In: Statistics and Probability
300 air conditioners were tested and it was found that the average power usage per hour during the summer months (for around 9 hours per day) is 4200 W. Find the 95% confidence interval if sigma = 100 W.
Solution :
Given,
= 4200
= 100
n = 300
Note that, Population standard deviation() is known..So we use z distribution.
Our aim is to construct 95% confidence interval.
c = 0.95
= 1- c = 1- 0.95 = 0.05
/2 = 0.05 2 = 0.025 and 1- /2 = 0.975
Search the probability 0.975 in the Z table and see corresponding z value
= 1.96
The margin of error is given by
E = /2 * ( / n )
= 1.96 * ( 100/ 300 )
= 11.316
Now , confidence interval for mean() is given by:
( - E ) < < ( + E)
( 4200 - 11.316 ) < < ( 4200 + 11.316 )
4188.684 < < 4211.316
Required 95% confidence interval is ( 4188.684 , 4211.316 )