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a)
To determine which chi-square test is appropriate for given setting.
a)
Answer to Problem 49E
We should use Chi-square test for independence.
Explanation of Solution
Given:
| Gender | |||
| Female | Male | ||
| Living location | Parent's home | 923 | 986 |
| Another person's home | 144 | 132 | |
| Own place | 1294 | 1129 | |
| Group quarters | 127 | 119 | |
First, we need to know which test should be apply for a specific scenario. There are three tests:
Chi-square goodness-of-fit test, chi-square test for homogeneity, chi-square test for independence.
We will know in details about these all tests.
When we are interested in the distribution of one variable, then we will use a chi-square goodness-of-fit test. When we are interested in the distribution of two variables in which there are multiple independent samples, then in this scenario we have to use a chi-square test for homogeneity. When we are interested in the distribution of two variables and if there is a single sample, then we would like to use a chi-square test for independence.
For given setting, we are given in two variables: Gender and living location There is one random sample. Therefore, we should use Chi-square test for independence.
b)
To state the null and alternative hypotheses.
b)
Answer to Problem 49E
Explanation of Solution
Given:
| Gender | |||
| Female | Male | ||
| Living location | Parent's home | 923 | 986 |
| Another person's home | 144 | 132 | |
| Own place | 1294 | 1129 | |
| Group quarters | 127 | 119 | |
Calculation:
The null and alternative hypotheses:
c)
To verify the conditions for inference.
c)
Answer to Problem 49E
All conditions satisfied.
Explanation of Solution
Given:
Calculation:
The conditions of chi square test are: Random, Independence and Large counts.
The sample is randomly selected, so condition of random sample is satisfied.
The
The expected count of each survey is at least 5 so, condition of large count is satisfied.
d)
To explain conclusion.
d)
Answer to Problem 49E
There is convincing evidence that the association between gender and living location.
Explanation of Solution
Given:
Calculation:
The p-value = 0.012. Therefore, if there is no association between the two variables, then the probability of getting a similar or more extreme sample as the given sample is 0.012 or 1.2%.
Decision: P-value < 0.05, reject H0.
Conclusion: There is convincing evidence that the association between gender and living location.
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