Question

In: Statistics and Probability

A random sample of 47 four year olds attended day care centers provided a yearly tuition...

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.

Solutions

Expert Solution

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)

orchestra answered 2 years ago

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