In: Statistics and Probability
In a random sample of 100 measurements from a population with known standard deviation 200, the average was found to be 50. A 95% confidence interval for the true mean is
Solution :
Given that,
Point estimate = sample mean = = 50
Population standard deviation = =200
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 * ( 200 / 100 )
E= 39.2
At 95% confidence interval estimate of the population mean
is,
- E < < + E
50 - 39.2 <
< 50+ 39.2
10.8<
< 89.2
( 10.8, 89.2)