Question

Salaries of 39 college graduates who took a statistics course in college have a​ mean, x​, of $ 61,600. Assuming a standard​ deviation, sigma​, of ​$17,362​, construct a 95​% confidence interval for estimating the population mean, μ.

Answer 1

Answer:  (56150.92, 67049.08)

Step-by-step explanation:

The confidence interval for population mean is given by :-

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

Given : Sample size : $n=39$

Sample mean = $\overline{x}=\ \ 61,600$

Standard deviation : $\sigma=\ \ 17,362$

Significance level : $1-0.95=0.05$

Critical value = $z_{\alpha/2}=1.96$

Now, the  95​% confidence interval for estimating the population mean $\mu$will be :-

$61600\pm (1.96)\dfrac{17362}{\sqrt{39}}\\\\\approx61600\pm5449.08\\\\=(61600-5449.08,61600+5449.08)\\\\=(56150.92,\ 67049.08)$

Hence, the 95​% confidence interval for estimating the population mean = (56150.92, 67049.08)