Questions: Use the sample data below to answer the following question(s). Use a 0.05 significance level and the observed frequencies of 144 drownings at the beaches of a randomly selected coastal state to test the claim that the number of drownings for each month is equally likely. Jan Feb Mar Apr May June July Aug Sept Oct Nov Dec 1 3 2 7 14 20 37 33 16 6 2 3 Determine the value of the χ^2 test statistic.

Solution Steps

The observed frequencies of drownings for each month are given as follows:

\[ O = [1, 3, 2, 7, 14, 20, 37, 33, 16, 6, 2, 3] \]

The total number of drownings is:

\[ \text{Total} = 1 + 3 + 2 + 7 + 14 + 20 + 37 + 33 + 16 + 6 + 2 + 3 = 144 \]

Since we are testing the claim that the number of drownings for each month is equally likely, the expected frequency for each month is:

\[ E = \frac{\text{Total}}{12} = \frac{144}{12} = 12 \]

Thus, the expected frequencies are:

\[ E = [12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12] \]

The Chi-Square test statistic is calculated using the formula:

\[ \chi^2 = \sum_{i=1}^{n} \frac{(O_i - E_i)^2}{E_i} \]

Substituting the observed and expected frequencies, we find:

\[ \chi^2 = 141.1667 \]

For a Chi-Square test with \( df = n - 1 = 12 - 1 = 11 \) degrees of freedom at a significance level of \( \alpha = 0.05 \), the critical value is:

\[ \chi^2(0.95, 11) = 19.6751 \]

The p-value associated with the test statistic is:

\[ P = P(\chi^2 > 141.1667) = 0.0 \]

Final Answer