A study was done to investigate whether a relationship existed between HPV (human papillomavirus) status and HIV infection status. HPV status and HIV infection status were obtained for 97 women.
Table of hpv by hiv |
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hpv |
hiv |
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Frequency |
Seropositive/Symptomatic |
Seropositive/Asymptomatic |
Seronegative |
Total |
Positive |
23 |
5 |
10 |
38 |
Negative |
10 |
14 |
35 |
59 |
Total |
33 |
19 |
45 |
97 |
Statistic |
DF |
Value |
Prob |
---|---|---|---|
Chi-Square |
2 |
19.6478 |
<.0001 |
Likelihood Ratio Chi-Square |
2 |
19.8286 |
<.0001 |
Mantel-Haenszel Chi-Square |
1 |
17.0081 |
<.0001 |
Phi Coefficient |
0.4501 |
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Contingency Coefficient |
0.4104 |
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Cramer's V |
0.4501 |
Fisher's Exact Test |
|
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Table Probability (P) |
<.0001 |
Pr <= P |
<.0001 |
Question 1
Which test is appropriate for these data, chi-square or Fisher's Exact test?
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Question 2
What are the test statistic, degrees of freedom, and p-value for the test you chose (chi-square or Fisher's Exact test)?
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Question 3
How would you present your conclusion to a "layperson" (i.e., someone who has no knowledge of statistics)?
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Question 4
What is the correct calculation of the degrees of freedom (d.f. = 2) for this contingency table?
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1) Here we will use chi-square test as here we want to test homogeneity also the variables here are categorical.
Option (a) is correct.
Q2)
From the table test statistics is 19.6478
degrees of freedom is 2
and p value <0.0001
So, Option (C) is correct.
Q3)
Since, our p-value<0.0001<0.05. So, we reject the null hypothesis and conclude that there is significant association between the HPV and HIV status.
Option (B) is correct.
Q4)
In Chi-square test degrees of freedom is given as
(r-1)*(c-1) = (2-1)(3-1) = 1*2
= 2
Option (C) is correct.
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