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Chapter 6: Analysing the Data |
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Example 4So, for an example with SPSS, let us use the first twenty cases in the data base given for Assignment II.
Figure 6.12 Data for SPSS Chi-square example. Where female = 0 and male = 1. Alccode = 0 if the person reported using alcohol less than once or twice per month (a 0, 1, 2, or 3 on the original scale) and alccode = 1 if the person reported using alcohol once or twice per month or more (a 4, 5, or 6 on the original scale). See the SPSS Screens and output booklet for how to select the right options for chi-square. Output 6.8 Chi square contingency table.Crosstabs
The above output indicates: Case Processing Summary Check here that the correct variables have been selected and that the right N appears. Check whether the number of missing cases is as you would expect (here zero is right). SEX * ALCCODE Crosstabulation Here are the Observed and Expected frequencies as determined form the data. In The Total column you can see that there were 9 females and 11 males. In the Total for ALCCODE you can see that there were 5 no alcohol users and 15 alcohol users. When you look at the individual cell frequencies you see that there were 4 female no alcohol users and 5 female alcohol users whereas there was only one male no alcohol user and 10 male alcohol users. On first inspection this seems quite a large difference. The male ratio of alcohol users to non alcohol users is 10:1!! Whereas the female ratio is 0.8:1. This appears a very large difference in support of the alternative hypothesis that there is an association between the two variables. Note the expected frequencies can be calculated as before. For female non-alcohol users
Which equals 2.3 when rounded But let us check the statistics. Chi-Square Tests The top line in this table is the one we want. Here it says that The main problem here is the small sample size. A total of 20 people classified into four categorises means that a small change in one of those frequencies will alter the final chi-square quite considerably. Notice that beneath this table, SPSS prints the number of cells that do not meet the minimum expected frequency requirement and then cites the actual minimum expected frequency that occurred in the table. Howell discusses this issue on pp. 151-152 and it is worth reading this. |
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2(1) = 3.300, p = .069. That is, the result is not significant and we have no evidence to reject the null hypothesis!