An economics professor randomly selected 100 millionaires in the United States. The average age of these millionaires was 54.8 years. If the standard deviation of the entire
Population of millionaires is 7.9 years, find the 95% confidence interval for the mean age
Of all United States millionaires.
A) $53.3

Question

An economics professor randomly selected 100 millionaires in the United States. The average age of these millionaires was 54.8 years. If the standard deviation of the entire
Population of millionaires is 7.9 years, find the 95% confidence interval for the mean age
Of all United States millionaires. 
A) $53.3 < \mu < 56.3$
B) $52.8 < \mu < 56.8$
C) $54.0 < \mu < 55.6$
D) $53.5 < \mu < 56.1$

Quizplus · Accepted Answer

The 95% confidence interval for the mean age of all United States millionaires can be calculated using the formula for the confidence interval of the mean: $\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $Z$ is the Z-score corresponding to the confidence level (for 95%, $Z = 1.96$), $\sigma$ is the population standard deviation, and $n$ is the sample size. Plugging in the given values: $54.8 \pm 1.96 \frac{7.9}{\sqrt{100}}$, we get $54.8 \pm 1.55$, which simplifies to the interval $53.25 < \mu < 56.35$, rounding to one decimal place gives $53.3 < \mu < 56.3$.

An Economics Professor Randomly Selected 100 Millionaires in the United 53.3<μ<56.353.3 < \mu < 56.353.3<μ<56.3

An Economics Professor Randomly Selected 100 Millionaires in the United $53.3 < \mu < 56.3$