Ap Statistics 2021 Free Response Questions Answers

The AP Statistics exam can be a daunting challenge, but mastering the free-response questions (FRQs) is key to achieving a high score. The 2021 AP Statistics exam, administered in various formats due to the pandemic, presented unique challenges. In this comprehensive guide, we will delve into the 2021 AP Statistics FRQs, providing detailed solutions, explanations, and insights into the grading criteria. Understanding these questions and their solutions will not only help you review essential concepts but also prepare you for future exams by understanding the types of questions commonly asked.

Introduction

The AP Statistics exam evaluates your understanding of statistical concepts and your ability to apply them in real-world scenarios. The FRQs are designed to test your problem-solving skills, communication of statistical reasoning, and ability to draw conclusions from data. The 2021 exam was unique because it included multiple administrations to accommodate various testing environments. Each set of FRQs covered different topics and required a deep understanding of statistical principles. Let’s explore these questions in detail.

Question 1: Focus on Probability and Simulation

Context: This question presents a scenario involving a manufacturing process and assesses your ability to apply probability concepts and simulation techniques.

Question:

A company manufactures খেলনা and ships them in boxes. The company claims that 80% of all খেলনা are flawless. Assume that the খেলনা are independent.

(a) Suppose a box contains 20 খেলনা. What is the probability that more than 15 খেলনা in the box are flawless?

(b) Consider randomly selecting 50 boxes, each containing 20 খেলনা. Let X be the number of boxes with more than 15 flawless খেলনা. Would you expect the mean of X to be greater than 10? Justify your answer.

(c) The company is concerned that the manufacturing process may not be producing 80% flawless খেলনা. To investigate, they will select a random sample of 500 খেলনা. Let p̂ be the proportion of flawless খেলনা in the sample. Describe a simulation to estimate the probability that p̂ is less than or equal to 0.75, assuming that the true proportion of flawless খেলনা is 0.80.

(d) Suppose the simulation in part (c) is conducted, and the estimated probability is 0.001. Based on this result, is there convincing statistical evidence, at a significance level of α = 0.05, that the true proportion of flawless খেলনা is less than 0.80? Explain.

Solution:

(a) To find the probability that more than 15 খেলনা are flawless, we need to calculate P(X > 15), where X follows a binomial distribution with n = 20 and p = 0.8. This means we need to find P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20).

The probability mass function for a binomial distribution is: P(X = k) = (n choose k) * p^k * (1-p)^(n-k)

• P(X = 16) = (20 choose 16) * (0.8)^16 * (0.2)^4 ≈ 0.2182
• P(X = 17) = (20 choose 17) * (0.8)^17 * (0.2)^3 ≈ 0.2054
• P(X = 18) = (20 choose 18) * (0.8)^18 * (0.2)^2 ≈ 0.1369
• P(X = 19) = (20 choose 19) * (0.8)^19 * (0.2)^1 ≈ 0.0576
• P(X = 20) = (20 choose 20) * (0.8)^20 * (0.2)^0 ≈ 0.0115

P(X > 15) = 0.2182 + 0.2054 + 0.1369 + 0.0576 + 0.0115 ≈ 0.6296

Therefore, the probability that more than 15 খেলনা in the box are flawless is approximately 0.6296.

(b) Let X be the number of boxes with more than 15 flawless খেলনা. Since we are selecting 50 boxes, and each box has a probability of 0.6296 (from part a) of having more than 15 flawless খেলনা, X follows a binomial distribution with n = 50 and p = 0.6296.

The expected value (mean) of a binomial distribution is E(X) = n * p. E(X) = 50 * 0.6296 ≈ 31.48

Yes, we would expect the mean of X to be greater than 10 because the expected value is approximately 31.48, which is significantly larger than 10.

(c) Simulation to estimate the probability that p̂ is less than or equal to 0.75:

1. Generate Random Samples: For each of the 500 খেলনা, generate a random number between 0 and 1.
2. Simulate Flawless খেলনা: If the random number is less than or equal to 0.80, consider the খেলনা flawless; otherwise, consider it flawed.
3. Calculate Sample Proportion: Calculate the sample proportion p̂ by dividing the number of flawless খেলনা by the total number of খেলনা (500).
4. Repeat: Repeat steps 1-3 a large number of times (e.g., 10,000 times).
5. Estimate Probability: Count the number of times p̂ is less than or equal to 0.75 and divide by the total number of simulations. This gives the estimated probability.

(d) Based on the simulation result of 0.001, we have strong evidence to suggest that the true proportion of flawless খেলনা is less than 0.80. Since the estimated probability (0.001) is less than the significance level (α = 0.05), we reject the null hypothesis that the true proportion is 0.80. The result indicates that observing a sample proportion as low as or lower than 0.75 is very unlikely if the true proportion were indeed 0.80.

Question 2: Focus on Experimental Design

Context: This question assesses your understanding of experimental design principles, including randomization, control groups, and potential confounding variables.

Question:

A researcher wants to investigate whether a new fertilizer increases the yield of tomato plants. The researcher has 20 tomato plants available for the experiment.

(a) Describe a completely randomized design that the researcher could use to compare the yield of tomato plants grown with the new fertilizer to the yield of tomato plants grown without the new fertilizer.

(b) Explain how the design described in part (a) could be improved by using a blocking variable. What would be a reasonable blocking variable in this context?

(c) The researcher implements the blocked design described in part (b). At the end of the growing season, the researcher collects data on the weight of tomatoes produced by each plant. Explain how the researcher could use these data to determine whether the new fertilizer increases the yield of tomato plants.

Solution:

(a) Completely Randomized Design:

1. Random Assignment: Number each of the 20 tomato plants from 1 to 20.
2. Randomization: Use a random number generator to select 10 unique numbers between 1 and 20. These plants will be assigned to the treatment group (new fertilizer). The remaining 10 plants will be assigned to the control group (no new fertilizer).
3. Treatment Application: Apply the new fertilizer to the treatment group according to the manufacturer’s instructions. The control group receives no fertilizer.
4. Data Collection: At the end of the growing season, measure the weight of tomatoes produced by each plant.
5. Comparison: Compare the average yield of the treatment group to the average yield of the control group to determine if there is a significant difference.

(b) Improvement with Blocking:

Blocking involves grouping experimental units (tomato plants in this case) into blocks based on a characteristic that might affect the response variable (tomato yield). This reduces variability within each block, making it easier to detect a treatment effect.

A reasonable blocking variable in this context could be the initial size or height of the tomato plants. Plants of similar size are likely to have similar yields, so blocking by size can reduce variability.

1. Blocking: Divide the 20 tomato plants into 10 pairs based on their initial size (height). Pair plants that are most similar in size.
2. Random Assignment within Blocks: Within each pair, randomly assign one plant to the treatment group (new fertilizer) and the other to the control group (no new fertilizer).
3. Treatment Application: Apply the new fertilizer to the treatment group plants according to the manufacturer’s instructions. The control group plants receive no fertilizer.
4. Data Collection: At the end of the growing season, measure the weight of tomatoes produced by each plant.

By blocking, we ensure that each treatment is applied within similar initial conditions, reducing the noise and making it easier to see the fertilizer’s effect.

(c) Determining the Effect of Fertilizer:

1. Calculate Paired Differences: For each pair of plants (block), calculate the difference in tomato yield between the treatment plant (fertilizer) and the control plant (no fertilizer). This is the treatment effect within each block.
2. Calculate the Mean Difference: Calculate the mean of these paired differences. This is the average treatment effect across all blocks.
3. Statistical Test: Perform a paired t-test (also known as a matched pairs t-test) to determine if the mean difference is significantly different from zero. The null hypothesis is that there is no difference in yield between the fertilizer and no-fertilizer treatments (mean difference = 0). The alternative hypothesis is that there is a difference (mean difference ≠ 0).
4. Interpretation: If the p-value from the paired t-test is less than a predetermined significance level (e.g., α = 0.05), then we reject the null hypothesis and conclude that there is a statistically significant difference in yield between the plants grown with the new fertilizer and those grown without it. If the mean difference is positive, this indicates that the new fertilizer increases the yield of tomato plants.

Question 3: Focus on Inference

Context: This question assesses your ability to conduct and interpret the results of a hypothesis test.

Question:

A local community is concerned about the level of lead in its drinking water. A random sample of 100 homes is selected, and the lead level is measured in each home. The sample mean lead level is 11.5 parts per million (ppm), and the sample standard deviation is 4.0 ppm.

(a) Construct a 95% confidence interval for the mean lead level in the community’s drinking water.

(b) Based on the confidence interval constructed in part (a), is there convincing statistical evidence that the mean lead level in the community’s drinking water is greater than 10 ppm? Explain.

(c) Suppose it is later learned that the lead levels in the sample are not normally distributed, but are heavily skewed to the right. Would the confidence interval constructed in part (a) still be valid? Explain.

Solution:

(a) Constructing a 95% Confidence Interval:

Since the sample size is large (n = 100), we can use the t-distribution to construct a confidence interval for the population mean. The formula for a confidence interval is:

CI = x̄ ± t*(s / √n)

Where:

• x̄ is the sample mean (11.5 ppm)
• s is the sample standard deviation (4.0 ppm)
• n is the sample size (100)
• t* is the critical t-value for a 95% confidence level with n-1 degrees of freedom (df = 99)

For a 95% confidence level and 99 degrees of freedom, the t-value is approximately 1.984.

CI = 11.5 ± 1.984 * (4.0 / √100) CI = 11.5 ± 1.984 * (4.0 / 10) CI = 11.5 ± 1.984 * 0.4 CI = 11.5 ± 0.7936

The 95% confidence interval is (10.7064, 12.2936) ppm.

(b) Evidence that Mean Lead Level is Greater than 10 ppm:

The 95% confidence interval for the mean lead level is (10.7064, 12.2936) ppm. Since the entire interval lies above 10 ppm, there is convincing statistical evidence that the mean lead level in the community’s drinking water is greater than 10 ppm. The interval does not contain 10 ppm, indicating that 10 ppm is an unlikely value for the population mean.

(c) Validity of the Confidence Interval:

Even though the lead levels in the sample are heavily skewed to the right, the Central Limit Theorem (CLT) can still apply because the sample size is large (n = 100). The CLT states that the sampling distribution of the sample mean approaches a normal distribution as the sample size increases, regardless of the shape of the population distribution.

Since the sample size is large enough, the sampling distribution of the sample mean is approximately normal, and the t-confidence interval is still valid. However, it is important to note that the validity of the interval relies on the CLT.

Question 4: Focus on Chi-Square Test

Context: This question examines your ability to apply the chi-square test for independence.

Question:

A survey was conducted to investigate the relationship between education level and employment status. A random sample of adults was classified according to their highest level of education (high school, bachelor’s degree, graduate degree) and their employment status (employed, unemployed). The data are summarized in the table below:

	Employed	Unemployed
High School	200	50
Bachelor’s Degree	300	25
Graduate Degree	150	25

(a) State the null and alternative hypotheses for testing whether there is an association between education level and employment status.

(b) Calculate the expected counts for each cell in the table, assuming the null hypothesis is true.

(c) Calculate the chi-square test statistic.

(d) Determine the degrees of freedom for the test and the p-value. Based on the p-value, is there convincing statistical evidence of an association between education level and employment status at a significance level of α = 0.05? Explain.

Solution:

(a) Hypotheses:

• Null Hypothesis (H0): There is no association between education level and employment status.
• Alternative Hypothesis (Ha): There is an association between education level and employment status.

(b) Expected Counts:

To calculate the expected counts for each cell, use the formula: Expected Count = (Row Total * Column Total) / Grand Total

First, calculate the row and column totals:

	Employed	Unemployed	Total
High School	200	50	250
Bachelor’s Degree	300	25	325
Graduate Degree	150	25	175
Total	650	100	750

Now, calculate the expected counts:

• High School, Employed: (250 * 650) / 750 = 216.67
• High School, Unemployed: (250 * 100) / 750 = 33.33
• Bachelor’s Degree, Employed: (325 * 650) / 750 = 281.67
• Bachelor’s Degree, Unemployed: (325 * 100) / 750 = 43.33
• Graduate Degree, Employed: (175 * 650) / 750 = 151.67
• Graduate Degree, Unemployed: (175 * 100) / 750 = 23.33

Here's the table with expected counts:

	Employed	Unemployed
High School	216.67	33.33
Bachelor’s Degree	281.67	43.33
Graduate Degree	151.67	23.33

(c) Chi-Square Test Statistic:

The chi-square test statistic is calculated as: χ² = Σ [(Observed - Expected)² / Expected]

χ² = [(200 - 216.67)² / 216.67] + [(50 - 33.33)² / 33.33] + [(300 - 281.67)² / 281.67] + [(25 - 43.33)² / 43.33] + [(150 - 151.67)² / 151.67] + [(25 - 23.33)² / 23.33]

χ² ≈ [1.28 + 8.33 + 1.17 + 7.51 + 0.02 + 0.12] ≈ 18.43

(d) Degrees of Freedom and P-Value:

The degrees of freedom (df) for the chi-square test are calculated as: df = (Number of Rows - 1) * (Number of Columns - 1) df = (3 - 1) * (2 - 1) = 2 * 1 = 2

Using a chi-square distribution table or calculator with χ² = 18.43 and df = 2, the p-value is approximately 0.0001.

Since the p-value (0.0001) is less than the significance level (α = 0.05), we reject the null hypothesis. There is convincing statistical evidence of an association between education level and employment status.

Tips & Expert Advice

• Understand Key Concepts: A solid grasp of probability, distributions, hypothesis testing, and experimental design is crucial.
• Practice Regularly: Work through a variety of problems to build your skills and confidence.
• Show Your Work: Always show your steps clearly and logically. Partial credit is often awarded for correct methodology.
• Interpret Results: Be able to explain what your calculations mean in the context of the problem.
• Review Past Exams: Familiarize yourself with the format and types of questions asked on previous AP Statistics exams.
• Use Technology: Become proficient with using a calculator or statistical software to perform calculations and simulations.
• Manage Your Time: Practice pacing yourself so you can complete all FRQs within the allotted time.
• Communicate Clearly: Use precise statistical language and clearly explain your reasoning.

FAQ (Frequently Asked Questions)

Q: What is the best way to prepare for the AP Statistics FRQs? A: Practice with previous FRQs, understand the underlying statistical concepts, and work on communicating your reasoning clearly.

Q: How is the AP Statistics exam graded? A: The exam consists of two sections: multiple choice and free response. Each section is worth 50% of the total score. The FRQs are graded by AP readers based on a rubric that assesses statistical knowledge, problem-solving skills, and communication.

Q: What are common mistakes students make on the AP Statistics exam? A: Common mistakes include misinterpreting the question, not showing work, using incorrect formulas, and failing to provide context in interpretations.

Q: Can I use a calculator on the AP Statistics exam? A: Yes, you are allowed to use a graphing calculator with statistical capabilities on the exam.

Q: How important is it to understand the context of the problem? A: Understanding the context is crucial for interpreting results and communicating statistical reasoning effectively.

Conclusion

The 2021 AP Statistics free-response questions provide valuable insights into the types of problems you can expect on the exam. By thoroughly reviewing these questions, understanding the solutions, and practicing regularly, you can improve your chances of achieving a high score. Remember to focus on understanding key concepts, showing your work, and interpreting your results in context.

How do you plan to incorporate these insights into your study routine? Are you ready to tackle future AP Statistics challenges?