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For this written assignment I have attached a data set to this announcement for you to use. In order to complete this assignment, I strongly recommend that you use Vassar Stats for the chi-square analysis. Your challenge will be to select the appropriate chi-square for the problem described in the attachment. In your write-up, make sure to provide an APA style interpretation of the chi-square results, in addition to answering the other related questions.

What is Chi Square?

As discussed above, chi square, or c2, is what we call a non-parametric test.  Non-parametric statistics do not make assumptions about the shape of your distribution as inferential statistics do (assumed normal curve).  Non-parametric statistics are generally used when we have data that is not continuous.  It may be either ranked (ordinal) or categorical (nominal) data.  Because data of these types have no variability, they need different tests.

The c2 is very similar to the Pearson product-moment correlation that we studied in Week 10 in that it looks for a relationship between two variables.  What differs is the use of variables that are nominal or ordinal rather than continuous.

Computing a c2 Test for Independence

To compute a c2, you need to know two things:  The number of times a construct is observed (O), and the number of times you would expect to see that construct observed in the population (E).  By comparing these, you can find whether or not your cells are independent.  If they are independent, there is no relationship between your variables.  If they are not independent, you do have a relationship.

Start by making a table of your observations.  Perhaps you are interested in whether or not there is a relationship between owning a dog and home ownership.  You collect data on people who do and do not own dogs, and ask those people whether or not they own a home.  Your results are as follows:

Own Home

Do not Own Home

Own Dog

23

15

Do not Own Dog

7

25

Once you have collected O, you can calculate E.  First, compute your column and row totals, then compute the percentage of total observations you have in each row.  i.e., 23 + 15 = 38.  38/70 = 54%

Own Home

Do not Own Home

Totals

Own Dog

23

15

38

54%

Do not Own Dog

7

25

32

46%

Totals

30

40

70

100%

Next, for each cell, multiply the row’s percentage by the column total.  Your expected frequencies should add up to the same totals as your observed frequencies.  i.e., 16.29 + 14.01 ≈ 30 and 16.29 + 21.72≈ 38 (you can expect rounding errors).

Own Home

Do not Own Home

Totals

Own Dog

23(O)

30*.543 = 16.29(E)

15(O)

40*.543 = 21.72(E)

38

54.3%

Do not Own Dog

7(O)

30*.467 = 14.01(E)

25(O)

40*.467 = 18.68(E)

32

46.7%

Totals

30

40

70

100%

Finally, you can compute your c2, which is ∑(O-E)2 / E.  In this case,

c2  = ((23-16.29)2/16.29) + ((15-21.72)2 / 21.72) + ((7-14.01)2 / 14.01) + ((25-18.68)2 / 18.68)

= (45.02/16.29) + (45.16 / 21.72) + (49.14 / 14.01) + (39.94 / 18.68)

= 2.76 + 2.08 + 3.51 + 2.14

= 10.79

Is c2 = 10.79 significant?  We have a table in the appendix on p. 672 (Table A-4) that will give us the cutoff we need.  First, we need degrees of freedom, defined as number of columns – 1 * number of rows – 1.  As this is a 2 x 2 design (2 rows and 2 columns), our df = (2-1)(2-1) = 1*1 = 1.

If we look at significance level .05 for 1 df, we find a cut-off of 3.841.  Since 10.79 is greater than 3.841, we can determine that the cell means are not independent, meaning that there is a significant relationship between owning a dog and owning a home (c2(1) = 10.79, p < .05).  Looking at our table of observations we see that of 38 people who own dogs, 23 also own their home, while out of 32 people who do not own dogs, only 7 own their home.