Question

Celebrity Net Worth Listed below are the amounts of net worth (in millions of dollars) of these ten wealthiest celebrities: Tom Cruise, Will Smith, Robert De Niro, Drew Carey, George Clooney, John Travolta, Samuel L. Jackson, Larry King, Demi Moore, and Bruce Willis. Construct a 98% confidence interval. What does the result tell us about the population of all celebrities? Do the data appear to be from a normally distributed population as required?

250 200 185 165 160 160 150 150 150 150

Short answer

At confidence interval, the result indicate that themean amount of all the given celebrities lies between $143.3 and $200.7.

No, since the values deviate from the straight line, the data is non-normal.

Step-by-step solution

Given information

The amount of net worth of the 10 wealthiest celebrities is recorded. The confidence interval is $98\%$.

Calculate the mean 

Let $n=$10 be the number of wealthiest celebrities.

Themean value is given below:

$$\begin{gathered} \overline{x}=\frac{{\sum }_{i=1}^n{x}_i}{n} \\ =\frac{250+200+....+150+150}{10} \\ =\frac{1720}{10} \\ =172 \end{gathered}$$

The mean value of the amount is 172 million dollars.

Calculate the standard deviation

The standard deviation of the given amount is as follows:

$$\begin{gathered} s_{}=\sqrt{\frac{{\sum }_{i=1}^n{(x_i-\overline{x})}^2}{n-1}} \\ =\sqrt{\frac{{\left(250-172\right)}^2+{\left(200-172\right)}^2+...+{\left(150-172\right)}^2+{\left(150-172\right)}^2}{10-1}} \\ =\sqrt{\frac{9310}{9}} \\ =32.1628 \end{gathered}$$

The standard deviation is 32.1628 million dollars.

Compute the degrees of freedom

Assume that the population is normally distributed with an unknown standard deviation.

Thus, t-distribution would be used in this case.

The degrees of freedom are as follows:

$$\begin{gathered} \text{df}=n-1 \\ =10-1 \\ =9 \end{gathered}$$

Compute the critical value

At confidence interval and with 9 degrees of freedom, the critical value is obtained using the t-table.

$$\begin{gathered} t_{crit}=t_{\frac{\alpha }{2},df} \\ =t_{\frac{0.02}{2},9} \\ =2.82 \end{gathered}$$

Compute the margin of error

The margin of error is calculated by multiplying the critical value with the standard error$\left(\text{SE}=\frac{s}{\sqrt{n}}\right)$. It is denoted by E.

Therefore,

$$\begin{gathered} E=t_{crit}\times \frac{s}{\sqrt{n}} \\ =2.82\times \frac{32.1628}{\sqrt{10}} \\ =28.6962 \end{gathered}$$

Compute the confidence interval

The formula for the confidence interval is given as follows:

$$\begin{gathered} CI=\overline{x}-E<\mu <\overline{x}+E \\ =\left(172-28.6962<\mu <172+28.6962\right) \\ =(143.3038<\mu <200.6962) \\ =\left(143.3<\mu <200.7\right) \end{gathered}$$

The result indicates that with 98% confidence level it can be expressed that the mean amount of all the given celebrities lies between ($143.3 and $200.7).

Check normality

The normal Q-Q plot is sketched using the following steps:

1. Draw two axes, horizontal and vertical.
2. Mark z-scores corresponding to observations on the change by means of scaling the axes.
3. Thus, the relevant graph along with a tentative line is shown below.

From the graph, it is clear that the values are not close to the straight line.

Since the points deviate from the straight line, the population does not follow a normal distribution. Hence, it does not meet the requirement of the population.