A recent national report states the marital status distribution of the male population age 18 or older is as follows: Never Married (32.7\%), Married (54.5\%), Widowed (2.8\%), Divorced (10\%). The table below shows the results of a random sample of 1676 adult men from California. Test the claim that the distribution from California is as expected at the $\alpha=0.01$ significance level. a. Complete the table by filling in the expected frequencies. Round to the nearest whole number: Frequencies of Marital Status \begin{tabular}{|c|l|l|} \hline Outcome & Frequency & Expected Frequency \\ \hline Never Married & 567 & \\ \hline Married & 898 & \\ \hline Widowed & 60 & \\ \hline Divorced & 151 & \\ \hline \end{tabular} b. What is the correct statistical test to use? Select an answerv c. What are the null and alternative hypotheses? $H_{0}$ : Marital status and residency are dependent. The distribution of marital status in California is the same as it is nationally. The distribution of marital status in California is not the same as it is nationally. Marital status and residency are independent.

Step 1 ：a. To find the expected frequencies, we need to multiply the total number of the sample by the national percentage for each category. The total number of the sample is 1676. For "Never Married", the expected frequency is 0.327 * 1676 = 548 (rounded to the nearest whole number). For "Married", the expected frequency is 0.545 * 1676 = 913. For "Widowed", the expected frequency is 0.028 * 1676 = 47. For "Divorced", the expected frequency is 0.1 * 1676 = 168. So, the completed table is: \begin{tabular}{|c|l|l|} \hline Outcome & Frequency & Expected Frequency \\ \hline Never Married & 567 & 548 \\ \hline Married & 898 & 913 \\ \hline Widowed & 60 & 47 \\ \hline Divorced & 151 & 168 \\ \hline \end{tabular} b. The correct statistical test to use in this case is the Chi-Square Goodness of Fit Test. This test is used to determine if there is a significant difference between the observed frequencies and the expected frequencies. c. The null hypothesis (H0) and the alternative hypothesis (H1) are as follows: H0: The distribution of marital status in California is the same as it is nationally. This means that marital status and residency are independent. H1: The distribution of marital status in California is not the same as it is nationally. This means that marital status and residency are dependent. The null hypothesis assumes that there is no significant difference between the observed and expected frequencies, while the alternative hypothesis assumes that there is a significant difference.