\pm 1.08\\\\=(13.03-1.08,\ 13.03+1.08)\\\\ =(11.95,\ 14.11)[/tex]
A 90% confidence interval for the mean number of years of education = (11.95, 14.11)
