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Applied Univariate, Bivariate, and Multivariate Statistics. Daniel J. Denis
Читать онлайн.Название Applied Univariate, Bivariate, and Multivariate Statistics
Год выпуска 0
isbn 9781119583011
Автор произведения Daniel J. Denis
Жанр Математика
Издательство John Wiley & Sons Limited
The chi‐square goodness‐of‐fit test is one such statistical method that utilizes the chi‐square test statistic to evaluate the tenability of a null hypothesis. Recall that such a test is suitable for categorical data in which counts (i.e., instead of means, medians, etc.) are computed within each cell of the design. The goodness‐of‐fit test is given by
Table 2.1 Contingency Table for 2 × 2 Design
| Condition Present (1) | Condition Absent (0) | Total | |
|---|---|---|---|
| Exposure yes (1) | 20 | 10 | 30 |
| Exposure no (2) | 5 | 15 | 20 |
| Total | 25 | 25 | 50 |
where Oi and Ei represent observed and expected frequencies, respectively, summed across r rows and c columns.
As a simple example, consider the hypothetical data (Table 2.1), where the frequencies of those exposed to something adverse are related to whether a condition is present or absent. If you are a clinical psychologist, then you might define exposure as, perhaps, a variable such as combat exposure, and condition as posttraumatic stress disorder (if you are not a psychologist, see if you can come up with another example).
The null hypothesis is that the 50 counts making up the entire table are more or less randomly distributed across each of the cells. That is, there is no association between condition and exposure. We can easily test this hypothesis in SPSS by weighting the relevant frequencies by cell total:
| exposure | condition | freq |
| 1.00 | 0.00 | 10.00 |
| 1.00 | 1.00 | 20.00 |
| 2.00 | 0.00 | 15.00 |
| 2.00 | 1.00 | 5.00 |
WEIGHT BY freq. CROSSTABS /TABLES=condition BY exposure /FORMAT=AVALUE TABLES /STATISTICS=CHISQ /CELLS=COUNT /COUNT ROUND CELL.
The output follows in which it is first confirmed that we set up our data file correctly:
| Exposure * Condition Crosstabulation | ||||
|---|---|---|---|---|
| Count | ||||
| Condition | Total | |||
| 1.00 | 0.00 | |||
| Exposure | 1.00 | 20 | 10 | 30 |
| 2.00 | 5 | 15 | 20 | |
| Total | 25 | 25 | 50 |
We focus on the Pearson chi‐square test value of 8.3 on a single degree of freedom. It is statistically significant (p = 0.004), and hence we can reject the null hypothesis of no association between condition and exposure group.
| Chi‐square Tests | |||||
| Value | df | Asymp. Sig. (two‐sided) | Exact Sig. (two‐sided) | Exact Sig. (one‐sided) | |
| Pearson chi‐square | 8.333a | 1 | 0.004 | ||
| Continuity correctionb | 6.750 | 1 | 0.009 | ||
| Likelihood ratio | 8.630 | 1 | 0.003 | ||
| Fisher's exact test | 0.009 | 0.004 | |||
| Linear‐by‐linear association | 8.167 | 1 | 0.004 | ||
| No. of valid cases | 50 | ||||
a 0 cells (0.0%) have expected count less than 5. The minimum expected count is 10.00.
b Computed only for a 2 × 2 table.
In R, we can easily perform the chi‐square test on this data. We first build the matrix of cell counts, calling it diag.table :
> diag.table <- matrix(c(20, 5, 10, 15), nrow = 2) > diag.table [,1] [,2] [1,] 20 10 [2,] 5 15 > chisq.test(diag.table, correct = F) Pearson's Chi-squared test data: diag.table X-squared = 8.3333, df = 1, p-value = 0.003892
We see that the result in R agrees with what we