Question

The data on hospitalizations for swine flu are summarized in the table below.

	Age Group
	Under 5	5 -14	15-29	30-44	45-60	Over 60	Total
Hospitalized, Yes	8	12	13	15	0	1	49
Hospitalized, No	47	183	262	57	37	11	597
Total	55	195	275	72	37	12	646

Fill in the table below with the expected counts if Age and Hospitalization are not associated.

	Age Group
	Under 5	5 -14	15-29	30-44	45-60	Over 60	Total
Hospitalized, Yes
Hospitalized, No
Total

Do we meet the requirements for a test for association?

Complete the following table to combine categories from the first table in this problem.

	Age Group
	Under 30	Over 30	Total
Hospitalized, Yes
Hospitalized, No
Total

Find the expected counts if age and hospitalization are not associated in this new table.

	Age Group
	Under 30	Over 30	Total
Hospitalized, Yes
Hospitalized, No
Total

Do we meet the requirements for a test for association?

Write out the hypotheses.

Find the test statistic.

Find the p-value.

p=

State your conclusion in the context of this question using a significance level of 0.05.

Answer 1

First of all we obtain expected counts as :

For Hospitalized and age group under 5 , the expected count wil be = 49 * 55 / 646 = 4.172

For Hospitalized and age group 5-14 , the expected count wil be = 49 * 195 / 646 = 14.8

For Hospitalized and age group under 15-29 , the expected count wil be = 49 * 275 / 646 = 20.86

and so on

So, we obtain the following table for expected counts :

	Age group
	Under 5	5-14	15-29	30-44	45-60	Over 60	Total
Hospitalized, Yes	4.17	14.79	20.86	5.46	2.81	0.91	49
Hospitalized, No	50.83	180.21	254.14	66.54	34.19	11.09	597
Total	55	195	275	72	37	12	646

We do not meet the the requirements of test for association because there are some cells in which the counts are still less than 5. hence, we combine the categories.

The following table is obtained by combining categories from the first table in this problem :

	Age Group
	Under 30	Over 30	Total
Hospitalized, Yes	33	16	49
Hospitalized, No	492	105	597
Total	525	121	646

Now, we obtain the following table for expected counts :

	Age Group
	Under 30	Over 30	Total
Hospitalized, Yes	39.82	9.18	49
Hospitalized, No	485.18	111.82	597
Total	525	121	646

Now, We meet the the requirements of test for association because there are no cells in which the counts are less than 5.

Null hypothesis : Age and Hospitalization are not associated.

Alternative hypothesis : Age and Hospitalization are associated.

Test statistic is given by :

where,

Oi = Observed counts and Ei = Expected counts

So, using the above formula

Test statistic = 6.7466

p-value , p = 0.0094

Since , P-value = 0.0094 < 0.05 ( significance level ) , we reject null hypothesis and conclude that Age and Hospitalization are associated.