# At a local health clinic, 300 women were screened for HIV infection; 100 women who were HIV positive were compared to 200 women who were HIV negative. All women were interviewed with respect to...

At a local health clinic, 300 women were screened for HIV infection; 100 women who were HIV positive were compared to 200 women who were HIV negative. All women were interviewed with respect to their recent history of sexual partners: 60 of the infected women had a sexual partner within the last 2 years who was an intravenous drug user, while 25 of the uninfected women had a sexual partner in the last 2 years who was an intravenous drug user.

The odds ratio for this ques is 10.5. Calculate the 95% confidence interval for this. In the study is the association between HIV infection and having a partner who is an iv drug user statistically significant.

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### 1 Answer

First calculate the odds ratio from the 2x2 contingency table

HIV+

___Y____N___ |____

Exposure Y | 60 25 | 85

N |__40___175___|_ 215

100 200 | 300

Odds of being HIV+ given exposure is `O_1 = 60/25 ` and the odds of being HIV+ given lack of exposure is `O_2 = 40/175 ` . The *odds ratio *then of being HIV+ given exposed versus not exposed is

`OR = O_1/O_2 = ("60/25")/("40/175") = ("12/5")/("8/35") = 12/("8/7") = "84/8" = 10.5 ` as required.

Now the sampling distribution of the odds ratio is very skewed to the right. Taking the natural logarithm of the odds ratio negates this problem substantially giving a more Normal (Gaussian) sampling distribution. Hence, we can calculate a 95% confidence interval for the *log odds ratio *and transform it back to the *odds ratio *scale.

Here, the log odds ratio is simply `"ln"(OR) = "ln"(10.5) = 2.351375 `

To calculate the *standard error *of this estimate we use the formula

`"SE"("ln"(OR)) = sqrt(1/a + 1/b + 1/c + 1/d) `

where here `a = 60, b = 25, c = 40, d = 175 ` giving

`"SE"(ln(OR)) = sqrt(1/60 + 1/25 + 1/40 + 1/175) = 0.2956027 `

A 95% confidence interval for `"ln"(OR) ` can then be calculated as

`"ln"(OR) pm 1.96 times "SE"("ln"(OR)) `

which here is

`2.351375 pm 1.96 times 0.2956027 = [1.771944, 2.930757] `

Converting this back the the *odds ratio *scale, by exponentiating it, we get a 95% confidence interval for the `OR ` of

`[ 5.88, 18.74]`

Given that this interval does not include 1 (an odds ratio indicating exposure to drug-using partners has no effect on disease status of HIV), and is significantly greater than 1, the positive association between being HIV+ and interacting with drug-using partners in the previous 2 years is statistically significant at the 5% level, with indications that the person is `OR = 2.35` times more likely to be HIV+ given exposure versus no exposure.

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