A two-sample t-test for the difference between two population means is used to determine if there is statistically significant evidence of a difference between two independent groups. The test requires checking conditions including random assignment, sample size ≥30 for normality, and appropriate degrees of freedom. The test statistic is calculated as the difference between sample means divided by the standard error, and the P-value is compared to the significance level (typically 0.05) to make a conclusion. In this 2026 AP Statistics FRQ #4, a farmer compared orange yields between trees fertilized with brand C versus brand N, finding convincing evidence (P ≈ 0.03) that the mean number of oranges differs between the two fertilizer brands.
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2026 AP Statistics FRQ #4 Walkthrough | Full Solution & Explanation for the Free Response QuestionAdded:
What's up my stats stars? In this video, we're going to take a look at the fourth free response question from the 2026 AP statistics exam. Now, this question was an inference based question where we were required to use a two sample T test for the difference between two population means. We were looking at the number of oranges grown on orange trees and some of them use fertilizer and some of them didn't. So, we're trying to compare those two different groups, the group that used fertilizer and the group that didn't. But, to do this problem, you did need to use a two sample T test for the difference between two population means. Let's dive into this problem right now and take a look at everything that's required for you to get full credit.
So, this year question number four was our inference based question. Here it was. A farmer wants to know whether there's a difference in the mean number of oranges that grow on trees fertilized with the current fertilizer brand C or a new fertilizer brand N. The farmer randomly assigns 50 trees to be treated with brand C and another 50 to be treated with brand N and the number of oranges from each tree is recorded. Now, the summary statistics for the two different groups are below and they want us to determine at the 5% significance level, do the data provide convincing statistical evidence that the mean number of oranges on trees with brand C is different than the mean number of trees fertilized with brand N. Now, right away, I recognize that I'm dealing with quantitative data, the number of oranges on the trees, which means this is going to be working with means and I clearly see the word mean throughout the whole problem. So, I know I'm going to be needing a T distribution. Next up, there are two separate samples for sure.
We got oranges or trees with brand C fertilizer and another separate sample with brand N. And I also recognize I do not need a confidence interval here because it A doesn't ask for a 99 or 95 or 90% confidence interval. It asks for convincing evidence and a test is what's going to be required for you to do that.
So, this type of question is going to be graded in three sections. The first section is define the parameter, name the procedures, and giving the hypotheses. The second section is going to be checking the conditions and getting your test statistic and your p-value and then the third section is going to be making your conclusion in context. So, let's look at section one right now.
So, a two-sample t-test for the difference between the mean number of oranges grown on similar trees using fertilizer N and the mean number of oranges grown on similar trees using fertilizer C. So, right away I'm naming the procedure a two-sample t-test for the difference between the two population means and then I'm actually going to define what these are. Mu sub N is the mean number of oranges grown on similar trees with fertilizer N and mu C is the mean number of oranges grown on similar trees with fertilizer C. So, I'm representing the fact that I know what the problem is about. I'm defining my two things in the problem, the mean for N and the mean for C, and I'm giving the name of it all in context. Really, really important there.
Now, the hypothesis starts off with the null being there's no difference. The mean number of oranges with fertilizer N is the exact same as the mean number of oranges with fertilizer C. And then the alternative is that there is a difference. Now, it never said, if you go back and read the problem, it never said that the new fertilizer produces more oranges or the old fertilizer produces less oranges. It just said is there evidence of a difference. Now, the other thing you could have is you could subtract this over and you could have mu N minus mu C equals zero. That's the same thing as saying that they're equal, saying that there's no difference. And then down here you could have mu N minus mu C is not equal to zero, meaning that there is in fact a difference. All right, that's it for section one. Section two, we have to start off by checking those conditions. The first is that there must be random assignment of the trees to the two different treatment groups. Now, we did not select the trees at random. So, this is an experiment. This is not just an observational study where we were selecting trees at random to look at. We were doing an experiment here. So, we must have random assignment to the two different treatment groups and it did say we had that. Then we have to make sure that our samples are big enough so that we can assume normality. Well, both samples of 58 are larger than 30. Both the N and the C over 58, both of those are larger than 30, so normality can be assumed. Now, I want you to notice I did not check the less than 10% condition because in an experiment based problem, that condition is not supposed to be checked because we're not sampling from the population, we had random assignment. So, if you're not sampling from the population, we don't have to check that under 10% condition.
All right, next up for section two is the test statistic and the Z score. Now, here is me showing the work, but I want to make a huge point here. You do not have to show any of this work, but I think it's valuable to show it. So, the T score is going to be taking our difference of seven and subtracting zero cuz we assumed there was no difference, right? That was the null. Dividing it by the standard error for the two samples and that's our formula right there and I got 2.202.
Then you could use a T distribution.
Now, the problem here is degrees of freedom, right? We know that we had 58 in one and 58 in the other. Now, the correct degrees of freedom formula is extremely complicated. So, what most kids are going to do, at least what I recommend to do is you could either use 57 degrees of freedom, that is the degrees of freedom from each sample, or you could take 57 plus 57 or add your samples together and subtract two to get 114 degrees of freedom. Now, technically both of those estimates of the degrees of freedom are incorrect, but they're close enough to get you a good enough answer. Then what you could do is you could use a T distribution and you're going to be looking greater than this T score right here with the proper number of degrees of freedom, again either 57 or you could use 114, and they're going to get your P value. But, I want to make sure that you realize you do not have to do any of that cuz you could use Desmos or your calculator to get the T score and the P value for you with absolutely no work needed.
Now, one more thing I want to make sure I mention here. Could you have done the problem with 141 minus 148? Yes, you would therefore get a negative seven difference and your T score would be negative. So, when you go to use your T distribution, you'd want to look at below that value cuz it was negative.
But, the whole point that I want to show you right now is if you know how to use Desmos, you don't have to do any of that work to get your T score and your P value. So, on Desmos, I'm going to click that little plus sign and choose inference and then I'm going to choose for quantitative data a T test. Now, I want to click this little tab right here that says stats because I don't actually have the data, I have the stats. And I'm going to enter in the first sample, 58, mean 148, standard deviation 119. Then, click the little plus sign right here to add a second sample, 58, 141, and 15.
Again, you could do that in the other order, it would be completely okay.
Then, I'm going to hit create test and click on this little arrow right down here because we're not doing a confidence interval, we're doing a significance test. So, it should default to this being a zero, which is what we had it set at. And then we want to make sure both is selected because remember, our alternative was it could be greater than or less than, we just care that it was something different. It wasn't a left-sided, it wasn't a right-sided, it was both. So, as long as that's selected, your P value is right here, about 0.03, and here's the correct degrees of freedom, by the way, about a 108.2, and then here's our T score. Now, if you click on these three little dots right here, it'll give you all of that with more decimals. So, the degrees of freedom, the T score, and the P value.
So, at the end of the day, all you have to have for section two is first off the conditions checked, and then you just have to have your T score and your P value written down and identified. Don't just write two numbers down, write down T score equals 2.02202, and the P value is 0.02977.
All right, now for section three, making our conclusion. Our P value is low, although it's not super low, but it is below 5%, so we are going to reject the null. So, since the P value of 0.02977 is less than 0.05, the null would be rejected. That needs to be done. You must compare your P value to your significance level to get the full points here. Then, we have to give it the context. The data does provide convincing statistical evidence that the mean number of oranges on trees fertilized with brand C is different from the mean number of oranges on trees fertilized with brand N for all trees similar to those in the study. Now, I want to let you in on a big secret. I literally copied all of this from the problem. That's exactly what you should have done as well. Don't mince words.
Don't try to shortcut it. We have concluded that we could reject the null, which means there is evidence that what they asked us to find evidence for is true. So, just copy it from the problem, giving yourself a huge advantage in not having to worry about what do I need to write. Now, make sure you do not write that the alternative is 100% definitely true. No, no test can conclude that. We just have evidence. We have evidence.
That's what tests are all about, showing if there is evidence or lack of evidence. And in this problem, since our P value is low, there is evidence that, well, the mean number of oranges on trees fertilized with brand C is different than the mean number of trees uh for oranges on trees fertilized with brand N for all trees similar to those in the study. And that last piece is actually really, really important because we don't want to say this is true for all orange trees cuz we did not randomly select the orange trees from the population of all orange trees. We just randomly assigned the ones that we had, and that means that we can only conclude that it's true for trees similar to those in our study. Again, another reason why you're going to want to copy that sentence from the question and use it in your answer. All right, that's it for question number four.
Hopefully, you enjoyed it. I hope you did really, really well on it.
And now for part D, interpret the standard deviation calculated in part C II in context. So, they want us to interpret the standard deviation that we just calculated, but the funny thing is it's really hard to interpret standard deviation if you don't also incorporate the mean.
So, here's what I said. The number of games Ben needs to attend until the first time the team song is over 120 seconds, typically differs from the mean by about 3.53 games. So, we expect it to take an average of 4.065 games until the first time the song is over 120 seconds, but that number of games could vary by 3.53 games. So, the idea is the mean is the number of games we expect it to take until we get a song over 120 seconds, and that's 4.065. But, that number can certainly vary, and that's where the standard deviation comes in. The number of games that that number could vary by is 3.53 games. All right, hopefully that made sense. Definitely a tough question that I bet a lot of kids are not going to get right this year, but no worries.
If you're studying for 2027 or further, you're actually never going to see a question like this come up, so good for you. All right, see you in the next video.
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