AP Stats Practice FRQ Responses & Feedback (Unit 6)

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!

On the AP Statistics Exam, Section II contains 6 free-response questions: 5 questions in Part A (65 minutes total) and 1 investigative task in Part B (25 minutes). AP Statistics free-response questions are scored using question-specific scoring guidelines. Students earn credit for correctly selecting and carrying out methods, showing required statistical reasoning, and clearly communicating conclusions in context when the rubric calls for it. For many inference FRQs, strong responses define parameters, identify an appropriate procedure, verify required conditions, carry out the mechanics correctly, and interpret the result in context. However, the exact required components depend on the specific question and its scoring guidelines.

The Practice FRQ Prompt

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?

FRQ Writing Samples and Teacher Feedback

FRQ Practice Submission 1

P(yes|new) = 88/105 = 0.838

P(yes|used) = 85/120 = 0.708

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

margin of error = +/- 0.108

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.

Teacher Feedback

A strong AP Statistics significance-test response will usually include:

1. Defining the parameters in context
2. Stating null and alternative hypotheses
3. Choosing an appropriate procedure by name
4. Checking the conditions for the chosen procedure
5. Carrying out the mechanics correctly
6. Interpreting the result in context

FRQ Practice Submission 2

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

2-sample z test for p_1 - p_2

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

Conditions:

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 - for a 2-sample z test, use the pooled proportion under the null hypothesis: p̂_c = (88 + 85)/(105 + 120) = 173/225 = 0.7689. Then 105(0.7689) = 80.73, 105(0.2311) = 24.27, 120(0.7689) = 92.27, and 120(0.2311) = 27.73, all of which are at least 10, so the sampling distribution of p̂_1 - p̂_2 is approximately normal.

p-hat_1 = 88/105 = 0.838, n_1=105

p-hat_2 = 85/120 = 0.708, n_2=120

z* = 2.303

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.

Teacher Feedback

FRQ Practice Submission 3

Test type- two sample proportional difference hypothesis z-test

• 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

Conditions

1. Since we are dealing with a two-sample proportional difference hypothesis test, we will have to pool/combine our proportions.
   • p_c = (88 + 85) / (105 + 120) = 173/225 = 0.7689
   • q_c = 1 - 0.7689 = 0.2311
   • pooled large counts: 225(0.7689) = 173.0 ≥ 10 and 225(0.2311) = 52.0 ≥ 10

Conditions for new cars:

• 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.

Conditions for old cars:

• simple random sample: stated in the problem- “the company randomly selects…120 customers…”
• independence: 10(120)=1200 Assume that the population of used 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.

Solve

• z=.84-.708/(sqrt.(.2311x.7689)/105 + (.2311x.7689)/120) = 2.28 --> *I looked at table z to find the p-value of -2.28, p-value=.0129(2)=.0258

Conclusion

• (.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.

Teacher Feedback

FRQ Practice Submission 4

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.

Ho= p1-p2=0 Ha= p1-p2 does not equal 0

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

Conditions:

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- using the pooled proportion for the significance test, p̂_c = (88 + 85)/(105 + 120) = 173/225 = 0.7689. Then 105(0.7689)=80.73, 105(0.2311)=24.27, 120(0.7689)=92.27, and 120(0.2311)=27.73, all of which are at least 10.

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.

Teacher Feedback

FRQ Practice Submission 5

Hypotheses

H_o: p_1 = p_2

H_a: p_1 ≠ p_2

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.

Assumptions

• 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:- for a hypothesis test, use the pooled proportion p̂_c = (88 + 85) / (105 + 120) = 173/225 = 0.7689.- 105(0.7689) = 80.73 ≥ 10- 105(0.2311) = 24.27 ≥ 10- 120(0.7689) = 92.27 ≥ 10- 120(0.2311) = 27.73 ≥ 10- Since all 4 are greater than 10, the sampling distribution is approximately normal.

Calculations

p_hat_combined = (88 + 85) / (105 + 120) = 173/225 = 0.7689.

z = ((0.8381 - 0.7083) - 0) / sqrt[(0.7689)(1 - 0.7689)/105 + (0.7689)(1 - 0.7689)/120] = 2.3036.

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

alpha = 0.05

p-value<alpha

Conclusion

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.

Teacher Feedback

FRQ Practice Submission 6

H null: p1-p2=0

H alternative: p1-p2 doesn’t equal 0.

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).

Conditions:

1. For a two-sample z test, the large-counts condition should be checked using the pooled proportion under H0: p1 = p2. Here, p̂c = (88 + 85)/(105 + 120) = 173/225 = 0.7689, so the counts are 105(0.7689) = 80.73, 105(0.2311) = 24.27, 120(0.7689) = 92.27, and 120(0.2311) = 27.73, all at least 10.
2. Because the customers in each group were randomly selected and each sample is less than 10% of its respective population, observations within each sample can be treated as approximately independent.
3. The two samples come from different groups (new-vehicle customers and used-vehicle customers), so comparing the two sample proportions is appropriate. Z= (p1 hat - p2 hat)/(phat *(1-phat)(1/n1 + 1/n2))^.5 where phat is the pooled proportion = 173/235 = 0.7361

p1 hat = 0.8381

p2 hat = 0.7083

Z=0.1298/((0.7361)(0.2639)(0.0179))^.5

Z=2.19

pvalue = 2(1-0.9857) = 0.0286.

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.

Teacher Feedback

FRQ Practice Submission 7

STATE

Ho: p1=p2

Ha: p1=/p2 (=/ means not equal to)

Alpha = 0.05

p1 = the true proportion of all 2018 customers who bought a new vehicle who would answer “yes” to the recommendation question.

p2 = the true proportion of all 2018 customers who bought a used vehicle who would answer “yes” to the recommendation question.

2 sample z test for proportions

PLAN

Random: met, since stated in the problem that company selects customers randomly.

10%: 105<= 1/10 all customers who bought new vehicle

120<= 1/10 all customers who bought used vehicle

Large Counts: For a significance test, use the pooled proportion.

Since, Large counts conditions are met for both we can assume that the distribution is approximately normal.

DO

p_c = (88 + 85) / (105 + 120) = 173/225 = 0.7689.

z = ((88/105) - (85/120)) / sqrt[(0.7689)(1 - 0.7689)(1/105 + 1/120)] = 2.3039.

p-value = 2P(Z > 2.3039) ≈ 0.0212.

CONCLUDE Since, the p value is 0.0212 and the alpha level is 0.05, the p value is smaller than alpha. Therefore, we reject the null hypothesis of p1=p2. We have convincing evidence to say that there is a difference between the proportion of customers who would answer “Yes” to the survey question is different from new vs used vehicle sales.

Teacher Feedback

FRQ Practice Submission 8

2-sample z test for proportion

Ho: p1 = p2

Ha: p1 =/= p2

Conditions:

• 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.

Calculation: Pc = 88+85/105+120 = 173/225

z= (88/105-85/120)-0/ sqrt(173/225)(52/225) x sqrt(1/105+1/120) = 2.304

p-value: .0212

Interpret: Because or p-value (.0212) is below the alpha (0.05) we reject the Ho. There is significant evidence that the proportion of customers who would answer “yes” to the survey question is different for new vs used vehicle sales.

Teacher Feedback

FRQ Practice Submission 9

pnew= proportion of customers who bought a new car and would recommend to a friend.

pused= proportion of customers who bought a used car and would recommend to a friend.

H0=pnew−pold=0 Ha=pnew−pold≠0

Conditions:

1. Random: “The company randomly selects 105 customers who bought a new vehicle and 120 customers who bought a used vehicle”
2. Normal: For a two-sample z test, use the pooled proportion under the null hypothesis. p̂_c = (88 + 85)/(105 + 120) = 173/225 = 0.7689, so 105(0.7689)=80.73, 105(0.2311)=24.27, 120(0.7689)=92.27, and 120(0.2311)=27.73, all at least 10.
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. We will be conducting a 2 sample z-test for difference of population proportions.

Calculations:

p=0.021

Conclusion: Since our p-value of 0.021<0.05 , we can reject the H0 . We have convincing statistical evidence that there is a difference between population proportion of the people who would recommend to a friend between the people who bought new and old cars.

Teacher Feedback