AP Stats Unit 7 Practice FRQ #2

An experiment is designed to test a new material being used in the creation of baseball bats. A group of 20 volunteers with baseball-playing experience are recruited to participate in the experiment. Each of the volunteers is randomly assigned a bat made up of either the new material or an existing old material, and hits a ball off a tee 25 times, with the mean distance that the ball travels calculated at the end of the trials. After an appropriate period of rest, the volunteers then use the other bat.

A table summarizing the results of the experiment are shown below:
Unit 7 a

Histograms summarizing the results are also shown here:
Unit 7 b

The researchers wish to determine if the new material resulted in the average distance traveled by the batted balls was greater than the average from the old material.

a. For the experiment described above, define the hypotheses that the researcher wishes to test and name test that they should conduct. You do not need to execute the stated test.

b. For the test named in part (a), have the conditions for running the test been met? Explain why or why not.

c. The results of the experiment result in a statistically significant conclusion. Based on the design of the experiment, can the manufacturer conclude that the average distance traveled by batted balls is greater for the new material than the old material, for players similar to those in the experiment? Justify your response.

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Welcome - thanks for posting! On to the feedback:

Up until last year, your response in part (a) would be sufficient. You choose the correct test and write correct null/alternative hypotheses using standard notation (“mu”) with clear subscripts. However, starting last year, an emphasis has been placed on defining what “mu” or “p” stand for in the context of the specific scenario; that is, for a question like this, you’d need to mention “mean distance traveled by the balls” or something similar. A lot of students lost points on last year’s test for not defining what “p” stood for in their hypotheses (it was a 2-sample z-test). A way you can guarantee credit is by using full sentences instead of the standard symbols that you’ve used in the past.

In part (b), you correctly check the “random” condition by referring to the “random assignment.” You make a small error in checking the “normal” condition by referring to the given graphs as “sampling distributions.” Unfortunately, the histograms given are sample distributions and not sampling distributions, because they reflect data only from these 20 individuals in the experiment, and not repeated samples of 20 individuals. Yes, the experiment contained those 20 individuals repeating the batted-ball process 25 times each, so I recognize how it can be confusing. I may actually need to re-write this scenario to reflect that, as I’m guessing exam writers wouldn’t put us in a situation like that. My apologies there.

I can comment on the 10% condition though - we actually don’t need to check that here! The 10% condition should be checked in contexts of random sampling without replacement (i.e. surveys or observational studies) to ensure independence among observations. In an experiment, there is no sampling being done, so we do not need to check this condition.

Finally, in part ( c ), you come to the correct conclusion and give a clear reason why. Nice job!

Thanks again for participating, and I hope you try some other problems soon!

a) mu_diff = true mean difference in distance travelled by the baseballs made from the new material and the baseballs made from the old material

H_0: mu_diff = 0
H_a: mu_diff > 0

1-sample t test for mu_diff

b) The conditions for running the test have been met. The random condition is met, as each volunteer is randomly assigned a bat made up of either the new or an existing old material.
The Normal/Large Counts condition is met, as the the histograms of the new and old material does not show strong skewness or any outliers, and therefore it is reasonable to assume that the samples were taken from a population with an approximately normal distribution. This implies that the sampling distribution of x-bar is normal, regardless of sample size.

c) The manufacturer can conclude that the average distance travelled by batted balls is greater than the new material than the old material for players simlar to those in the experiment because this was a matched pairs experiment that allows inference about cause and effect.

Nice job top to bottom on this one - I was going to comment on a lack of labeling of “matched pairs” in part (a) but you clearly describe that in part ( c ), and it is a one-sample test for a mean difference when we do matched-pairs, so I think you’re all good on this! Full credit all around.


thank you for 10 minutes straight of feedback :smiley:

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H_o: mu_d = 0
H_a: mu_d > 0
Where mu_d is the true mean difference between average distance in feet traveled by baseballs with the old material versus the new material.

b. Yes
-We have a random sample of baseballs batted.
-Population of baseballs batted is at least 200.
-Since the histograms have no outliers and show no skewness, it can be assumed that the sample is taken from a population that is distributed approximately normally.

c. Yes
The manufacturer can conclude that the average distance traveled by the batted balls is greater for new material than old material. This test is an example of matched pair t-procedure, in which the average distance traveled by each sample of the old material is paired with the average distance traveled by each sample of the new material. Thus, this test shows that the difference in distance traveled was due to the fact that the material of the batted ball was changed from old to new.

Nice job once again! Be sure to read carefully - you choose the correct test (matched-pairs t-test), but don’t name it in part (a) like you were asked to. Your hypotheses are correct, though.

Why is this a matched pairs vs. two sample t test?

I checked a and think I have that one down but here are b and c :slight_smile:

b) random: problem says they are “randomly assigned to a bat” so random condition is met
independence: we can assume there are greater than 20(10) 200 bats in the total population
normal: we’d have to construct a dot plot of the mean differences in new-old bats to check is the data have any outliers or strong skew to assume normality.

c) Since the data provides convincing evidence, and the p-value is less than alpha, we reject the null hypothesis and there is evidence that the average distance traveled by batted balls is greater for the news bats than for the old bats.

I may be wrong, but I think it is because there are not 2 samples. It is the same 20 volunteers who try out the old bat and the new bat. It is comparing 2 measurements from the same sample. Sorry if this is completely off base!

Even though there are two “groups” of data (balls hit with new bat and balls hit with old bat), every person in the experiment used both bats, so the data is paired in that way. If the experiment had been done where half of the players used the new bat, half of the players used the old bat, and there was no overlap, we’d do a two-sample t-test.

You nailed it! I also replied to that user because I hadn’t read this yet :slight_smile:

Solid answer! The order that you phrased part ( c ) in is a little unorthodox though: the reason the data provides convincing evidences is because p-value is less than alpha; you make it sound like it’s the other way around. A typical structure is “since the p-value is less than alpha, we reject Ho and conclude that [Ha in context]”

H0= mew sub alpha=0
HA= mew sub alpha =/ 0
They wish to conduct the 2 Sample Paired Means Test. (Idk if you need to put this, but we know this because they have a 3rd column indicating the difference of means, which gives it away)

Random: It was given that the volunteers were given the bat at random.
Since we have satisfied this conditions, we can assume independence for the pairs
The distributions of both old material and new material looks to be normal, no skewedness or anything irregular.
Since we satisfied normality, we can satisfy large enough samples.
This means that the conditions have been met.

c) Yes , based off the design of this experiment the manufacturer can conclude that the average distance traveled by batters balls is greater for the new material than the older ones because all the A/C’s were satisfied and if it is statistically significant, that means we can reject the null hypothesis in favor of the alternate hypothesis which said that their is a clear difference between old and new material in relation to the ball distance.

Side Note: Thank you for providing these questions and feedback, it has been a great help to prepare for Friday!

I’m glad you’re finding them helpful! :slight_smile: I’ll admit that I had fun making them. (Why yes, I’m that kind of teacher)

Your work is solid once again - but be careful on part (a). You call this a “2-sample paired means test.” That’s… not a thing. There are 2-sample t-tests for a difference in means, and there are matched-pairs t-tests for a difference in means (which are really done as one-sample tests). So part (a) would be marked partially correct. This is a matched-pairs t-test, by the way.


  • Ho: µn=µo
    Ha: µn>µo
  • Paired t confidence interval for mean difference
    ?I don’t really get why this is not a 2 sample t test for difference in Means could you explain?



  • Random: We are told Random Assignment was used.
    ?Why isn’t it suspect since they say they are using volunteers?(I saw that u said it was random.)
  • Normal: The sampling distribution is approximately normal because provided graphs of the differences are not strongly skewed and have no outliers.
  • 10% Condition: The population is greater than 10*20

Yes because the fact that it is statistically significant tells us that 0 isn’t on its interval thus there is a statistically significant difference between distance batted with new and old bats for like individuals.

Good questions!

The test is paired because each volunteer uses both bats, and therefore the two groups of data aren’t actually two independent groups, but rather two sets of data from the same group of individuals. This prevents us from using a 2-sample procedure, since that requires the groups to be independent of each other (i.e. involving separate participants. For example, had we had 20 volunteers only use the new bat and a different 20 volunteers only use the old bat, we’d use a 2-sample t-test).

Then for conditions:
“random” doesn’t actually always mean “random sample.” When we do experiments, random assignment helps to ensure that we’ve avoided confounding variables and made the groups as balanced as they can be. So the “random” condition can be met from random sampling in surveys/observational studies, but in experiments it is met through random assignment of treatments. Volunteers are OK in experiments, as long as random assignment then happens. Furthermore, if random assignment is done, we don’t check the 10% condition, because that’s there to protect against sampling without replacement. We don’t need to worry about that when the sampling isn’t what we’re concerned with.

One last thing: part ( c ) was about recognizing that since this is an experiment, we can infer cause-and-effect relationships based on treatments when we get a statistically significant result.

a.) Paired t confidence interval for mean difference
Ho: μn - μo = 0
Ha: μn -μo > 0

Random: We are told it is “randomly assigned”
Normal: It is approx. normal because there are no extreme outliers or strong skews.
10% condition: There are more than 20 x 10 baseball players

Yes, because it is statistically significant, this means that zero is not on the interval and would provide a difference of more than 0 which is what the alternative hypothesis wanted.

Solid work. In part (a), you choose the correct test and your hypotheses are valid, though you do not define what your symbols stand for in this context (aka you didn’t define the parameter of interest). That is becoming a more common scoring component, so practice defining parameters before tomorrow.

In part (b), you check the conditions correctly - though you did not need to check the 10% condition, because this was an experiment. The 10% condition is only for scenarios involving sampling without replacement.

Part ( c ) was more of an experiment-design question: since this was an experiment, a cause-and-effect relationship can be inferred when there is a statistically significant result.

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