Answer to Question #161292 in Quantitative Methods for Sunny

Question

Based on below data, test that Brand preference and Education qualifications are independent, at 5% level of significance.

(No. OF PERSONS PREFERRING BRANDS)

Education Brand A Brand B Brand C

qualification

SSC 22 34 35

HSC 28 35 28

Graduate 30 44 27

Post graduate 55 42 28

Answer 1

The Chi-Square Test of Independence:

$H_0$ : Brand preference is independent of Education qualifications

$H_1$ : Brand preference is dependent of Education qualifications

$\Chi^2=\displaystyle\sum^R_{i=1}\displaystyle\sum^C_{j=1}\frac{(a_{ij}-e_{ij})^2}{e_{ij}}$

$e_{ij}=\frac{row(i)total\cdot col(j)total}{grandtotal}$

$grandtotal=408$

$e_{11}=\frac{135\cdot91}{408}=30.11$ , $e_{12}=\frac{155\cdot91}{408}=34.57$ , $e_{13}=\frac{118\cdot91}{408}=26.32$

$e_{21}=\frac{135\cdot91}{408}=30.11$ , $e_{22}=\frac{155\cdot91}{408}=34.57$ , $e_{23}=\frac{118\cdot91}{408}=26.32$

$e_{31}=\frac{135\cdot101}{408}=33.42$ , $e_{32}=\frac{155\cdot101}{408}=38.37$ , $e_{33}=\frac{118\cdot101}{408}=29.21$

$e_{41}=\frac{135\cdot125}{408}=41.36$ , $e_{42}=\frac{155\cdot125}{408}=47.49$ , $e_{43}=\frac{118\cdot125}{408}=36.15$

$\Chi^2=2.18+0.01+2.86+0.15+0.01+0.11+0.35+0.83+0.17+$

$+4.50+0.63+1.84=13.64$

Degrees of freedom:

$df=$ (number of columns – 1)(number of rows – 1)

$df=2\cdot3=6$

p-value$=P(\Chi^>13.64)=0.038$

$\alpha=0.05$ , $\alpha>$ p-value

Since α > p-value, reject H₀. This means that the factors are not independent.