# Solved (Free): In the United States, approximately 19% of adult deaths are caused by cancer, 47% by heart disease, and the remaining 34% by other causes

#### ByDr. Raju Chaudhari

Apr 3, 2021

In the United States, approximately 19% of adult deaths are caused by cancer, 47% by heart disease, and the remaining 34% by other causes. A study was done on causes of death for 650 smokers, with the results given below.

Cause of death Cancer Heart disease Other
Number of adults 135 310 205

Use a 0.05 significance level to test the claim that causes of death for smokers do not fit the distribution given above.

#### Solution

The observed data is

Cause Obs. Freq.$(O)$ Prop.
Cancer 135 0.19
Heart Disease 310 0.47
Other 205 0.34
##### Step 1 Setup the hypothesis

The null and alternative hypothesis are as follows:

$H_0:p_{\text{cancer}} = 0.19, p_{\text{Heart disease}}=0.47, p_{\text{Other}}=0.34$

$H_1:$ Causes of death for smokers do not fit the specified distribution.

##### Step 2 Test statistic

The test statistic for testing above hypothesis is

 $$\begin{equation*} \chi^2= \sum \frac{(O_{i} -E_{i})^2}{E_{i}} \sim \chi^2_{(k-1)}\\ \end{equation*}$$

##### Step 3 Level of Significance

The level of significance is $\alpha =0.05$.

##### Step 4 Critical value of $\chi^2$

The level of significance is $\alpha =0.05$. Degrees of freedom $df=k-1=3-1 =2$.

The critical value of $\chi^2$ for $df=2$ and $\alpha=0.05$ level of significance is $\chi^2 =5.9915$.

##### Step 5 Test Statistic

The expected frequencies can be calculated as

 $$\begin{equation*} E_{i} =N*p_i \end{equation*}$$

For example, $E_{1}$ is given by

 $$\begin{eqnarray*} E_{1} & = &N*p_1\\ &=& 650*0.19\\ &=&123.5. \end{eqnarray*}$$

Cause Obs. Freq.$(O)$ Prop. $p_i$ Expe.Freq.$(E)$ $(O-E)^2/E$
Cancer 135 0.19 123.5 1.071
Heart Disease 310 0.47 305.5 0.066
Other 205 0.34 221 1.158

The test statistic is

 $$\begin{eqnarray*} \chi^2&=& \sum \frac{(O_{i} -E_{i})^2}{E_{i}} \sim \chi^2_{(k-1)}\\ &=&\frac{(135-123.5)^2}{123.5}+\cdots + \frac{(205-221)^2}{221}\\ &=& 1.071 +\cdots + 1.158\\ &=& 2.295. \end{eqnarray*}$$

##### Step 6 Decision (Traditional approach)

The test statistic is $\chi^2 =2.295$ which falls $outside$ the critical region bounded by the critical value $5.9915$, we $\textit{fail to reject}$ the null hypothesis.

OR

##### Step 6 Decision ($p$-value approach)

The p-value is $P(\chi^2_{2}>2.295) =0.31743$.

As the p-value $0.3174$ is $\textit{greater than}$ the significance level of $\alpha = 0.05$, we $\textit{fail to reject}$ the null hypothesis.

There is no sufficient evidence to support the claim that causes of death for smokers do not fit the specified distribution.