6.6.1. Contingency Table
This procedure assumes that the data is in the form of frequency counts and entered in a table format. Any number of data columns can be selected as the columns of a Contingency Table, provided that their lengths are equal. If this condition is not met, then the program will not proceed.
After selecting the variables, an Output Options Dialogue asking for the score method to be used (see 6.6.2.0. Scores) and containing four check boxes (with an [Opt] button to their left) will be displayed. Each one of these options has further options. The full output from the Contingency Table procedure is large and computations are demanding. For a full discussion of these options see section 6.6.2. CrossTabulation.
Example 1
Example 23.1 on p. 490 from Zar, J. H. (2010). The null hypothesis “Human hair colour is independent of sex in the population sampled” is tested.
Open TABLES, select Statistics 1 → Tables → Contingency Table and select Black, Brown, Blond and Red (C1 to C4) as [Variable]s. From the Tables options select only the Frequency and from R x C Table Statistics select only the Chisquare Tests output options. Go back to Step 2 Output Options Dialogue and click [Finish] to obtain the following results:
Contingency Table
2 Rows x 4 Columns
Frequency

Black 
Brown 
Blond 
Red 
Row Sum 
R1 
32.0000 
43.0000 
16.0000 
9.0000 
100.0000 
R2 
55.0000 
65.0000 
64.0000 
16.0000 
200.0000 
Column Sum 
87.0000 
108.0000 
80.0000 
25.0000 
300.0000 
Chisquare Tests

ChiSquare Statistic 
Degrees of Freedom 
RightTail Probability 
Pearson 
8.9872 
3 
0.0295 
LikelihoodRatio 
9.5121 
3 
0.0232 
+ Yates Correction 



# Linearbylinear 
2.6155 
1 
0.1058 
~ McNemarBowker 



+ Reported for 2 x 2 tables.
# Table scores
~ Reported for 3 x 3 or larger square tables.
Cells with expected count < 5 = 0 ( 0.00%)
Minimum expected count = 8.3333
Phi = 
0.1731 
Contingency Coefficient = 
0.1705 
Cramer’s V = 
0.1731 
In this example Zar only reports Pearson’s chisquare statistic and its tail probability.
Example 2
Example 23.4 on p. 499 from Zar, J. H. (2010). The null hypothesis “the ability of snails to resist the current is no different between the two species” is tested. Open TABLES, select Statistics 1 → Tables → Contingency Table, select Resisted and Yielded (C10 and C11) as [Variable]s. Leave output option selections as in the previous example.
Contingency Table
2 Rows x 2 Columns
Frequency

Resisted 
Yielded 
Row Sum 
R1 
12.0000 
7.0000 
19.0000 
R2 
2.0000 
9.0000 
11.0000 
Column Sum 
14.0000 
16.0000 
30.0000 
Chisquare Tests

ChiSquare Statistic 
Degrees of Freedom 
RightTail Probability 
Pearson 
5.6622 
1 
0.0173 
LikelihoodRatio 
6.0162 
1 
0.0142 
+ Yates Correction 
3.9993 
1 
0.0455 
# Linearbylinear 
5.4734 
1 
0.0193 
~ McNemarBowker 



# Table scores
~ Reported for 3 x 3 or larger square tables.
Cells with expected count < 5 = 0 ( 0.00%)
Minimum expected count = 5.1333
Phi = 
0.4344 
Contingency Coefficient = 
0.3985 
Cramer’s V = 
0.4344 
Zar reports the chisquare statistic with Yates correction and its tail probability.
Example 3
Example 8.4 on p. 231 from Armitage & Berry (2002). The effects of PAS and streptomycin in the treatment of pulmonary tuberculosis are given. The null hypothesis “the probabilities for rows (columns) to fall into different columns (rows) are the same” is tested.
Open TABLES, select Statistics 1 → Tables → Contingency Table and include Negative smear, NegSmrPosCult and NegSmrNegCult (C7 to C9) as [Variable]s. Include only Frequency, Expected and Chisquare Statistics output options to obtain the following results:
Contingency Table
3 Rows x 3 Columns
Frequency

Negative smear 
NegSmrPosCult 
NegSmrNegCult 
Row Sum 
R1 
56.0000 
30.0000 
13.0000 
99.0000 
R2 
46.0000 
18.0000 
20.0000 
84.0000 
R3 
37.0000 
18.0000 
35.0000 
90.0000 
Column Sum 
139.0000 
66.0000 
68.0000 
273.0000 
Expected

Negative smear 
NegSmrPosCult 
NegSmrNegCult 
R1 
50.4066 
23.9341 
24.6593 
R2 
42.7692 
20.3077 
20.9231 
R3 
45.8242 
21.7582 
22.4176 
Chisquare Tests

ChiSquare Statistic 
Degrees of Freedom 
RightTail Probability 
Pearson 
17.6284 
4 
0.0015 
LikelihoodRatio 
17.7770 
4 
0.0014 
+ Yates Correction 



# Linearbylinear 
11.4263 
1 
0.0007 
McNemarBowker 
14.9937 
3 
0.0018 
+ Reported for 2 x 2 tables.
# Table scores
Cells with expected count < 5 = 0 ( 0.00%)
Minimum expected count = 20.3077
Phi = 
0.2541 
Contingency Coefficient = 
0.2463 
Cramer’s V = 
0.1797 