Question

Construct 90%, 95%, and 99% confidence intervals to estimate μ from the following data. State the point estimate. Assume the data come from a normally distributed population.

12.1 11.6 11.9 12.6 12.5 11.4 12.0 11.7 11.8 12.1

(Round the intermediate values to 4 decimal places. Round your answers to 2 decimal places.)

90% confidence interval: _____ ≤ μ ≤ _____

95% confidence interval: _____ ≤ μ ≤ _____

99% confidence interval: _____ ≤ μ ≤ _____

The point estimate is _____ .

Answer 1

Solution :

We are given a data of sample size n = 10

12.1,11.6,11.9,12.6,12.5,11.4,12.0,11.7,11.8,12.1

Using this, first we find sample mean() and sample standard deviation(s).

=   

= (12.1 + 11.6.......+ 12.1)/10

= 11.97

Now ,

s=   

Using given data, find Xi - for each term.Take square for each.Then we can easily find s.

s = 0.3773

Note that, Population standard deviation() is unknown..So we use t distribution.

Our aim is to construct 90% confidence interval.   

c = 0.90

= 1- c = 1- 0.90 = 0.10

  /2 = 0.10 2 = 0.05

Also, d.f = n - 1 = 10 -1 = 9

    =    =  _0.05,9 = 1.8331

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

The margin of error is given by

E =  /2,d.f. * ( / n)

= 1.8331* ( 0.3773/ 10 )

= 0.2187

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

( - E ) <   <  ( + E)

( 11.97 - 0.2187 )   <   <  ( 11.97 + 0.2187 )

11.75 <   < 12.19

Required 90% Confidence interval is ( 11.75 , 12.19 )

Construct 95% confidence interval.

c = 0.95

= 1- c = 1- 0.95 = 0.05

  /2 = 0.05 2 = 0.025

Also, d.f = n - 1 = 10 - 1 = 9

    =    =  _0.025,9 = 2.2622

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

The margin of error is given by

E =  /2,d.f. * ( / n )

= 2.2622* ( 0.3773/ 10 )

= 0.2699

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

( - E ) <   <  ( + E)

( 11.97 - 0.2699 )   <   <  ( 11.97 + 0.2699 )

11.70 <   < 12.24

Required 95% confidence interval is ( 11.70 , 12.24 )

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 = 10 - 1 = 9

    =    =  _0.005,9 = 3.2498

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

The margin of error is given by

E =  /2,d.f. * ( / n )

= 3.2498* ( 0.3773/ 10 )

= 0.3877

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

( - E ) <   <  ( + E)

( 11.97 - 0.3877 )   <   <  ( 11.97 + 0.3877)

11.58 <   < 12.36

Required 99% confidence interval is ( 11.58 , 12.36 )