11.1 Chi-Square Test for Independence

Remember, qualitative data is where you collect data on individuals that are categories or names. Then you would count how many of the individuals had particular qualities. An example is that there is a theory that there is a relationship between breastfeeding and autism. To determine if there is a relationship, researchers could collect the time period that a mother breastfed her child and if that child was diagnosed with autism. Then you would have a table containing this information. Now you want to know if each cell is independent of each other cell. Remember, independence says that one event does not affect another event. Here it means that having autism is independent of being breastfed. What you really want is to see if they are not independent. In other words, does one affect the other? If you were to do a hypothesis test, this is your alternative hypothesis and the null hypothesis is that they are independent. There is a hypothesis test for this and it is called the Chi-Square Test for Independence. Technically it should be called the Chi-Square Test for Dependence, but for historical reasons it is known as the test for independence. Just as with previous hypothesis tests, all the steps are the same except for the assumptions and the test statistic.

Now you want to find out how many individuals you expect to be in a certain cell. To find the expected frequencies, you just need to multiply the probability of that cell times the total number of individuals. Do not round the expected frequencies.

Expected frequency $(cellA\text{ and }B)=E(A\text{ and }B)$

$$\begin{array}{c}=n(\frac{\text{ row total }}{n}·\frac{\text{ column total }}{n})\\ =\frac{\text{ row total -column total }}{n}\end{array}$$

If the variables are independent the expected frequencies and the observed frequencies should be the same. The test statistic here will involve looking at the difference between the expected frequency and the observed frequency for each cell. Then you want to find the “total difference” of all of these differences. The larger the total, the smaller the chances that you could find that test statistic given that the assumption of independence is true. That means that the assumption of independence is not true. How do you find the test statistic? First find the differences between the observed and expected frequencies. Because some of these differences will be positive and some will be negative, you need to square these differences. These squares could be large just because the frequencies are large, you need to divide by the expected frequencies to scale them. Then finally add up all of these fractional values. This is the test statistic.

Test Statistic:

The symbol for Chi-Square is ${\chi }^2$

$${\chi }^2=\sum \frac{(O-E{)}^2}{E}$$

where O is the observed frequency and E is the expected frequency

Distribution of Chi-Square

${\chi }^2$ has different curves depending on the degrees of freedom. It is skewed to the right for small degrees of freedom and gets more symmetric as the degrees of freedom increases (see Figure $1$). Since the test statistic involves squaring the differences, the test statistics are all positive. A chi-squared test for independence is always right tailed.

p-value:

Using the TI-83/84: $\chi \text{ cdf (lower limit, }1E99,df)$

Using R: $1-\text{ pchisq }(x^2,df)$

Where the degrees of freedom is $$df=(\#\text{ of rows }-1)*(\#\text{ of columns }-1)$$

4. Conclusion

This is where you write reject $H_o$ or fail to reject $H_o$. The rule is: if the p-value < $\alpha$, then reject $H_o$. If the p-value $\ge \alpha$, then fail to reject $H_o$.

5. Interpretation

This is where you interpret in real world terms the conclusion to the test. The conclusion for a hypothesis test is that you either have enough evidence to show $H_A$ is true, or you do not have enough evidence to show $H_A$ is true.

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