Chi Square Test
We will now use Excel to run an example of a chi square test.Chi square test is checking the independence of two variables. Ourexample will test if taking hormonal pills and being overweight arerelated. We will test the independence on 200 random patients.Thus, N=200. They will be divided first into two groups, those whotake hormonal pills and those who do not. Second, they will bedivided into three groups based on weight, not overweight,overweight and obese. All data is in this table
Observed frequency table Not overweight overweight obese total
Not taking hormonal pills 35 36 49 120
Take hormonal pills 33 32 15 80
Total 68 68 64 200
We will start in Excel by making the above table in regionA1-E4, first five columns and first four rows. That is, in celll B1you will type Not overweight, in cell A2 Not taking hormonal pills,etc
Next we construct the expected table. Let's make it in the regionA8-E11. Type Expected frequency table in cell A8, not overweight incell B8 etc. Data in the table is calculated in this fashion. CellB10 corresponds to the take hormonal pills row and not overweightcolumn. Thus in cell B10 we type =B4*E3/E4. In cell D10 we type=D4*E3/E4. Using that strategy complete the expected frequencytable.
Next we check if chi square test will work for this example. Whenyou remove total from the expected frequency table, you have a 2x3table with 6 entries. To run chi square we should first have nozero entries out of those 6. In cell A13 type zero entries. In cellB13 type the actual value of how many zero entries you have inexpected frequency table. Second, you should have at most 20%entries that are less than 5. In cell A14 type percentage ofentries less than 5. In cell B14 calculate the actual value ofpercents of entries in expected frequency table that are less than5.
Now let's evaluate chi square parameters. In cell A16 type df. Incell B16 evaluate df. In cell A20 type chi square. We will evaluatechi square in cell B20. In cell B20 type=(B2-B9)^2/B9+(B3-B10)^2/B10+(C2-C9)^2/C9+(C3-C10)^2/C10+(D2-D9)^2/D9+(D3-D10)^2/D10.In cell A22 type table chi square and then find the table value onpage 416 with .05 level of significance and degrees of freedom dffrom B16. Put that value in cell B22.
Now we do testing. In cell A24 type H0 and in cell B24 state thenull hypothesis. In cell A25 type H1 and in cell B25 state thealternate hypothesis.
Now compare the values in cells B20 and B22. State if we reject ordo not reject the null hypothesis in cell A26. Explain how youobtained your conclusion in cell B26.
Next we will test it another way, with asymptotic significance(probability).
In cell A28 type Asymp. Sig. (probability). We will evaluate Sig.in cell B28. We will use an Excel command for finding sig. in a chisquare test. In cell B28 type =CHITEST(B2:D3,B9:D10).
Compare the sig. in cell B28 with the significance level of .05 andusing that comparison, state in cell A31 if we reject or do notreject the null hypothesis. Explain how you have reached yourstatement in cell B31.
