An economics professor randomly selected 100 millionaires in the United States. The average age of these...

An economics professor randomly selected 100 millionaires in the United States. The average age of these millionaires was 54.8 years. The population standard deviation is known to be 7.9 years. What is the 95% confidence interval for the mean age, of all United States millionaires? Interpret the confidence interval.

Point Estimate: ______

Critical Value:

Margin of Error: ________

Confidence Interval:______

Interpretation:

Solutions

Expert Solution

Solution :

Given that,

Point estimate = sample mean = = 54.8

Population standard deviation = = 7.9

Sample size = n =100

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96 ( Using z table )

Margin of error = E = Z/2 * ( / $\sqrt$ n)

= 1.96* ( 7.9/ $\sqrt$ 100 )

= 1.548
At 95% confidence interval
is,

- E < < + E

54.8- 1.548 < < 54.8 + 1.548

53.252 < < 56.348

lower bound 53.252

upper bound 56.348

orchestra answered 1 year ago

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