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

`52.1-1.984(12.3)/(√(100))=49.66`

`52.1 +1.984(12.3)/(√(100))=54.54`

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

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)(s)/(√(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(12.3)/(√(100))=49.66`

`52.1 +1.984(12.3)/(√(100))=54.54`

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