Solution :
In this case we have to use [tex]$\text{chi square test}$[/tex] for the goodness of fit.
Null hypothesis is : [tex]$H_0 :$[/tex] Data follows the given distribution
Alternate hypothesis is : [tex]$H_a:$[/tex] Data do not follow the given distribution
The level of significance, [tex]$\alpha = 0.05$[/tex]
The test statistics is given by :
Chi square = [tex]$\sum\frac{(O-E)^2}{E}$[/tex]
Here O is the observed frequencies
E is the expected frequencies
So we have, N = number of the categories = 4
df = degrees o freedom = N - 1
= 4 - 1 = 3
Critical value = [tex]$7.814727764$[/tex]
Calculating the table for the test statistics are given as :
Category prop. O E [tex]$\frac{(O-E)^2}{E}$[/tex]
A 0.2 14 20 1.80
C 0.3 28 30 0.13
M 0.1 6 10 1.60
P 0.4 52 40 3.60
Total 1 100 100 7.13
The test statics = chi square = [tex]$\sum\frac{(O-E)^2}{E}$[/tex] = 7.13
[tex]$X^2$[/tex] statistics = 7.13
P-value = 0.067767248
P -value > [tex]$\alpha = 0.05$[/tex]
Therefore, we do not reject the [tex]$\text{null hypothesis}$[/tex]. There is sufficient evidence to conclude that the data follows the given distribution. So there is sufficient evidence to conclude the given claim is true.
