Question

2. Multinomial Data: A multinomial experiment with k=3 cells and n-320 produced the data shown in the accompanying table. Do these data provide sufficient evidence to contradict the null hypothesis that p1 0.25, p2-0.25 and p3-0.5? Assume a Type I error rate of 0.05. Cell ni 78 60 182

Answer 1

The expected frequencies here are computed as:

E₁ = 0.25*320 = 80
E₂ = 0.25*320 = 80
E₃ = 0.5*320 = 160

The chi square test statistic here is computed as:

$\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$

$\chi^2 = \frac{(78 - 80)^2}{80} + \frac{(60 - 80)^2}{80} + \frac{(182 - 160)^2}{160}$

$\chi^2 = 8.075$

For n - 1 = 2 degrees of freedom, we get from the chi square tables: we get:

$p = P(\chi^2_2 > 8.075 ) = 0.0176$

As the p-value here is 0.0176 < 0.05, therefore the test is not significant and we can reject the null hypothesis here and conclude that data provide sufficient to contradict the null hypothesis here