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.

Respuesta :

Answer: (2823, 3417)

Explanation:

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

[tex]\mu\ \pm E[/tex], where E is the margin of error.

Formula for Margin of error :-

[tex]z_{\alpha/2}\times\dfrac{\sigma}{\sqrt{n}}[/tex]

Given : Significance level : [tex]\alpha=1-0.95=0.05[/tex]

Sample size = 20

[tex]\mu=3120[/tex]

[tex]\sigma=677[/tex]

Critical value : [tex]z_{\alpha/2}=z_{0.025}=\pm1.96[/tex]

Margin of error : [tex]E=1.96\times\dfrac{677}{\sqrt{20}}\approx 297[/tex]

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

[tex]3120\ \pm\ 297 =(3120-297\ ,\ 3120+297)=(2823,\ 3417)[/tex]