# Chi-Square Test

• Nov 18th 2009, 04:15 PM
statmajor
[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: $\displaystyle Q = \Sigma \frac{(X_i - np_i)^2}{np_i} - \frac{(Y_i - np_i)^2}{np_i}$.

Since it's a multinomial distribution, I know that the expectation is np_i, but how would I find p_i?
• Nov 19th 2009, 01:09 AM
mr fantastic
Quote:

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: $\displaystyle Q = \Sigma \frac{(X_i - np_i)^2}{np_i} - \frac{(Y_i - np_i)^2}{np_i}$.

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 $\displaystyle n_{ij}$ for a contigency table is $\displaystyle \frac{r_i \cdot c_j}{n}$ where $\displaystyle r_i$ is the row total (for row i) and $\displaystyle c_j$ is the column total (for column j) and n is the total frequency.
• Nov 19th 2009, 06:47 AM
statmajor
Got it; thanks.