Chi-square and ordinal tests
The ASTA team
Contingency tables
A contingency table
The dataset popularKids, we study the association between the factors Goals and Urban.Rural:
Urban.Rural: The students were selected from urban, suburban, and rural schools.
Goals: The students indicated whether good grades, athletic ability, or popularity was most important to them.
Based on a sample we make a cross tabulation of the factors and we get a so-called contingency table (krydstabel).
popKids <- read.delim("https://asta.math.aau.dk/datasets?file=PopularKids.dat")
library(mosaic)
tab <- tally(~Urban.Rural + Goals, data = popKids, margins = TRUE)
tab
## Goals
## Urban.Rural Grades Popular Sports Total
## Rural 57 50 42 149
## Suburban 87 42 22 151
## Urban 103 49 26 178
## Total 247 141 90 478
A conditional distribution
- Another representation of data is the probability distribution of
Goals for each level of Urban.Rural, i.e. the sum in each row of the table is \(1\) (up to rounding):
## Goals
## Urban.Rural Grades Popular Sports Sum
## Rural 0.383 0.336 0.282 1.000
## Suburban 0.576 0.278 0.146 1.000
## Urban 0.579 0.275 0.146 1.000
## Total 0.517 0.295 0.188 1.000
Here we will talk about the conditional distribution of Goals given Urban.Rural.
An important question could be:
- Are the goals of the kids different when they come from urban, suburban or rural areas? I.e. are the rows in the table significantly different?
There is (almost) no difference between urban and suburban, but it looks like rural is different.
Independence
Recall, that two factors are independent, when there is no difference between the population’s distributions of one factor given the levels of the other factor.
Otherwise the factors are said to be dependent.
If we e.g. have the following conditional population distributions of Goals given Urban.Rural:
## Goals
## Urban.Rural Grades Popular Sports
## Rural 0.5 0.3 0.2
## Suburban 0.5 0.3 0.2
## Urban 0.5 0.3 0.2
Then the factors Goals and Urban.Rural are independent.
We take a sample and “measure” the factors \(F_1\) and \(F_2\). E.g. Goals and Urban.Rural for a random child.
The hypothesis of interest today is: \[H_0: F_1 \mbox{ and } F_2 \mbox{ are independent, } \quad H_a: F_1 \mbox{ and } F_2 \mbox{ are dependent.}\]
The Chi-squared test for independence
- Our best guess of the distribution of
Goals is the relative frequencies in the sample:
tab <- tally(~Urban.Rural + Goals, data = popKids)
n <- margin.table(tab)
pctGoals <- round(margin.table(tab, 2) / n, 3)
pctGoals
## Goals
## Grades Popular Sports
## 0.517 0.295 0.188
If we assume independence, then this is also a guess of the conditional distributions of Goals given Urban.Rural.
The corresponding expected counts in the sample are then:
## Goals
## Urban.Rural Grades Popular Sports Sum
## Rural 77.0 (0.517) 44.0 (0.295) 28.1 (0.188) 149.0 (1.000)
## Suburban 78.0 (0.517) 44.5 (0.295) 28.4 (0.188) 151.0 (1.000)
## Urban 92.0 (0.517) 52.5 (0.295) 33.5 (0.188) 178.0 (1.000)
## Sum 247.0 (0.517) 141.0 (0.295) 90.0 (0.188) 478.0 (1.000)
Calculation of expected table
## Goals
## Urban.Rural Grades Popular Sports Sum
## Rural 77.0 (0.517) 44.0 (0.295) 28.1 (0.188) 149.0 (1.000)
## Suburban 78.0 (0.517) 44.5 (0.295) 28.4 (0.188) 151.0 (1.000)
## Urban 92.0 (0.517) 52.5 (0.295) 33.5 (0.188) 178.0 (1.000)
## Sum 247.0 (0.517) 141.0 (0.295) 90.0 (0.188) 478.0 (1.000)
Chi-squared (\(\chi^2\)) test statistic
- We have an observed table:
## Goals
## Urban.Rural Grades Popular Sports
## Rural 57 50 42
## Suburban 87 42 22
## Urban 103 49 26
- And an expected table, if \(H_0\) is true:
## Goals
## Urban.Rural Grades Popular Sports Sum
## Rural 77.0 44.0 28.1 149.0
## Suburban 78.0 44.5 28.4 151.0
## Urban 92.0 52.5 33.5 178.0
## Sum 247.0 141.0 90.0 478.0
If these tables are “far from each other”, then we reject \(H_0\). We want to measure the distance via the Chi-squared test statistic:
\(X^2=\sum\frac{(f_o-f_e)^2}{f_e}\): Sum over all cells in the table
\(f_o\) is the frequency in a cell in the observed table
\(f_e\) is the corresponding frequency in the expected table.
We have: \[X^2_{obs} = \frac{(57-77)^2}{77} + \ldots + \frac{(26-33.5)^2}{33.5}=18.8\]
Is this a large distance??
\(\chi^2\)-test template.
We want to test the hypothesis \(H_0\) of independence in a table with \(r\) rows and \(c\) columns:
We take a sample and calculate \(X^2_{obs}\) - the observed value of the test statistic.
p-value: Assume \(H_0\) is true. What is then the chance of obtaining a larger \(X^2\) than \(X^2_{obs}\), if we repeat the experiment?
This can be approximated by the \(\mathbf{\chi^2}\)-distribution with \(df=(r-1)(c-1)\) degrees of freedom.
For Goals and Urban.Rural we have \(r=c=3\), i.e. \(df=4\) and \(X^2_{obs}=18.8\), so the p-value is:
1 - pdist("chisq", 18.8, df = 4)
## [1] 0.0008603303
- There is clearly a significant association between
Goals and Urban.Rural.
The function chisq.test
- All of the above calculations can be obtained by the function
chisq.test.
tab <- tally(~ Urban.Rural + Goals, data = popKids)
testStat <- chisq.test(tab, correct = FALSE)
testStat
##
## Pearson's Chi-squared test
##
## data: tab
## X-squared = 18.828, df = 4, p-value = 0.0008497
## Goals
## Urban.Rural Grades Popular Sports
## Rural 76.99372 43.95188 28.05439
## Suburban 78.02720 44.54184 28.43096
## Urban 91.97908 52.50628 33.51464
- The frequency data can also be put directly into a matrix.
data <- c(57, 87, 103, 50, 42, 49, 42, 22, 26)
tab <- matrix(data, nrow = 3, ncol = 3)
row.names(tab) <- c("Rural", "Suburban", "Urban")
colnames(tab) <- c("Grades", "Popular", "Sports")
tab
## Grades Popular Sports
## Rural 57 50 42
## Suburban 87 42 22
## Urban 103 49 26
##
## Pearson's Chi-squared test
##
## data: tab
## X-squared = 18.828, df = 4, p-value = 0.0008497
The \(\chi^2\)-distribution
Summary
For the the Chi-squared statistic, \(X^2\), to be appropriate we require that the expected values have to be \(f_e\geq 5\).
Now we can summarize the ingredients in the Chi-squared test for independence.
Residual analysis
If we reject the hypothesis of independence it can be of interest to identify the significant deviations.
In a given cell in the table, \(f_o-f_e\) is the deviation between data and the expected values under the null hypothesis.
We assume that \(f_e\geq 5\).
If \(H_0\) is true, then the standard error of \(f_o - f_e\) is given by \[se=\sqrt{f_e (1 - \textbf{row proportion}) (1 - \textbf{column proportion})}\]
The corresponding \(z\)-score \[z=\frac{f_o-f_e}{se}\] should in 95% of the cells be between \(\pm 2\). Values above 3 or below -3 should not appear.
In popKids table cell Rural and Grade we got \(f_e = 77.0\) and \(f_o = 57\). Here column proportion \(=0.517\) and row proportion \(=149/478=0.312\).
We can then calculate \[z = \frac{57-77}{\sqrt{77(1-0.517)(1-0.312)}} = -3.95\]
Compared to the null hypothesis there are way too few rural kids who find grades important.
In summary: The standardized residuals allow for cell-by-cell (\(f_e\) vs \(f_o\)) comparision.
Residual analysis in R
- In
R we can extract the standardized residuals from the output of chisq.test:
tab <- tally(~ Urban.Rural + Goals, data = popKids)
testStat <- chisq.test(tab, correct = FALSE)
testStat$stdres
## Goals
## Urban.Rural Grades Popular Sports
## Rural -3.9508449 1.3096235 3.5225004
## Suburban 1.7666608 -0.5484075 -1.6185210
## Urban 2.0865780 -0.7274327 -1.8186224
Cramér’s V
To measure the strength of the association, the Swedish mathematician Harald Cramér developed a measure which is estimated by \[
V = \sqrt{\frac{X^2}{n\cdot \text{min}(r-1,c-1)}}
\] where r and c are the number of columns and rows in the contingency table and n is the sample size.
Property:
- Cramér’s V lies between 0(no association) and 1(complete association)
In the situation with the factors Goals and Urban.Rural from the dataset popularKids we get \[
V = \sqrt{\frac{X^2}{n\cdot \text{min}(r-1,c-1)}} = \sqrt{\frac{18.8}{479\cdot \text{min}(3-1,3-1)}} = 0.14,
\] which indicates a weak (but significant) association.
The function CramerV in the package DescTools gives you the value and a confidence interval
library(DescTools)
CramerV(tab, conf = 0.95, type = "perc")
## Cramer V lwr.ci upr.ci
## 0.14033592 0.06014641 0.19419139
Ordinal variables
Association between ordinal variables
| < 15k |
1 |
3 |
10 |
6 |
| 15-25k |
2 |
3 |
10 |
7 |
| 25-40k |
1 |
6 |
14 |
12 |
| > 40k |
0 |
1 |
9 |
11 |
We might do a chi-square test to see whether Q1 and Q2 are associated, but the test does not exploit the ordinality.
We shall consider a test that incorporates ordinality.
Gamma coefficient
Consider a pair of respondents, where respondent 1 is below respondent 2 in relation to Q1.
Let:
\(C=\) the number of concordant pairs in our sample.
\(D=\) the number of disconcordant pairs in our sample.
We define the estimated gamma coefficient \[\hat{\gamma} = \frac{C-D}{C+D} = \underbrace{\frac{C}{C+D}}_{concordant~prop.} - \underbrace{\frac{D}{C+D}}_{discordant~prop.}\]
Gamma coefficient
Properties:
Gamma lies between -1 og 1
The sign tells whether the association is positive or negative
Large absolute values correspond to strong association
The standard error \(se({\hat{\gamma}})\) on \({\hat{\gamma}}\) is complicated to calculate, so we leave that to software.
We can now determine a 95% confidence interval: \[{\hat{\gamma}}\pm 1.96 se({\hat{\gamma}})\] and if zero is contained in the interval, then there is no significant association, when we perform a test with a 5% significance level.
Example
- First, we need to install the package
vcdExtra, which has the function GKgamma for calculating gamma. It also has the dataset on job satisfaction and income built-in:
## satisfaction
## income VeryD LittleD ModerateS VeryS
## < 15k 1 3 10 6
## 15-25k 2 3 10 7
## 25-40k 1 6 14 12
## > 40k 0 1 9 11
GKgamma(JobSat, level = 0.90)
## gamma : 0.221
## std. error : 0.117
## CI : 0.028 0.414
- A positive association. Marginally significant at the 10% level, but not so at the 5% level.
Validation of data
Goodness of fit test
You have collected a sample and want to know, whether the sample is representative for people living in Hirtshals.
E.g. whether the distribution of gender, age, or profession in the sample do not differ significantly from the distribution in Hirtshals.
Actually, you know how to do that for binary variables like gender, but not if you e.g. have 6 agegroups.
Example
As an example we look at \(k\) groups, where data from Hjørring kommune tells us the distribution in Hirtshals is given by the vector \[\pi = (\pi_1, \ldots, \pi_k),\] where \(\pi_i\) is the proportion which belongs to group number \(i,\ \
i=1,2\ldots,k\) in Hirtshals.
Consider the sample represented by the vector: \[O=(O_1, \ldots, O_k),\] where \(O_i\) is the observed number of individuals in group number \(i,\ \ i = 1, 2, \ldots, k\).
The total number of individuals: \[n = \sum_{i = 1}^k O_i.\]
The expected number of individuals in each group, if we have a sample from Hirtshals: \[ E_i = n \pi_i,\ i = 1, 2, \ldots, k \]
Goodness of fit test
- We will use the following measure to see how far away the observed is from the expected:
\[X^2=\sum_{i=1}^k\frac{(O_i-E_i)^2}{E_i}\]
- If this is large we reject the hypothesis that the sample has the same distribution as Hirtshals. The reference distribution is the \(\chi^{2}\) with \(k-1\) degrees of freedom.
Example
- Assume we have four groups and that the true distribution is given by:
k <- 4
pi_vector <- c(0.3, 0.2, 0.25, 0.25)
- Assume that we have the following sample:
O_vector <- c(74, 72, 40, 61)
- Expected number of individuals in each group:
n <- sum(O_vector)
E_vector <- n * pi_vector
E_vector
## [1] 74.10 49.40 61.75 61.75
Xsq = sum((O_vector - E_vector)^2 / E_vector)
Xsq
## [1] 18.00945
p_value <- 1 - pchisq(Xsq, df = k-1)
p_value
## [1] 0.0004378808
Test in R
Xsq_test <- chisq.test(O_vector, p = pi_vector)
Xsq_test
##
## Chi-squared test for given probabilities
##
## data: O_vector
## X-squared = 18.009, df = 3, p-value = 0.0004379
- As the hypothesis is rejected, we look at the standardized residuals (\(z\)-scores):
## [1] -0.01388487 3.59500891 -3.19602486 -0.11020775
- We conclude that group 1 and 4 is close to true distribution in Hirtshals, but in groups 2 og 3 we have a significant mismatch.