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The data on heart attack patient's quality of life in Canada and the United States is given below

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 independent and the samples can be at most $10\%$of the population.

$\text{Expected Count}=\frac{\text{row total}{}^{*}\text{coloumn total}}{\text{Total}}$

$\text{Formula used:}{蠂}^2=鈭�\left(\frac{(\text{Observed}-\text{Expected}{)}^2}{\text{Expected}}\right)$

$\text{Degree of freedom}=(\text{no. of rows}-1)*(\text{no. of columns}-1)$

The data came from separate random samples of $2600$U.S and $400$Canadian 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 the Chi-square test for homogeneity.

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

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

The row total and column total of the table is calculated as shown below

$\begin{array}{ccc}\text{Quality of life}& \text{Canada}& \text{U.S}\\ \text{Much better}& \frac{{616}^{脳}311}{2476}=77.37& \frac{616脳2165}{2476}=538.63\\ \text{Somewhat better}& \frac{{569}^{脳}311}{2476}=71.47& \frac{{569}^{脳}2165}{2476}=497.53\\ \text{About the same}& \frac{{875}^{脳}311}{2476}=109.91& \frac{{875}^{脳}2165}{2476}=765.09\\ \text{Somewhat worse}& \frac{{332}^{脳}311}{2476}=41.70& \frac{{332}^{脳}2165}{2476}=290.30\\ \text{Much worse}& \frac{{84}^{脳}311}{2476}=10.55& \frac{{84}^{脳}2165}{2476}=73.45\end{array}$

The test statistic is calculated as shown below

$$\begin{gathered} {蠂}^2=鈭�\left(\frac{(\text{Observed}-\text{Expected}{)}^2}{\text{Expected}}\right) \\ =\frac{(75鈭�77.37{)}^2}{77.37}+\frac{(71鈭�71.47{)}^2}{71.47}+\frac{(96鈭�109.91{)}^2}{109.91}+..+\frac{(65鈭�73.45{)}^2}{73.45} \\ =11.72548 \end{gathered}$$

$$\begin{gathered} Degreeoffreedom=(5-1)*(2-1) \\ =4 \end{gathered}$$

The $p$-value for $4$degrees 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.