The type of household for the U.S. population and for a random sample of 411 households from a community in Montana are shown below.
Type of Household | Percent of
U.S. Households |
Observed
Number of Households in the Community |
Married with children | 26% | 105 |
Married, no children | 29% | 120 |
Single parent | 9% | 28 |
One person | 25% | 91 |
Other (e.g., roommates, siblings) | 11% | 67 |
Use a 5% level of significance to test the claim that the distribution of U.S. households fits the Dove Creek distribution.
(a) What is the level of significance?
State the null and alternate hypotheses.
H_{0}: The distributions are the same.
H_{1}: The distributions are the same.
H_{0}: The distributions are different.
H_{1}: The distributions are
different.
H_{0}: The distributions are the same.
H_{1}: The distributions are different.
H_{0}: The distributions are different.
H_{1}: 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 the
test statistic to three decimal places.)
What sampling distribution will you use?
normal
chi-square
binomial
uniform
Student's t
What are the degrees of freedom?
(c) Find or estimate the P-value of the sample test
statistic. (Round your answer to three decimal places.)
(d) Based on your answers in parts (a) to (c), will you reject or
fail to reject the null hypothesis that the population fits the
specified distribution of categories?
Since the P-value > ?, we fail to reject the null hypothesis.
Since the P-value > ?, we reject the null hypothesis.
Since the P-value ? ?, we reject the null hypothesis.
Since the P-value ? ?, we fail to reject the null hypothesis.
(e) Interpret your conclusion in the context of the
application.
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.
At the 5% level of significance, the evidence is insufficient to conclude that the community household distribution does not fit the general U.S. household distribution.
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) | O_{i} | E_{i}=total*p | R^{2}_{i}=(O_{i}-E_{i})^{2}/E_{i} | |
married with children | 0.260 | 105.000 | 106.860 | 0.032 | |
married,no children | 0.290 | 120.000 | 119.190 | 0.006 | |
single parent | 0.090 | 28.000 | 36.990 | 2.185 | |
no parent | 0.250 | 91.000 | 102.750 | 1.344 | |
Others | 0.110 | 67.000 | 45.210 | 10.502 | |
total | 1.000 | 411 | 411 | 14.069 |
chi-square statistic =14.069
What sampling distribution will you use --Chi square
degrees of freedom =categories-1=4
c)
p value =0.007
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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