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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.

Short Answer

Expert verified

pvalue<a, the null hypothesis is rejected.tSothe data does not fit the distribution.

Step by step solution

01

Given Information

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

The expected marital status frequency:

Marital StatusPercentExpected Frequencynever married31.3married56.1widowed2.5divorced/separated10.1

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

The frequency of marital status :

Marital StatusExpected Frequencynever married140married238widowed2divorced/separated20

Therefore, the total frequency is400

02

Calculation of Expected Frequency

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

Test statistic χ2=k(O-E)2E

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

Marital statusPercentExpected frequencyNever married31.331.3100×400=125.20Married56.156.1100×400=224.40Widowed2.52.5100×400=10.00Divorced/separated10.110.1100×400=40.40

H0=the data fits the distribution

Ha=the data does not fit the distribution

The degree of freedom will be:

df=4-1=3

03

Calculation of Test statistic

OE(O-E)(O-E)2(O-E)2E140125.2014.80219.041.75238224.4013.60184.960.82210.00-8646.402040.40-20.40416.1610.304(O-E)2E19.27

Therefore, the test statistic will be:

χ2=19.27

α=0.05

p=0.0002

Sincepvalue<a, the null hypothesis is rejected.tSothe data does not fit the distribution.

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