Question

In: Statistics and Probability

For each exercise, answer the following along with any additional questions.  Select and justify the...

For each exercise, answer the following along with any additional questions.  Select and justify the best test(s). The chi-square, Phi, Yates, or Lambda (or even a combination) might be best for a problem given the data and research question. Do not assume the independent is always on the row.  Provide the null and alternative hypotheses in formal and plain language for the appropriate test at the 0.05 significance level.  Do the math and reject/retain null at a=.05. State your critical value.  Explain the results in plain language.

3. The Nevada state chapter for women’s issues non-profit wonders if perceptions of sexual harassment in casinos varies by gender. They get a random sample of 100 employees and ask “How common is sexual harassment in your workplace?” (C15PROB3.SAV).

  Prevalence of Sexual Harassment

High Medium Low/None

Male 15 15 23

Female 21 15 11

Solutions

Expert Solution

Based on the observed and expected values, the squared distances can be computed according to the following formula: (O-E)2/E. The table with squared distances is shown below:

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

H0: The two variables are independent

Ha: The two variables are dependent

This corresponds to a Chi-Square test of independence.

Rejection Region

Based on the information provided, the significance level is α=0.05 , the number of degrees of freedom is df=(2−1)×(3−1)=2, so then the rejection region for this test is R={χ2:χ2>5.991}.

Test Statistics

The Chi-Squared statistic is computed as follows:

Decision about the null hypothesis

Since it is observed that χ^2=4.893≤χc2​=5.991, it is then concluded that the null hypothesis is not rejected.

Conclusion: It is concluded that the null hypothesis Ho is not rejected. Therefore, there is NOT enough evidence to claim that the two variables are dependent, at the 0.05 significance level.


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