- 1 11 25 21 1 89 Solution Here We Use Chi Square Test Of Independence We Have Given 2 X 5 Contingency Table Number O 1 (56.65 KiB) Viewed 22 times
- 1 11 25 21 1 89 Solution Here We Use Chi Square Test Of Independence We Have Given 2 X 5 Contingency Table Number O 2 (56.65 KiB) Viewed 22 times
1.11 25 21 1.89 Solution: Here we use "Chi-square test" of independence. We have given 2 x 5 contingency table. Number of rows = r = 2 Number of clumn = c = 5 Here we take level of significance a = 0.05. Test of hypothesis: Ho : There is no association between Academic Level and focused VS Hy: There is association between Academic Level and focused. The table of observed and Expected Count: Observed Oi Expected Ei (Oi-Ei) (0;-E;) (Oi-Ei)2/Ei 22 17.751 4.25 18.0625 1.01760563 23.89 1.2321 0.05157388 19.11 3.5721 0.18692308 -4.82 23.2324 3.40651026 3.41 -2.41 5.8081 1.70325513 8.25 -4.25 18.0625 2.18939394 10 11.1 -1.1 1.21 0.10900901 8.881 -1.88 3.5344 0.39801802 3.17 23.3 7.3 744 14 1.58 2.42 5.8564 3.70658228 Total =104 1041 Total =20.128 Test statistics The Chi-square statistics is Σ (0, - E.) calculated X1 E ......Last column total calculated X = 20.128 P-value: Degrees of freedom = (r - 1)(c - 1) = (2-1)(5-1) = 1*4 =4 The P-vlaue at df =4 is, P-value = 0.000942 ..........using chi-square table The required P-value is 0.000942. 2 6.82 1 14 7