Question

The age of U.S. millionaires is normally distributed with a standard deviation o = 13.0years. A group of 36 millionaires is randomly selected and their mean age is & = 58.5years old. Construct a 95% confidence interval for the mean age of U.S. millionaires.

Answer 1

a) Parameter, population standard deviation σ = 13.0

__

Sample statistics n=36 x = 58.5

b)The margin of error E is 4.2467

c)The confidence interval is (54.2533, 62.7467)

d) we are 95% confident that the mean age of US millionaires is between (54.2533 , 62.7467)

EXPLANATION

From the given question;

a) The given Parameter or (statistics)

Parameter, population standard deviation σ = 13.0

__

Sample statistics n=36 x =58.5

b) The margin of Error E can be calculated using the formula below:

$M.E=Z_{\frac{\propto}{2}}\times\frac{\sigma}{\sqrt[]{n}}$

Substitute the the values into the formula and simplify.

$=1.96\times\frac{13}{\sqrt[]{36}}$

$=1.96\times\frac{13}{6}$

$=\frac{25.48}{6}$

$=4.2467$

Hence, the margin of error E is 4.2467

c) The confidence interval can be calculated using the formula below:

$C.I=\bar{x}\text{ }\pm Z_{\propto\text{ /2}}\times\frac{\sigma}{\sqrt[]{n}}$

$This\text{ implies; C.I = mean age }\pm\text{ margine error}$

Substitute the values and simplify.

$C\mathrm{}I=58.5\pm4.2467$

$C.I=(54.2533,\text{ 62.7467)}$

Hence, the confidence interval is (54.2533, 62.7467)

d)Conclusion

Hence, we are 95% confident that the mean age of US millionaires is between (54.2533 , 62.7467)