Question

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

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 What is the Expected Frequency
never married 137 ?
married 240 ?
widowed 3 ?
divorced/separated 20 ?

Part (a)

State the null hypothesis. Choose 1 or 2

1. The data fit the distribution of marital status for the U.S. adult population.

2. The data do not fit the distribution of marital status for the U.S. adult population.

Part (b)

State the alternative hypothesis.

1. The data fit the distribution of marital status for the U.S. adult population.

2. The data do not fit the distribution of marital status for the U.S. adult population.

Part (c)

What are the degrees of freedom? (Enter an exact number as an integer, fraction, or decimal.)

________

Part (d)

State the distribution to use for the test.

t3

?23

OR

t4

?24

What is the Test STatistic?

What is the PValue?

null hypothesis: 1. The data fit the distribution of marital status for the U.S. adult population.

alternative hypothesis: 2. The data do not fit the distribution of marital status for the U.S. adult population.

c)

degrees of freedom =categories-1=4-1=3

d)?23

applying chi square test:

 observed Expected Chi square category Probability(p) Oi Ei=total*p R2i=(Oi-Ei)2/Ei never married 0.313 137.000 125.20 1.112 married 0.561 240.000 224.40 1.084 widowed 0.025 3.000 10.00 4.900 dvorced/seperated 0.101 20.000 40.40 10.301 total 1 400 400 17.398

Test STatistic =17.398

p value =0.0006

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