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.

Question

Quizplus · Accepted Answer

The 90% confidence interval for the mean can be calculated using the formula: $\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $t_{\alpha/2}$ is the t-score for the desired confidence level (which depends on the degrees of freedom, $n-1$), $s$ is the sample standard deviation, and $n$ is the sample size. Given $\bar{x} = 13.03$, $s = 3.22$, and $n = 26$, and knowing that the degrees of freedom $df = 25$, for a 90% confidence interval, the t-score is approximately 1.708. Plugging these values into the formula gives a margin of error of approximately 1.08, leading to a confidence interval of (12.0, 14.1).

A Survey Asked 26 Adults How Many Years of Education