Finding relative risk and odds ratio for clinical trials

Intro

Howdy!  I'm Professor Curtis of Aspire Mountain Academy here with more statistics homework help.  Today we're going to learn how to find the relative risk and odds ratio for clinical trials.

Here's our problem statement: In a clinical trial of 2091 subjects treated with a certain drug, 29 reported headaches.  In a control group of 1721 subjects given a placebo, 26 reported headaches.  Denoting the proportion of headaches in the treatment group by p-sub-t and denoting the proportion of headaches in the control (or placebo) group by p-sub=c, the relative risk is p-sub-t divided by p-sub-c.  The relative risk is a measure of the strength of the effect of the drug treatment.  Another such measure is the odds ratio, which is the ratio of the odds in favor of a headache for the treatment group to the odds in favor of a headache for the control (or placebo) group found by evaluating p-sub-t over 1 minus p-sub-t all over p-sub-c over 1 minus p-sub-c.  The relative risk and odds ratios are commonly used in medicine and epidemiological studies. (What a tongue-twister!)  Find the relative risk and odds ratio for the headache data.  What do the results suggest about the risk of a headache from the drug treatment?

Find the relative risk

OK, so first we're asked to find the relative risk for the headache treatment.  Now we know from the problem statement that the relative risk is defined as p-sub-t over p-sub-c.  Because this is going to get a little complicated with fractions within fractions, from this point on I'm going to simply denote these two quantities as PT and PC.

So PT divided by PC is the relative risk that's defined here in the problem statement.  Now how do we find PT and PC?  Well, PT is defined here as the proportion of headaches in the treatment group.  And the proportion of headaches in the treatment group will simply be the part of that group that had headaches divided by the total number (the whole) of that group.  So 29 out of 2091 is PT.  We know this is true because of this first sentence in our problem statement: “In a clinical trial we have 2091 subjects treated with a certain drug and 29 reported headaches.”  So this is PT.  We can do the same thing to find PC.  Note that we have the part over the whole.  We’ll do the same thing to find PC.  PC is going to be 26 over 1721.  Note again we're taking the part of that group that had headaches and dividing it by the whole of that group.

A fraction within a fraction can be evaluated; we simply take the reciprocal of the fraction in the denominator and multiply it by the fraction in the numerator.  So 26 over 1721 gets flipped over to become 1721 over 26, and then we multiply that by the fraction in the numerator, which is 29 over 2091.  This gives us 49909 over 54366, which we divide out to get approximately 0.91802.  Note our problem statement says we want to round to three decimal places, so taking this number and rounding it to three decimal places gives us 0.918.  So that's the number that I'm going to put in the answer field.  Nice work!

Find the odds ratio

Now we're asked to find the odds ratio.  Well, the odds ratio is found similarly by using the formula that they give you in the problem statement.  So we start by writing out that equation.  We can use the same values for PT and PC here that we used before in calculating relative risk.  So everywhere I see PT, I'm just going to substitute in 29 over 2091, and everywhere I see PC, I'm going to substitute in 26 over 1721.  This looks really awful, but if we take it piece by piece, we can actually simplify this so that it comes out really nice.

The first thing we're going to do is take care of that one 29 over 2091 that we see in the big fraction inside our numerator.  In order to subtract a whole number and a fraction, we need to first convert the whole number to a fraction.  So our common denominator will be 2091.  1 is 2091 over 2091, and then we just subtract the numerators.  So 2091 minus the 29 gives me 2062.  So now I've simplified this into a fraction.  I can do the same thing in the mega-fraction that we have on the bottom.  So again the same process — taking that one and converting it with the common denominator.  In this case it will be 1721over 1721.  1721 minus 26 gives 7 — excuse me — 1695.

So now we have fraction within a fraction within a fraction.  Gosh, this thing looks awful!  But again, we take it step by step, and we'll get to where we need to be.  Let's just move that over to the left so we have some more room to work.  In order to evaluate a fraction within a fraction, we need to take the reciprocal of what's in the bottom and multiply it by what's in the top.  But here we have layers of fractions, so we have to perform that procedure within each of the layers of our fractions.

So first, the two fractions that are on the top of the mega-fraction — I'm going to take 29 over 2091 and multiply it by the reciprocal of the fraction just beneath it.  So 2062 over 2091 flipped over becomes 2091 over 2062.  Then I multiply that by 29 over 2091.  I can do the same thing in the bottom of my mega-fraction, and you get what you see here.  Notice that we can simplify this even further.  Up here, the top part of my fraction, I've got 2091 on top and 2091 on the bottom, so they'll actually cancel out.  So when I multiply through I get 29 over 2062.  I can perform the same operation in the bottom part of my mega-fraction.  Notice 1721 on top and 1721on the bottom.  They cancel each other out leaving me with 26 over 1695.

So now we've gotten rid of a layer of our fractions, but we still have a fraction within a fraction.  So we have to apply the procedure once more.  Take the reciprocal of what's on the bottom and multiply by what's on the top.  When we do, we get 29 over 2062 times 1695 over 26.  Multiply top by top and bottom by bottom, and we get 49155 divided by 53612.  This then yields approximately 0.91687, or rounded to three decimal places 0.917.  So that’s what I'm going to put here.  Excellent!

Interpret the results

And now the final part of our problem: What do the results suggest about the risk of a headache from the drug treatment?  Well let's review what we've actually uncovered from our calculations.  If we take the relative risk and the odds ratio and expand those out to five decimal places so we can get a better idea of what's going on, and then we also recalculate PT and PC so we have that here for our reference — now, let's look at our answer options.

Option A says, “The drug appears to pose a risk of headaches because the odds ratio is greater than 1.0.”  Well the odds ratio is not greater than 1.0.  You can see here that the odds ratio is actually less than 1, not greater than 1.  So Answer A cannot be correct.

Answer Option B says, “The drug does not appear to pose a risk of headaches because PT is slightly less than PC.”  If we compare PT and PC, we see that, yes indeed, PT is slightly less than PC.  What are these quantities representing?  Well, PT is the proportion of those in the treatment group who suffered headaches, and PC is the proportion of those in the control group — who weren't given the treatment — how many of them received headaches.  So we can see the drug has some minor effect in reducing the number of headaches.  So it doesn't appear to impose a risk of headaches, meaning there's no increase in the number of headaches, because the proportion among those who took the treatment went down.

So Answer Option B is very viable.  Let's look at the other answer options just to be sure before we select.

Answer Option C says, “The drug has no risk because the relative risk and odds ratios are almost equal.”  Well, if we look here, we see that the relative risk and the odds ratios are indeed approximately equal to each other.  However, this does not signify that the drug has no risk; the relative risk is almost a certainty —  0.918!  You've got to be kidding me that that's not saying there's no risk to it.  So that part of Answer Option C just doesn't pan out, so we can't select that as our answer.

Finally, Answer Option D says, “The drug appears to pose a risk of headaches because PT is greater than PC.  Well, again we see here PT is not greater than PC.  PT is actually less than PC.  This means we are indeed going to select Answer Option B for our answer.  Well done!

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