1. Hypothesis testing. Chi-square test
Georgi Iskrov, MBA, MPH, PhD
Department of Social Medicine

2. Parametric and non-parametric tests
Parametric test – to estimate at least one population parameter from sample statistics
Assumption: the variable we have measured in the sample is normally distributed in the population to which we plan to generalize our findings
Non-parametric test – distribution free, no assumption about the distribution of the variable in the population

3. Choosing a statistical test
Choice of a statistical test depends on:
Level of measurement for the dependent and independent variables
Number of groups or dependent measures
Number of units of observation
Type of distribution
The population parameter of interest (mean, variance, differences between means and/or variances)

4. Parametric and non-parametric tests

5. Normality test
Normality tests are used to determine if a data set is modeled by a normal distribution and to compute how likely it is for a random variable underlying the data set to be normally distributed.
In descriptive statistics terms, a normality test measures a goodness of fit of a normal model to the data – if the fit is poor then the data are not well modeled in that respect by a normal distribution, without making a judgment on any underlying variable.
In frequentist statistics statistical hypothesis testing, data are tested against the null hypothesis that it is normally distributed.

6. Normality test Graphical methods
An informal approach to testing normality is to compare a histogram of the sample data to a normal probability curve. The empirical distribution of the data (the histogram) should be bell-shaped and resemble the normal distribution. This might be difficult to see if the sample is small.

7. Normality test Frequentist tests
Tests of univariate normality include the following:
D'Agostino's K-squared test
Jarque–Bera test
Anderson–Darling test
Cramér–von Mises criterion
Lilliefors test
Kolmogorov–Smirnov test
Shapiro–Wilk test
Etc.

8. Normality test Kolmogorov–Smirnov test
K–S test is a nonparametric test of the equality of distributions that can be used to compare a sample with a reference distribution (1-sample K–S test), or to compare two samples (2-sample K–S test).
K–S statistic quantifies a distance between the empirical distribution of the sample and the cumulative distribution of the reference distribution, or between the empirical distributions of two samples.
The null hypothesis is that the sample is drawn from the reference distribution (in the 1-sample case) or that the samples are drawn from the same distribution (in the 2-sample case).

9. Normality test Kolmogorov–Smirnov test
In the special case of testing for normality of the distribution, samples are standardized and compared with a standard normal distribution. This is equivalent to setting the mean and variance of the reference distribution equal to the sample estimates, and it is known that using these to define the specific reference distribution changes the null distribution of the test statistic.

10. Is there an association?
Chi-square test
Chi-square test is used to check for an association between 2 categorical variables.
H0: There is no association between the variables.
HA: There is an association between the variables.
If two categorical variables are associated, it means the chance that an individual falls into a particular category for one variable depends upon the particular category they fall into for the other variable.
Is there an association?

11. Assumptions A large sample of independent observations
All expected counts should be ≥ 1 (no zeros)
At least 80% of expected counts should ≥ 5

12. Chi-square test
The following table presents the data on place of birth and alcohol consumption.
The two variables of interest, place of birth and alcohol consumption, have r = 4 and c = 2, resulting in 4 x 2 = 8 combinations of categories.
Place of birth
Alcohol
No alcohol
Big city
620 75
Rural
240 41
Small town
130 29
Suburban
190 38

13. Chi-square test E11 = (695 x 1180) / 1363 E12 = (695 x 183) / 1363
Place of birth
Alcohol
No alcohol
Total
Big city
O11 = 620
O12 = 75
R1 = 695
Rural
O21 = 240
O22 = 41
R2 = 281
Small town
O31 = 130
O32 = 29
R3 = 159
Suburb
O41 = 190
O42 = 38
R4 = 228
C1 = 1180
C2 = 183
n=1363
E11 = (695 x 1180) / E12 = (695 x 183) / 1363
E21 = (281 x 1180) / E22 = (281 x 183) / 1363
E31 = (159 x 1180) / E32 = (159 x 183) / 1363
E41 = (228 x 1180) / E42 = (228 x 183) / 1363

14. Chi-square test
The test statistic measures the difference between the observed the expected counts assuming independence.
If the statistic is large, it implies that the observed counts are not close to the counts we would expect to see if the two variables were independent. Thus, 'large' χ2 gives evidence against H0, and supports HA.
The p-value of the χ2 test is the probability that the χ2 statistic is as large or larger than the value we obtained if H0 is true.
To get this probability we need to use a chi-square distribution with (r-1) x (c-1) df.

15. Association is not causation.
Beware!
Association is not causation.
The observed association between two variables might be due to the action of a third, unobserved variable.

16. Special case
In a lot of cases the categorical variables of interest have two levels each. In this case, we can summarize the data using a contingency table having 2 rows and 2 columns (2x2 table):
In this case, the χ2 statistic has a simplified form:
Under the null hypothesis, χ2 statistic has chi-square distribution with (2-1) x (2-1) = 1 degrees of freedom.
Column 1
Column 2
Total
Row 1 A B R1
Row 2 C D R2 C1 C2 n

17. Special case Gender Alcohol No alcohol Total Male 540 52 592 Female
325 31
356
865 83
948

18. Limitations No categories should be less than 1
No more than 1/5 of the expected categories should be less than 5
To correct for this, can collect larger samples or combine your data for the smaller expected categories until their combined value is 5 or more
Yates Correction*
When there is only 1 degree of freedom, regular chi-test should not be used
Apply the Yates correction by subtracting 0.5 from the absolute value of each calculated O-E term, then continue as usual with the new corrected values

19. Fisher’s exact test This test is only available for 2 x 2 tables.
For small n, the probability can be computed exactly by counting all possible tables that can be constructed based on the marginal frequencies. Thus, the Fisher exact test computes the exact probability under the null hypothesis of obtaining the current distribution of frequencies across cells, or one that is more uneven.

20. Fisher’s Exact Test Gender Dieting Non-dieting Total Male 1 11 12
Female 9 3 10 14 24
Gender
Dieting
Non-dieting
Total
Male a c
a + c
Female b d
b + d
a + b
c + d
a + b + c + d

21. Fisher’s Exact Test Gender Dieting Non-dieting Total Male 1 11 12
Female 9 3 10 14 24

22. Ordinal data independent samples. Mann-Whitney test
Null hypothesis: Two sampled populations are equivalent in location (they have the same mean ranks).
The observations from both groups are combined and ranked, with the average rank assigned in the case of ties.
If the populations are identical in location, the ranks should be randomly mixed between the two samples.

23. Ordinal data independent samples. Kruskal-Wallis test
Null hypothesis:
K sampled populations are equivalent in location.
The observations from all groups are combined and ranked, with the average rank assigned in the case of ties.
If the populations are identical in location, the ranks should be randomly mixed between the K samples.

24. Ordinal data 2 related samples. Wilcoxon signed rank test
Two related variables. No assumptions about the shape of distributions of the variables.
Null hypothesis:
Two variables have the same distribution.
Takes into account information about the magnitude of differences within pairs and gives more weight to pairs that show large differences than to pairs that show small differences.
Based on the ranks of the absolute values of the differences between the two variables.