Question

The marital status distribution of the U.S. male population, age 15 and older, is as shown below.

Marital Status	Percent
never married	31.3
married	56.1
widowed	2.5
divorced/separated	10.1

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 at the 5% level. Calculate the frequency one would expect when surveying 400 people. Fill in the table below, rounding to two decimal places.

Marital Status	Frequency	Expected Frequency
never married	138
married	239
widowed	3
divorced/separated	20

A. What are the degrees of freedom? (Enter an exact number as an integer, fraction, or decimal.)
B. State the distribution to use for the test.

C. What is the test statistic? (Round your answer to two decimal places.)
D. What is the p-value? (Round your answer to four decimal places.)
E. Alpha (Enter an exact number as an integer, fraction, or decimal.)
α =

Answer 1

We are given with table which shows the % of distribution to each class

Marital Status	Probability
never married	0.313
married	0.561
widowed	0.025
divorced/separated	0.101

So we will now calculate the Expected frequnecy.

Marital status	Probability	observed	Expected
never married	0.313	138	= 400 * 0.313 = 125.2
married	0.561	239	= 400 * 0.561 = 224.4
widowed	0.025	3	= 400 * 0.025 = 10
divorced/seprated	0.101	20	= 400 * 0.101 = 40.4

A)

Here degree of freedoms are given by no of statuses - 1

i.e., 4 - 1 = 3

So df = 3

B)

Here we will use distribution.

i.e, here we are checking goodness of fit so here we will use chi square test.

and here under null hypothesis test statistic will follow with 3 df.

C)

Here test statistic is given by:

D)

Here as we know that this test statistic will follow chi square distribution with 3 df

So

It is calculated in R

code is attached herewith.

E)

Here we are given that alpha level is 5%

so

Here as we can see that p value =  0.0005684475 <

which implies that we have enough evidence to reject null hypothesis.