Question

Construct a 95% confidence interval for the population mean, μ. Assume the population has a normal distribution. A sample of 20 part-time workers had mean annual earnings of $3120 with a standard deviation of $677. Round to the nearest dollar.

Answer 1

Answer: (2823, 3417)

Explanation:

The confidence interval for the population mean is given by :-

$\mu\ \pm E$, where E is the margin of error.

Formula for Margin of error :-

$z_{\alpha/2}\times\dfrac{\sigma}{\sqrt{n}}$

Given : Significance level : $\alpha=1-0.95=0.05$

Sample size = 20

$\mu=3120$

$\sigma=677$

Critical value : $z_{\alpha/2}=z_{0.025}=\pm1.96$

Margin of error : $E=1.96\times\dfrac{677}{\sqrt{20}}\approx 297$

Now, the 95% confidence interval for the population mean will be :-

$3120\ \pm\ 297 =(3120-297\ ,\ 3120+297)=(2823,\ 3417)$