In: Statistics and Probability
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:
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
* (
/n)
= 1.96* ( 7.9/ 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