AP Stats Mixed Units Practice FRQ #2 & Feedback

📊AP Statistics Review

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 Mixed Units FRQ Practice Prompt

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 a 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 for road repairs?” Of those who reply, 170 say yes; the other 280 say no.

Explain why 0.378 (170/450) may be a biased estimate of the proportion of all city residents who would reply yes to the question.
Another survey is commissioned; in it, a random sample of 300 city residents is contacted. They are also asked if they would support an increase in city taxes for the purpose of providing additional funding for road repairs. In the second survey, 161 of those surveyed reply “yes.” The city council plans to use the results of this survey to construct a 95% confidence interval to estimate p, the proportion of all city residents who would reply “yes” to the question.
(a) Explain the meaning of the 95% confidence level in the context of this problem.
(b) The 95% confidence interval constructed from the survey is (0.480, 0.593). Does this interval provide convincing statistical evidence that the proportion of all city residents who would say “yes” to the question is at least 50%? Justify your answer.
One of the conditions that must be met for the confidence interval in (2) to be constructed is that at least 10 residents must have said “yes” and at least 10 residents must have said “no.” Explain why this condition is necessary.

Student Samples and Teacher Feedback

Student Response 1

Those who responded yes probably make a lot of money during their job. People who have a well paid job usually travel daily on roads to arrive at work everyday. This makes it seem like that most of the city residents travel long distances on roads so would like to increase city taxes for road repairs.
If I took many many samples and generated confidence intervals on them, 95% of city residents would contain the true proportion of all city residents who support an increase in city taxes to help fund road repairs.
No, since the interval ranges from .48-.593 there is a possibility that the true mean of city residents who support an increase in city taxes for road repairs is not at least 50% since .48-.49999 is less than 50%.
This condition is necessary as it makes sure that the samples are independent from one another meaning that one city resident whose decision supports an increase in city taxes for road repairs does not affect another city resident whose decision supports an increase in city taxes.

Teacher Feedback

In #1, you give a plausible reason for bias, but the explanation needs to be tied more clearly to the sample proportion 0.378. To explain why 0.378 may be biased, identify the source of bias (nonresponse) and explain that people who respond may differ from those who do not respond, so the sample proportion may not represent all residents. If you can reasonably justify a likely direction (overestimate or underestimate), include it, but a direction should not be presented as universally required when the scenario does not establish one. On AP Statistics FRQs, precision of terminology matters and unclear wording can cost credit, depending on the specific scoring guidelines for that question.

In part 2, your sentence is close, but “95% of city residents” is not what the confidence level describes. It should be “95% of the confidence intervals constructed from many random samples” or similar wording. You have something similar in part 3, where you say “true mean” instead of “true proportion.” On AP Statistics FRQs, using the wrong statistical term can matter, depending on the scoring guidelines for that question.

Finally, in part 4, the idea of independence is important, but that comes from the 10% condition, not from the “at least 10 yes and 10 no” condition. The reason for the “at least 10 successes/failures” condition is to ensure an approximately normal sampling distribution of $\hat{p}$ , which is what allows us to construct the confidence interval.

Student Response 2

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.
(A) 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. (B) 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.

Teacher Feedback

In part #1, your response correctly identifies nonresponse bias as the source. To make the explanation stronger, you should add why that creates bias: the residents who replied may differ systematically from the residents who did not reply, so the 37.8% from respondents may not represent the proportion for all city residents. If you can justify a likely direction, you can include it, but you should not treat “underestimate” as automatic here. In this scenario, the information given does not force one direction of bias. On AP Statistics FRQs, precision of explanation matters and incomplete reasoning can cost credit, depending on the scoring guidelines for that question.

Additionally, getting responses from fewer people than expected does not automatically produce an underestimate. “Underestimate” specifically means the sample proportion is lower than the population proportion. A biased sample could produce either an underestimate or an overestimate depending on who responds.

In part #2, we have a little reviewing to do. In part (a), you correctly describe the idea of a confidence interval, but a confidence level is a long-run capture rate. A correct version would sound like: “If we were to take many, many random samples of 300 city residents and construct a confidence interval from each sample, about 95% of those intervals would capture the true value of p, the proportion of all city residents who would respond yes to the question.”

For #2 part (b), the conclusion should be no, not yes. Because 0.50 is inside the interval, values below 0.50 are also plausible, so the interval does not provide convincing evidence that the true population proportion is at least 50%. We could only say that if the entire interval were at or above 0.50.

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

Student Response 3

The result may be a biased estimate of the proportion of all city residents who would reply yes to the question because only 450 of those 1000 people replied. This could lead to under coverage of the population and non-response bias: people who do not reply might have different opinions about the plan.
Part a and b:
a. To be 95% confidence in the answer ye of the residents means that if many random samples of size 300 residents are conducted and many confidence intervals are created, then 95% of the interval would contain the true proportion of city residents who would reply “yes” to the question.
b. No. The interval (0.480, 0.593) does not provide convincing statistical evidence that the proportion of all city residents who would say “yes” to the question is at least 50% because 50% is contained in the interval. It means one could get a result of a random sample of 300 city residents that 50% say yes just by random chance alone.
The Success/ Failures must be met for the confidence interval because to make sure the sample is large enough - the sampling distribution of the proportion is approximately Normal.

Teacher Feedback

For part 1, you correctly point to nonresponse bias and note that people who do not reply might have different opinions. That core idea is exactly what needs to be present: the respondents may differ systematically from the nonrespondents, so 0.378 may not represent all residents. You do not always have to force a direction of bias if the scenario does not establish one. If you can reasonably justify overestimate or underestimate, you may include it, but identifying the source of bias and explaining why it could distort the estimate is the key idea here. Also, “undercoverage” is not really the best label for this scenario; the main issue is nonresponse.

Your response in part 2a is strong and shows a clear understanding of what a confidence level represents. In part 2b, you give the correct answer (“no”) with a correct reason (“50% is contained in the interval”), but the last sentence is a little off. A more direct statement would be: “This means that 50% is a plausible value for the proportion of all residents who would say yes, so we do not have statistical evidence that the proportion is greater than or equal to 50%.”

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

Student Response 4

Since this is a survey that was sent out to a 1000 people but only 450 chose to reply, this may lead to a bias where the people who are strongly opinionated about the topic of raising a city tax to provide funding for road repairs, may choose to respond compared to others, who did not respond, may have a different opinion/ response or didn’t feel strongly enough to respond. This will lead to a over representation of the strong opinionated people.
Part a and b
a) using the method, if we sampled repeatedly, 95% of the intervals created would contain the true proportion of people who would reply yes to the survey
b) no because 50 is included in the interval of plausible values of the 95% confidence interval. This means that 50 is a plausible value for the proportion of all adults who would say yes and therefore we do not have statistical evidence that the proportion is greater
The reason of the rule is to ensure that we have a large enough sample and that we have an approximately normal sampling distribution.

Teacher Feedback

In part (a), you clearly identify the possibility of nonresponse bias and explain it well: people who feel more strongly may be more likely to respond, and the nonrespondents may have different opinions. That is the essential reasoning. If you can reasonably justify a likely direction of bias, you can add it, but a direction is not universally required when the situation does not make one clear. On AP Statistics FRQs, precise terminology and a clear explanation of the source of bias matter most.

In parts (b) and (c), you’ve provided correct answers with appropriate context.

Student Response 5

In this scenario, there is a non-response bias because only 450 people out of the 1000 who were mailed the survey chose to respond to the survey. It is reasonable to assume that only people who were very opinionated on the subject chose to reply. People who saw an increase of taxes as a threat may have been more likely to respond than people who saw the increase as a good thing. This results in an underestimation of the true number of individuals who would respond yes to the survey.
Part a and b
a) If the city was to repeat the survey multiple times, about 95% of the intervals would contain the true proportion of residents who would reply yes to the question.
b) This interval does not provide convincing evidence that the true proportion is at least 50% because the interval contains numbers that are less than 50.
This is the large counts condition and is necessary to ensure that there is a approximately normal sampling distribution.

Teacher Feedback

Great work! All three parts are complete. In part (a), you named the source of bias, explained how it might affect who responds, and connected that to the proportion we were trying to estimate. Your direction of bias is also reasonably justified here, though notice that direction is only supportable when you give a believable explanation for why respondents would lean one way more than nonrespondents.

In part (b), you correctly interpret both parts, and in part (c) you give the correct reason for checking that condition.

Student Response 6

This is an example of Voluntary Response Bias. When given the option to respond, the people that are inclined to do it follow through with it; in this context that may be citizens that checked their mail, or citizens with strong opinions over whether or not to raise the city tax, based on their economic status or their history with road conditions or other factors.
Part a and b:
a. 95% of samples of 161 people capture the true proportion of city residents that would reply “yes” to the question.
b. The interval implies that we are 95% confident that the true proportion of city residents that respond yes falls between 0.48 and 0.593. 50% falls within this interval, so this is a likely outcome, albeit not the only possibility because I can be less than 50%.
The success/failure condition indicates that the sampling distribution is approximately Normal.

Teacher Feedback

First, an important terminology correction: this is better described as nonresponse bias, not voluntary response bias. The city first selected a random sample of 1,000 residents, but only 450 responded. Voluntary response bias would occur if people chose themselves to enter the sample, such as responding to an open online poll. Here, the bias comes from selected individuals not responding.

To explain why 0.378 may be biased, identify the source of bias (nonresponse) and explain that people who respond may differ from those who do not respond, so the sample proportion may not represent all residents. If you can reasonably justify a likely direction (overestimate or underestimate), include it, but a direction should not be treated as universally required when the scenario does not establish one.

In part 2, your confidence level interpretation needs adjustment: it should refer to many random samples and the confidence intervals produced from them, not “95% of samples of 161 people.” Also, the sample size in the second survey is 300, not 161; 161 is the number who said yes. In part (b), you reach the right general conclusion that 50% is plausible because it lies in the interval, so we do not have convincing evidence that the true proportion is at least 50%.

Part 3 has the correct rationale for the 10 successes/failures condition.

Final Tips for This Kind of FRQ

For a bias question, start by naming the source of bias correctly. In this prompt, the issue is nonresponse bias.
Then explain why that could affect the estimate: respondents may differ from nonrespondents, so the sample proportion may not represent the whole population.
If the context clearly supports a likely direction of bias, include it. But don’t assume you must always force an overestimate/underestimate claim when the information given doesn’t justify one.
For confidence level, think long-run: many random samples, many confidence intervals, about 95% of those intervals capturing the true proportion.
For evidence from a confidence interval, a claim like “at least 50%” is supported only if the entire interval is at or above 0.50.
For the 10 successes/10 failures condition, the purpose is to justify using an approximately normal sampling distribution for $\hat{p}$ .