Question

Use the following data to create the contingency tables.

AGE

Male

          16        17        17        19        19        19        18        17        18        17                                16        19            19        19        17        16        17        16        19        19                                24        31        23        44            21        42        23        43        43        33                                30        41        35        40        24        43            22        30        25        32

            43        51        55        80        61        58        65        52        67        75                                90        63            71        74

Female

             17        16        17        19        19        18        17        19        16        18                                19        17            19        17        18        19        19        16        33        23                                46        46        23        21            46        47        48        47        48        30                                35        24        48        49        47        25            84        54        77        63                                51        72        90        57        69        81

1. In the first table, use gender (male and female) as your row variable and age (<20, 20-50, and >50) for

     your column variable. Run a Chi-square test of independence and find the test statistic, p-value, and

     degrees of freedom.

2. In the second table, use gender (male and female) as your row variable and age (<18, 18-25, 26-45,    

     and >45) for your column variable. Run a Chi-square test of independence and find the test statistic,  

     p-value, and degrees of freedom.

3. Compare the results and comment on problems that may occur when categorizing continuous   

     variables.

Answer 1

H0: The variables gender and age are independent.

H1: The variables gender and age are not independent.

(H0: Null hypothesis; H1: Alternative hypothesis)

1.

Contingency table (Observed frequencies):

	<20	20-50	>50	Total
Male	20	21	13	54
Female	18	18	10	46
Total	38	39	23	N =100

Expected frequency, E =(Corresponding Row Total* Corresponding Column Total)/N

Contingency table (Expected frequencies):

	<20	20-50	>50	Total
Male	54*38/100 =20.52	54*39/100 =21.06	54*23/100 =12.42	54
Female	46*38/100 =17.48	46*39/100 =17.94	46*23/100 =10.58	46
Total	38	39	23	N =100

Test statistic(​​​​​​):

Sl. No:	Observed frequency: O	Expected frequency: E	(O-E)2/E
1.	20	20.52	(20-20.52)2/20.52 =0.0132
2.	18	17.48	(18-17.48)2/17.48 =0.0155
3.	21	21.06	0.0002
4.	18	17.94	0.0002
5.	13	12.42	0.0271
6.	10	10.58	0.0318
Total	100	100	0.088

The test statistic, = =0.088

Degrees of freedom, df =(r-1)(c-1) =(2-1)(3-1) =1(2) =2

For the test statistic, =0.088 and at df =2, the p-value = 0.956954

p-value is very high (>0.01; >0.05 and >0.10 significance levels) indicating that we cannot reject the null hypothesis that says "the variables gender and age are independent".

2.

Contingency table for observed frequencies:

	<18	18-25	26-45	>45	Total
Male	10	17	14	13	54
Female	8	15	3	20	46
Total	18	32	17	33	N =100

Contingency table for expected frequencies:

	<18	18-25	26-45	>45	Total
Male	9.72	17.28	9.18	17.82	54
Female	8.28	14.72	7.82	15.18	46
Total	18	32	17	33	N =100

Test statistic(​​​​​​):

Sl. No.	Observed frequency: O	Expected frequency: E	(O-E)2/E
1.	10	9.72	0.0081
2.	8	8.28	0.0095
3.	17	17.28	0.0045
4.	15	14.72	0.0053
5.	14	9.18	2.5308
6.	3	7.82	2.9709
7.	13	17.82	1.3037
8.	20	15.18	1.5305
Total	100	100	8.3633

The test statistic, =8.3633

Degrees of freedom, df =(r-1)(c-1) =(2-1)(4-1) =1(3) =3

For the test statistic, =8.3633 and at df =3, the p-value = 0.039071

p-value is low (<0.05; <0.10) indicating that we can reject the null hypothesis that says "the variables gender and age are independent".

(However at 0.01 significance level, p-value of 0.039071 > 0.01 indicating that we cannot reject the null hypothesis).

3.

The results of 1. and 2. are significantly different with higher and lower p-values that resulted in opposite conclusions at 5% and 10% significance levels.

This is because of the problem of different cut points when categorizing the continuous variable (age in this case).

When categorizing continuous variables, cut points are a major problem. How can one decide where to cut? It depends purely on the researcher and what he wants to determine but it's not that simple to decide where to cut and different cut points may result in contradictory conclusions as above.

Another problem is the loss of information.