Higher Engineering Mathematics

612 STATISTICS AND PROBABILITY

Table 63.2

12 3 456

Class Class z-value for Area from Area for Expected

mid-point boundaries,x class boundary 0 toz class frequency

4.009 −2.56 0.4948

4.011 0.0255 13

4.013 −1.87 0.4693

4.015 0.0863 43

4.017 −1.19 0.3830

4.019 0.1880 94

4.021 −0.51 0.1950

4.023 0.2628 131

4.025 0.17 0.0678

4.027 0.2345 117

4.029 0.85 0.3023

4.031 0.1347 67

4.033 1.53 0.4370

4.035 0.0494 25

4.037 2.21 0.4864

Total: 490

To determine theχ^2 -value

Theχ^2 -value is calculated using a tabular method
as shown below.

Diameter Observed Expected,

of rivets frequency,o frequency,e

4.011 12 13

4.015 47 43

4.019 86 94

4.023 123 131

4.027 107 117

4.031 97 67

4.035 28 25

o−e (o−e)^2

(o−e)^2

e

− 1 1 0.0769

4 16 0.3721

− 8 64 0.6809

− 8 64 0.4885

− 10 100 0.8547

30 900 13.4328

3 9 0.3600

χ^2 =

∑

{

(o−e)^2

e

}

= 16. 2659

To test the significance of theχ^2 -value

The number of degrees of freedom is given by
N− 1 −M, whereMis the number of estimated

parameters in the population. Both the mean and the

standard deviation of the population are based on

the sample value,M=2, henceν= 7 − 1 − 2 =4.

From Table 63.1, theχ^2 p-value corresponding toχ^20. 95

andν 4 is 9.49.Hence the null hypothesis that the

diameters of the rivets are normally distributed

is rejected.Forχ^20. 10 ,ν 4 , theχ^2 p-value is 1.06, hence

the fit is not ‘too good’. Since the null hypothesis

is rejected, the second significance test need not be

carried out.

Now try the following exercise.

Exercise 227 Further problems on fitting

data to theoretical distributions

1. Test the null hypothesis that the observed data
   given below fits a binomial distribution of the
   form 250(0. 6 + 0 .4)^7 at a level of significance
   of 0.05.
   Observed
   frequency 8 27 62 79 45 24 5 0
   Is the fit of the data ‘too good’ at a level of
   confidence of 90%?
   ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣
   Expected frequencies:
   7, 33, 65, 73, 48, 19, 4, 0;
   χ^2 -value= 3 .62,χ^20. 95 ,
   ν 7 = 14 .1, hence hypothesis
   accepted.χ^20. 10 ,ν 7 = 2 .83,
   hence data is not ‘too good’

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦