The Situation:

• We will be considering problems involving count data that are classified according to two variables (say, A and B ) and are summarized in a two-way frequency table.
• Each variable will have 2 or more non-overlapping categories and each response must fall into one of the categories.
• The rows of the two-way table correspond to the categories for variable A and the columns of the two-way table correspond to the categories for variable B .
• Each row and column category defines a cell in the table. If there are R rows and C columns then there are $R \times C$ cells in the table.

Example

• Failures of an electronic device are classified by the manufacturer by type of failure and location of failure. There are two types of failure: mechanical and electronic. There are three primary locations within the device that failures occur which will denote as locations 1, 2, and 3. Thus,

  Variable A is ``type of failure''.

  Variable B is ``location of failure''.

• The data (once collected) can be summarized by filling in a two-way table with the observed counts. For the electronic device example, the data are displayed:
  
  
  
• There are R=2 rows and C=3 columns so there are $2 \times 3 = 6$ cells in this table.
  • The 6 cells in the electronic device example can be represented as a pair (Type of Failure, Location of Failure). The 6 cells are (Mechanical,1), (Mechanical,2), (Mechanical,3), (Electronic,1), (Electronic,2), (Electronic,3).
• Also shown are marginal totals. Thus there were 50+16+31 = 97 mechanical failures and there were 16+26=42 failures (both mechanical and electrical) at location 2.
• The counts alone can give a misleading picture of the relationship between type of failure and location of failure. For example we see that there were 50 mechanical failures at location 1 and only 31 at location 2 but $(50/111)100 = 45.0\%$ of the failures at location 1 were mechanical while $(31/47) = 66.0\%$ of the failures at location 3 were mechanical. In general percentages give a more accurate picture of the relationship.

Marginal Distributions

• The marginal distribution of the row variable summarizes the percentage of items having each of the possible values of that variable.
• To compute the marginal row distribution, compute the percentage of each row based on the grand total (the total sample size).
  • The marginal row distribution for Type of Failure in the above example is
    Type of Failure
    
    
    
  • The percentages were computed as shown below.
    • Percent of Mechanical Failures: (97/200)100=48.5 %.
    • Percent of Electrical Failures: (103/200)100=51.5 %.
• The marginal column distribution of the column variable summarizes the percentage of items having each of the possible values of that variable.
• To compute the marginal column distribution, compute the percentage of each row based on the grand total (the total sample size).
• the marginal column distribution for Location of Failure in the above example is
  
  
  
• The percentages were computed as shown below.
  • Percent of failures at Location 1: (111/200)100=55.5 %.
  • Percent of failures at Location 2: (42/200)100=21.0 %.
  • Percent of failures at Location 3: (47/200)100=23.5 %.

Conditional Distributions

• The conditional distribution of the row variable, given the column variable, is found by expressing the counts in the frequency table as percentages of the column totals.
  • The distribution of Type of Failure given Location of Failure is shown below.
    
    
    
• The percentages were computed as follows.
  • Percent of failures at Location 1 that were mechanical: (50/111)100=45.0 %.
  • Percent of failures at Location 2 that were mechanical: (16/42)100=38.1 %.
  • Percent of failures at Location 1 that were electrical: (61/111)100=55.0 %.
  • You should make sure you know how to compute the other percentages shown in the table.
• The conditional distribution of the column variable, given the row variable, is found by expressing the counts in the frequency table as percentages of the row totals.
  • the conditional distribution of Location of Failure given Type of Failure is shown below.
    
    
    
• The percentages were computed as follows.
  • Percent of mechanical failures occurring at Location 1: (50/97)100=51.5 %.
  • Percent of mechanical failures occurring at Location 2: (16/97)100=16.5 %.
  • Percent of electrical failures occurring at Location 1: (61/103)100=59.2 %
  • You should make sure you know how to compute the other percentages shown in the table.
• It can help to draw bar graphs displaying the conditional distributions.

Simpson's Paradox

• An association that holds for all of several groups can reverse direction when the data are combined to form a single group. This reversal is called Simpson's Paradox.
• Below is table of applicants to Upper Wabash Tech's 2 professional schools categorized by gender and admission decision. There were 1200 applicants
  
  
  
• Note that
  • (490/700)100=70.0 % of Males were admitted.
  • (280/500)100=56.0 % of Females were admitted.
  There appears to be evidence of gender discrimination, with Males being favored over Females.
• Upper Wabash Tech has 2 professional schools, Business and Law. There were 800 applicants for the Business School and 400 for the Law School. Below are 2 frequency tables showing the relationship between gender and admission decision for each school separately.
  
  
  
  
  
  

• Note that
  • (480/600)100=80.0 % of Males were admitted to the Business School and (180/200)100=90.0 % of Females were admitted to the Business School.
  • (10/100)100=10.0 % of Males were admitted to the Law School and (100/300)100=33.3 % were admitted to the Law School.
• There is no longer any evidence of gender discrimination in favor of Males. In fact it appears that Females may be favored over Males.
• What happened? The Law School is harder to get into than the Business School. Only (110/400)100=27.5 % of applicants to the Law School were admitted whereas 82.5 % of the applicants to the Business School were admitted. Of the 500 Female applicants, (300/500)100=60.0 % applied to the more selective Law School whereas only 14.3 % of the men applied to the Law School. The original frequency table showed a favoritism towards Males because it failed to account for the particular school to which applicants had applied.
• Numbers will lie to you if you are not careful.

Testing for Significant Associations in Two-Way Tables: Chapter 13, Section 2

• We will be looking at two types of tests: a test of homogeneity and a test of independence. Fortunately both tests use exactly the same test statistic.
• Method: The chi-square test. The approach taken is:
  • For both tests the data consist of observed counts, the number of observations that fall into each cell of a frequency table.
  • The counts we expect to see under H₀ will be computed. These are referred to as expected counts. We will see how to compute these below.
  • If H₀ is true the observed counts (O ) and expected counts (E ) should be close to one another. We will measure how close using a chi-square test statistic.
• The test statistic, denoted X² , compares the set of observed counts with the set of expected counts. The formula for X² is $$X^2=\displaystyle{\sum_{\rm all \ cells} \frac{ \left(O - E \right)^2}{E}}.$$
• Large values of X² are evidence against H₀ . The question remains ``How large is too large?''. To answer this question we refer to a new distribution called the chi-square distribution. In Greek letter notation, we write $\chi^2$distribution.
• With each $\chi^2$ distribution is an associated degrees of freedom. We use the notation $\chi^2(df)$ to denote a $\chi^2$ distribution with df degrees of freedom.
• For an $R \times C$ table, the test statistic X² will approximately follow a $\chi^2$ distribution with $\left(R-1\right) \left(C-1\right)$ degrees of freedom when H₀ is true.
• Properties of the $\chi^2$ distribution:
  • The distribution is skewed to the right.
  • The values are nonnegative.
  • There is a different $\chi^2$ distribution for different degrees of freedom.
  • The mean of a $\chi^2$ distribution is equal to its degrees of freedom.
  • The variance of a $\chi^2$ distribution is equal to twice its degrees of freedom.
• Tabled values of $\chi^2$ distributions can be found in Table VI on page 662.

Testing Hypotheses about Two-Way Tables

• There are 2 tests of interest.

  1.

  Test of Homogeneity.

  2.

  Test of Independence.

• To determine which test is appropriate you need to know how the data were collected.
• For a test of homogeneity there are C independent random samples from each of the C populations under study. The response is a qualitative variable with R possible categories. The data are summarized in a two-way table with the columns referring to the populations and the rows to the categories of the response.
  • The test is designed to see if the C populations are homogeneous with respect to the discrete distribution of the response, i.e. the test is designed to see if the marginal column distributions are the same. The null hypothesis is that the populations are homogeneous with respect to the response and the alternative is that the populations are not homogeneous with respect to the response.
  • Example: A survey was designed to study how the operations of a group of businesses vary with their size. Companies were classified as small, medium, and large. SRS's of size 200 were taken from each size group. Not all the questionnaires were returned and it was decided to examine whether or not the response rate varied with the size of the business. There are C=3 populations. The response variable is Response Rate and there are R=2 categories (Responded and Did Not Respond). The data are shown below.
    
    
    

    The hypotheses are:
    H₀ : The distributions of the Response Rate are the same for the 3 business sizes.
    H₁ ; The distributions of the Response Rate are not the same for the 3 business sizes.
    The null hypothesis is saying that the 3 marginal column distributions are exactly the same.

• For a test of independence there is just one SRS from a single population. For each unit two qualitative responses are recorded. One response has R categories and the other has C categories.
  • The test is designed to see if there is a relationship between the two variables in the population. The null hypothesis is that there is no relationship (i.e. the two variables are independent) and the alternative is that there is a relationship.
  • Example: The first example given in this handout, failures of an electronic device works here. A SRS of 200 failures was taken from the population of failures. For each failure 2 responses were recorded: Type of Failure with R=2 categories and Location of Failure with C=3 categories. The data are summarized in the following table.
    
    
    

    The hypotheses are:
    H₀ : Location of Failure is independent of Type of Failure
    H₁ ; Location of Failure and Type of Failure are dependent.

• Fortunately both hypotheses are tested in exactly the same way.
  The $\chi^2$ Test
• To test the null hypothesis in $R \times C$ tables, we compare the observed cell counts with the expected cell counts. The expected cell counts are calculated under the assumption that H₀ is true.
• The test statistic, denoted X² , compares the set of observed counts with the set of expected counts. We calculate X² as follows:

  1.

  Calculate each expected cell count. The expected count for a cell is $$E \;\; = \;\; \frac{\left( {\rm row \ total} \right) \times \left({\rm column \ total} \right)}{\rm{ overall \ total}}.$$

  2.

  Calculate $\ds \;\;\; \frac{(O \; - \; E)^2}{E} \;\;\;$ for each cell in the table.

  3.

  The test statistic X² is the sum of all of the values calculated in step 2. That is, $$X^2 = \sum \frac{(O - E)^2}{E}$$

• Back to the electronic device example.
  • For cell (i,j) the expected cell counts are: $${\rm For} \; (1,1): \;\; \frac{(97)(111)}{200} = 53.835 \hsp... ...4in} {\rm For} \; (2,1): \;\; \frac{(103)(111)}{200} = 57.165$$$${\rm For} \; (1,2): \;\; \frac{(97)(\;42)}{200} = 20.370 \hs... ...in} {\rm For} \; (2,2): \;\; \frac{(103)(\;42)}{200} = 21.630$$$${\rm For} \; (1,3): \;\; \frac{(97)(\;47)}{200} = 22.795 \hs... ...in} {\rm For} \; (2,3): \;\; \frac{(103)(\;47)}{200} = 24.205$$
  • Then the (O-E)²/E values for cell (i,j) are: $${\rm For} \; (1,1): \;\; \frac{(50-53.835)^2}{53.835} = 0.27... ...{\rm For} \; (2,1): \;\; \frac{(61-57.165)^2}{57.165} = 0.2573$$$${\rm For} \; (1,2): \;\; \frac{(16-20.370)^2}{20.370} = 0.93... ...{\rm For} \; (2,2): \;\; \frac{(26-21.630)^2}{21.630} = 0.8829$$$${\rm For} \; (1,3): \;\; \frac{(31-22.795)^2}{22.795} = 2.95... ...{\rm For} \; (2,3): \;\; \frac{(16-24.205)^2}{24.205} = 2.7813$$
  • Then $\ds \;\; X^2_{OBS} \; = \; 0.2732+0.9375+2.9534+0.2573+0.8829+2.7813 \; = \; 8.0856$.
• Regardless of which of the 2 tests is being conducted if H₀ is true then the observed and expected cell frequencies should be similar. When these frequencies are similar, the X² statistic will tend to be small.
• On the other hand, if H₀ is false then we will tend to find observed and expected cell frequencies that differ significantly. When these frequencies do differ significantly, the X² statistic will tend to be large.
• For the hypothesis test, we can get bounds on the p -value from Table VI (page 662) by doing the following:

  1.

  Go to the (R-1)(C-1) row in Table VI .

  2.

  Scan the row until you find two consecutive values that X² falls between.

  3.

  Look at the column headings corresponding to tail probabilities. These are used to bound the p -value. For example if the column headings are X²_0.95 and X²_0.975 then the p -value is between 0.025 and 0.05. (If the X² falls outside the first or last column we say the p -value is $\gt.995$ or $<.005$, respectively.)

• At this point, we follow the decision rule: Reject H₀ if p -value $\leq \alpha$ or Fail to Reject H₀ if p -value $\gt \alpha$.
• If a significant association is detected, it is recommended to view the bar charts described earlier to gain some understanding of the nature of the association.
• Example For the electronic devices example, the test statistic was X²_OBS=8.0856 . Suppose $\alpha=0.10$. If the null hypothesis is true, i.e. location of failure is independent of type of failure then the test statistic is approximately distributed as $\chi^2$ with (R-1)(C-1)=(1)(2)=2 degrees of freedom. Using Table VI we see that the p -value is bounded between 0.01 and 0.025. Thus we would reject H₀ and conclude that Type of Failure depends on Location of Failure.
• Note that all this test tells us is the two variables are dependent on one another. It tells us nothing about the nature of the dependence. You should always look at the appropriate marginal/conditional distributions to get some idea of the nature of the dependence.
• The example was for a test of independence but exactly the same procedure is followed for a test of homogeneity. Of course, statement of the null and alternative hypotheses will differ as will your conclusions but the mechanical part of carrying out the test is identical.