Question

An economics professor randomly selected 100 millionaires in the U.S. The average age of these millionaires was 52.1 years with a standard deviation of 12.3 years. What is a 95% confidence interval for the mean age, μ, of all U.S. millionaires?

Answer 1

Answer:

$52.1-1.984\frac{12.3}{\sqrt{100}}=49.66$    

$52.1 +1.984\frac{12.3}{\sqrt{100}}=54.54$    

The 95% confidence interval would be given by (49.66;54.54)    

Step-by-step explanation:

Information given

$\bar X= 52.1$ represent the sample mean

$\mu$ population mean (variable of interest)

s=12.3 represent the sample standard deviation

n=100 represent the sample size  

Confidence interval

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

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

The degrees of freedom are given by:

$df=n-1=100-1=99$

The Confidence is 0.95 or 95%, and the significance would be $\alpha=0.05$ and $\alpha/2 =0.025$, the critical value for this case is: $t_{\alpha/2}=1.984$

Replacing the info given we got:

$52.1- 1.984\frac{12.3}{\sqrt{100}}=49.66$    

$52.1 +1.984\frac{12.3}{\sqrt{100}}=54.54$    

The 95% confidence interval would be given by (49.66;54.54)