Question

Twenty-five adult citizens of the United States were asked to estimate the average income of all U.S. households. The mean estimate was x¯=$65,000 and s=$15,000 . (Note: The actual average household income at the time of the study was about $79,000 .) Assume the 25 adults in the study can be considered an SRS from the population of all adult citizens of the United States. A 95% confidence interval for the mean estimate of the average income of all U.S. households is

Answer 1

Answer:

$65000-2.06 \frac{15000}{\sqrt{25}}=58820$    

$65000+2.06 \frac{15000}{\sqrt{25}}=71180$    

We are confident that the true mean for the average income of all U.S. households is between (58820;71180)    

Step-by-step explanation:

Information provided

$\bar X=65000$ represent the sample mean for the average income of all US households

$\mu$ population mean

s=15000 represent the sample standard deviation

n=25 represent the sample size  

Confidene interval

The confidence interval for the true mean of interest is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

The degrees of freedom are given by:

$df=n-1=25-1=24$

The Confidence level is is 0.95 or 95%, then the significance is $\alpha=0.05$ and $\alpha/2 =0.025$, the critical value for this case would be $t_{\alpha/2}=2.06$

Replacing in formula (1) we got:

$65000-2.06 \frac{15000}{\sqrt{25}}=58820$    

$65000+2.06 \frac{15000}{\sqrt{25}}=71180$    

We are confident that the true mean for the average income of all U.S. households is between (58820;71180)