# Chi-Squared Test

1. Test the null hypothesis that death and location of death are independent variables.

2. For the "over 74" group, estimate the conditional proportion of those who are in acute care, with 90% confidence. (Note: the estimator and its standard error are based on the "sample" consisting just of the "over 74" group.)

3. For the "over 74" group, estimate the difference in conditional proportions for those in acute care and those in chronic care, with 90% confidence.

See attached file for full problem description.

© BrainMass Inc. brainmass.com June 3, 2020, 7:39 pm ad1c9bdddfhttps://brainmass.com/statistics/chi-squared-test/chi-squared-test-106823

#### Solution Preview

a) We're going to use a chi-squared test to test the null hypothesis that location and age of death are independent.

The test statistic is:

X2 = Σ (O - E)2/E where O is the observed number and E is the expected.

df = (number of rows - 1)(number of columns - 1) = (3)(2) = 6

The observed values are in the table in the question. You can calculate the expected values by calculating the row and column totals, then for each cell of the table, multiple the row column by the column total, then dividing by the total number of cases. For example, the expected number of deceased 15-54 year olds who died at home is:

(94 + 418 + 23)(94 + 116 + 156 + 138)/2989 = (535)(504)/2989 = 90.211

Find the expected value for each of the 12 cells in the table.

Then calculate X2:

X2 = (94 - 90.2)2/90.2 + (418 ...

#### Solution Summary

The solution uses a chi-squared test to test the null hypothesis that the two variables are independent of each other. Then, confidence intervals are created for (1) a proportion and (2) a difference of proportions. Complete, step-by-step explainations are provided.