Question

You are to construct a 99% confidence interval of a normally distributed population; the population standard deviation is known to be 25. A random sample of size 28 is taken; (i) the sample mean is found to 76 and (ii) the sample standard deviation was found to be 30. Construct the Confidence interval. Clearly name the standard distribution you use

Answer 1

Solution:

1) Given,

= 76

= 25   

n  = 28

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 * ( 25/ 28 )

= 12.170

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

( - E ) <   <  ( + E)

( 76 - 12.170 )   <   <  ( 76 + 12.170)

63.83 <   < 88.17

Required 99% confidence interval is ( 63.83 , 88.17 )

2) Solution =

Given that,

n = 28

   = 76

s = 30

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

Also, d.f = n - 1 = 27  

    =    =  0.005,27 = 2.771

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

The margin of error is given by

E =  /2,d.f. * ( / n )

= 2.771* ( 30/ 28 )

= 15.708

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

( - E ) <   <  ( + E)

( 76 - 15.708 )   <   <  ( 76 + 15.708 )

60.292 <   < 91.708

Required.99% confidence interval is ( 60.292 , 91.708 )