In: Statistics and Probability
A random sample of 47 four year olds attended day care centers provided a yearly tuition average of $3966 and the population standard deviation of $636.find the 99% confidence interval of the true mean.round your answer to the nearest whole number.
Solution:
Given,
= 3966
= 636
n = 47
Note that, Population standard deviation() is known..So we use z distribution.
Our aim is to construct 99% confidence interval.
c = 0.99
= 1- c = 1- 0.99 = 0.01
/2 = 0.01 2 = 0.005 and 1- /2 = 0.995
Search the probability 0.995 in the Z table and see corresponding z value
= 2.576
The margin of error is given by
E = /2 * ( / n )
= 2.576 * (636 / 47)
= 238.975866711
= 239
Now , confidence interval for mean() is given by:
( - E ) < < ( + E)
(3966 - 239) < < (3966 + 239)
3727 < < 4205
Required 99% confidence interval is (3727 , 4205)