In: Statistics and Probability
A survey is conducted to estimate the average household income in a large metropolitan area. A random sample of 150 households in this area yielded an average household income of $55,000 with a standard deviation of $13,200. Construct a 95% confidence interval for the mean household income in this area. (Choose the BEST answer)
Solution :
Given that,
Point estimate = sample mean = = 55000
Population standard deviation = = 13200
Sample size n =150
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 * ( 13200 / 150)
= 2112.4
At 95% confidence interval
is,
- E < < + E
55000 - 2112.4 <
<55000 + 2112.4
52887.6 <
< 57112.4