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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.