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.

(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 :

Sample standard deviation : s = 3.22

Significance level :

Degree of freedom: df = n-1 = 25

Critical t-value for and df = 25 will be

90% confidence interval for mean:

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

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