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It is given that out of 450 weather-related deaths, 320 were males.

Letp is the population proportion of males who died.

The null hypothesis is as follows:

The proportion of male deaths is equal to 0.50.

Symbolically,

\({H_o}:p = 0.50\)

The alternative hypothesis is as follows:

The proportion of male deaths is not equal to 0.50.

Symbolically,

\({H_o}:p \ne 0.50\)

It is a two-tailed test.

Let\(\hat p\)denote the sample proportion of male deaths.

The value of\(\hat p\)is computed as shown below:

\(\begin{aligned}{c}\hat p = \frac{{{\rm{Number}}\;{\rm{of}}\;{\rm{male}}\;{\rm{deaths}}}}{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{deaths}}}}\\ = \frac{{320}}{{450}}\\ = 0.711\end{aligned}\)

Here, p = 0.50.

Thus,

\[\begin{aligned}{c}q = 1 - p\\ = 1 - 0.50\\ = 0.50\end{aligned}\]

Since the sample size (n) equal to 450 is large, the value of the z-score is computed as follows:

\[\begin{aligned}{c}z = \frac{{\hat p - p}}{{\sqrt {\frac{{pq}}{n}} }}\;\;\; \sim N\left( {0,1} \right)\\ = \frac{{0.711 - 0.50}}{{\sqrt {\frac{{\left( {0.50} \right)\left( {0.50} \right)}}{{450}}} }}\\ = 8.957\end{aligned}\]

Thus, the test statistic, z is 8.957.

Refer to the standard normal table,

The critical value of z at 0.05 level of significance for a two-tailed test is equal to 1.96.

The corresponding p-value obtained using the test statistic is equal to 0.000.

Since the absolute value of the z-score is greater than the critical value and the p-value is less than 0.05, the null hypothesis is rejected.

There is enough evidence to conclude that the percentage of male deaths out of the given sample ofweather-related deaths is not equal to 50%.

It can be said that a comparatively greater number of males participate in outdoor activities like golfing, fishing and surfing during the months of May, June and July. Thus, their proportion is greater than 0.5.