Unit 3 test

AP Statistics Unit 3 Test

15 multiple-choice questions and 1 free-response question, every one from Unit 3: Inference for Categorical Data: Proportions, written in the format of the AP Statistics Exam.

Take the multiple choice on its own, timed at the exam's pace or in practice mode with each answer checked as you give it. Or sit the whole test, free response included, with each section on its own clock; the free-response scoring guide opens when that section ends.

AP Statistics Unit 3 Test

15 multiple-choice and 1 free-response question, a shorter set laid out like the AP Statistics Exam. Each section runs on its own clock and ends when time runs out.

Section	Questions	Time
Section I: Multiple Choice	15 of the exam's 42	32 minutes 9 seconds
Section II: Free Response	1 of the exam's 4	22 minutes 30 seconds

A graphing calculator with statistical capabilities is expected on both sections.

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Section I: Multiple Choice

15 questions · 32 minutes 9 seconds

A county's records show that 44% of the 30,000 property tax bills it sent last year were paid online. Before those records were tallied, a clerk had selected a random sample of 300 of the bills, and 41% of the sampled bills were paid online. A commissioner says the gap shows that the sample proportion from a random sample is a biased estimator of the proportion of all bills paid online. Which response is best?
Answer: D.
1. A
  One estimate that misses is not bias. Bias describes where an estimator's values center over many samples, and any single random sample can land above or below the parameter by chance.
2. B
  Sample size affects how much the estimates vary, not where they center. The sample proportion from a random sample centers on the population proportion at every sample size, so a larger sample would only make the misses smaller.
3. C
  The conclusion is right but the reason is not. Being close on one sample does not make an estimator unbiased, and a biased method could also land close on one try. Unbiased means the estimator's values average out to the parameter across repeated samples.
4. D
  Correct. An estimator is unbiased when, over many random samples, its values neither overestimate nor underestimate the parameter on average. The sample proportion from a random sample has that property. With $n = 300$ and $p = 0.44$ the standard deviation of $\hat{p}$ is about 0.029, so missing by 0.03 is an ordinary result from one sample.
About 12% of the bats in a cave colony of roughly 9,000 bats carry a leg band from an earlier study. Biologists will capture a random sample of 150 bats from the colony and find $\hat{p}$ , the proportion of the sampled bats that carry a band. Over all possible random samples of 150 bats, the values of $\hat{p}$ have mean 0.12 and standard deviation about 0.0265. Which statement correctly interprets the standard deviation?
Answer: A.
1. A
  Correct. The standard deviation of the sampling distribution, $\sqrt{0.12 (0.88) / 150} \approx 0.0265$ , describes how far $\hat{p}$ typically lands from its mean, 0.12, across many random samples of 150 bats.
2. B
  One standard deviation either side of the mean is not a guarantee. Many samples land farther than 0.0265 from 0.12; for a roughly normal distribution only about two thirds land within one standard deviation.
3. C
  The proportion of the colony with a band is a fixed number, 0.12. It is the sample proportion that changes from sample to sample, and the standard deviation describes that variation.
4. D
  This mixes up a count with a proportion. The value 0.0265 is on the proportion scale. The number of banded bats in a sample varies from 18 by about $150 (0.0265) \approx 4$ bats, not by a fraction of a bat.
A public health clinic selected a random sample of 240 of the 5,000 adults on its patient list and found that 66 of them were overdue for a tetanus booster. Assume the conditions for inference hold. Which interval estimates, with 95% confidence, the share of the clinic's adult patients who are overdue for a tetanus booster?
Answer: B.
1. A
  This puts 0.5 in the standard error, $\sqrt{0.5 (0.5) / 240} \approx 0.0323$ . That value is for planning a sample size before any data exist. The interval uses the sample proportion, $\hat{p} = 0.275$ .
2. B
  Correct. $\hat{p} = 66 / 240 = 0.275$ and $SE = \sqrt{0.275 (0.725) / 240} \approx 0.0288$ . The margin of error is $1.96 (0.0288) \approx 0.0565$ , so the interval is $0.275 \pm 0.0565$ , or about $(0.219, 0.331)$ .
3. C
  This uses $z^{*} = 1.645$ , which leaves 5% in one tail. A 95% interval needs the middle 95% of the standard normal distribution, so $z^{*} = 1.96$ , leaving 2.5% in each tail.
4. D
  This uses the standard error, 0.0288, as the margin of error. The margin of error is the critical value times the standard error, $1.96 (0.0288) \approx 0.0565$ .
A weather app selected a random sample of 900 of its users and found that 549 of them have the app's dark mode turned on. From these data, a 95% confidence interval for the proportion of all the app's users who have dark mode turned on is $(0.578, 0.642)$ . Which statement correctly describes what the 95% confidence level means?
Answer: C.
1. A
  This reads the confidence level as the share of results that land inside this one interval. Sample proportions center on the true proportion, which is unknown, not on 0.61, so there is no reason 95% of them would fall between 0.578 and 0.642.
2. B
  Once the interval is computed, the proportion of all users either is in it or is not. The 95% describes the method over many samples, not a probability about this one interval.
3. C
  Correct. The confidence level describes the method: if many random samples of 900 users were taken and a 95% interval built from each in the same way, about 95% of those intervals would contain the proportion of all the app's users who have dark mode on.
4. D
  Each new sample gives a different sample proportion and so a different interval. The intervals change from sample to sample, and the 95% is about how many of them capture the true proportion.
A school district says that 5% of the textbooks students return at the end of the year are too damaged to use again. A board member suspects the proportion is higher this year. From the 38,000 textbooks returned this year, she selects a random sample of 180 and finds that 16 are too damaged to use again. She plans a test of $H_{0} : p = 0.05$ against $H_{a} : p > 0.05$ . Which statement about the normality condition for this test is correct?
Answer: C.
1. A
  Observed counts are the check for a confidence interval. A test is carried out assuming the null hypothesis is true, so its normality condition uses the expected counts under $H_{0}$ .
2. B
  There is no rule of 30 for proportions. The normality condition asks for at least 10 expected successes and 10 expected failures, which depends on $p_{0}$ as well as on $n$ .
3. C
  Correct. For a test the normality condition uses the null value: $n p_{0} = 180 (0.05) = 9$ and $n (1 - p_{0}) = 171$ . Both must be at least 10, and 9 is not, so the condition is not met. The randomization and 10% conditions are fine (180 is far less than 10% of 38,000).
4. D
  How far $\hat{p}$ is from $p_{0}$ is what the test measures; it is not a condition. The conditions are checked before looking at whether the data disagree with $H_{0}$ .
A statewide health survey found that 40% of the state's adults could name at least three warning signs of a stroke. A county health officer wants to know whether the proportion in her county is different. In a random sample of 500 of the county's adults, 222 (44.4%) could name at least three warning signs. With $p$ standing for the proportion of all the county's adults who can name at least three warning signs, she tests $H_{0} : p = 0.40$ against $H_{a} : p \neq 0.40$ and gets $z \approx 2.01$ and a p-value of about 0.045. Which statement correctly interprets the p-value?
Answer: A.
1. A
  Correct. The p-value is computed assuming $H_{0}$ is true, that $p = 0.40$ . Because $H_{a}$ is two-sided, it counts results as extreme as 0.444 in both directions: sample proportions of 0.444 or more and of 0.356 or less. Together they have probability about 0.045.
2. B
  This describes only one tail. The alternative is two-sided, so the p-value counts results this far from 0.40 in either direction. The upper tail alone has probability about 0.022.
3. C
  A p-value is not the probability that the null hypothesis is true. It is computed by assuming the null hypothesis is true and asking how unusual the sample result would be.
4. D
  A p-value is computed assuming the null hypothesis is true, not the alternative. If the proportion were something other than 0.40, the probability of such a result would depend on what that value was.
Last year, 72% of the seniors in a large school district completed the federal financial aid application. This year the district's counselors ran evening workshops on the form. A counselor selected a random sample of 300 of the district's 4,100 seniors and found that 234 (78%) had completed it. With the conditions for inference met, the counselor tests $H_{0} : p = 0.72$ against $H_{a} : p > 0.72$ , with $p$ standing for the proportion of this year's seniors in the district who completed the form. The test's p-value is 0.010. Using $α = 0.05$ , which conclusion is correct?
Answer: D.
1. A
  Rejecting the null hypothesis does not prove the alternative or show that the null is false. A small p-value gives convincing evidence, and a Type I error is still possible.
2. B
  This reverses the decision rule. The p-value of 0.010 is less than $α = 0.05$ , so the counselor rejects $H_{0}$ .
3. C
  The test is about the proportion of seniors this year, and the data come from a sample, not an experiment. Nothing was randomly assigned, so the test cannot say the workshops caused the change; other things may have changed since last year too.
4. D
  Correct. Because $0.010 < 0.05$ , reject $H_{0}$ . There is convincing statistical evidence that the proportion of all this year's seniors in the district who completed the financial aid form is greater than 0.72.
On a meditation app, 25% of new users are still active one month after they sign up. The app's designers rewrote the first week of sessions. Letting $p$ be the share of all new users who would still be active after a month with the new sessions, they will test $H_{0} : p = 0.25$ against $H_{a} : p > 0.25$ with $α = 0.05$ , using a random sample of 340 new users. If the true value is $p = 0.31$ , the power of the test is about 0.80. If $p = 0.25$ , how likely is the test to make a Type I error, and if $p = 0.31$ , how likely is it to make a Type II error?
Answer: B.
1. A
  The significance level is the probability of a Type I error only. A Type II error happens when $H_{0}$ is false, and its probability is 1 minus the power.
2. B
  Correct. The probability of a Type I error, rejecting $H_{0}$ when $p = 0.25$ , is $α = 0.05$ . The probability of a Type II error when $p = 0.31$ is $1 - power = 1 - 0.80 = 0.20$ .
3. C
  This is the power, the probability of correctly rejecting a false null hypothesis. A Type II error is failing to reject it, so its probability is 1 minus the power, 0.20.
4. D
  The two errors are swapped. A Type I error rejects a true null hypothesis, and its probability is $α = 0.05$ ; the 0.20 belongs to the Type II error.
Biologists know that 30% of the walleye in Arden Lake and 18% of the walleye in Birch Lake carry a certain parasite. Each lake holds many thousands of walleye. The biologists will take independent random samples of 120 walleye from Arden Lake and 150 from Birch Lake and compute ${\hat{p}}_{A} - {\hat{p}}_{B}$ . Across all possible pairs of samples, what are the mean and the standard deviation of the values of ${\hat{p}}_{A} - {\hat{p}}_{B}$ ?
Answer: C.
1. A
  This is the variance, $0.00175 + 0.00098 \approx 0.0027$ . The standard deviation is its square root.
2. B
  This subtracts the variances, $\sqrt{0.00175 - 0.00098}$ . The variability of a difference of independent sample proportions comes from both samples, so the variances add.
3. C
  Correct. The mean is $0.30 - 0.18 = 0.12$ , and the standard deviation is $\sqrt{0.30 (0.70) / 120 + 0.18 (0.82) / 150} = \sqrt{0.00175 + 0.00098} \approx 0.0523$ .
4. D
  This adds the two standard deviations, $0.0418 + 0.0314$ . Variances add for independent samples, and standard deviations do not, so take the square root of the sum of the variances.
A school district wants to estimate how much the proportion of its ninth graders who attended at least one school dance this year differs from the proportion of its twelfth graders who did. It selects a random sample of 150 ninth graders and a separate random sample of 150 twelfth graders from its schools. Which procedure and parameter fit the district's goal?
Answer: D.
1. A
  Combining the two grades into one proportion loses the comparison the district wants. The goal is the difference between two population proportions.
2. B
  A test decides whether the proportions differ; it does not say by how much. The district wants an estimate of the difference, which calls for a confidence interval.
3. C
  The sample proportions are known exactly once the data are in, so there is nothing to estimate about them. The interval estimates the difference between the population proportions.
4. D
  Correct. Two independent random samples and a goal of estimating a difference call for a two-sample $z$ -interval for $p_{9} - p_{12}$ , the difference between the proportions of all the district's ninth graders and all its twelfth graders who attended at least one dance.
A conservation group placed 300 nest boxes along a woodland trail and randomly assigned 150 to a new design and 150 to the standard design. By the end of spring, 87 of the new boxes and 78 of the standard boxes had been used by nesting birds. Let $p_{1}$ and $p_{2}$ be the proportions of boxes like these that would be used with the new and with the standard design. From the data, the group built a 95% confidence interval, $(- 0.052, 0.172)$ , for the difference $p_{1} - p_{2}$ . The group's newsletter wants to report that the new design is used more often. Which statement is best?
Answer: A.
1. A
  Correct. The interval contains 0, so no difference between the designs is a plausible value. The data do not give convincing evidence that boxes like these are used more often with the new design, even though the new boxes were used more often in this study.
2. B
  The share of the interval that is positive does not matter. As long as 0 is inside the interval, a difference of zero is plausible, and the claim is not supported.
3. C
  A plausible value is not a proven one. Values as large as 0.172 are also in the interval, so the interval cannot show the designs are the same; it only fails to show they differ.
4. D
  Random assignment of the two designs is what lets the group compare them, and the conclusion applies to boxes like these. Random selection would matter only for extending it to all nest boxes, which the claim does not need.
A county elections office wants to know whether the proportion of mail ballots rejected for a signature problem differs between first-time voters and returning voters. It selects a random sample of 400 mail ballots from first-time voters and a separate random sample of 400 from returning voters. In the samples, 4.5% of the first-time voters' ballots and 2.0% of the returning voters' ballots were rejected for a signature problem. Let $p_{F}$ and $p_{R}$ be the proportions of all mail ballots from first-time and from returning voters that are rejected for a signature problem. Which pair of hypotheses should the office test?
Answer: B.
1. A
  Hypotheses are about population proportions. The sample proportions, 0.045 and 0.020, are already known, so there is nothing to test about them.
2. B
  Correct. The null hypothesis is no difference, $p_{F} = p_{R}$ , and the office asked whether the proportions differ, with no direction in mind, so the alternative is two-sided, $p_{F} \neq p_{R}$ .
3. C
  The direction of the alternative comes from the question asked before the data, not from which sample proportion turned out larger. The office asked whether the proportions differ, so the alternative is two-sided.
4. D
  These are reversed. The null hypothesis states no difference, the claim assumed true unless the data give convincing evidence against it, and the alternative holds the claim being investigated.
A county health department randomly assigned 415 adults who called its quit-smoking line to one of two programs: phone coaching alone, or phone coaching plus daily text messages. Six months later, 62 of the 210 adults given coaching plus texts and 44 of the 205 given coaching alone said they had not smoked in the past 30 days. For a test of $H_{0} : p_{T} = p_{C}$ against $H_{a} : p_{T} > p_{C}$ , where $p_{T}$ and $p_{C}$ are the proportions of callers like these who would stop smoking with coaching plus texts and with coaching alone, which gives the test statistic and p-value?
Answer: A.
1. A
  Correct. ${\hat{p}}_{T} = 62 / 210 \approx 0.2952$ , ${\hat{p}}_{C} = 44 / 205 \approx 0.2146$ and the combined proportion is ${\hat{p}}_{c} = 106 / 415 \approx 0.2554$ . Then $z = 0.0806 / \sqrt{0.2554 (0.7446) (1 / 210 + 1 / 205)} \approx 0.0806 / 0.0428 \approx 1.88$ , and the area above 1.88 under the standard normal curve is about 0.030.
2. B
  This doubles the tail area, which is the p-value for a two-sided alternative. Here the alternative is one-sided, $p_{T} > p_{C}$ , so the p-value is the area above 1.88 only.
3. C
  This is the area below $z = 1.88$ . The alternative says $p_{T} > p_{C}$ , so results in the upper tail are the evidence, and the p-value is the area above 1.88.
4. D
  This uses one sample size for the whole study, $\sqrt{0.2554 (0.7446) / 415} \approx 0.0214$ . The standard error for a difference needs $1 / 210 + 1 / 205$ , which makes it twice as large.
A recipe app selected a random sample of 1,000 of its users and recorded each user's household size (1 person, 2 people, or 3 or more) and the kind of recipe they save most often (quick weeknight meals, desserts, vegetarian dishes, or slow-cooker meals). The app wants to know whether household size and kind of recipe are related among its users. Which hypotheses should the app's analyst state?
Answer: D.
1. A
  These are reversed. The null hypothesis of a chi-square test for independence is no association, and the alternative is the association being looked for.
2. B
  Hypotheses describe the population, all of the app's users. Whether the two variables are related in the 1,000 sampled users can be seen directly from the table, with no test needed.
3. C
  This looks at only one kind of recipe, desserts. The question is about the whole distribution of kind of recipe across household sizes, so the hypotheses involve all four kinds at once.
4. D
  Correct. One random sample classified by two categorical variables calls for a chi-square test for independence. The null hypothesis is that household size and kind of recipe are independent among all the app's users, and the alternative is that they are associated.
A city's civic engagement office selected independent random samples of adult residents from three of its wards and asked each one whether they could name their own city council member. The results are shown.
$\begin{matrix} Could name & Could not name & Total \\ Ward 1 & 46 & 74 & 120 \\ Ward 2 & 72 & 78 & 150 \\ Ward 3 & 37 & 93 & 130 \\ Total & 155 & 245 & 400 \end{matrix}$
The conditions for inference are met. The office tests whether the distribution of responses is the same in the three wards. Which gives the test statistic and p-value?
Answer: B.
1. A
  This adds $| observed - expected | / expected$ without squaring the differences. Each term of the chi-square statistic is $(observed - expected)^{2} / expected$ .
2. B
  Correct. The expected counts are row total times column total over 400: 46.5 and 73.5 for Ward 1, 58.13 and 91.88 for Ward 2, 50.38 and 79.63 for Ward 3. Summing $(observed - expected)^{2} / expected$ over the six cells gives $χ^{2} \approx 11.21$ , and with $df = (3 - 1) (2 - 1) = 2$ the area to the right of 11.21 is about 0.004.
3. C
  This uses 5 degrees of freedom, the number of cells minus 1. For a two-way table the degrees of freedom are $(3 - 1) (2 - 1) = 2$ .
4. D
  This is the area to the left of $χ^{2} \approx 11.21$ . Large values of the statistic are the evidence against $H_{0}$ , so the p-value is the area to the right.

Section II: Free Response

1 question · 22 minutes 30 seconds

Question 3: Inference (Hypothesis Test or Confidence Interval) 10 points

A city sends text messages about severe weather, flooding and road closures to residents who sign up for its emergency alerts. Two years ago, 35% of the city's adult residents were signed up. Since then the city has run a sign-up drive at libraries, schools and community events, and the city's emergency manager believes that more than 35% of adult residents are now signed up. The city has about 84,000 adult residents. An analyst randomly selected 450 of them and found that 179 were signed up for the alerts. The results are shown in the table.

Alert sign-ups among 450 randomly selected adult residents
Signed up for alerts Number of adults

Yes 179
No 271
Total 450

At the 0.05 level of significance, complete the appropriate inference procedure to determine whether there is convincing statistical evidence that more than 35% of the city's adult residents are now signed up for the emergency alerts.

Use the given information to respond to parts A, B, C, and D. Label any subparts (e.g., i and ii) that may be present.
1. A.
  
  i. Identify the appropriate inference procedure.
  
  ii. Define the parameter of interest in context.
  Question 3, part A: your response. It stays in this browser tab and is never sent anywhere.
  Score part A
  - 1 point Identifies the procedure as a one-sample $z$ -test for a population proportion. A correct formula, $z = \frac{\hat{p} - p_{0}}{\sqrt{p_{0} (1 - p_{0}) / n}}$ , also earns the point.Earns it: "one-sample $z$ -test for a proportion"; " $z = \frac{\hat{p} - 0.35}{\sqrt{0.35 (0.65) / 450}}$ "Does not: "one-sample $z$ -interval for a proportion" (the question asks for a decision about a claim, which calls for a test); " $t$ -test" (the response variable is categorical, signed up or not, so the procedure is for a proportion); "two-sample $z$ -test" (there is one sample from one population; 35% is a value from two years ago, not a second sample)
  - 1 point Defines $p$ as the proportion of all the city's adult residents who are now signed up for the emergency alerts. The definition must name the proportion, the response (signed up for the alerts) and the population (the city's adult residents now).Earns it: " $p$ = the proportion of all adult residents of the city who are now signed up for the emergency text alerts"Does not: " $p$ = the proportion of the 450 adults who are signed up" (that is the sample proportion, $\hat{p}$ ); " $p$ = the proportion signed up" (the population is not named); " $p$ = 179" (a count from the sample, not a parameter)
  Model response
  i. A one-sample $z$ -test for a population proportion.
  
  ii. Let $p$ be the proportion of all the city's adult residents who are now signed up for the emergency text alerts.
2. B.
  
  Complete the inference procedure, including calculating the appropriate statistics.
  Question 3, part B: your response. It stays in this browser tab and is never sent anywhere.
  Score part B
  - 1 point Hypotheses: states $H_{0} : p = 0.35$ and $H_{a} : p > 0.35$ , with $p$ as defined in part A or here.Earns it: " $H_{0} : p = 0.35$ , $H_{a} : p > 0.35$ "Does not: " $H_{0} : \hat{p} = 0.35$ , $H_{a} : \hat{p} > 0.35$ " (hypotheses are about the population proportion, not the sample proportion); " $H_{a} : p \neq 0.35$ " (the emergency manager believes the proportion is greater, so the alternative is one-sided); " $H_{0} : p = 0.398$ " (the null value is the claim being tested, 0.35, not the sample result)
  - 1 point Randomization and 10% conditions: states that the 450 adults were randomly selected, and that 450 is less than 10% of the city's 84,000 adult residents (450 is at most 8,400). Both are required.Earns it: "Random: the analyst randomly selected the 450 adults. 10%: 450 is less than 10% of 84,000 = 8,400."Does not: "Random: met. 10%: met." (no justification for either); "The adults were randomly assigned." (this is a sample survey; nothing was assigned); "450 is at least 10% of the population." (the condition is that the sample is at most 10% of the population)
  - 1 point Normality condition: checks the expected numbers of successes and failures using the null value, $n p_{0} = 450 (0.35) = 157.5$ and $n (1 - p_{0}) = 450 (0.65) = 292.5$ , both at least 10. A check that uses the observed counts, 179 and 271, does not earn the point, because the test assumes $H_{0}$ is true.Earns it: " $450 (0.35) = 157.5 \geq 10$ and $450 (0.65) = 292.5 \geq 10$ , so the sampling distribution of $\hat{p}$ is approximately normal."Does not: "179 and 271 are both at least 10." (observed counts are the check for a confidence interval; a test uses $p_{0}$ ); " $n = 450 \geq 30$ " (not the condition for a proportion); "Large counts: met." (no values shown)
  - 1 point Test statistic: reports $\hat{p} = 179 / 450 \approx 0.398$ and $z \approx 2.12$ (2.13 or 2.14 from a rounded $\hat{p}$ ), computed with the standard error $\sqrt{0.35 (0.65) / 450} \approx 0.0225$ . An arithmetic or transcription error can be ignored if correct work is shown.Earns it: " $\hat{p} = 0.398$ , $z = \frac{0.398 - 0.35}{\sqrt{0.35 (0.65) / 450}} \approx 2.13$ "Does not: " $z = \frac{0.398 - 0.35}{\sqrt{0.398 (0.602) / 450}} \approx 2.07$ " (the standard error for a test uses $p_{0}$ , not $\hat{p}$ ); " $z = 2.12$ " with no work and no sample proportion (the value cannot be checked)
  - 1 point p-value: reports a p-value of about 0.017, the area above the test statistic, consistent with the value found. From a table, a p-value between 0.0162 and 0.0170 also earns the point.Earns it: "p-value $= P (z \geq 2.12) \approx 0.017$ "Does not: "p-value = 0.034" (doubles the tail area; the alternative is one-sided); "p-value = 0.983" (the area below the test statistic)
  Model response
  $H_{0} : p = 0.35$ and $H_{a} : p > 0.35$ , where $p$ is the proportion of all the city's adult residents who are now signed up for the emergency text alerts.
  
  The randomization condition is met because the analyst randomly selected the 450 adults. The 10% condition is met because 450 is less than 10% of the city's 84,000 adult residents, which is 8,400. The normality condition is met because, assuming $H_{0}$ is true, the expected numbers are $450 (0.35) = 157.5$ adults signed up and $450 (0.65) = 292.5$ not signed up, both at least 10.
  
  $\hat{p} = 179 / 450 \approx 0.398$ , and
  $z = \frac{0.398 - 0.35}{\sqrt{\frac{0.35 (0.65)}{450}}} \approx 2.12.$
  The p-value is $P (z \geq 2.12) \approx 0.017$ .
3. C.
  
  Justify a conclusion in context.
  Question 3, part C: your response. It stays in this browser tab and is never sent anywhere.
  Score part C
  - 1 point Compares the p-value with the significance level and makes a formal decision about the null hypothesis: because $0.017 < 0.05$ , reject $H_{0}$ . The p-value used must be the one from part B.Earns it: "The p-value of 0.017 is less than $α = 0.05$ , so reject $H_{0}$ ."Does not: "p is small, so reject." (the values are not compared); "0.017 < 0.05, so $H_{0}$ is false." (a test gives evidence against $H_{0}$ ; it does not prove it false, and no formal decision is stated)
  - 1 point States that there is convincing statistical evidence that the proportion of all the city's adult residents who are now signed up for the alerts is greater than 0.35. The conclusion must be in terms of the alternative hypothesis, use non-definitive language, and refer to the population.Earns it: "There is convincing statistical evidence that more than 35% of all the city's adult residents are now signed up for the emergency alerts."Does not: "The sign-up drive raised the proportion above 35%." (no random assignment, so the test cannot say what caused the change, and the wording is definitive); "There is convincing evidence that more than 35% of the 450 adults are signed up." (the sample proportion is known; the conclusion is about all adult residents); "This proves that 39.8% of adults are signed up." (definitive, and the sample proportion is an estimate)
  Model response
  Because the p-value of 0.017 is less than $α = 0.05$ , reject $H_{0}$ . There is convincing statistical evidence that the proportion of all the city's adult residents who are now signed up for the emergency text alerts is greater than 0.35.
4. D.
  
  Based on your conclusion in part C, identify the type of error that could have been made. Describe that error in context, and describe one consequence it could have for the city.
  Question 3, part D: your response. It stays in this browser tab and is never sent anywhere.
  Score part D
  - 1 point Identifies a Type I error, describes it in context (concluding that more than 35% of the city's adult residents are signed up when the true proportion is 35%), and gives a consequence that follows from that mistake, such as the city scaling back or ending the sign-up drive while the proportion has not actually risen. The type of error must be consistent with the decision in part C, and all three elements are required.Earns it: "Type I error: deciding that more than 35% of adults are now signed up when really only 35% are. The city might stop the sign-up drive too early, and fewer residents would get warnings about floods and severe weather than the city thinks."Does not: "Type II error: failing to find that more than 35% are signed up." (the null hypothesis was rejected, so a Type II error is not possible); "Type I error: rejecting a true null hypothesis. The city could waste money." (not described in context, and the consequence does not follow from this error); "The analyst might have counted some adults twice." (a data-collection mistake, not a Type I or Type II error)
  Model response
  Because $H_{0}$ was rejected, a Type I error could have been made. That would mean concluding that more than 35% of the city's adult residents are now signed up for the alerts when, in fact, the proportion is still 35%. One consequence: believing the drive has worked, the city might end it, so fewer residents would be signed up to get warnings about floods and severe weather than city officials think.
Your score: 0 of 10, from the boxes you ticked.

Alert sign-ups among 450 randomly selected adult residents
Signed up for alerts	Number of adults
Yes	179
No	271
Total	450

AP Statistics Unit 3 Test

Section I: Multiple Choice

Section II: Free Response

Question 3: Inference (Hypothesis Test or Confidence Interval) 10 points

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