Chi square test for Goodness of fit Â
 Â
expected frequncy,E = expected proportions*total
frequency Â
total frequency=Â Â 4048
|
observed frequencey, O |
expected proportion |
expected frequency,E |
(fi - ei) |
|
(O-E)²/E |
| 2226 |
0.333 |
1349.333 |
876.667 |
|
569.573 |
| 891 |
0.333 |
1349.333 |
-458.333 |
|
155.684 |
| 931 |
0.333 |
1349.333 |
-418.333 |
|
129.696 |
chi square test statistic,X² = Σ(O-E)²/E = Â
854.9531Â Â Â Â Â Â Â Â
  Â
       Â
     Â
level of significance, α=  0.05 Â
        Â
Degree of freedom=k-1=Â Â 3Â Â -Â Â
1Â Â =Â Â 2
       Â
     Â
P value =Â Â 0.0000Â Â [ excel function:
=chisq.dist.rt(test-stat,df) ]Â Â Â Â Â
  Â
Decision: P value < α, Reject Ho    Â
        Â
There is significant evidence that current, former, and
never smokers are not equally represented among adults diagnosed
with gum disease.
