Refer to the previous exercise about the study of different HIV intervention pro
ID: 3049882 • Letter: R
Question
Refer to the previous exercise about the study of different HIV intervention programs:
Researches Jemmott III e al. (Journal of American Medical Association, 1998) reported a sstudy in which a group of inner-city African American adolescents from Philadelphia were randomly divided into three groups as part of an experiment to assess the effectiveness of different HIV prevention programs. One group was given "abstinence HIV intervention," the next group was given safer-sex HIV intervention," and the third group was given "health promotion intervention," which was to serve as a control. One outcome that was measured was whether subjects who were sexually active during a three-month period reported consistent condom use. The 8.2.2A table shows the summarized data.
a. State the appropriate null and alternative hypotheses in words.
b. Explain why it would be okay to use the theory-based method to find a p-value for this study.
c. Use an appropriate applet (for example, the Multiple Proportions applet) to find and report the following from the data)
- The name of the statistic and its numeric value
- Te theory-based p-value
d. Comment on the strength of evidence based on the p-value.
e. Comment on causation and generalization. Be sure to explain how you are arriving at your conclusion.
f. Explain why it is not particularly useful to look at the confidence intervals for the parameter(s) of interest in this study.
Abstinence Sater-sex intervention intervention Contro Total 14 20 34 20 12 32 21 20 41 Yes Consistent 52 condom use? No 107 Total EXERCISE 8.2.2A: Observed countsExplanation / Answer
Given table data is as below
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calculation formula for E table matrix
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expected frequecies calculated by applying E - table matrix formulae
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calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above
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set up null vs alternative as
null, Ho: exists no relation between the diffrent group resluts obtained
alternative, H1: exists no relation between the diffrent group resluts obtained
level of significance, = 0.05
from standard normal table, chi square value at right tailed, ^2 /2 =5.991
since our test is right tailed,reject Ho when ^2 o > 5.991
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o = 3.002
critical value
the value of |^2 | at los 0.05 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 3 - 1 ) = 1 * 2 = 2 is 5.991
we got | ^2| =3.002 & | ^2 | =5.991
make decision
hence value of | ^2 o | < | ^2 | and here we do not reject Ho
^2 p_value =0.223
ANSWERS
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a.
null, Ho: exists no relation between the diffrent group resluts obtained
alternative, H1: exists no relation between the diffrent group resluts obtained
b.
we have success, failure cases and the data is an observation data from the theory, hence this is suitable to conduct a theory based approach to find out the p -value and conculde the result in following
c.
chi square test for independence test
test statistic: 3.002
critical value: 5.991
d.
p-value:0.223
decision: do not reject Ho
MATRIX col1 col2 col3 TOTALS row 1 14 20 21 55 row 2 20 12 20 52 TOTALS 34 32 41 N = 107Related Questions
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