Applied Univariate, Bivariate, and Multivariate Statistics. Daniel J. Denis

Applied Univariate, Bivariate, and Multivariate Statistics - Daniel J. Denis


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with a number of tests on model coefficients in a variety of statistical methods. The multivariate analog to the chi‐square distribution is that of the Wishart distribution (see Rencher, 1998, p. 53, for details).

      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

equation
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.

      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 1 0.004
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

      > 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


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