In: Math
I have draw a random sample of 20 of my neighbors. I ask them their income (What can I say? I'm a nosy neighbor). My sample average (x bar) is $41,000. I want to create a 95% confidence interval around x bar.
My estimated standard deviation is $5,000.
What is the 95% confidence internal for the average income in my neighborhood? $38,652 to $43.347, $38,663 to $43,336, $38,809 to $43,191, or not enough information?
Solution :
Given that,
= $41000
=$5000
n = 20
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
Margin of error = E = Z/2
* (
/n)
= 1.96 * ( 5000 / 20
)
= 2191
At 95% confidence interval estimate of the population mean is,
- E < < + E
41000- 2191 <
< 41000+ 2191
38809 <
<43191
( $38,809 to $43,191 )