Chapter 11.2, Problem 4.2CYU

To determine

To test if there is a significant difference between the two distribution of quality of life ratings at significance level 0.01.

Answer to Problem 4.2CYU

Explanation of Solution

Given:

The data of heart attack patients quality of life for Canada and United States is given below

  

The level of significance = 0.01

Concept used:

The conditions to be met to use a Chi-square test for homogeneity.

• The data should be chosen randomly
• The sample size should be large, so that the expected counts are not less than 5.
• There should be independence and the samples can be at most 10% of the population.

Formula used: $\begin{array}{l}ExpectedCount=\frac{rowtotal*coloumntotal}{Total}\\ {\chi }^2={\displaystyle \sum \left(\frac{{(Observed-Expected)}^2}{Expected}\right)}\\ Degreeoffreedom=(no.ofrows-1)*(no.ofcolumns-1)\end{array}$

The data came from separate random samples of 2600 U.S and 400 Canada heart attack patients. The sample size is large enough so that the expected counts are greater than 5.

The samples are taken from two different countries, so they are independent and we can safely assume that there will be more than 26000 and 4000 heart patients in U.S and Canada respectively. Hence all conditions are met to carry out Chi-square test for homogeneity.

Null hypothesis : There is no significant difference in the distribution of quality of life of heart attack patients of Canada and U.S.

Alternate hypothesis : There is a significant difference in the distribution of quality of life of heart attack patients of Canada and U.S.

Calculation:

The row total and column total are calculated as shown below

  

The expected values are calculated as shown below

  $\begin{array}{ccc}Qualityoflife& Canada& U.S\\ Muchbetter& \frac{616*311}{2476}=77.37& \frac{616*2165}{2476}=538.63\\ Somewhatbetter& \frac{569*311}{2476}=71.47& \frac{569*2165}{2476}=497.53\\ Aboutthesame& \frac{875*311}{2476}=109.91& \frac{875*2165}{2476}=765.09\\ Somewhatworse& \frac{332*311}{2476}=41.70& \frac{332*2165}{2476}=290.30\\ Muchworse& \frac{84*311}{2476}=10.55& \frac{84*2165}{2476}=73.45\end{array}$

The test statistic is calculated as shown below

  $\begin{array}{l}{\chi }^2={\displaystyle \sum \left(\frac{{(Observed-Expected)}^2}{Expected}\right)}\\ {\chi }^2=\frac{{\left(75-77.37\right)}^2}{77.37}+\frac{{\left(71-71.47\right)}^2}{71.47}+\frac{{\left(96-109.91\right)}^2}{109.91}+\mathrm{..}+\frac{{\left(65-73.45\right)}^2}{73.45}\\ {\chi }^2=11.72548\end{array}$

Degree of freedom = (5-1) *(2-1) = 4

The p value for 4 degree of freedom and test statistic 11.72 is 0.019476.

The p value is more than the level of significance, so we have insufficient evidence at 0.01 level of significance to reject the null hypothesis.

Conclusion:

At 0.01 level of significance there is no significant difference in the distribution of quality of life of heart attack patients in Canada and U.S.