Question

A survey asked 26 adults how many years of education they had. The sample mean was 13.03 with a standard deviation of 3.22. It is reasonable to assume that the population is approximately normal. Construct a 90% confidence interval for the mean number of years of education. (I saw this one on g when I tried to socratic it.

Answer 1

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 : $\overline{x}=13.03$

Sample standard deviation : s = 3.22

Significance level : $\alpha=100\%-90\%=10\%=0.1$

Degree of freedom: df = n-1 = 25

Critical t-value for $\alpha=0.1$ and df = 25 will be

$t_{\alpha/2, df}=t_{0.05,25}=1.7081$

90% confidence interval for mean:

$\overline{x}\pm t_{\alpha/2,df}\dfrac{s}{\sqrt{n}}\\\\=13.03\pm (1.7081)\dfrac{3.22}{\sqrt{26}}\\\\=13.03\pm (1.7081)\dfrac{3.22}{5.09902}\\\\=13.03\pm 1.08\\\\=(13.03-1.08,\ 13.03+1.08)\\\\ =(11.95,\ 14.11)$

A 90% confidence interval for the mean number of years of education = (11.95, 14.11)