AP Stats Practice FRQ Responses & Feedback (Unit 6)

The Practice FRQ Prompt

FRQ Writing Samples and Teacher Feedback

FRQ Practice Submission 1

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.

1. Stating null/alternative hypotheses
2. Defining the parameters in the null/alternative hypotheses
3. Choosing an appropriate test/interval by name
4. Checking the conditions to run the chosen test/interval
5. Writing the results from the chosen test/interval
6. Correctly interpreting the results from the chosen test/interval in terms of whether we do or don’t have evidence for the alternative hypothesis.

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 not 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.

FRQ Practice Submission 2

1. Random - Stated that the company “randomly selects” customers for the survey
2. 10% Condition for Independence - satisfied since it is safe to assume that there are at least 105(10) = 1050 customers who bought a new vehicle at the car company, and at least 120(10) = 1200 customers who bought a used vehicle at the car company.
3. Large Counts Condition - satisfied since
   1. n_1p-hat_1 = 1050.838 = 87.99 >=10
   2. n_1*(1-p-hat_1) = 1050.162 = 17.01 >=10
   3. n_2p-hat_2 = 1200.708 = 84.96 >= 10
   4. n_2(1-p-hat_2) = 120*0.292 = 35.04 >=10
   5. With Large Counts Condition satisfied, the sampling distribution of p-hat_1 - p-hat_2 is approximately normal.

Strong execution from top to bottom, presented clearly. One thing that you’re going to facepalm about: you defined the parameters p1 and p2 as the exact same thing. “p2” should say “used”

FRQ Practice Submission 3

• H0= p1-p2=0
• Ha=p1-p2 does not = 0
• p1=the proportion of all customers from a list of 2018 vehicle sales who bought a new vehicle and would recommend our company to a friend
• p2=the proportion of all customers from a list of 2018 vehicle sales who bought a used vehicle and would recommend our company to a friend

1. Since we are dealing with a two-sample proportional difference hypothesis test, we will have to pool/combine our proportions.
   1. pc=88+85/105+120=0.75
   2. qc=1-0.75=0.25
      1. *we have to check for the independence of our pooled data --> .75(225)=168.75 >=10 & .25(225)=56.25 >= 10

• 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.

• z=.84-.708/(sqrt.(.25x.75)/105 + (.25x.75)/120) = 2.28 --> *I looked at table z to find the p-value of -2.28, p-value=.0129(2)=.0258

• (.0258<.05 our significance level) --> Reject the H0 in favor of the Ha. We have significant evidence that the proportion of all customers from a list of 2018 vehicle sales who bought a new vehicle and said they would recommend the company is different than the proportion of all customers from a list of 2018 vehicle sales who bought a used vehicle and said that they would recommend our company to a friend.

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

FRQ Practice Submission 4

1. Random- A random sample of 105 customers who bought a new vehicle and 120 customers who bought a used vehicle is taken
2. Normal-
   1. Sample of new cars: np = 105 * 0.838= 88 is greater than or equal to 10.
      1. n(1-p)= 105(0.162)= 17 is greater than or equal to 10.
   2. Sample of used cars: np= 120* 0.708= 85 is greater than or equal to 10.
      1. n(1-p)= 120(0.292)= 35 is greater than or equal to 10.

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

FRQ Practice Submission 5

• Independence:
  • We have 2 independent random samples of customers from 2018 vehicle sales.
  • Population of new vehicle customers is at least 1050 and the population of used vehicle customers is at least 1200.
• Normality:
  • n_1 * p-hat_1 = 105 * 0.8381 = 88 ≥ 10
  • n_1 * (1-p-hat_1) = 105* (0.1619) = 16.9995 ≥ 10
  • n_2 * p-hat_2 = 120 * 0.7083 = 84.996 ≥ 10
  • n_2 * (1-p-hat_2) = 120 * 0.2917 = 35.004 ≥ 10
  • Since all 4 are greater than 10, the sampling distribution is approximately normal.

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

FRQ Practice Submission 6

1. np1 hat and nq1 hat both greater than 10 (or 15). np2 hat and nq2 hat both greater than 10 or (15). Calculations omitted, but it is clear that both groups both have more than 10 successes and failures.
2. Samples are random and independent. This is met as the question specifies that the selection of customers from both groups was random, and it is clear that the selection of a customer from one group does not affect the selection of a customer from the other group.
3. It is also safe to assume that the number of customers who bought new and used cars is at least 10510 = 1050 and 12010 - 1200 people, respectively.

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.

FRQ Practice Submission 7

Nicely done! You’ve done all components of a hypothesis test correctly (and shout-out to the state-plan-do-conclude method that’s in the textbook I use as well).

FRQ Practice Submission 8

• Random: We are told that the customers are “randomly selected”
• Normal: It is approximately normal because:
  • (88/105)(105)=88 ≥10 (17/105)(105) = 17 ≥ 10
  • (85/120)(120)=85 ≥10 (35/120)(120) = 35 ≥10
• 10% condition: There are more than 10 x 120 and 10 x 105 customers. The two samples are independent from each other.

The only part you need more on is your hypotheses; you don’t define what you mean by “p1” and “p2”, so you’d only get partial credit there (you didn’t define the parameter). Every other part is done appropriately and would earn full credit.

FRQ Practice Submission 9

1. Random: “The company randomly selects 105 customers who bought a new vehicle and 120 customers who bought a used vehicle”
2. Normal: For the new vehicle population, there is 88≥10 successes and 105−88=17≥10 failures. For the old vehicle population, there is 85≥10 successes and 120−85=35≥10 failures.
3. Independent: It is reasonable to assume that the population of the people who bought new and old cars is at least 1050 and 1200 respectively.

Very good work. Little thing: on “calculations” part, it is typical to show both the test statistic (in this case, your z), as well as the p-value.