In 2000 there were almost 300,000 Canadians with more than 1 million dollars in financial assets! If the average age of one hundred randomly selected millionaires was 54.8 years with a standard deviation of 7.9 years, what is the 95% confidence interval for the mean age, μ, of all Canadian millionaires?

Question

Quizplus · Accepted Answer

We can use the formula:

CI = X̄ ± z*(σ/√n)

Where:
X̄ = sample mean age = 54.8
σ = population standard deviation = 7.9
n = sample size = 100
z = z-value for 95% confidence level = 1.96

So:

CI = 54.8 ± 1.96*(7.9/√100)
CI = 54.8 ± 1.55
CI = (53.25, 56.35)

Therefore, the 95% confidence interval for the mean age of all Canadian millionaires is (53.25, 56.35) years. Choice B is the correct answer. We can use the formula for a confidence interval for a population mean:

CI = x̄ ± z*(σ/√n)

where x̄ is the sample mean, σ is the population standard deviation (which we don't know, so we use the sample standard deviation as an estimate), n is the sample size, and z is the critical value for the desired level of confidence (in this case, 95%).

First, we need to find the critical value, z. Since we want a 95% confidence interval, we want to leave 2.5% in each tail of the normal distribution. Using a table or calculator, we find that the z-score for the 97.5th percentile is 1.96.

Next, we plug in the given values:

x̄ = 54.8
σ = 7.9 (from the standard deviation given in the question)
n = 100
z = 1.96 (from the critical value for 95% confidence)

CI = 54.8 ± 1.96*(7.9/√100)
CI = 54.8 ± 1.55
CI = (53.25, 56.35)

So the 95% confidence interval for the mean age of all Canadian millionaires is (53.25, 56.35) years. Choice B is the only option that matches this interval.

In 2000 There Were Almost 300,000 Canadians with More Than