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

1)

The Chi-Square test of independence is used to determine if there is a significant relationship between two factors.

For variables Gender and Age

The Chi-Square test of independence is performed in following steps,

Step 1: The hypothesis is defined as,

Null hypothesis, Ho:There is no association between two variables.

Alternative hypothesis, Ha There is an association present between the two variables.

Step 2: The significance level for the test is,

Step 3: The Chi-Square test statistic is obtained as follow,

The observed values are,

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

Step 4: The expected values are obtained using the formula,

The expected values are,

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

Step 5: Now the Chi-Square Value is obtained using the formula,

Observed,	Expected,
20	20.52	-0.5200	0.2704	0.0132
21	21.06	-0.0600	0.0036	0.0002
13	12.42	0.5800	0.3364	0.0271
18	17.48	0.5200	0.2704	0.0155
18	17.94	0.0600	0.0036	0.0002
10	10.58	-0.5800	0.3364	0.0318
			Sum	0.0879

The P-value is obtained from chi square distribution table for degree of freedom = (r-1)(c-1)=(2-1)(3-1)=2

Since the P-value is greater than 0.05 at 5% significance level, the null hypothesis is not rejected.

2)

The Chi-Square test statistic is obtained as follow,

The observed values are,

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

the expected values are obtained using the formula,

The expected values are,

	<18	18-25	26-45	>45	Total
Male	9.6364	17.1313	9.1010	17.6667	53
Female	8.3636	14.8687	7.8990	15.3333	46
Total	18	32	17	33	100

Now the Chi-Square Value is obtained using the formula,

Observed,	Expected,
10	9.72	0.2800	0.0784	0.0081
17	17.28	-0.2800	0.0784	0.0045
14	9.18	4.8200	23.2324	2.5308
13	17.82	-4.8200	23.2324	1.3037
8	8.28	-0.2800	0.0784	0.0095
15	14.72	0.2800	0.0784	0.0053
3	7.82	-4.8200	23.2324	2.9709
20	15.18	4.8200	23.2324	1.5305
			Sum	8.3632

The P-value is obtained from chi square distribution table for degree of freedom = (r-1)(c-1)=(2-1)(4-1)=3

Since the P-value is less than 0.05 at 5% significance level, the null hypothesis is rejected.

3)

The null hypothesis is not rejected in first part while null hypothesis is rejected in second part of the problem. The main cause of this happen was due to change in the age categorization.

The main problem with the categorization is to choosing the number of cut points in for the continuous variable.

.