The type of household for the U.S. population and for a random sample of 411 households...

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The type of household for the U.S. population and for a randomsample of 411 households from a community in Montana are shownbelow.

Type of HouseholdPercent of U.S.
Households
Observed Number
of Households in
the Community
Married with children26%        104            
Married, no children29%        102            
Single parent9%        38            
One person25%        103            
Other (e.g., roommates, siblings)11%        64            

Use a 5% level of significance to test the claim that thedistribution of U.S. households fits the Dove Creekdistribution.

(a) What is the level of significance?


State the null and alternate hypotheses.

H0: The distributions are different.
H1: The distributions are the same.

H0: The distributions are different.
H1: The distributions aredifferent.    

H0: The distributions are the same.
H1: The distributions are different.

H0: The distributions are the same.
H1: The distributions are the same.


(b) Find the value of the chi-square statistic for the sample.(Round the expected frequencies to two decimal places. Round thetest statistic to three decimal places.)


Are all the expected frequencies greater than 5?

Yes

No    


What sampling distribution will you use?

uniform

chi-square    

binomial Student's t

normal


What are the degrees of freedom?


(c) Find or estimate the P-value of the sample teststatistic. (Round your answer to three decimal places.)


(d) Based on your answers in parts (a) to (c), will you reject orfail to reject the null hypothesis that the population fits thespecified distribution of categories?

Since the P-value > ?, we fail to rejectthe null hypothesis.

Since the P-value > ?, we reject the nullhypothesis.    

Since the P-value ? ?, we reject the nullhypothesis.

Since the P-value ? ?, we fail to reject thenull hypothesis.


(e) Interpret your conclusion in the context of theapplication.

At the 5% level of significance, the evidence is sufficient toconclude that the community household distribution does not fit thegeneral U.S. household distribution.

At the 5% level of significance, the evidence is insufficient toconclude that the community household distribution does not fit thegeneral U.S. household distribution.

Answer & Explanation Solved by verified expert
3.6 Ratings (326 Votes)

a)level of significance =0.05

H0: The distributions are the same.
H1: The distributions are different.

b)

applying chi square test:

observed Expected Chi square
category Probability(p) Oi Ei=total*p R2i=(Oi-Ei)2/Ei
married with children 0.260 104.000 106.860 0.077
married no children 0.290 102.000 119.190 2.479
single parent 0.090 38.000 36.990 0.028
one person 0.250 103.000 102.750 0.001
other 0.110 64.000 45.210 7.809
total 1.000 411 411 10.393

chi-square statistic =10.393

Are all the expected frequencies greater than 5? --Yes

What sampling distribution will you use? --chi square

What are the degrees of freedom? - categories-1 =5-1=4

c)

p value =0.034

d)

Since the P-value ? ?, we reject the null hypothesis.

e)

At the 5% level of significance, the evidence is sufficient to conclude that the community household distribution does not fit the general U.S. household distribution.


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