Question

Weather-Related Deaths For a recent year, the numbers of weather-related U.S. deaths for each month were 28, 17, 12, 24, 88, 61, 104, 32, 20, 13, 26, 25 (listed in order beginning with January). Use a 0.01 significance level to test the claim that weather-related deaths occur in the different months with the same frequency. Provide an explanation for the result.

Short answer

There is enough evidence to conclude thatweather-related deaths do not occur with equal frequency in different months.

The months of May, June, and July tend to have a disproportionately higher number of weather-related mortalities, which is due to the increased number of vacations and outdooractivities during those months.

Step-by-step solution

Given information

The number of weather-related deaths that occurred in different months of a year is provided.

Hypotheses

The null hypothesis is as follows:

Weather-related deaths occur equally frequently in different months.

The alternative hypothesis is as follows:

Weather-related deaths do not occur equally frequently in different months.

It is considered a right-tailed test.

If the test statistic value is greater than the critical value, then the null hypothesis is rejected, otherwise not.

Determine the observed and expected frequencies

Let the months of the year be denoted by 1, 2, …..,12 starting from January.

Observed Frequency:

The frequencies provided in the table are the observed frequencies. They are denoted by\(O\).

The following table shows the observed frequency for each month:

\({O_1}\)=28	\({O_2}\)=17	\({O_3}\)=12	\({O_4}\)=24	\({O_5}\)=88	\({O_6}\)=61
\({O_7}\)=104	\({O_8}\)=32	\({O_9}\)=20	\({O_{10}}\)=13	\({O_{11}}\)=26	\({O_{12}}\)=25

Expected Frequency:

It is given that the deaths should occur with the same frequency each month.

Thus, the expected frequency for each of the 12 months is equal to:

\(E = \frac{n}{k}\)

where

n is the total frequency

k is the number of months

Here, the value of n is equal to:

\(\begin{aligned}{c}n = 28 + 17 + ..... + 25\\ = 450\end{aligned}\)

The value of k is equal to 12.

Thus, the expected frequency is computed below:

\(\begin{aligned}{c}E = \frac{n}{k}\\ = \frac{{450}}{{12}}\\ = 37.5\end{aligned}\)

The table below shows the necessary calculations:

O	E	O-E	\({\left( {O - E} \right)^2}\)	\(\frac{{{{\left( {O - E} \right)}^2}}}{E}\)
28	37.5	-9.5	90.25	2.407
17	37.5	-20.5	420.25	11.207
12	37.5	-25.5	650.25	17.340
24	37.5	-13.5	182.25	4.860
88	37.5	50.5	2550.25	68.007
61	37.5	23.5	552.25	14.727
104	37.5	66.5	4422.25	117.927
32	37.5	-5.5	30.25	0.807
20	37.5	-17.5	306.25	8.167
13	37.5	-24.5	600.25	16.007
26	37.5	-11.5	132.25	3.527
25	37.5	-12.5	156.25	4.167

Test statistic

The test statistic is computed as shown below:

\[\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}\;\;\; \sim } {\chi ^2}_{\left( {k - 1} \right)}\\ = 2.407 + 11.207 + ....... + 4.167\\ = 269.1467\end{aligned}\]

Thus, \({\chi ^2} = 269.1467\).

Critical vale, p-value and conclusion of the test

The degrees of freedom are computed below:

\(\begin{aligned}{c}df = \left( {k - 1} \right)\\ = \left( {12 - 1} \right)\\ = 11\end{aligned}\)

The critical value of\({\chi ^2}\)for 11 degrees of freedom at 0.01 level of significance for a right-tailed test is equal to 24.725.

The corresponding p-value is approximately equal to 0.000.

Since the value of the test statistic 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 thatweather-related deaths do not occur equally frequently in different months.

Possible Explanation

The months of May, June, and July tend to have a disproportionately higher number of weather-related mortality, which is due to the increased number of vacations and outdoor activities during those months.