Question

Random sample of size 7 is drawn from a normal population with variance 2.5 the sample observations are 9, 16, 10, 14, 8, 13, 14. find 99% confidence interval for the population mean what would be the confidence interval if the variance is unknown.

Answer 1

Solution:

Given ,

Population variance = ² = 2.5

Sample size = n = 7

9, 16, 10, 14, 8, 13, 14

1) Population variance is known .

So we use z distribution.

² = 2.5

So , =  2.5 = 1.58113883008

Using calculator , we find sample mean

=(9+16+10+14+8+13+14)/7 = 84/7 = 12

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 * (1.58113883008 / 7)

= 1.539

Now , confidence interval for mean() is given by:

( - E ) <   <  ( + E)

(12 - 1.539)   <   <  (12 + 1.539)

10.461 <   < 13.539

(10.461 , 13.539)

2)

Suppose that the population variance ² is unknown.

So we use t distribution.

We need to find the sample standard deviation.

Let s be the sample standard deviation.

s=   

Using calculator , we can find s directly.

s = 3

Our aim is to construct 99% confidence interval.   

c = 0.99

= 1 - c = 1 - 0.99 = 0.01

  /2 = 0.01 2 = 0.005

Also, d.f = n - 1 = 7 - 1 = 6  

    =    =  _0.005,6 = 3.707

( use t table or t calculator to find this value..)

The margin of error is given by

E =  /2,d.f. * ( / n )

= 3.707 * (3 / 7)

= 4.203

Now , confidence interval for mean() is given by:

( - E ) <   <  ( + E)

(12 - 4.203)   <   <  (12 + 4.203)

7.797 <   < 16.203

(7.797 , 16.203)