Question

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. Our example will test if taking hormonal pills and being overweight are related. We will test the independence on 200 random patients. Thus, N=200. They will be divided first into two groups, those who take hormonal pills and those who do not. Second, they will be divided 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 region A1-E4, first five columns and first four rows. That is, in celll B1 you will type Not overweight, in cell A2 Not taking hormonal pills, etc
Next we construct the expected table. Let's make it in the region A8-E11. Type Expected frequency table in cell A8, not overweight in cell B8 etc. Data in the table is calculated in this fashion. Cell B10 corresponds to the take hormonal pills row and not overweight column. Thus in cell B10 we type =B4*E3/E4. In cell D10 we type =D4*E3/E4. Using that strategy complete the expected frequency table.
Next we check if chi square test will work for this example. When you remove total from the expected frequency table, you have a 2x3 table with 6 entries. To run chi square we should first have no zero entries out of those 6. In cell A13 type zero entries. In cell B13 type the actual value of how many zero entries you have in expected frequency table. Second, you should have at most 20% entries that are less than 5. In cell A14 type percentage of entries less than 5. In cell B14 calculate the actual value of percents of entries in expected frequency table that are less than 5.
Now let's evaluate chi square parameters. In cell A16 type df. In cell B16 evaluate df. In cell A20 type chi square. We will evaluate chi 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 on page 416 with .05 level of significance and degrees of freedom df from B16. Put that value in cell B22.
Now we do testing. In cell A24 type H0 and in cell B24 state the null hypothesis. In cell A25 type H1 and in cell B25 state the alternate hypothesis.
Now compare the values in cells B20 and B22. State if we reject or do not reject the null hypothesis in cell A26. Explain how you obtained 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 chi square test. In cell B28 type =CHITEST(B2:D3,B9:D10).
Compare the sig. in cell B28 with the significance level of .05 and using that comparison, state in cell A31 if we reject or do not reject the null hypothesis. Explain how you have reached your statement in cell B31.

Answer 1

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Explanation solution

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