Question

Find the requested confidence interval. The data below consists of the test scores of 32 students. Construct a 95.44% confidence interval for the population mean. 88 70 67 95 99 72 86 60 57 92 90 88 70 73 88 97 109 63 76 71 97 88 70 79 54 80 83 92 91 66 90 90

Answer 1

Solution:

Given a sample of size n = 32.

88 70 67 95 99 72 86 60 57 92 90 88 70 73 88 97 109 63 76 71 97 88 70 79 54 80 83 92 91 66 90 90

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

=   

= (80 + 70 + 67 + .......+ 90)/32

=  80.71875

Now , sample variance

Sample variance s² =

Using this formula we can easily find the sample variance.

s² = 184.01512096774

So ,

s = 13.565217321066

Our aim is to construct 95.44% confidence interval.

c = 0.9544

= 1- c = 1- 0.9544 = 0.0456

  /2 = 0.0456 2 = 0.0228

Now , df = n - 1 = 32 - 1 = 31

Using t value calculator ,

     =  _0.0228,31 = 2.083

The margin of error is given by

E =  /2,d.f. * ( / n )

= 2.083 * (13.565217321066 / 32)

= 4.9951

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

( - E ) <   <  ( + E)

(80.71875 - 4.9951)   <   <  (80.71875 + 4.9951)

75.72365 <   < 85.713813764

Required 95.44% confidence interval is (75.72 , 85.71)