Practicing with FRQs is a great way to prep for the AP exam! Work through this FRQ from Unit 6, then review sample student responses and corresponding feedback from Fiveable teacher Jerry Kosoff!

After completing a sale, a car company likes to send a follow-up survey where customers can indicate their level of satisfaction with their experience. One of the questions in the survey asks “would you recommend our company to a friend looking to purchase a vehicle?” The company wonders if people would answer the question differently based on whether they bought a new or used vehicle. From a list of all 2018 vehicle sales, the company randomly selects 105 customers who bought a new vehicle 120 customers who bought a used vehicle. 88 of the customers who bought new vehicles answered “yes,” while 85 of the customers who bought used vehicles answered “yes.”

At the significance level of 0.05, do the data provide convincing statistical evidence that the proportion of customers who would answer “yes” to the survey question is different for new vs used vehicle sales?

margin of error = +/- (1.960)sqrt[(0.838(0.162)/105)+(0.708(0.292)/120)]

confidence interval = 0.05 +/- 0.108 = (-0.058, 0.113)

No, the data do not provide convincing statistical evidence that the proportion of customers who would answer “yes” to the survey question is different for new vs used vehicle sales. Since 0 is captured in the 95% confidence interval of -0.058 and 0.113, the data shows that the true difference in proportions could be 0.

I’ll give feedback on your work below, but I want to start with noticing that you used a 2-sample confidence interval to answer the question. That is a totally valid strategy for a situation like this, but only because the alternative hypothesis was “different”; had the scenario asked “higher” or “lower” the confidence interval would not work in the same way. Typically, when given a significance level, and asked if there is “convincing statistical evidence” of something, we should be running a hypothesis test. That said, you will still be scored for your work with the confidence interval. 

The scoring for a “convincing statistical evidence…” scenario includes:

Given that list (some parts are scored together to create a question with 3-4 scoring components), you can likely see that your work doesn’t have enough there to be earning much of the available credit. You calculate the appropriate margin of error, and therefore obtain a confidence interval, but never name the interval, check conditions (random samples, approximately normal sampling distribution [at least 10 successes/at least 10 failures], 10% condition), or write hypotheses. Additionally, you used “0.05” in the interval, instead of using (0.838 - 0.708 = 0.13) as your difference of proportions to add/subtract the margin of error. That would have led you to a different confidence interval where 0 was 

 included. Given that your interval did include 0 though, your conclusion that we do not have convincing evidence would get scored as correct, because you interpreted the answer you got correctly. Unfortunately, you would not get credit for the other components of the question.

p_1 = the proportion of customers who bought a new vehicle and answered yes to the survey question

p_2 = the proportion of customers who bought a new vehicle and answered yes to the survey question

H_0: p_1 - p_2 = 0, H_a: p_1 - p_2 not equal 0

P-val = P(z>=2.303 or z<=-2.303) = 0.021247

Since 0.021247 < alpha of 0.05, we reject the null hypothesis, because there is convincing statistical evidence that the proportion of customers who would answer “yes” to the survey question is different for new vs used vehicle sales.

Strong execution from top to bottom, presented clearly. One thing that you’re going to facepalm about: you defined the parameters 

 as the exact same thing. “p2” should say “used”

Test type- two sample proportional difference hypothesis z-test

simple random sample: stated in the problem- “the company randomly selects 105 customers…”

independence: 10(105)=1050 Assume that the population of new cars purchased in 2018 is greater than 1050.

normal: .84(105)=(88 >= 10), .162(105)=(17 >= 10)
All of the conditions for new cars are met.

simple random sample: stated in the problem- “the company randomly selects…120 customers…”

independence: 10(120)=1200 Assume that the population of new cars purchased in 2018 is greater than 1200.

normal: .708(120)=(85 >= 10), .292(120)=(35 >= 10)

All of the conditions for old cars are met.

This is about as thorough a response as I’ve seen! Very well done - you’ve nailed all of the components.

p1= Proportion of customers who bought a new car and answers “yes” to the survey question.

p2= Proportion of customers who bought a used car and answered “yes” to the survey question.

We are interested in conducting a 2 sample z test for a difference in population proportions.

Calculator: 2-Prop Z Test {x1=88, n1=105, x2=85, n2=120, p1 does not equal p2} = p: 0.0212

Since the p-value of 0.0212 is less than our alpha level of 0.05, we have convincing statistical evidence to reject the null hypothesis. The proportion of customers who would answer “yes” to the survey question is different for new vs. used vehicle sales.

Nice job! You’ve defined parameters, checked conditions, named the test, obtained appropriate test statistic and p-value, and made an appropriate conclusion. 

Where p_1 is the true proportion of customers who bought a new vehicle and answered “yes” to the survey question.

Where p_2 is the true proportion of customers who bought a used vehicle and answered “yes” to the survey question.

p_hat_combined = 105(0.8381) + 120(0.7083) / 105+102 = 0.7689

z = (0.8381 - 0.7083) - 0 / sqrt((0.7689 * (1-0.7689) / 105) + (0.7869 * (1-0.7869) / 120) = 2.3036

p-value = 2*normalcdf(2.3036, 1E99, 0, 1) = 0.0212

Since the p-value<alpha, we reject the H_o. There is sufficient evidence to suggest that the proportion of customers who bought a new vehicle and answered “yes” to the survey question is different from the proportion of customers who bought a used vehicle and answered “yes” to the survey question.

Well done from top to bottom - you’ve got parameters, conditions, appropriate calculations, and appropriate conclusions. You’re ready!

Anything with subscript 1 pertains the new vehicle group while any thing with subscript 2 pertains to the used vehicle group.

The test we will be using is a two sample z-test (for difference in proportion).

Z= (p1 hat - p2 hat)/(phat *(1-phat)(1/n1 + 1/n2))^.5 where phat is the pooled proportion = 173/235 = 0.7361

Conclusion: Since the pvalue of 0.0286 is less than the alpha level of 0.05, we reject the null hypothesis, leadings us to the conclusion that the data does provide convincing statistical evidence that the proportion of customers who would answer yes to the survey question is different for new vs used vehicle sales.

Good work! You’ve done all required parts of a hypothesis test and answered appropriately. The only place where you might not receive full credit is in your hypotheses: you’ve used symbols, and named which group is which, but did not define the parameter in context. The parameter here was “proportion of people who would say yes to the survey”, and then can be differentiated between new/used sales.

p1: The proportions of customers that answered “yes” to “would you recommend our company to a friend looking to purchase a vehicle?”from buying the new vehicle

p2: The proportions of customers that answered “yes” to “would you recommend our company to a friend looking to purchase a vehicle?”from buying the used vehicle

AP Stats Practice FRQ Responses & Feedback (Unit 6)

Practicing with FRQs is a great way to prep for the AP exam! Work through this prompt from Unit 3, then review student responses and corresponding feedback from Fiveable teacher Jerry Kosoff!

A large group of voters in two nearby counties were surveyed about which political party they prefer between parties X, Y, and Z. The relative frequencies in the table below summarize the results of the survey.

a. If one surveyed person is selected at random: 

b. In the survey, are the events “Lives in County B” and “Prefers Party X” independent? Explain why or why not. 

c. Assume that the distribution of party preference shown in the table holds true in the entire state. If 5 voters from the state are selected at random, what is the probability that exactly 2 of them prefer Party Y?

a) i. P(County A and Party Z) = 0.152. There is a 0.152 probability that the person lives in County A and prefers Party Z.

ii) P(County A or Party Z) = P(County A) + P(Party Z) - P(County A and Party Z)

P(County A or Party Z) = 0.4 + 0.31 - 0.152 = 0.558. There is a 0.558 probability that the person lives in County A or prefers Party Z.

iii) P(County A|Party Z) = P(County A and Party Z) / P(Party Z) = 0.152 / 0.31 = 0.49. There is a 0.49 probability that the person lives in County A, given that they prefer Party Z.

b) If independent, P(Lives in County B and Prefers Party X) = P(Lives in County B) * P(Prefers Party X).

The events “Lives in County B” and “Prefers Party X” are independent, as the two events satisfy the multiplication rule for independent events.

 Binary - Party Y or not Party Y, Independent - reasonable to assume preferences of 5 voters are independent from each other, Number of Trials = 5 voters are selected at random (n=5), S = probability of selecting a voter who prefers Party Y is the same for every voter selected (p = 0.51)

X = the number of voters who prefer Party Y when the distribution of X is binomial with p = 0.51 and n = 5

P(X=2) = 0.306 using binomialpdf with p = 0.51, n = 5, and successes = 2.

There is a 0.306 probability that exactly two voters prefer Party Y if 5 voters from the state are selected at random.

Well done! All three parts show strong command of statistical notation, and clear communication. Your answers are all correct and would all earn credit.

a.) i. P(County A AND Party Z)= 0.152. There is a 15.2% chance that the person lives in County A and prefers Party Z.  

ii.) P(County A OR Party Z)=0.400+0.310-0.152=0.558. There is a 55.8% chance that the person lives in County A or prefers Party Z

iii.) P(County A|Party Z)=0.152/0.310=0.490. There is a 49% chance that the person lives in County A given that they prefer Party Z.

b.) P(County B)= 0.600, P(County B|Party X)=0.108/0.180=0.60. Because the two probabilities are the same, the events “Live in County B” and “Prefers Party X” are independent.

c.) This scenario has each trial ends is either a success or failure, they are independent, the 2 selected are set, and the probability for those who prefer Party Y is the same at 0.510. Therefore, this is a binomial probability.

P(X=2)= BinomPDF(n=5, p=0.510, x=2)=0.306

There is a 30.6% chance that if 5 voters from the state are selected at random, exactly 2 of them will prefer Party Y.

In all three parts, your answers are correct and presented with accompanying work and appropriate notation. You would get full credit for each part! Part ( c ) provides a good contrast to your work on Unit 4 FRQ #2 - by saying “the probability for those who prefer Party Y is the same at 0.510,” you provide an 

 reason for the scenario being binomial. Well done!

A i. P(County A and Party Z)=.152/1=.152 

ii. P(County A or Party Z)= .400+.310-.152=.558

B. Yes, “lives in county B” and “prefers party x” are independent because:

P(County B/Party X)= P(County B): P(.6)= P(.600)

P(Party X/County B)= P(Party X): P(.18)= P(.180)

To find out I first check the binomial distribution requirements:

Each trial has 2 possible outcomes (Party Y or not party Y)

Probability of success is the same for each trial (p=.510)

I then proceeded to put on my calculator binompdf( 5, .510, 2). I knew to use the function binompdf because the problem is asking to find exactly 2 voters.

All calculations and explanations are correct, with appropriate work shown. However, in part (b), you make a small notation error. P(County B | Party X) = P(County B) = 0.6. The probabilities themselves should 

 be inside the P( ) notation. Think of it like input/output function notation - inside the P( ) is what probability you’re finding (in words or symbols), while outside on the other side of the equals sign is the numerical answer.

a i) The probability that a person lives in Country A and prefers Party Z is 0.152.

ii) The probability that the person lives in County A or prefers Party Z is 0.558.

P(AUB) = P(person lives in county A) + P(Person prefers Party Z) - P(person who lives in Country A and prefers Party Z)= 0.558

iii) P(Person lives in County A | Person prefers Party Z) = 0.152/0.310= 0.490

b) P(Living in County B)= 0.600 P(Living in County B | Prefers Party X)=0.108/ 0.180= 0.6

The events “Lives in County B” and "Prefers Party X"are independent as the probability of Living in County B equals the probability of living in county B given preferring Party X.

Binary: Prefer Party Y or not prefer Party Y.

Independent: Sample size of 5 voters is less that 10% of the population(voters from the entire state)

Number of trials: There are a set number of trials (5) n=5

Calculator: P(x=2)= Binom (n=5, p=0.50, x=2) = 0.306

If 5 voters from the state are selected at random, the probability that exactly 2 of them prefer Party Y is 0.306.

Small thing: for part a-ii, you provide the correct answer and use correct notation. However, you don’t show the 

 that go into the calculation, so on some rubrics that would be considered a “naked” answer in that you didn’t show the numbers along the way. Taking the extra step to show calculations with probability scenarios is always helpful. Everything else should receive full credit.

a) i. P(county A and prefers party z)=.152 --> There’s a 15.2% chance that a person lives in county A and prefers party z.

ii. P(county A or prefers party z)= P(county A) + P(party z) - P(county A and party z) --> .4 + .31 - .152=.558 or 55.8% --> There’s a 55.8% chance that a person lives in county A or prefers party Z.

iii. P(county A|party z)= P(county A and party z)/P(party z) --> .152/.31=.49 --> There’s a 49% chance that the person lives in county A given that they prefer party Z.

b) P(Lives in county B|party x) --> P(county b and party x)/P(party x) = P(county B) <-- if this is true, the two events are independent

P(.108)/P(.18)=.6 which equals P(.6). This statement is true, which shows that these two events are independent of each other.

c) This is a binomial random variable because there is 2 outcomes for each trial – prefer party y (.51) or don’t prefer party y (.49). There is a set number of trials – 5 randomly selected voters from this state. The trials are independent of each other – seen in part b. Lastly, the probability of a success will not change in each trial (p=.51).

P(exactly 2/5 of randomly selected voters from the state prefer party y) = (5 2) = 5!/2!(5-2)! --> 10(.51)^2(.49)^3 = .306 --> There’s a 30.6% chance that exactly 2/5 of the randomly selected voters from the state prefer party Y.

Nice job on your calculations (and super shout-out for going by hand - or at least writing it all out by hand - on the binomial calculations). Be really careful with your notation on part (b). Probability notation is kind of like function notation f(x) from algebra. The “stuff” inside the parentheses is like an input - and should represent what we’re finding the probability of. This will include variables and equality/inequality statements in some cases, but not just “plain numbers”, since those are the “output” in this situation. So P(.108) isn’t a thing - it should be P(B and X) = .108. That is, you already wrote the notations correctly - just put the actual numbers on your second line for part (b), and you’ll be good to go to say yes for independence.

a) i) the probability that a person lives in country A and prefers party Z is 0.152

ii) the probability that a person lives in country A or prefers party Z is 0.4+0.31-0.152 = 0.558

iii) the probability that they live in country A given that they prefer party Z is 0.152/0.31 = 0.49032

b) for the events of living in country B and preferring party X to be independent the probability that a person lives in country B and the probability. that. they live in B given that. they prefer party X has to be equal. P(B) = 0.6 P(BㅣX) = 0.6549
since the probabilities are not equal, the two events are not independent.

binompdf (trials : 5, p : 0.51, x value : 2) = 0.306

On part (a) and part ( c ), you’ve done all calculations correctly. For part (b), you’ve defined what we need to find to show independence… but your calculation P(B | X) is incorrect. You should get 0.108 / 0.18 = 0.6, so it 

 be independent. Double check what you did. Given what probability you got, you made an appropriate conclusion, so you’d still be eligible for partial credit.

Z)=P(A)+P(Z)-P(A∩Z)=0.400+0.310-0.152=0.558

b)When the situation is Independent P(A∩B)=P(A)×P(B)

So the events “Lives in County B” and “Prefers Party X” are Independent.

c) We know this is a Binomial Distribution since it’s 

b. Yes they are independent because P(BandX)=.108 and P(B)*P(X)=.108. Since these two numbers are the same then the events are independent.

c. This is a binomial distribution because there is a success or failure as to whether the voters prefer Party Y or not, each voter chosen is independent of one another, Number of trials is fixed at 5, and there is a set number of probabilities for each choice. Binompdf(trials=5,p=.510,x=2)=.306

There is a .306 probability that exactly 2 of the 5 voters prefer Party Y.

AP Stats Practice FRQ Responses & Feedback (Unit 4)

Practicing with FRQs is a great way to prep for the AP exam! Review student responses for a FRQ combining multiple units and corresponding feedback from Fiveable teacher Jerry Kosoff.

A researcher in a city with a large public train system wondered if the rent prices of one-bedroom apartments was related to the distance from the nearest train station. From a list of 250 similarly-sized one-bedroom apartments in the city, the researcher selected a simple random sample of 20 apartments. The researcher then measured the walking distance, in minutes, to the nearest train station and created a scatterplot comparing the walking distances to the advertised weekly rent, in dollars. A scatterplot and the output from computer regression software are shown below.

a. Explain a procedure by which the researcher may have selected the simple random sample.

b. Describe the association between walking distance from the nearest train station and weekly rent for the apartments included in the sample.

c. Interpret the value of the coefficient of determination (r-squared) in the context of this problem.

d. Before examining the data, a second researcher makes a prediction that for each additional minute of walking distance, the weekly rent will decrease by approximately $2. A 95% confidence interval for the slope of the regression line is constructed from the data, and is found to be (-2.471, -1.845). Does the confidence interval contradict the researcher’s claim? Justify your answer.

e. The second researcher wants to conduct a similar study to the first researcher. However, in the second researcher’s study, a mixture of one, two, and three-bedroom apartments were selected. Do you expect the value of r-squared for the second study to be greater than, less than, or equal to the value of r-squared in the first study? Justify your response.

a. The researcher could have assigned each apartment in the sample a number from 1-250 and used a random number generator to choose 20 apartments, taking out any repeats.

b. There is a strong, negative linear correlation between the weekly rent and distance from the nearest train station. (should I say correlation or association?)

c. 92.1% of the variability of in the weekly rent dollars can be accounted for by the variability in the distance from nearest train station.

d. No it does not, because $2 falls within the given confidence interval.

e. I expect the r-squared value to be less than the first study. This is because r will decrease since there will be lower residual because there will be an increased variability in the weekly rent since it can differ depending on the number of bedrooms. If r decreases, r squared decreases as well.

Little things:
in (a), you must specify that the 20 numbers should be contained within the interval of 1-250 (you’re leaving yourself open to other numbers the way you phrase it).

In (b), either word is ok - correlation or association. I tend to use the language that is included in the question to play it safe.

c, d, and e give fully appropriate explanations!

a. The researcher could have numbered the 250 similarly-sized apartments from 1-250 and wrote the numbers on an equally sized piece of paper. Then, the researcher could put these slips in a hat and she it very well. Then the researcher can pick 20 slips of paper which would represent the 20 apartments that will be used in the sample.

b. The association between walking distances from the nearest train station and weekly rent for the apartments included in the sample is strong, negative, and linear with no suspected outliers. It is strong as the correlation coefficient= square root of r^2 = square root of .921 is .9596 which shows us that the data points follow a strong pattern and linear pattern. We know the data points have a negative association as the slope of the linear regression line is -2.158.

c. The coefficient of determination, says that 92.1% of the variation in weekly rent is accounted for by the regression line when x= distance from nearest train station.

d. Since the confidence interval contains -2, the researchers claim is not contradicted.

e. We expect the r-squared to be less than the first study. This is because we will have more variability in the weekly rent since there are more types of apartments which vary in weekly rent. This leads us to have less of an association between the weekly rent and the train station.

AP Stats Mixed Units Practice FRQ #4 & Feedback

A group of randomly-selected students at a large high school were given a survey about their transportation to school. One of the questions asked students for their primary mode of transportation; another asked students for the typical number of minutes it takes them to travel to school in the morning. Two groups of students are singled out for further comparison: those that said they walk to school as their primary mode of transportation, and those that said they drive to school as their primary mode of transportation. The two histograms below show the distribution of travel times, in minutes, for the two groups - though the group labels are not present. There are a similar number of students in each group. 

Histogram K represents the group of students who walk to school because the mean is pulled towards outliers. Since Histogram K is skewed left, the mean would be pulled towards the skewness while the median would stay put as it is resistant to outliers. Histogram J however, shows the mean being greater than the median as it is being pulled to the right.

The resulting histogram of the combination of the two groups of students who walk to school and time it takes for school would be normally distributed.

Name of test: 2 sample t-test. We want to test a claim about the difference between mu1 (the true mean travel time it takes to travel to school for 50 students) and mu2 (the true mean travel time it takes to travel to school for 50 students.

Your answer for part 1 is very good - you pick the correct histogram and give a clear reason for doing so (and clearly explain why it WOULDN’T be the other histogram). Full credit. For part 2, you say “normal,” which implies a unimodal graph with a peak in the center of the distribution. That won’t be the case here: if you look carefully at the x-axis for each distribution the peaks will end up separated (one around 8 minutes and the other around 40 minutes), whereas the center of the combined number line (around 24 minutes) has very few observations. Thus, 

 is the description we’d be looking for here.

For part 3, your work is good - you’ve checked conditions, cited the use of the CLT, and correctly calculated the parameters of the sampling distribution. One thing - you never mention the shape (just saying “normality” as a condition isn’t equivalent to saying “the sampling distribution would be approximately normal.”). Therefore, you’d get partial instead of full credit. Advice - after citing the CLT, state that this means the sampling distribution is approximately normal.

Finally, in part 4, you’d also receive partial credit, if any. A 2-sample t-test is the correct test. However, you do not write fully correct statements for Ho and Ha. Your Ho would be that the mean travel times for the two groups are the same; Ha would be that the mean travel times for the groups are different. You only mention the ‘difference’ and do not cite it as the alternative hypothesis. Additionally, you define mu1 and mu2 as representing 50 students - but they would represent ALL students and ALL teachers at the school (since “mu” is a population parameter). I’m working on the assumption that you meant for one of your parameters to mention teachers. Be careful when typing/writing responses!

All of your answers are correct and give appropriate reasoning. A tip for part (a): when asked to make a choice in AP stats, you should explain not only why your choice is correct, but why the other option is incorrect. You could add “since histogram J is right skewed, it does not match the description” or something similar. All of your other answers are thorough, in context, and show strong understanding. Well done!

The students and teachers were randomly selected.

50 and 40 are both greater than 30, which satisfies the Central Limit Theorem, indicating that the distributions are approx. normal.

There are at least 501 students and 401 teachers, so the 10% condition for independence is satisfied.

Nice job! One small thing: for #2, you’re correct in calling the shape “bimodal,” but the rubric for a similar question only gave full credit for identifying where the clusters/peaks were centered. So you’d get partial credit. Full credit: “bimodal, with peaks at around 8 minutes and 40 minutes”

AP Stats Mixed Units Practice FRQ #3 & Feedback

Practicing FRQs is a great way to prep for the AP exam! Review student responses for a Mixed Units FRQ and corresponding feedback from Fiveable teacher Jerry Kosoff. 

The city council of a large city is considering raising a city tax to provide funding for road repairs. The council wishes to gauge citizen interest in the plan. The council mails a survey to random sample of 1,000 city residents, of whom 450 reply. The survey asks “should we increase city taxes in order to provide additional funding to road repairs?” Of those who reply, 170 say yes; the other 280 say no.

In #1, you give a plausible reason for bias (people travel a lot and would like to see the taxes increase), but do not 

 describe the impact this has on the estimate they got (37.8%). Saying “it seems like most of the city residents would like to increase city taxes” is not the same as clearly stating that you believe this the survey resulted in an 

 of the true proportion. I know it seems harsh, but that’s how the rubrics go with describing bias: (1) explain HOW the bias comes to be, (2) explain WHY the bias happens, (3) give a specific DIRECTION of the bias (over/under estimate of the true ____). Your answer only does #1 and #2.

In part 2, your sentence looks great - except you said “95% of city residents” instead of “95% of the constructed intervals” or something similar. Read your sentence back, and I think you’ll see what I mean. That would ding your answer from full to partial credit unfortunately. You have something similar in part 3 - you say “true MEAN of city residents” instead of “true PROPORTION of city residents”. Mixing up mean & proportion there would also knock you down a scoring level, even though the rest of your answer is on point.

Finally, in part 4, you are correct in that we need the 

 (not “samples”) within the sample to be independent, but that comes from the “10% condition” (that the sample size is less than 10% of the overall population size). The reason for the “at least 10 successes/failures” condition is to ensure an approximately normal sampling distribution of p-hat, from which we can calculate the confidence interval.

The 0.378 may be a biased estimate of the proportion of all city residents who would reply yes to the question because the survey has non-response bias present. According to the information present in the passage, the survey indicated that not everyone responded to the survey which means that the proportion is not an accurate estimate of all city resident individuals who would reply yes to the question because the proportion is an underestimate of the true proportion who would reply yes to the question.

In the context of this problem, a 95% confidence level means that we are 95% confident that the interval captures the true portion of all city residents who would reply “yes” to the question.


This interval does provide convincing statistical evidence that the proportion of all city residents who would say “yes” to the question is at least 50% since 50% is included in the interval 0.480 to 0.593.

This condition is necessary in order to ensure that the sampling distribution of p-hat is approximately normal.

In part #1, you would likely earn partial credit. When discussing bias on the AP exam, you typically have to do 3 things: (1) explain the source of the bias (“how” it happens), (2) explain the reason for that source existing (“why” it happens), and (3) explain the impact on the result (“what” happens). When reading your response, I see evidence for #1 and #3 - you mention “not everyone responded to the survey” (#1 - how) and that this will probably “underestimate the true proportion” (#3 - the impact). To my eyes, though, your response does not address 

 the non-response will lead to an underestimate, which would imply that the 37.8% is lower than the true proportion if we actually asked everyone (and it would be maybe more like 50% or something like that). You would need to make an argument for *

why the people who responded to the survey are more likely to say no and thus produce an underestimate

 * - perhaps they are strongly opposed to taxes of any kind, or the wording of the question made them feel like their money could be better spent elsewhere. Whatever you decide is the case, you should present and defend why it impacts the responses. For nonresponse to turn into nonresponse bias, the people who 

 participate must be more likely to answer a certain way than the people who 

Additionally, while I’m not assuming this is the case, I often have students misunderstand that getting responses from fewer people than you expect does not automatically produce an 

estimate. “Underestimate” specifically refers to the proportion/mean/whatever-statistic-is-being-measured being lower than would be reflected in the population. A small and biased sample can produce an 

estimate just as easily as an underestimate - perhaps in this scenario we ask a small group of people who live near roads with lots of potholes what they think. They would be likely to support the city’s proposal more than others, and therefore produce an overestimate. [OK, thanks for coming to my TED Talk about bias. On to the next part…]

In part #2, we have a little bit of reviewing to do. In part (a), you correctly interpret what a 95% confidence 

 is, but that is not the same as a confidence 

 represents a “long-run capture rate” that is then reflected in each specific confidence interval. You can check out an overview from a previous stream 

 - it’s time-stamped to the part you’d need. The correct answer in this case would sound something like “if we were to take many, many random samples of 300 city residents and ask them the question, about 95% of the confidence intervals we constructed would capture the correct value for 

, the proportion of all city residents who would respond yes to the question.”

For #2 part (b), you’ve also committed a relatively common error, in that while it is true that 50% is in fact in the interval, the presence of other, smaller values in the interval provides evidence against the claim that at least 50% of residents support the proposal. It’s just as plausible that 48.5%, or 49%, or 49.9% would say “yes”. And since 

 values within a confidence interval are considered “reasonable” values for 

, we cannot say with confidence that the true population proportion is at least 50%. We could only say that if the 

 interval is 50% of higher - for example, (0.512, 0.592).

In part ( c ), you give the correct rationale for the “large counts” condition - short and to the point! This would earn full credit.

For part 1, you would likely earn partial credit. When discussing bias on the AP exam, you typically have to do 3 things: (1) explain the source of the bias (“how” it happens), (2) explain the reason for that source existing (“why” it happens), and (3) explain the impact on the result (“what” happens). When reading your response, I see evidence for (1) and (2) when you describe “only 450 of those 100 people replied” and cite the possibility that those who replied “might have different opinions about the plan.” From there, you should “take a stand” per se and make a conjecture as to 

 those people would differ in their opinions from the general population, and if that will produce an overestimate (more likely to say yes) or underestimate (less likely to say yes) of the true proportion. You can actually justify either direction here, as long as your explanation is clear.

Your response in part 2a is strong, and shows a clear understanding of what a confidence 

 represents. In part 2b, you give the correct answer (“no”) with a correct reason (“50% is contained in the interval”), but you lose me a bit with your last sentence. In theory, we could get 

 proportion in a sample of 300 residents just by random chance alone. A more direct statement may be something like “this means that 50% is a plausible value for the proportion of all adults who would say yes, and we therefore do not have statistical evidence that the proportion is greater”

Your response in part 3 is on the money - the approximately normal sampling distribution is why we check that condition!

In part (a), you clearly identify the possibility of non-response bias. However, when discussing bias on the exam, it’s important to pick a 

 of the bias - you’re correct that we may end up with an over-representation of strongly opinionated people, but you should “pick a side” as to how those people will land (either more or less in favor of the proposal than the general public), resulting in either an under or over-estimate of the true parameter. In most cases, you can defend either side, as long as you give a reasonable possibility. In parts (b) and ( c ), you’ve provided correct answers with appropriate context.

Great work! All three parts are complete. In part (a), you named the source of bias, explained 

 it might impact people’s responses, 

 connected that to the proportion we were trying to estimate. In part (b), you correctly interpret both parts, and in part ( c ) you give the correct reason for checking that condition.

When discussing bias like in part (a), you need to take it step further and explicitly comment on whether you think the sample results in this case are too high (an overestimate of 

). In a case like this where it’s not obvious, it’s OK to “pick a side” and just go with it: for example, “citizens who are concerned about tax increases may be more likely to respond and say no, producing an underestimate of the proportion of all citizens who would support the proposal”

In part 2, your confidence level interpretation is well done (though I think you should say “95% of samples of 161 

 that…”, and you reach the correct conclusion in part b. 3 has the correct rationale for the 10 successes/failures condition.

AP Stats Mixed Units Practice FRQ #2 & Feedback

Practicing FRQs is a great way to prep for the AP exam! Review student responses for an FRQ prompt and corresponding feedback from Fiveable Teacher Jerry Kosoff. 

AP Statistics Exam Prep 2020-21📊

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Unit 1

Exploring One-Variable Data

Unit 2

Exploring Two-Variable Data

Unit 3

Collecting Data

Unit 4

Probability, Random Variables, and Probability Distributions

Unit 5

Sampling Distributions

Unit 6

Inference for Categorical Data: Proportions

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Inference for Quantitative Data: Slopes

Free Reviews 2020

Free Response Questions (FRQs)

Blogs

Jenni MacLean

Jerry Kosoff

Josh Argo

Shane Durkan

Bob Amar

Karen Knight

Harrison Burnside

Peter Cao

Sander Owens

Brianna Bukowski

Kanya Shah

Jackson Schuster

Jacob Jeffries

Lusine Ghazaryan

John Le

Tyler Fazenbaker

Krish Gupta

Derek Schiltz

Tyler Fazenbaker