Q 8.19RP. - Elementary Statistics

Question

Millionaires. Dr. Thomas Stanley of Georgia State University has surveyed millionaires since $1973$ Among other information, Stanley obtains estimates for the mean age, $\mu$, of all U.S. millionaires. Suppose that $36$ randomly selected U.S. millionaires are the following ages, in years. 

Determine a $95\%$ confidence interval for the mean age $\mu$of all U.S. millionaires. Assume that the standard deviation of ages of all U.S. millionaires is $13.0$ years. (Note: The mean of the data is $58.53$ years)

Step-by-Step Solution

Answer

We are $95\%$ confident that mean age, $\mu$ of all us. Millionaires, lies some where between $54.28$ years and $62.78$ years

Step 1: Given information

The figure is given: 

31	45	79	64	48	38	39	68	52
59	68	79	42	79	53	74	66	66
71	61	52	47	39	54	67	55	71
77	64	60	75	42	69	48	57	48

Step 2: Concept

The formula used: $\left(\overline{x}-z_{\alpha /2}\frac{\sigma }{\sqrt{n}},\overline{x}+z_{\alpha /2}\frac{\sigma }{\sqrt{n}}\right)$

Step 3: Calculation

Let the population mean age be $\mu$

Population S.D $\sigma =13$ years

Sample size $n=36$

Sample mean $\overline{x}=58.53$

$\overline{x}$ is roughly normally distributed because the sample is huge.

The $100(1-\alpha )\%\mathrm{Cl}$ of $\mu$ is calculated using the $z-$interval technique.

$\left(\overline{x}-z_{\alpha /2}\frac{\sigma }{\sqrt{n}},\overline{x}+z_{\alpha /2}\frac{\sigma }{\sqrt{n}}\right)$

Here confidence level $=95\%=100\times 0.95\%$

$$\begin{gathered} \therefore 1-\alpha =0.95 \\ \Rightarrow \alpha =1-0.95 \\ \Rightarrow \alpha =0.05 \\ \Rightarrow \alpha /2=0.025 \\ \therefore z_{\alpha /2}=z^{125}=1.96 \end{gathered}$$

Step 4: Calculation

$$\begin{gathered} \therefore 95\%\mathrm{Cl}\text{ of }\mu \\ =\left(\overline{x}-Z_{0.025}\frac{\sigma }{\sqrt{n}},\overline{x}+z_{0.025}\frac{\sigma }{\sqrt{n}}\right) \\ =\left(58.53-1.96\times \frac{13}{\sqrt{36}},58.53+1.96\times \frac{13}{\sqrt{36}}\right) \\ =(58.53-4.25,58.53+4.25) \\ =(54.28,62.78) \end{gathered}$$

i.e., we are $95\%$ certain that the average age of all of us is $\mu$ Millionaires are born between the ages of $54.28$ and $62.78$