91Ó°ÊÓ

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 ${\mathrm{χ}}^2=\underset{k}{∑}\frac{(O-E{)}^2}{E}$

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

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

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

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

The degree of freedom will be:

$df=4-1=3$

$\begin{array}{|lllll|}\hline O& E& (O-E)& (O-E)2& \frac{(O-E{)}^2}{E}\\ 140& 125.20& 14.80& 219.04& 1.75\\ 238& 224.40& 13.60& 184.96& 0.82\\ 2& 10.00& -8& 64& 6.40\\ 20& 40.40& -20.40& 416.16& 10.30\\ \underset{4}{∑}\frac{(O-E{)}^2}{E}& & 19.27& & \\ \hline\end{array}$

Therefore, the test statistic will be:

${\mathrm{χ}}^2=19.27$

$\mathrm{Î\pm }=0.05$

$p=0.0002$

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