Lesson 16: Chi-squared test

TB sections 8.3-8.4

Meike Niederhausen and Nicky Wakim

2024-12-02

Learning Objectives

  1. Understand the Chi-squared test and the expected cell counts under the null hypothesis distribution.

  2. Determine if two categorical variables are associated with one another using the Chi-squared test.

Where are we?

Last time

  • We looked at inference for a single proportion
  • We looked at inference for a difference in two independent proportions

 

  • If there are two groups, we could see if they had different proportions by testing if the difference between the proportions were the same (null) or different (alternative, two-sided, \(\neq\))

 

  • What happens when we want to compare two or more groups’ proportions?

    • Can no longer rely on the difference in proportions
    • Need a new method to make inference (Chi-squared test!)

Learning Objectives

  1. Understand the Chi-squared test and the expected cell counts under the null hypothesis distribution.
  1. Determine if two categorical variables are associated with one another using the Chi-squared test.

From Lesson 4: Example: hypertension prevalence (1/2)

  • US CDC estimated that between 2011 and 20141, 29% of the population in America had hypertension

 

  • A health care practitioner seeing a new patient would expect a 29% chance that the patient might have hypertension
    • However, this is only the case if nothing else is known about the patient

From Lesson 4: Example: hypertension prevalence (2/2)

  • Prevalence of hypertension varies significantly with age
    • Among adults aged 18-39, 7.3% have hypertension
    • Adults aged 40-59, 32.2%
    • Adults aged 60 or older, 64.9% have hypertension

 

  • Knowing the age of a patient provides important information about the likelihood of hypertension

    • Age and hypertension status are not independent - Can we back up this claim??

  • While the probability of hypertension of a randomly chosen adult is 0.29…

    • The conditional probability of hypertension in a person known to be 60 or older is 0.649

 

Question: Is there an association between age group and hypertension?

From Lesson 4: Contingency tables

  • We can start looking at the contingency table for hypertension for different age groups
    • Contingency table: type of data table that displays the frequency distribution of two or more categorical variables
Table: Contingency table showing hypertension status and age group, in thousands.
Age Group Hypertension No Hypertension Total
18-39 yrs 8836 112206 121042
40-59 yrs 42109 88663 130772
60+ yrs 39917 21589 61506
Total 90862 222458 313320

Test of General Association + Hypotheses

  • General research question: Are two variables (both categorical, nominal) associated with each other?

General wording for hypotheses

Test of “association” wording

  • \(H_0\): There is no association between the two variables

  • \(H_A\): There is an association between the two variables

Test of “independence” wording

  • \(H_0\): The variables are independent

  • \(H_A\): The variables are not independent

Hypotheses test for example

Test of “association” wording

  • \(H_0\): There is no association between age and hypertension

  • \(H_A\): There is an association between age and hypertension

Test of “independence” wording

  • \(H_0\): The variables age and hypertension are independent

  • \(H_A\): The variables age and hypertension are not independent

\(H_0\): Variables are Independent (under the null)

  • Recall from Chapter 2, that events \(A\) and \(B\) are independent if and only if

\[P(A \cap B)=P(A)P(B)\]

  • If age and hypertension are independent variables, then theoretically this condition needs to hold for every combination of levels, i.e.

\[\begin{align} P(18-39 \cap \text{hyp}) &= P(18-39)P(\text{hyp})\\ P(18-39 \cap \text{no hyp}) &= P(18-39)P(\text{no hyp})\\ P(40-59 \cap \text{hyp}) &= P(40-59)P(\text{hyp})\\ P(40-59 \cap \text{no hyp}) &= P(40-59)P(\text{no hyp})\\ P(60+ \cap \text{hyp}) &= P(60+)P(\text{hyp})\\ P(60+ \cap \text{no hyp}) &= P(60+)P(\text{no hyp})\\ \end{align}\]

Age Group Hypertension No Hypertension Total
18-39 yrs 8836 112206 121042
40-59 yrs 42109 88663 130772
60+ yrs 39917 21589 61506
Total 90862 222458 313320

\[\begin{align} P(18-39 \cap \text{hyp}) &= \frac{121042}{313320}\cdot\frac{90862}{313320}\\ & ...\\ P(60+ \cap \text{no hyp}) &=\frac{61506}{313320}\cdot\frac{222458}{313320} \end{align}\]

With these probabilities, for each cell of the table we calculate the expected counts for each cell under the \(H_0\) hypothesis that the variables are independent

Expected counts (if variables are independent)

  • The expected counts (if \(H_0\) is true & the variables are independent) for each cell are
    • \(np = \text{total table size} \cdot \text{probability of cell}\)
    • \(\text{expected count} = \dfrac{\text{column total}\cdot \text{row total}}{\text{table total}}\)

Expected count of 40-59 years old and hypertension:

\[\begin{align} \text{expected count} &= \dfrac{\text{column total}\cdot \text{row total}}{\text{table total}} \\ &= \dfrac{\text{90862}\cdot \text{130772}}{\text{313320}} \\ &= 37923.55 \end{align}\]

Age Group Hypertension No Hypertension Total
18-39 yrs 8836 112206 121042
40-59 yrs 42109 88663 130772
60+ yrs 39917 21589 61506
Total 90862 222458 313320

 

  • If age group and hypertension are independent variables
    • (as assumed by \(H_0\)),
  • then the observed counts should be close to the expected counts for each cell of the table
  • Test to see how likely is it that we observe our data given the null hypothesis (no association)

Observed vs. Expected counts

  • The observed counts are the counts in the 2-way table summarizing the data

 

Age Group Hypertension No Hypertension
18-39 yrs 8836 112206
40-59 yrs 42109 88663
60+ yrs 39917 21589
  • The expected counts are the counts the we would expect to see in the 2-way table if there was no association between age group and hypertension
Age Group Hypertension No Hypertension
18-39 yrs 35101.87 85940.13
40-59 yrs 37923.55 92848.45
60+ yrs 17836.58 43669.42

 

Expected count for cell \(i,j\) :

\[\textrm{Expected Count}_{\textrm{row } i,\textrm{ col }j}=\frac{(\textrm{row}~i~ \textrm{total})\cdot(\textrm{column}~j~ \textrm{total})}{\textrm{table total}}\]

Poll Everywhere Question 2

Using R for expected cell counts

  • R calculates expected cell counts using the expected() function in the epitools package

  • Make sure dataset is in matrix form using as.matrix()

hyp_data2
          Hypertension No_Hypertension
18-39 yrs         8836          112206
40-59 yrs        42109           88663
60+ yrs          39917           21589
library(epitools)
expected(hyp_data2)
          Hypertension No_Hypertension
18-39 yrs     35101.87        85940.13
40-59 yrs     37923.55        92848.45
60+ yrs       17836.58        43669.42

Learning Objectives

  1. Understand the Chi-squared test and the expected cell counts under the null hypothesis distribution.
  1. Determine if two categorical variables are associated with one another using the Chi-squared test.

Reference: Steps in a Hypothesis Test

  1. Check the assumptions

  2. Set the level of significance \(\alpha\)

  3. Specify the null ( \(H_0\) ) and alternative ( \(H_A\) ) hypotheses

    1. In symbols
    2. In words
    3. Alternative: one- or two-sided?

  4. Calculate the test statistic.

  5. Calculate the p-value based on the observed test statistic and its sampling distribution

  6. Write a conclusion to the hypothesis test

    1. Do we reject or fail to reject \(H_0\)?
    2. Write a conclusion in the context of the problem

Step 1: Check assumptions

  • Independence
    • All individuals are independent from one another
      • In particular, observational units cannot be represented in more than one cell
      • For example, someone cannot be in two differnt age groups
  • Sample size
    • In order for the distribution of the test statistic to be appropriately modeled by a chi-squared distribution we need
    • 2 \(\times\) 2 table
      • expected counts are at least 10 for each cell
    • Larger tables
      • No more than 20% of expected counts are less than 5
      • All expected counts are greater than 1

 

 

 

expected(hyp_data2)
          Hypertension No_Hypertension
18-39 yrs     35101.87        85940.13
40-59 yrs     37923.55        92848.45
60+ yrs       17836.58        43669.42

All expected counts > 5

Step 2 and 3: Significance level and Hypotheses

  • Set \(\alpha = 0.05\)

 

Hypotheses test for example

Test of “association” wording

  • \(H_0\): There is no association between age and hypertension

  • \(H_A\): There is an association between age and hypertension

Test of “independence” wording

  • \(H_0\): The variables age and hypertension are independent

  • \(H_A\): The variables age and hypertension are not independent

Step 4: Calculate the \(\chi^2\) test statistic (1/2)

Test statistic for a test of association (independence):

\[\chi^2 = \sum_{\textrm{all cells}} \frac{(\textrm{observed} - \text{expected})^2}{\text{expected}}\]

  • When the variables are independent, the observed and expected counts should be close to each other

Step 4: Calculate the \(\chi^2\) test statistic (2/2)

\[\begin{align} \chi^2 =& \sum\frac{(O-E)^2}{E} \\ =& \frac{(8836-35101.87)^2}{35101.87} + \frac{(112206-85940.13)^2}{85940.13} + \\ & \ldots + \frac{(21589-43669.42)^2}{43669.42} \\ =& 66831 \end{align}\]

Is this value big? Big enough to reject \(H_0\)?

Observed:

Age Group Hypertension No Hypertension
18-39 yrs 8836 112206
40-59 yrs 42109 88663
60+ yrs 39917 21589

 

Expected:

Age Group Hypertension No Hypertension
18-39 yrs 35101.87 85940.13
40-59 yrs 37923.55 92848.45
60+ yrs 17836.58 43669.42

Poll Everywhere Question 2

Step 5: Calculate the p-value

The \(\chi^2\) distribution shape depends on its degrees of freedom

  • It’s skewed right for smaller df,
    • gets more symmetric for larger df
  • df = (# rows-1) x (# columns-1)

  • The p-value is always the area to the right of the test statistic for a \(\chi^2\) test

 

  • We can use the pchisq function in R to calculate the probability of being at least as big as the \(\chi^2\) test statistic:
pv <- pchisq(66831, df = 2, 
       lower.tail = FALSE)
pv
[1] 0

Step 4-5: Calculate the test statistic and p-value

  • Data need to be in a matrix or table: use as.matrix() or table()
    • Use matrix if data already in contingency table form
    • Use table if data are two columns with each row for each observation (tidy version)
  • Notice that age groups are rownames! Age does not have its own column
hyp_data2
          Hypertension No_Hypertension
18-39 yrs         8836          112206
40-59 yrs        42109           88663
60+ yrs          39917           21589
  • Run chisq.test() in R
chisq.test(x = hyp_data2)

    Pearson's Chi-squared test

data:  hyp_data2
X-squared = 66831, df = 2, p-value < 2.2e-16

Step 6: Conclusion

Recall the hypotheses to our \(\chi^2\) test:

  • \(H_0\): There is no association between age and hypertension

  • \(H_A\): There is an association between age and hypertension

 

  • Recall the \(p\)-value = 0.0402
  • Use \(\alpha\) = 0.05
  • Do we reject or fail to reject \(H_0\)?

Conclusion statement:

  • There is sufficient evidence that there is an association between age group and hypertension (p-value < 0.0001`)

Warning!!

If we fail to reject, we DO NOT say variables are independent! We can say that we have insufficient evidence that there is an association.

Chi-squared test: Example all together

  1. Check expected cell counts threshold
expected(hyp_data2)
          Hypertension No_Hypertension
18-39 yrs     35101.87        85940.13
40-59 yrs     37923.55        92848.45
60+ yrs       17836.58        43669.42

All expected cells are greater than 5.

  1. \(\alpha = 0.05\)

  2. Hypothesis test:

    • \(H_0\) : There is no association between age group and hypertension
    • \(H_1\) : There is an association between age group and hypertension

4-5. Calculate the test statistic and p-value for Chi-squared test in R

chisq.test(x = hyp_data2)

    Pearson's Chi-squared test

data:  hyp_data2
X-squared = 66831, df = 2, p-value < 2.2e-16
  1. Conclusion

We reject the null hypothesis that age group and hypertension are not associated (\(p<2.2\cdot10^{-16}\)). There is sufficient evidence that age group and hypertension are associated.

Lesson 16 Slides