Lightning Deaths The graph in Cumulative Review Exercise 5 was created by using data consisting of 232 male deaths from lightning strikes and 55 female deaths from lightning strikes. Assume that these data are randomly selected lightning deaths and proceed to test the claim that the proportion of male deaths is greater than 1/2. Use a 0.01 significance level. Any explanation for the result?

There are 232 male deaths and 55 female deaths due to lightning strikes. It is claimed that the proportion of male deaths is greater than $\frac{1}{2}$ or 0.5.

The null hypothesis is written as follows:

The proportion of male deaths is equal to 0.5.

$H_0:p=0.5$

The alternative hypothesis is written as follows:

The proportion of male deaths is greater than 0.5.

$H_1:p>0.5$

The test is right-tailed.

The sample size is computed below:

$$\begin{gathered} n=232+55 \\ =287 \end{gathered}$$

The sample proportion of male deaths is equal to:

$$\begin{gathered} \hat{p}=\frac{\text{Number}\text{of}\text{male}\text{deaths}}{\text{Sample}\text{size}} \\ =0.808 \end{gathered}$$

The population proportion of male deaths is equal to 0.5.

The value of the test statistic is computed below:

$$\begin{gathered} z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}} \\ =\frac{0.808-0.5}{\sqrt{\frac{0.5\left(1-0.5\right)}{287}}} \\ =10.44 \end{gathered}$$

Thus, z=10.44.

Referring to the standard normal table, the critical value of z at for a right-tailed test is equal to 2.3263.

Referring to the standard normal table, the p-value for the test statistic value of 10.44 is equal to 0.000.

Since the p-value is less than 0.01, the null hypothesis is rejected.

There is enough evidence to support the claim that the proportion of male deaths is greater than 0.5.

Since a greater number of males are engaged in outdoor activities like construction and fishing, there is a comparatively large number of male deaths due to lightning strikes.