# Thread: Chi-Square Test

1. ## [SOLVED] Chi-Square Test

Two different teaching procedures were used in two different groups of students. Each group had 100 students of similiar ability. These are the results:

Group A B C D F
I 15 25 32 17 11
II 9 18 29 28 16

Data is independent from two respective multinomial distribution with k = 5. Test at the 5% level the hypothesis that the two teaching methods are equally effective.

So let X be the students in group 1, and Y students in group 2.

So the test statistic would be: $Q = \Sigma \frac{(X_i - np_i)^2}{np_i} - \frac{(Y_i - np_i)^2}{np_i}$.

Since it's a multinomial distribution, I know that the expectation is np_i, but how would I find p_i?

2. Originally Posted by statmajor
Two different teaching procedures were used in two different groups of students. Each group had 100 students of similiar ability. These are the results:

Group A B C D F
I 15 25 32 17 11
II 9 18 29 28 16

Data is independent from two respective multinomial distribution with k = 5. Test at the 5% level the hypothesis that the two teaching methods are equally effective.

So let X be the students in group 1, and Y students in group 2.

So the test statistic would be: $Q = \Sigma \frac{(X_i - np_i)^2}{np_i} - \frac{(Y_i - np_i)^2}{np_i}$.

Since it's a multinomial distribution, I know that the expectation is np_i, but how would I find p_i?
Under the null hypothesis the maximum likelihood estimate of the expected value of cell frequency $n_{ij}$ for a contigency table is $\frac{r_i \cdot c_j}{n}$ where $r_i$ is the row total (for row i) and $c_j$ is the column total (for column j) and n is the total frequency.

3. Got it; thanks.