Perform a chi-square homogeneity test. An independent simple
random sample of residents in three regions gave the data on race
shown in the table. At the
1 %1% significance level, do the data provide sufficient evidence to conclude that a difference exists in race distributions among the three regions? |
|
What are the null and alternative hypotheses?
A.
Upper H 0H0:
The racial distribution is the same in at least two of the three regions.
Upper H Subscript aHa:
The racial distribution is not the same in at least two of the three regions.
B.
Upper H 0H0:
The racial distribution is the same in each of the three regions.
Upper H Subscript aHa:
The racial distribution is not the same in each of the three regions.
C.
Upper H 0H0:
The racial distribution is not the same in each of the three regions.
Upper H Subscript aHa:
The racial distribution is the same in each of the three regions.
D.
Upper H 0H0:
The racial distribution is not the same in at least two of the three regions.
Upper H Subscript aHa:
The racial distribution is the same in at least two of the three regions.
Find the test statistic.
chiχ2equals=nothing
(Round to three decimal places as needed.)
Find the P-value.
Pequals=nothing
(Round to four decimal places as needed.)
Choose the correct conclusion below.
A.
Do not rejectDo not reject
Upper H 0H0.
The data
do not providedo not provide
sufficient evidence that there is a difference.
B.
RejectReject
Upper H 0H0.
The data
provideprovide
sufficient evidence that there is a difference.
C.
RejectReject
Upper H 0H0.
The data
do not providedo not provide
sufficient evidence that there is a difference.
D.
Do not rejectDo not reject
Upper H 0H0.
The data
provideprovide
sufficient evidence that there is a difference.
E.
At least one assumption is violated. Therefore, the test cannot be conducted.
Click to select your answer(s).
>> The null and alternative hypotheses: Option(B)
H0: The racial distribution is the same in each of the three regions.
Ha: The racial distribution is not the same in each of the three regions.
>> The test statistic:
χ2 = 5.191
>> The P-value:
P = 0.2683
>> The correct conclusion: Option(A)
Do not reject H0. The data do not provide sufficient evidence that there is a difference.
STATISTICAL OUTPUTS:
Actual Values:
97 10 8
116 17 4
117 15 13
Warning: Actual Value less than 5.
Results not reliable.
Expected Values:
95.5919 12.1662 7.24181
113.879 14.4937 8.6272
120.529 15.34 9.13098
Chi-Squared Values:
0.0207406 0.385709 0.0793788
0.0395002 0.433397 2.4818
0.103325 0.00753806 1.6394
Chi-Square Test Statistic = 5.19079
Degrees of Freedom = 4
P-value = 0.268276
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