Questions: Listed below are the amounts of net worth (in millions of dollars) of the ten wealthiest celebrities in a country. Construct a 99% 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? 250 220 175 168 160 158 153 153 153 148 What is the confidence interval estimate of the population mean μ ? million < μ < million (Round to one decimal place as needed.)

Solution Steps

The mean (\(\mu\)) of the dataset is calculated as follows:

\[ \mu = \frac{\sum_{i=1}^N x_i}{N} = \frac{250 + 220 + 175 + 168 + 160 + 158 + 153 + 153 + 153 + 148}{10} = \frac{1738}{10} = 173.8 \]

Thus, the mean of the data is \(173.8\) million dollars.

The variance (\(\sigma^2\)) is calculated using the formula:

\[ \sigma^2 = \frac{\sum (x_i - \mu)^2}{n-1} \]

Calculating the variance gives:

\[ \sigma^2 = 1153.29 \]

The standard deviation (\(\sigma\)) is then:

\[ \sigma = \sqrt{1153.29} \approx 33.96 \]

Thus, the sample standard deviation is \(33.96\).

For a 99% confidence interval with unknown variance and a small sample size, we use the formula:

\[ \bar{x} \pm t \frac{s}{\sqrt{n}} \]

Where:

• \(\bar{x} = 173.8\)
• \(t\) (critical value for \(n-1 = 9\) degrees of freedom at 99% confidence) is approximately \(3.2\)
• \(s = 33.96\)
• \(n = 10\)

Calculating the margin of error:

\[ \text{Margin of Error} = t \frac{s}{\sqrt{n}} = 3.2 \cdot \frac{33.96}{\sqrt{10}} \approx 34.9 \]

Thus, the confidence interval is:

\[ (173.8 - 34.9, 173.8 + 34.9) = (138.9, 208.7) \]

Final Answer