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. ?_0^2= Find the p-value. p= State your conclusion in the context of this question using a significance level of 0.05.

Answer 1

For (i,j) cell, the expected frequency =( i th row total * j th column total)/646 for all (i,j). By this formula, the expected frequency is calculated as

Now, because there are many cells for which the cell frequency is less than 5, we need to combined the classes. So, combining the age group as under 30 and above 30, we get the new contingency table for observed and expected frequency as

Now, the chi-square test statistic is calculated as

where

By, this formula, we get the value of test statistics as

so, the p-value will be

Which means we will reject our null hypothesis of independence. We conclude that there is a significant evidence that age and hospitalization are associated