First we find the sum of each column and each row. Then we find the grand sum which is equal to 500.
Table 1:ui=5001∑j=1cOij where c — number of columns,Oij — observed values of the contingency table,i=1,…,r,r — number of rows.In our case r=c=3.vj=5001∑i=1rOijTable 2:Eij=500uivj — values of the table (expected values)Table 3:aij=Eij(Eij−Oij)2chio2=∑i=1r∑j=1caij=28.75 — observed value of χ2chicr2≈9.49 — critical value of χ2chicr2=chicr2(alpha;df)alpha=0.05 — significance leveldf=(c−1)(r−1)=4 — degrees of freedomchio2>chicr2. So we can say that age and grade points arenot independent.Using CHISQ.TEST() we find p-value. We can seethat p-value<alpha. So we can say that age and grade points arenot independent.
Comments
Dear pratishtha singh, please use the panel for submitting new questions.
50 students selected at random from 500 students enrolled in a computer crash programme were classified according to the age and grade points giving the following data: Age (in years) Grade Point Below 20 21-30 Above 30 Up to 5 - 3 5 2 5.1 to 7.5 - 8 7 5 7.6 to 10.0 - 4 8 8 Test at 5% level of significance that age and grade points are independent