Chi square

klore717 · #1 $Old$ July 3rd, 2009, 12:46 PM

at a=.01, is the frequency of disciplinary action independant of the abusers level of privilege.

level disciplined not disciplined total

low 20 11 31

medium 42 3 45

high 33 3 36

column total 95 17 112

abender · #2 $Old$ July 6th, 2009, 07:32 PM

This is a chi-square test of independence problem. We are given the following data:

$\begin{tabular}{c || c | c | c} \text{ } & \text{Disciplined} & \text{Not Disciplined} & \text{Total} \\ \hline\hline \text{Low} & 20 & 11 & 31 \\ \hline \text{Medium} & 42 & 3 & 45 \\ \hline \text{High} & 33 & 3 & 36 \\ \hline \text{Column Total} & 95 & 17 & 112 \end{tabular}$

The null hypothesis is the frequency of disciplinary action is independent of the abusers level of privilege. In other words, the level (low, medium, or high) does not have any influence on the frequency of disciplinary action. So, if the null hypothesis is true, then 95/112 = 84.821% is the expected frequency of disciplinary action, regardless of level.

The next step is computing the expected (under the null hypothesis) cell frequencies in the contingency table. For example, consider the (first) cell, corresponding to Low and Disciplined. The observed value here is 20. The expected value, on the other hand, is computed by taking the product of the total number of low cases (which is 31) and the expected frequency of disciplinary action under the null hypothesis (84.821%), which gives us 31*0.84821 = 26.29451.

The following table is the same as the contingency table above, but with expected values shown in parentheses:

$\begin{tabular}{c || c | c | c} \text{ } & \text{Disciplined} & \text{Not Disciplined} & \text{Total} \\ \hline\hline \text{Low} & 20 (26.29451) & 11 (4.70549) & 31 \\ \hline \text{Medium} & 42 (38.16945) & 3 (6.83055) & 45 \\ \hline \text{High} & 33 (30.53556) & 3 (5.46444) & 36 \\ \hline \text{Column Total} & 95 & 17 & 112 \end{tabular}$

Now we know enough to calculate the chi-square statistic, ${\chi}^2$ , which is given by ${\chi}^2 = \sum{\frac{{(E-O)}^2}{E}}$ . Below is our calculation:

$\begin{matrix} \text{ } & \text{Observed} & \text{expected} & (O-E) & {(O-E)}^2 & \frac{{(O-E)}^2}{E} \\ \hline \text{Low} & 20 & 26.29451 & -6.29451 & 39.62086 & 1.50681 \\ \hline \text{Medium} & 42 & 38.16945 & 3.83055 & 14.67311 & 0.38442 \\ \hline \text{High} & 33 & 30.53556 & -2.46444 & 6.07346 & 0.19800 \\ \hline \text{Total} & 95 & 95 & \text{ } & \text{ } & 2.08923 \end{matrix}$

So, we have ${\chi}^2=2.08923$ .

The degrees of freedom, $df$ , is computed by $df=(C-1)(R-1)$ , where C represents the total number of columns and R represents the total number of rows (from the original contingency table). Thus, $df = (2-1)(3-1) = 2$ .

We are also given that $\alpha = .01$ .

Refer to a Chi Square Distribution table (I googled one), and you will see that the critical value at $df=2$ and $\alpha = .01$ , is 9.2104.

What conclusion can you now draw from this?