The marital status distribution of the U.S. male population, ages 15 a nd older, is as shown in Table

Suppose that a random sample of 400 U.S. young adult males, 18 to 24 years old, yielded the following frequency distribution. We are interested in whether this age group of males fits the distribution of the U.S. adult population. Calculate the frequency one would expect when surveying 400 people. Fill in Table 11.35, rounding to two decimal places.

Table 1 shows the marital status distribution of the male population in the United States, aged 15 and up.

The expected marital status frequency:

$\begin{array}{|lll|}\hline \text{Marital Status}& \text{Percent}& \text{Expected Frequency}\\ \text{never married}& 31.3& \\ \text{married}& 56.1& \\ \text{widowed}& 2.5& \\ \text{divorced/separated}& 10.1& \\ \hline\end{array}$

The marital status of the male population in the United States, aged 18 to 24 years.

The frequency of marital status :

$\begin{array}{|ll|}\hline \text{Marital Status}& \text{Expected Frequency}\\ \text{never married}& 140\\ \text{married}& 238\\ \text{widowed}& 2\\ \text{divorced/separated}& 20\\ \hline\end{array}$

Therefore, the total frequency is$400$

Let's calculate the Degrees of freedom df = n - 1

Test statistic ${\chi }^2=\sum _k\frac{(O-E{)}^2}{E}$

If the die is fair, we should expect the same frequency on either side.

$\begin{array}{|lll|}\hline \text{Marital status}& \text{Percent}& \text{Expected frequency}\\ \text{Never married}& 31.3& \frac{31.3}{100}\times 400=125.20\\ \text{Married}& 56.1& \frac{56.1}{100}\times 400=224.40\\ \text{Widowed}& 2.5& \frac{2.5}{100}\times 400=10.00\\ \text{Divorced/separated}& 10.1& \frac{10.1}{100}\times 400=40.40\\ \hline\end{array}$

${\mathrm{H}}_0=$the data fits the distribution

${\mathrm{H}}_{\mathrm{a}}=$the data does not fit the distribution

The degree of freedom will be:

$df=4-1=3$

$\begin{array}{|lllll|}\hline O& E& (O-E)& (O-E)2& \frac{(O-E{)}^2}{E}\\ 140& 125.20& 14.80& 219.04& 1.75\\ 238& 224.40& 13.60& 184.96& 0.82\\ 2& 10.00& -8& 64& 6.40\\ 20& 40.40& -20.40& 416.16& 10.30\\ \sum _4\frac{(O-E{)}^2}{E}& & 19.27& & \\ \hline\end{array}$

Therefore, the test statistic will be:

${\chi }^2=19.27$

$\alpha =0.05$

$p=0.0002$

$\text{Since}p\text{value}<a\text{, the null hypothesis is rejected.}\text{tSothe data does not fit the distribution.}$