A researcher is comparing mean cholesterol levels across 4 diet plans (A, B, C, D) in a One-Way ANOVA test. If H 0 H_0 was rejected and the researcher were to use a Bonferroni Test, how many pairs of comparisons would they do?

Welcome back, everybody. So now that we've gotten a chance to see an entire Bonferroni test in full, let's go and take a look at another question that's a little bit more foundational. Hopefully, you had a chance to work it out on your own. Let's get started. So we have a researcher that's comparing mean cholesterol levels across four diet plans. They're just given letters a, b, c, d, and a one way ANOVA test. Now if the null hypothesis for that one way ANOVA was rejected and the research were to follow-up using a Bonferroni test, how many comparisons or how many pairs of comparisons would they do? Now this sounds kinda trivial at first, but remember, there's really important information conceptually that you need to know here. For example, the null hypothesis is rejected. That's the one requirement that we have to use a Bonferroni test. Remember, if the HO is rejected, it means that at least one of the means was different. The Bonferroni test basically allows us to pair off the categories and test them separately to figure out which one or which mean or means is different. Alright? So basically, in order to figure out how many pairs, given the fact that we have these four diet plans, how do we actually find this? I'm just gonna go ahead and show you how to do this really quick and easy on your graphing calculators. We're gonna use the combinations feature, which is this this sort of a function called n c r. So let's go ahead and get our our graphing calculators out. We're just gonna go over to the math menu. We're We're going to go over to the probability tab, and it's going to be that third option there, which is n c r. If you're using an older calculator, it may not look exactly like this. You'll just have to do the first number, and then followed by the second number. Now, in this case, what we're trying to do here is we're trying to arrange four possible diet plans into pairs of two. In other words, we're taking four options and we're trying to come up with unique combinations of two of them. So that's four c two. So you're going to go to your graphing calculators. You're going to hit the four button and then go over to the right side, hit two, and you should hit the equal sign. Now what you should see is that the answer is equal to six. Now you may see this written in another way in your textbook, so I just want to go ahead and point this out. You may also see this written using sort of like a big parenthesis, but this notation basically says how many different ways can you choose two things out of four possible or total of four possible choices. Alright? So both of these formats actually are saying the exact same thing, and the answer is just six. Alright? Now if you actually didn't have access to a graphing calculator, here's how you would actually just do this. I recommend you just do this stuff by hand if you don't have access to a calculator. So, basically, I'm just gonna list out all the possible letters. We've got a, b, c, and d for my different diet plans. All I'm gonna do is just I'm gonna draw some lines in between them to figure out how many unique pairs I can come up with. So for example, if I draw a line between a and b, right, that's one possible pairing. I'm gonna pair off a and b, but I can also pair off a and c, and I can also pair off a and d. Right? All those are unique combinations. When I get to the second letter, though, I can do b and c. That's unique. That's the fourth one. B and d, that's the fifth one. But I can't actually go backwards and go from b to a because I've already counted that pair. Right? I've already got a to b. B to a is the same exact thing. So, basically, what happens is as you go across the letters, you're gonna end up with less and fewer and fewer unique combinations here. Alright? So if we move on to the third letter, c, I can't do c and a because I already have that combination done. C and b is already done as well. The only unique one that I haven't counted already yet is c and d. So if you go ahead and just count up all the little arcs and lines that you've made, you should see that there are six of them. Those are the only possible combinations that you have. Alright? So So I recommend, if you don't have a graphing calculator, you can just sort of do this stuff by hand. The maximum amount of categories that you see is probably four or five. So you really won't encounter anything too crazy here. All right? That's really all there's to it. Let's move on, and thanks for watching.

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Multiple Comparisons: Bonferoni Test

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14. ANOVA

Multiple Comparisons: Bonferoni Test

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Multiple Choice

A researcher is comparing mean cholesterol levels across 4 diet plans (A, B, C, D) in a One-Way ANOVA test. If $H sub 0$ was rejected and the researcher were to use a Bonferroni Test, how many pairs of comparisons would they do?

$12$

$6$

$4$

$16$

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Verified step by step guidance

Identify the number of groups being compared in the One-Way ANOVA. Here, there are 4 diet plans: A, B, C, and D.

Understand that the Bonferroni test is a post-hoc test used to make pairwise comparisons between group means after rejecting the null hypothesis in ANOVA.

Calculate the number of unique pairs of groups to compare. The formula for the number of pairwise comparisons among k groups is given by: $\frac{k(k-1)}{2}$.

Substitute the number of groups (k = 4) into the formula: $\frac{4(4-1)}{2} = \frac{4 \times 3}{2}$.

Simplify the expression to find the total number of pairwise comparisons that the Bonferroni test will perform.