Module stikpetP.tests.test_mod_log_likelihood_ind

Functions

def ts_mod_log_likelihood_ind(field1, field2, categories1=None, categories2=None, cc=None)

Mod-Log Likelihood Test of Independence

This test is similar as a Pearson Chi-Square test of independence, but approaches it from a likelihood-ratio approach (see Monica, 2015).

If the significance of this test is below 0.05 (or another pre-defined threshold), the two nominal variables have a significant association.

The test compares the observed counts of the cross table with the so-called expected counts. The expected values are the number of respondents you would expect if the two variables would be independent. See the Pearson Chi-Square test of independence for more details on expected counts.

One problem though is that the test should only be used if not too many cells have a so-called expected count, of less than 5, and the minimum expected count is at least 1. So you will also have to check first if these conditions are met. Most often ‘not too many cells’ is fixed at no more than 20% of the cells. This is often referred to as 'Cochran conditions', after Cochran (1954, p. 420). Note that for example Fisher (1925, p. 83) is more strict, and finds that all cells should have an expected count of at least 5 .

Parameters

field1 : list or pandas series

the first categorical field

field2 : list or pandas series

the first categorical field

categories1 : list or dictionary, optional

order and/or selection for categories of field1

categories2 : list or dictionary, optional

order and/or selection for categories of field2

cc : {None, "yates", "pearson", "williams"}, optional

method for continuity correction

Returns

A dataframe with:

 

• n, the sample size
• n rows, number of categories used in first field
• n col., number of categories used in second field
• statistic, the test statistic (chi-square value)
• df, the degrees of freedom
• p-value, the significance (p-value)
• min. exp., the minimum expected count
• prop. exp. below 5, proportion of cells with expected count less than 5
• test, description of the test used

Notes

The formula used: $$G=2\times\sum_{i=1}^{r}\sum_{j=1}^c\left(E_{i,j}\times ln\left(\frac{E_{i,j}}{F_{i,j}}\right)\right)$$ $$df = \left(r - 1\right)\times\left(c - 1\right)$$ $$sig. = 1 - \chi^2\left(G,df\right)$$

With: $$n = \sum_{i=1}^r \sum_{j=1}^c F_{i,j}$$ $$E_{i,j} = \frac{R_i\times C_j}{n}$$ $$R_i = \sum_{j=1}^c F_{i,j}$$ $$C_j = \sum_{i=1}^r F_{i,j}$$

Unclear what the original source is of the formula, but the one shown is taken from Cressie and Read (1984, p. 441).

The Yates correction uses $F_{i,j}'$ instead of $F_{i,j}$, defined as (Yates, 1934, p. 222): $$F_{i,j}' = \begin{cases} F_{i,j}-\frac{1}{2} & \text{ if } F_{i,j}> E_{i,j} \\ F_{i,j}+\frac{1}{2} & \text{ if } F_{i,j}< E_{i,j} \\ F_{i,j} & \text{ if } F_{i,j}= E_{i,j} \end{cases}$$

The Williams correction, adjusts the Pearson chi-square value: $$\chi_{wil}^2 = \frac{G}{q}$$

With: $$q = 1+\frac{\left(n\times\left(\sum_{i=1}^r \frac{1}{R_i}\right) - 1\right)\times \left(n\times\left(\sum_{j=1}^c \frac{1}{C_i}\right) - 1\right)}{6\times n \times\left(r - 1\right)\times\left(c - 1\right)}$$

The formula is probably from Williams (1976, p. 36) but the one shown here is taken from McDonald (1976, p. 36).

The Pearson correction also adjusts the G value with (E.S. Pearson, 1947, p. 157): $$\chi_{epearson}^2 = \frac{n - 1}{n}\times G$$

References

Cochran, W. G. (1954). Some methods for strengthening the common χ2 tests. Biometrics, 10(4), 417. doi:10.2307/3001616

Fisher, R. A. (1925). Statistical methods for research workers. Oliver and Boyd.

McDonald, J. H. (2014). Handbook of biological statistics (3rd ed.). Sparky House Publishing.

Monica, gung-R. (2015, April 2). Answer to “Why do my p-values differ between logistic regression output, chi-squared test, and the confidence interval for the OR?” Cross Validated. https://stats.stackexchange.com/a/144608

Pearson, E. S. (1947). The choice of statistical tests illustrated on the Interpretation of data classed in a 2 × 2 table. Biometrika, 34(1/2), 139–167. doi:10.2307/2332518

Williams, D. A. (1976). Improved likelihood ratio tests for complete contingency tables. Biometrika, 63(1), 33–37. doi:10.2307/2335081

Yates, F. (1934). Contingency tables involving small numbers and the chi square test. Supplement to the Journal of the Royal Statistical Society, 1(2), 217–235. doi:10.2307/2983604

Author

Made by P. Stikker

Companion website: https://PeterStatistics.com
YouTube channel: https://www.youtube.com/stikpet
Donations: https://www.patreon.com/bePatron?u=19398076

Expand source code

def ts_mod_log_likelihood_ind(field1, field2, categories1=None, categories2=None, cc=None):
    '''
    Mod-Log Likelihood Test of Independence
    -------------------------------------------------
    This test is similar as a Pearson Chi-Square test of independence, but approaches it from a likelihood-ratio approach (see Monica, 2015).
    
    If the significance of this test is below 0.05 (or another pre-defined threshold), the two nominal variables have a significant association.
    
    The test compares the observed counts of the cross table with the so-called expected counts. The expected values are the number of respondents you would expect if the two variables would be independent. See the Pearson Chi-Square test of independence for more details on expected counts.
    
    One problem though is that the test should only be used if not too many cells have a so-called expected count, of less than 5, and the minimum expected count is at least 1. So you will also have to check first if these conditions are met. Most often ‘not too many cells’ is fixed at no more than 20% of the cells. This is often referred to as 'Cochran conditions', after Cochran (1954, p. 420). Note that for example Fisher (1925, p. 83) is more strict, and finds that all cells should have an expected count of at least 5 .
    
    Parameters
    ----------
    field1 : list or pandas series
        the first categorical field
    field2 : list or pandas series
        the first categorical field
    categories1 : list or dictionary, optional
        order and/or selection for categories of field1
    categories2 : list or dictionary, optional
        order and/or selection for categories of field2
    cc : {None, "yates", "pearson", "williams"}, optional
        method for continuity correction
    
    Returns
    -------
    A dataframe with:
    
    * *n*, the sample size
    * *n rows*, number of categories used in first field
    * *n col.*, number of categories used in second field
    * *statistic*, the test statistic (chi-square value)
    * *df*, the degrees of freedom
    * *p-value*, the significance (p-value)
    * *min. exp.*, the minimum expected count
    * *prop. exp. below 5*, proportion of cells with expected count less than 5
    * *test*, description of the test used
    
    Notes
    -----
    The formula used:
    $$G=2\\times\\sum_{i=1}^{r}\\sum_{j=1}^c\\left(E_{i,j}\\times ln\\left(\\frac{E_{i,j}}{F_{i,j}}\\right)\\right)$$
    $$df = \\left(r - 1\\right)\\times\\left(c - 1\\right)$$
    $$sig. = 1 - \\chi^2\\left(G,df\\right)$$
    
    With:
    $$n = \\sum_{i=1}^r \\sum_{j=1}^c F_{i,j}$$
    $$E_{i,j} = \\frac{R_i\\times C_j}{n}$$
    $$R_i = \\sum_{j=1}^c F_{i,j}$$
    $$C_j = \\sum_{i=1}^r F_{i,j}$$
    
    Unclear what the original source is of the formula, but the one shown is taken from Cressie and Read (1984, p. 441).
    
    The **Yates** correction uses \\(F_{i,j}'\\) instead of \\(F_{i,j}\\), defined as (Yates, 1934, p. 222):
    $$F_{i,j}' = \\begin{cases} F_{i,j}-\\frac{1}{2} & \\text{ if } F_{i,j}> E_{i,j} \\\\ F_{i,j}+\\frac{1}{2} & \\text{ if } F_{i,j}< E_{i,j} \\\\ F_{i,j} & \\text{ if } F_{i,j}= E_{i,j} \\end{cases} $$
    
    The **Williams** correction, adjusts the Pearson chi-square value:
    $$\\chi_{wil}^2 = \\frac{G}{q}$$
    
    With:
    $$q = 1+\\frac{\\left(n\\times\\left(\\sum_{i=1}^r \\frac{1}{R_i}\\right) - 1\\right)\\times \\left(n\\times\\left(\\sum_{j=1}^c \\frac{1}{C_i}\\right) - 1\\right)}{6\\times n \\times\\left(r - 1\\right)\\times\\left(c - 1\\right)}$$
    
    The formula is probably from Williams (1976, p. 36) but the one shown here is taken from McDonald (1976, p. 36).
    
    The **Pearson** correction also adjusts the G value with (E.S. Pearson, 1947, p. 157):
    $$\\chi_{epearson}^2 = \\frac{n - 1}{n}\\times G$$
    
    References
    ----------
    Cochran, W. G. (1954). Some methods for strengthening the common χ2 tests. *Biometrics, 10*(4), 417. doi:10.2307/3001616
    
    Fisher, R. A. (1925). *Statistical methods for research workers*. Oliver and Boyd.
    
    McDonald, J. H. (2014). *Handbook of biological statistics* (3rd ed.). Sparky House Publishing.
    
    Monica,  gung-R. (2015, April 2). Answer to “Why do my p-values differ between logistic regression output, chi-squared test, and the confidence interval for the OR?” Cross Validated. https://stats.stackexchange.com/a/144608

    Pearson, E. S. (1947). The choice of statistical tests illustrated on the Interpretation of data classed in a 2 × 2 table. *Biometrika, 34*(1/2), 139–167. doi:10.2307/2332518
    
    Williams, D. A. (1976). Improved likelihood ratio tests for complete contingency tables. *Biometrika, 63*(1), 33–37. doi:10.2307/2335081
    
    Yates, F. (1934). Contingency tables involving small numbers and the chi square test. *Supplement to the Journal of the Royal Statistical Society, 1*(2), 217–235. doi:10.2307/2983604
    
    Author
    ------
    Made by P. Stikker
    
    Companion website: https://PeterStatistics.com  
    YouTube channel: https://www.youtube.com/stikpet  
    Donations: https://www.patreon.com/bePatron?u=19398076
    
    
    '''
    testUsed = "Mod-Log Likelihood test of independence"
    if cc == "yates":
        testUsed = testUsed + ", with Yates continuity correction"
        
    #create the cross table
    ct = tab_cross(field1, field2, categories1, categories2, totals="include")
    
    #basic counts
    nrows = ct.shape[0] - 1
    ncols =  ct.shape[1] - 1
    n = ct.iloc[nrows, ncols]
    
    #determine the expected counts & chi-square value
    chi2Val = 0
    expMin = -1
    nExpBelow5 = 0    
    expC = pd.DataFrame()
    for i in range(0, nrows):
        for j in range(0, ncols):
            expC.at[i, j] = ct.iloc[nrows, j] * ct.iloc[i, ncols] / n
            
            #add or remove a half in case Yates correction
            if cc=="yates":
                if ct.iloc[i,j] > expC.iloc[i,j]:
                    ct.iloc[i,j] = ct.iloc[i,j] - 0.5
                elif ct.iloc[i,j] < expC.iloc[i,j]:
                    ct.iloc[i,j] = ct.iloc[i,j] + 0.5
            
            chi2Val = chi2Val + expC.iloc[i, j]*log(expC.iloc[i, j]/ct.iloc[i, j])
            
            #check if below 5
            if expMin < 0 or expC.iloc[i,j] < expMin:
                expMin = expC.iloc[i,j]            
            if expC.iloc[i,j] < 5:
                nExpBelow5 = nExpBelow5 + 1
    
    chi2Val= 2*chi2Val
    nExpBelow5 = nExpBelow5/(nrows*ncols)
    
    #Degrees of freedom
    df = (nrows - 1)*(ncols - 1)
    
    #Williams and Pearson correction
    if cc == "williams":
        testUsed = testUsed + ", with Williams continuity correction"
        rTotInv = 0
        for i in range(0, nrows):
            rTotInv = rTotInv + 1 / ct.iloc[i, ncols]
        
        cTotInv = 0
        for j in range(0, ncols):
            cTotInv = cTotInv + 1 / ct.iloc[nrows, j]
        
        q = 1 + (n * rTotInv - 1) * (n * cTotInv - 1) / (6 * n * df)
        chi2Val = chi2Val / q
    elif cc == "pearson":
        testUsed = testUsed + ", with E.S. Pearson continuity correction"
        chi2Val = chi2Val * (n - 1) / n
    
    #The test
    pvalue = chi2.sf(chi2Val, df)
    
    #Prepare the results
    colNames = ["n", "n rows", "n col.", "statistic", "df", "p-value", "min. exp.", "prop. exp. below 5", "test"]
    testResults = pd.DataFrame([[n, nrows, ncols, chi2Val, df, pvalue, expMin, nExpBelow5, testUsed]], columns=colNames)
    pd.set_option('display.max_colwidth', None)
    
    return testResults