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)