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14. Dr. Antonio believes that you can predict people’s choice of mate from their

ID: 3228410 • Letter: 1

Question

14. Dr. Antonio believes that you can predict people’s choice of mate from their personal inventory of characteristics (PIC; measured 1-15). He administers the PIC to 100 students on campus and has those students rate the likelihood of dating a group of 10 co-eds (mate choice is measured on a 1-100 scale; 10 pts for each co-ed).

Please be sure to answer all the questions #1-#10. Most people do not answer it all the way *************

1.What are the researcher’s independent and dependent variables?

a.Independent=______________________________

b.Dependent=_______________________________

2. What are the null and alternative hypotheses (use the appropriate symbols not words)?

a.H0=__________________

b.H1=__________________

3.What is the appropriate test for the researcher to use?__________________

4.Why is that statistic appropriate?

If applicable to the specific test:

1.What are the test’s assumptions? N/A

2.What are the test’s df? N/A

3.What is the appropriate critical value? N/A

4.If the research obtains a test statistic of ____ , would the researcher retain or reject H0? N/A

5.Should the researcher do any follow up tests or make any graphs/charts? N/A

6.Interpret the findings in APA style

Explanation / Answer

1.What are the researcher’s independent and dependent variables?

a. Independent variable =personal inventory of characteristics

b. Dependent variable =people’s choice

            2.

            The null hypothesis

                        Ho: personal inventory of characteristics and people’s choice are independent

                The alternative hypothesis

                        H1: personal inventory of characteristics and people’s choice are dependent

            3. the appropriate test for the researcher use is Chi –square test

            4. because the data is qualitative data .

            1. assumptions are a) each frequency >5

            2.degrees of freedom, df =n-1

                                                =10-1

                                                =9

            3. critical value   - Chi-square t(0.05, 99) =123.23

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