AP · thinka-original Practice Paper

2024 AP AP Statistics Practice Paper with Answers

Thinka May 2024 AP-Style Mock — AP Statistics

24 marks90 mins2024

An original Thinka practice paper modelled on the structure and difficulty of the May 2024 AP AP Statistics paper. Not affiliated with or reproduced from AP.

Section II, Part A

Answer Questions 1 through 5. Show all your work. Indicate clearly the methods you use, because you will be scored on the correctness of your methods as well as on the accuracy and completeness of your results and explanations.

5 Question · 20 marks

Question 1 · Free Response

4 marks

A city transportation department is considering adding sheltered bicycle facilities at two major commuter rail stations: North Station and Central Station. The department wants to investigate whether the proportion of all daily commuters who regularly cycle to the station differs between the two stations. An independent random sample of 150 daily commuters at North Station and an independent random sample of 200 daily commuters at Central Station were surveyed.

The survey revealed that 36 of the 150 sampled commuters at North Station and 68 of the 200 sampled commuters at Central Station regularly cycle to the station.

At a significance level of $\alpha = 0.05$, do the data provide convincing statistical evidence of a difference in the proportion of all daily commuters at North Station who regularly cycle to the station and the proportion of all daily commuters at Central Station who regularly cycle to the station? Complete the appropriate inference procedure to justify your response.

Show answer & marking scheme

Worked solution

### Step 1: State Hypotheses and Identify Procedure
Let $p_{\text{N}}$ represent the true proportion of all daily commuters at North Station who regularly cycle to the station, and let $p_{\text{C}}$ represent the true proportion of all daily commuters at Central Station who regularly cycle to the station.

The hypotheses to be tested are:
\[ H_0: p_{\text{N}} - p_{\text{C}} = 0 \quad \text{(or } p_{\text{N}} = p_{\text{C}}\text{)} \]
\[ H_{\text{a}}: p_{\text{N}} - p_{\text{C}} \neq 0 \quad \text{(or } p_{\text{N}} \neq p_{\text{C}}\text{)} \]

The appropriate inference procedure is a **two-sample $z$-test for a difference in population proportions**.

---

### Step 2: Check Conditions and Calculate Test Statistic
1. Random Condition: We are given that independent random samples of commuters were selected from North Station and Central Station.
2. 10% Condition (Independence):
- Sample size at North Station is $n_{\text{N}} = 150$, and it is reasonable to assume there are more than $10(150) = 1{,}500$ daily commuters at North Station.
- Sample size at Central Station is $n_{\text{C}} = 200$, and it is reasonable to assume there are more than $10(200) = 2{,}000$ daily commuters at Central Station.
3. Large Counts Condition (Normality):
- Pooled sample proportion:
\[ \hat{p}_c = \frac{36 + 68}{150 + 200} = \frac{104}{350} \approx 0.2971 \]
- Expected counts:
- $n_{\text{N}}\hat{p}_c = 150(0.2971) \approx 44.57 \ge 10$
- $n_{\text{N}}(1 - \hat{p}_c) = 150(0.7029) \approx 105.43 \ge 10$
- $n_{\text{C}}\hat{p}_c = 200(0.2971) \approx 59.43 \ge 10$
- $n_{\text{C}}(1 - \hat{p}_c) = 200(0.7029) \approx 140.57 \ge 10$
- (Alternatively, observed counts: 36, 114, 68, 132 are all at least 10.)
Because all expected counts are at least 10, the sampling distribution of $\hat{p}_{\text{N}} - \hat{p}_{\text{C}}$ is approximately normal.

Calculations:
- Sample proportions:
\[ \hat{p}_{\text{N}} = \frac{36}{150} = 0.24, \quad \hat{p}_{\text{C}} = \frac{68}{200} = 0.34 \]
- Standard error of the difference:
\[ \text{SE}_{\text{pooled}} = \sqrt{0.2971(1 - 0.2971)\left(\frac{1}{150} + \frac{1}{200}\right)} \approx 0.04936 \]
- Test statistic:
\[ z = \frac{\hat{p}_{\text{N}} - \hat{p}_{\text{C}}}{\text{SE}_{\text{pooled}}} = \frac{0.24 - 0.34}{0.04936} \approx -2.026 \approx -2.03 \]
- $p$-value:
\[ p\text{-value} = 2 \cdot P(Z \le -2.026) \approx 0.0428 \quad \text{(or } 0.0424 \text{ using Table A with } z = -2.03\text{)} \]

---

### Step 3: Conclusion
Because the $p$-value $(\approx 0.0428)$ is less than the significance level $\alpha = 0.05$, we reject the null hypothesis $H_0$.

There is convincing statistical evidence that the proportion of all daily commuters at North Station who regularly cycle to the station is different from the proportion of all daily commuters at Central Station who regularly cycle to the station.

Marking scheme

Scored in three sections (Section 1, Section 2, Section 3):

Section 1: Hypotheses & Procedure Identification
- Essentially Correct (E) if the response:
1. Identifies the two-sample $z$-test for a difference in proportions by name or formula.
2. States correct null and two-sided alternative hypotheses with proper parameters.
3. Defines parameters with sufficient context (referencing both commuter populations and cycling behavior).
- Partially Correct (P) if 2 of the 3 components are met.
- Incorrect (I) otherwise.

Section 2: Verification of Conditions & Calculation
- Essentially Correct (E) if the response:
1. Verifies random sampling and the 10% condition for both independent samples.
2. Verifies large counts condition showing expected counts (or observed counts) are $\ge 10$.
3. Reports the correct test statistic value $z \approx -2.03$ (or $+2.03$).
4. Reports the correct $p$-value $(\approx 0.0424 \text{ to } 0.0428)$ consistent with the test statistic.
- Partially Correct (P) if 2 or 3 of the 4 components are met.
- Incorrect (I) otherwise.

Section 3: Decision & Conclusion
- Essentially Correct (E) if the response:
1. Compares the $p$-value to $\alpha = 0.05$ and states an appropriate decision (reject $H_0$).
2. States a correct conclusion in context, in terms of the alternative hypothesis, using non-definitive language.
- Partially Correct (P) if only 1 component is met.
- Incorrect (I) otherwise.

Final Score Conversion:
- 4 points: EEE
- 3 points: EEP
- 2 points: EEI, EPP, or PPP
- 1 point: EPI, PPI, or EII
- 0 points: PII or III

Question 2 · Exploration

4 marks

A regional environmental education center offers three types of weekend workshops for adults: Plant Identification, Bird Watching, and Nature Photography. The center tracks participants at two locations: Valley Park and Ridge Reserve.

At Valley Park, the relative frequencies of participants in the three workshops were $0.50$ for Plant Identification, $0.30$ for Bird Watching, and $0.20$ for Nature Photography.

At Ridge Reserve, the total number of participants was four times the total number of participants at Valley Park. For Ridge Reserve, the proportion of participants was equal for all three workshop types.

(a) Determine the relative frequency of participants for each of the three workshops at Ridge Reserve.

(b) A coordinator at Valley Park claims that Valley Park had more participants in the Plant Identification workshop than Ridge Reserve did because the relative frequency of Plant Identification at Valley Park ($0.50$) is greater than the relative frequency of Plant Identification at Ridge Reserve ($\frac{1}{3} \approx 0.333$). Is the coordinator's claim correct? Explain your response.

(c) Two other nature reserves in the county, Lakeview and Mountain Crest, also offer these three workshops. The distribution of participants across the three workshops at these two reserves is shown in the table below.

$$\begin{array}{|l|c|c|c|c|}\hline\text{Reserve} & \text{Plant Identification} & \text{Bird Watching} & \text{Nature Photography} & \text{Total Participants} \\hline\text{Lakeview} & 120 & 60 & 20 & 200 \\hline\text{Mountain Crest} & 300 & 350 & 150 & 800 \\hline\end{array}$$

(i) Which of the two reserves sold a greater proportion of registrations for the Plant Identification workshop? Justify your answer.

(ii) In a mosaic plot of the data for Lakeview and Mountain Crest (where the column width for each reserve is proportional to its total number of participants and each column is segmented by relative frequency), which reserve has a larger rectangular area representing participants in the Plant Identification workshop? Justify your answer.

Show answer & marking scheme

Worked solution

Part (a):
Because the proportion of participants at Ridge Reserve was equal for all three workshop types, the relative frequency for each workshop is:
$$\text{Relative Frequency} = \frac{1}{3} \approx 0.3333$$
- Plant Identification: $\frac{1}{3} \approx 0.333$
- Bird Watching: $\frac{1}{3} \approx 0.333$
- Nature Photography: $\frac{1}{3} \approx 0.333$

Part (b):
No, the coordinator's claim is incorrect.
Let $x$ represent the total number of participants at Valley Park. The number of participants in Plant Identification at Valley Park is $0.50x$.
Since Ridge Reserve had 4 times as many total participants, its total number of participants is $4x$. The number of participants in Plant Identification at Ridge Reserve is:
$$\frac{1}{3}(4x) = \frac{4}{3}x \approx 1.333x$$
Because $\frac{4}{3}x > 0.50x$ (or $1.333x > 0.50x$) for any positive $x > 0$, Ridge Reserve had more participants in the Plant Identification workshop than Valley Park did.

*(Example: If Valley Park had $100$ total participants, it had $0.50(100) = 50$ participants in Plant Identification. Ridge Reserve had $4(100) = 400$ total participants and $\frac{1}{3}(400) \approx 133.3$ participants in Plant Identification. Since $133.3 > 50$, Ridge Reserve had more participants.)*

Part (c):
(i)
- Proportion at Lakeview: $\frac{120}{200} = 0.60$
- Proportion at Mountain Crest: $\frac{300}{800} = 0.375$

Lakeview had a greater proportion of participants in the Plant Identification workshop because $0.60 > 0.375$.

(ii)
Mountain Crest has the larger rectangular area representing participants in Plant Identification.
In a mosaic plot, the area of a segmented rectangle is proportional to the joint frequency (the actual number of individuals in that category). Mountain Crest had $300$ participants in Plant Identification, whereas Lakeview had only $120$ participants (or the area fraction for Mountain Crest is $0.375 \times \frac{800}{1000} = 0.30$, compared to $0.60 \times \frac{200}{1000} = 0.12$ for Lakeview). Because $300 > 120$ (or $0.30 > 0.12$), Mountain Crest has a larger area.

Marking scheme

Scoring Guidelines:

Each part—(a), (b), and (c)—is scored as Essentially Correct (E), Partially Correct (P), or Incorrect (I).

---

Part (a) is scored as:
- Essentially Correct (E) if the response correctly indicates that the relative frequency for each workshop is $\frac{1}{3}$ (or approximately $0.333$ or $33.3\%$).
- Partially Correct (P) if the response states that the three values are equal but gives an incorrect value, or correctly finds $\frac{1}{3}$ for at least one workshop but makes an arithmetic error.
- Incorrect (I) if the response does not meet the criteria for E or P.

---

Part (b) is scored as:
- Essentially Correct (E) if the response satisfies the following three components:
1. States that the coordinator's claim is incorrect (or "No").
2. Provides valid mathematical support showing that Ridge Reserve had more participants (e.g., comparing $0.50x$ to $\frac{4}{3}x$, or providing a valid numerical example).
3. Includes context (references the reserves/parks and workshop/participants).
- Partially Correct (P) if the response satisfies two of the three components.
- Incorrect (I) if the response does not meet the criteria for E or P.

---

Part (c) is scored as:
- Essentially Correct (E) if the response satisfies all four of the following components:
1. Correctly selects Lakeview in (c-i).
2. Justifies (c-i) by comparing the correct proportions ($0.60$ and $0.375$, or $\frac{120}{200}$ and $\frac{300}{800}$).
3. Correctly selects Mountain Crest in (c-ii).
4. Justifies (c-ii) by noting that area in a mosaic plot corresponds to the number/count of participants ($300 > 120$) or by calculating and comparing the relative areas ($0.30 > 0.12$).
- Partially Correct (P) if the response satisfies two or three of the four components.
- Incorrect (I) if the response satisfies fewer than two components.

---

Final Score:
- 4 Points (Complete Response): 3 parts essentially correct (EEE)
- 3 Points (Substantial Response): 2 parts essentially correct and 1 part partially correct (EEP)
- 2 Points (Developing Response): 2 parts essentially correct and 0 parts partially correct (EEI), OR 1 part essentially correct and 1-2 parts partially correct (EPI, EPP), OR 3 parts partially correct (PPP)
- 1 Point (Minimal Response): 1 part essentially correct and 0 parts partially correct (EII), OR 0 parts essentially correct and 2 parts partially correct (PPI)

Question 3 · Data Collection and Experimental Design

4 marks

An agricultural research team is investigating the effects of a new biochar soil supplement on the biomass yield of bell pepper plants. Higher biomass yield is considered desirable.

(a) In an initial investigation, the researchers visited 40 local farms and recorded the biomass yield of bell pepper crops from farms that chose to use the biochar supplement and from farms that did not use the supplement. Is this an observational study or an experiment? Justify your answer in context.

(b) The researchers decide to conduct a formal experiment using 60 potted bell pepper seedlings of the same variety and age grown in a greenhouse. They will compare two treatments: soil enriched with biochar and standard soil without biochar. Describe an appropriate method the researchers could use to randomly assign 30 seedlings to the biochar treatment and 30 seedlings to the standard soil treatment in a completely randomized design.

(c) After 10 weeks, the mean biomass yield of the plants in the biochar group was found to be statistically significantly higher than that of the plants in the standard soil group. The researchers want to conclude that adding biochar will increase the mean biomass yield of all varieties of bell pepper plants grown on commercial farms nationwide. Explain why this conclusion is not appropriate based on the design of the study, and state what change to the study would be necessary to allow generalization to all commercial bell pepper crops nationwide.

Show answer & marking scheme

Worked solution

(a) This is an observational study. The researchers did not impose any treatment on the farms; they simply observed and recorded the biomass yields from farms that had already chosen whether or not to apply the biochar supplement.

(b) An appropriate random assignment method is as follows:
1. Assign each of the 60 seedlings a unique integer label from 1 to 60.
2. Using a random number generator, generate 30 unique integers between 1 and 60 without replacement.
3. The seedlings corresponding to these 30 selected numbers are assigned to the biochar soil treatment.
4. The remaining 30 seedlings are assigned to the standard soil treatment without biochar.

(Alternative method):
Write each number from 1 to 60 on identical slips of paper. Place all 60 slips into a container and mix them thoroughly. Randomly draw 30 slips without replacement; the seedlings corresponding to those 30 numbers receive the biochar treatment, and the remaining 30 seedlings receive the standard soil treatment.

(c) The conclusion is not appropriate because the 60 seedlings used in the experiment were of a single variety, grown under controlled greenhouse conditions, and were not randomly selected from the population of all commercial bell pepper varieties and farms nationwide. Therefore, the findings cannot be generalized beyond the specific variety and growing conditions tested.

To allow generalization to all commercial bell pepper crops nationwide, the researchers would need to randomly select bell pepper farms and plant varieties across the nation to participate in the study.

Marking scheme

Scoring Guidelines

Each of parts (a), (b), and (c) is scored as Essentially Correct (E), Partially Correct (P), or Incorrect (I).

---

### Part (a)
Essentially Correct (E) if the response satisfies the following three components:
1. Identifies the study as an observational study.
2. Justifies the choice by stating that no treatment was imposed (or that farmers chose their own soil treatment).
3. Includes context (e.g., biochar, bell pepper plants, farms).

Partially Correct (P) if the response satisfies two of the three components.

Incorrect (I) if the response satisfies fewer than two components.

---

### Part (b)
Essentially Correct (E) if the response satisfies the following three components:
1. Creates unique labels for the 60 seedlings (or treatments).
2. Describes a valid random assignment procedure where every possible assignment is equally likely (e.g., random number generator without repeats or thoroughly mixed slips of paper drawn without replacement).
3. Ensures that exactly 30 seedlings are assigned to the biochar treatment and exactly 30 seedlings are assigned to the standard soil treatment.

Partially Correct (P) if the response satisfies two of the three components.

Incorrect (I) if the response satisfies fewer than two components.

---

### Part (c)
Essentially Correct (E) if the response satisfies the following three components:
1. Explains why generalization is not valid (seedlings were not randomly selected from the target population / only one variety and specific greenhouse conditions were used).
2. Identifies that random sampling/selection from the population of interest is required for generalization.
3. Provides sufficient context referring to commercial bell pepper farms/varieties nationwide.

Partially Correct (P) if the response satisfies two of the three components.

Incorrect (I) if the response satisfies fewer than two components.

---

### Composite Score
- 4 Points: All 3 parts are E.
- 3 Points: 2 parts are E and 1 part is P.
- 2 Points: 2 parts are E and 0 parts are P; OR 1 part is E and 1–2 parts are P; OR 3 parts are P.
- 1 Point: 1 part is E and 0 parts are P; OR 0 parts are E and 2–3 parts are P.
- 0 Points: Does not meet the criteria for 1 point.

Question 4 · free_response

4 marks

A wildlife research team monitors a protected wetland using automated cameras to study the presence of a rare marsh bird. The cameras record continuous 10-minute video clips. Based on historical data, the probability that a randomly selected 10-minute video clip captures bird activity is 0.15. Assume whether bird activity occurs in each 10-minute clip is independent.

(a) An intern reviews consecutive 10-minute video clips until one showing bird activity is found.

(i) Calculate the mean (expected value) of the distribution of the number of video clips the intern must review until bird activity is found. Show your work.

(ii) Calculate the standard deviation of the distribution of the number of video clips the intern must review until bird activity is found. Show your work.

(b) Another researcher, Marcus, has limited time each morning and will stop reviewing video clips after finding one with bird activity or after reviewing 3 clips, whichever comes first. Let the random variable $W$ represent the number of video clips Marcus reviews in a morning session. The table shows the partially completed probability distribution of $W$.

$$\begin{array}{|l|c|c|c|}
\hline
\text{Number of video clips reviewed, } w & 1 & 2 & 3 \\
\hline
\text{Probability, } P(W = w) & 0.15 & & \\
\hline
\end{array}$$

(i) Calculate $P(W = 2)$. Show your work.

(ii) Calculate $P(W = 3)$. Show your work.

(c) Consider the probability distribution of $W$ from part (b).

(i) Calculate the mean of the distribution of $W$. Show your work.

(ii) Interpret the mean of the distribution of $W$ in context.

Show answer & marking scheme

Worked solution

Part (a):
Let $X$ be the number of video clips reviewed until bird activity is first found. $X$ follows a geometric distribution with probability of success $p = 0.15$.

(i) The mean of a geometric distribution is:
$$\mu_X = \frac{1}{p} = \frac{1}{0.15} = \frac{20}{3} \approx 6.67 \text{ clips}$$

(ii) The standard deviation of a geometric distribution is:
$$\sigma_X = \frac{\sqrt{1 - p}}{p} = \frac{\sqrt{1 - 0.15}}{0.15} = \frac{\sqrt{0.85}}{0.15} \approx \frac{0.92195}{0.15} \approx 6.146 \text{ clips (or } 6.15 \text{ clips)}$$

---

Part (b):

(i) Marcus reviews exactly 2 clips if the first clip has no bird activity (failure) and the second clip has bird activity (success):
$$P(W = 2) = (1 - 0.15)(0.15) = (0.85)(0.15) = 0.1275$$

(ii) Marcus reviews 3 clips if bird activity is not found in the first 2 clips. Since the sum of the probabilities in the distribution must equal 1:
$$P(W = 3) = 1 - P(W = 1) - P(W = 2) = 1 - 0.15 - 0.1275 = 0.7225$$
*(Alternatively, $P(W = 3) = P(\text{first two clips have no activity}) = (0.85)^2 = 0.7225$, or $(0.85)^2(0.15) + (0.85)^3 = 0.108375 + 0.614125 = 0.7225$.)*

---

Part (c):

(i) The expected value (mean) of $W$ is calculated as:
$$E(W) = \sum w \cdot P(W = w) = 1(0.15) + 2(0.1275) + 3(0.7225)$$
$$E(W) = 0.15 + 0.255 + 2.1675 = 2.5725 \text{ clips}$$

(ii) Interpretation: Over a large number of repeated morning sessions (in the long run), the average number of video clips reviewed per session is approximately 2.5725 clips.

Marking scheme

Scoring Guidelines:

This question is scored in three parts: Part (a), Part (b), and Part (c). Each part is scored as Essentially Correct (E), Partially Correct (P), or Incorrect (I).

---

### Part (a):
Essentially Correct (E) if the response satisfies at least 3 of the following 4 components:
1. In (a-i), correctly calculates the mean ($\mu \approx 6.67$).
2. In (a-i), shows supporting work using the geometric mean formula $1/p$.
3. In (a-ii), correctly calculates the standard deviation ($\sigma \approx 6.15$).
4. In (a-ii), shows supporting work using the geometric standard deviation formula $\sqrt{1-p}/p$.

Partially Correct (P) if the response satisfies only 2 of the 4 components.
Incorrect (I) if the response satisfies 0 or 1 component.

Note: Rounding the mean in (a-i) to an integer (e.g., 6 or 7) without stating the unrounded value does not satisfy component 1.

---

### Part (b):
Essentially Correct (E) if the response satisfies at least 3 of the following 4 components:
1. In (b-i), correctly calculates $P(W = 2) = 0.1275$.
2. In (b-i), provides supporting work for $P(W = 2)$.
3. In (b-ii), correctly calculates $P(W = 3) = 0.7225$ (consistent with b-i).
4. In (b-ii), provides supporting work for $P(W = 3)$.

Partially Correct (P) if the response satisfies only 2 of the 4 components.
Incorrect (I) if the response satisfies 0 or 1 component.

---

### Part (c):
Essentially Correct (E) if the response satisfies both components 1 and 2 AND at least two of components 3–5:
1. In (c-i), states the correct mean of the distribution ($2.5725$), consistent with values from part (b).
2. In (c-i), shows correct calculation work using the formula $\sum w P(W=w)$.
3. In (c-ii), interpretation conveys the concept of repeated trials / long run / many sessions.
4. In (c-ii), interpretation refers to the mean or average.
5. In (c-ii), interpretation is set in context (number of video clips reviewed).

Partially Correct (P) if the response satisfies two or three of components 1–4, but not the full criteria for E.
Incorrect (I) if the response satisfies fewer than 2 components.

---

### Composite Score Conversion:
- 4 Points (Complete Response): All 3 parts (a, b, c) are E.
- 3 Points (Substantial Response): 2 parts are E and 1 part is P.
- 2 Points (Developing Response): 2 parts are E and 0 parts are P; OR 1 part is E and 1–2 parts are P; OR all 3 parts are P.
- 1 Point (Minimal Response): 1 part is E and 0 parts are P; OR 0 parts are E and 2 parts are P.
- 0 Points: Response does not meet criteria for 1 point.

Question 5 · free-response

4 marks

A regional environmental research organization surveyed a random sample of 600 households in a large metropolitan county to investigate energy consumption patterns. Researchers recorded each household's home structure type and primary heating energy source. The counts are summarized in the two-way table below:

$$\begin{array}{|l|c|c|c|c|}
\hline
\textbf{Home Structure Type} & \textbf{Electricity} & \textbf{Natural Gas} & \textbf{Solar / Other} & \textbf{Total} \\
\hline
\text{Single-Family Detached} & 75 & 180 & 45 & 300 \\
\hline
\text{Townhouse} & 50 & 80 & 20 & 150 \\
\hline
\text{Multi-Family Apartment} & 85 & 55 & 10 & 150 \\
\hline
\textbf{Total} & 210 & 315 & 75 & 600 \\
\hline
\end{array}$$

(a) If one household is selected at random from the sample, what is the probability that the household resides in a Single-Family Detached home and uses Natural Gas as its primary heating source? Show your work.

(b) Given that a randomly selected household from the sample resides in a Multi-Family Apartment, what is the probability that its primary heating source is Electricity? Show your work.

(c) The researchers want to determine whether there is an association between home structure type and primary heating energy source.

(i) Identify the appropriate inference procedure to investigate this relationship.

(ii) State the appropriate null and alternative hypotheses for the test identified in part (c)(i).

(d) The hypothesis test yielded a test statistic of $\chi^2 = 28.42$ with an associated $p$-value of $0.0001$. Assuming the conditions for inference are met, what conclusion should the researchers reach at a significance level of $\alpha = 0.05$? Justify your response in context.

Show answer & marking scheme

Worked solution

(a)

The probability that a randomly chosen household resides in a Single-Family Detached home and uses Natural Gas is the joint relative frequency:

\[ P(\text{Single-Family Detached} \cap \text{Natural Gas}) = \frac{180}{600} = 0.30 \]

(b)

The conditional probability that a household uses Electricity, given that it is a Multi-Family Apartment, is:

\[ P(\text{Electricity} \mid \text{Multi-Family Apartment}) = \frac{P(\text{Electricity} \cap \text{Multi-Family Apartment})}{P(\text{Multi-Family Apartment})} = \frac{85 / 600}{150 / 600} = \frac{85}{150} = \frac{17}{30} \approx 0.5667 \]

(c)

(i) The appropriate inference procedure is a chi-square test for independence (or chi-square test of association).

(ii)
- $H_0$: There is no association between home structure type and primary heating energy source for all households in the metropolitan county (i.e., home structure type and primary heating energy source are independent).
- $H_a$: There is an association between home structure type and primary heating energy source for all households in the metropolitan county (i.e., home structure type and primary heating energy source are not independent).

(d)

Because the $p$-value of $0.0001$ is less than the significance level $\alpha = 0.05$, we reject the null hypothesis $H_0$.

There is convincing statistical evidence that there is an association between home structure type and primary heating energy source among all households in the metropolitan county.

Marking scheme

Scoring Guidelines:

This question is scored in 4 parts: Part (a), Part (b), Part (c), and Part (d). Each part is scored as Essentially Correct (E), Partially Correct (P), or Incorrect (I).

---

### Part (a)
- Essentially Correct (E): The response satisfies both of the following:
1. Correctly calculates the probability as $\frac{180}{600} = 0.30$ (or equivalent fraction/percentage).
2. Shows supporting work from the table.
- Partially Correct (P): Calculates the correct probability without showing work, OR shows correct work but makes an arithmetic calculation error.
- Incorrect (I): Does not meet the criteria for E or P.

---

### Part (b)
- Essentially Correct (E): The response satisfies both of the following:
1. Correctly calculates the conditional probability as $\frac{85}{150} \approx 0.5667$ (or equivalent fraction $\frac{17}{30}$).
2. Shows supporting work using the appropriate row total as the denominator.
- Partially Correct (P): Calculates $\frac{85}{210}$ (reversing the condition) with work, OR provides the correct answer $\frac{85}{150}$ without work.
- Incorrect (I): Does not meet the criteria for E or P.

---

### Part (c)
- Essentially Correct (E): The response satisfies all three components:
1. Identifies the procedure by name as a "chi-square test for independence" (or "chi-square test of association").
2. States a correct null hypothesis of independence/no association AND a correct alternative hypothesis of dependence/association.
3. Hypotheses include context (both variables: home structure type and primary heating energy source) and reference to the population (all households in the county).
- Partially Correct (P): Satisfies 2 of the 3 components required for E.
- Incorrect (I): Satisfies at most 1 of the 3 components required for E.

---

### Part (d)
- Essentially Correct (E): The response satisfies both components:
1. Compares the $p$-value to $\alpha = 0.05$ ($0.0001 < 0.05$) and states a correct decision to reject $H_0$.
2. States a conclusion in context in terms of the alternative hypothesis using non-definitive language (e.g., "convincing statistical evidence of an association").
- Partially Correct (P): Satisfies only 1 of the 2 components required for E.
- Incorrect (I): Does not meet the criteria for E or P.

---

### Composite Score Determination:
- 4 (Complete Response): 4 parts essentially correct (4E)
- 3 (Substantial Response): 3 parts E and 1 part P, or 3 parts E and 0 parts P
- 2 (Developing Response): 2 parts E, or 1 part E and 2–3 parts P, or 4 parts P
- 1 (Minimal Response): 1 part E and 0–1 part P, or 2–3 parts P
- 0 (No Credit): Does not meet criteria for 1

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Section II, Part B

Answer Question 6 (Investigative Task). Show all work and explain reasoning clearly.

1 Question · 4 marks

Question 1 · Investigative Task

4 marks

A marine biologist is studying the mass, in grams, of a certain species of deep-sea snail. The distribution of masses of these snails in the population is known to be approximately normal with mean $\mu$ and standard deviation $\sigma$.

For a normal distribution, the theoretical interquartile range is related to the population standard deviation by $\text{IQR} \approx 1.35\sigma$. Therefore, an alternative estimator for the population standard deviation based on sample data is given by:

\[ \widehat{\sigma}_{\text{IQR}} = \frac{\text{IQR}}{1.35} \]

(a) A random sample of 16 snails was collected from the population, and their masses were recorded. The summary statistics for this sample are shown in the table below:

$$\begin{array}{|c|c|c|c|c|c|c|c|}\hline \text{Sample Size} & \text{Mean} & \text{Standard Deviation } (s) & \text{Minimum} & Q_1 & \text{Median} & Q_3 & \text{Maximum} \\hline 16 & 45.20\text{ g} & 6.40\text{ g} & 33.10\text{ g} & 40.50\text{ g} & 45.00\text{ g} & 49.50\text{ g} & 58.20\text{ g} \\hline\end{array}$$

(i) Calculate the interquartile range ($\text{IQR}$) of this sample.
(ii) Calculate the value of the alternative estimator $\widehat{\sigma}_{\text{IQR}}$ for this sample.

(b) Identify whether the sample standard deviation $s$ or the alternative estimator $\widehat{\sigma}_{\text{IQR}}$ is more resistant to extreme outliers. Explain your reasoning.

(c) An automated digital scale occasionally malfunctions and records a value that is ten times the true mass of a specimen. Suppose that for a second sample of 16 snails, one snail with an actual mass of $48.00\text{ g}$ was mistakenly recorded as $480.00\text{ g}$, while all other 15 measurements were recorded correctly (with all 15 remaining between $33.00\text{ g}$ and $58.00\text{ g}$).

(i) Describe the impact of this recording error on the sample standard deviation $s$. Justify your response.
(ii) Describe the impact of this recording error on the alternative estimator $\widehat{\sigma}_{\text{IQR}}$. Justify your response.

(d) A statistician conducted a computer simulation drawing 1,000 random samples of size $n = 16$ from a normal distribution with $\mu = 45.00$ and $\sigma = 6.00$ without any measurement errors. For each sample, both $s$ and $\widehat{\sigma}_{\text{IQR}}$ were calculated. The sampling distribution statistics for the 1,000 estimates are summarized below:

$$\begin{array}{|c|c|c|}\hline \text{Estimator} & \text{Mean of 1,000 Estimates} & \text{Standard Deviation of 1,000 Estimates} \\hline \text{Sample standard deviation } (s) & 5.91 & 1.10 \\hline \text{IQR-based estimator } (\widehat{\sigma}_{\text{IQR}}) & 5.98 & 1.65 \\hline\end{array}$$

(i) Based on the simulation results, which estimator exhibits greater precision (less variability from sample to sample) when sampling from an uncontaminated normal distribution? Justify your answer using the table.
(ii) Considering both resistance to measurement errors and precision under normal conditions, explain the circumstances under which a researcher should prefer using $\widehat{\sigma}_{\text{IQR}}$ instead of $s$ to estimate $\sigma$.

Show answer & marking scheme

Worked solution

Part (a):
(i) The interquartile range is calculated as:
\[ \text{IQR} = Q_3 - Q_1 = 49.50 - 40.50 = 9.00\text{ g} \]
(ii) Using the provided formula:
\[ \widehat{\sigma}_{\text{IQR}} = \frac{\text{IQR}}{1.35} = \frac{9.00}{1.35} \approx 6.67\text{ g} \]

Part (b):
The estimator $\widehat{\sigma}_{\text{IQR}}$ is more resistant to extreme outliers. The sample standard deviation $s$ involves deviations of every observation from the mean $\sum(x_i - \bar{x})^2$, so a single extreme value heavily inflates both $\bar{x}$ and $s$. In contrast, $\widehat{\sigma}_{\text{IQR}}$ is based on quartiles ($Q_1$ and $Q_3$), which are positional measures that are unaffected by the magnitude of extreme values in the tails of the distribution.

Part (c):
(i) The sample standard deviation $s$ will increase substantially because the error creates an extreme outlier at $480.00\text{ g}$, which greatly increases the distance of that point from the sample mean and consequently inflates $\sum(x_i - \bar{x})^2$.
(ii) The estimator $\widehat{\sigma}_{\text{IQR}}$ will have little to no change. Since $48.00\text{ g}$ was already above the median and near/above $Q_3$, changing its value to $480.00\text{ g}$ does not change the rank order of the observations, leaving the positions and values of $Q_1$ and $Q_3$ virtually unchanged (or completely unchanged).

Part (d):
(i) The sample standard deviation $s$ exhibits greater precision because the standard deviation of its sampling distribution ($1.10$) is smaller than the standard deviation of the sampling distribution for $\widehat{\sigma}_{\text{IQR}}$ ($1.65$), indicating less sample-to-sample variability.
(ii) A researcher should prefer $\widehat{\sigma}_{\text{IQR}}$ when there is a known possibility of data contamination, measurement errors, or severe outliers (such as intermittent equipment malfunction), because $\widehat{\sigma}_{\text{IQR}}$ is resistant to extreme values, whereas $s$ would produce heavily distorted estimates of the population spread.

Marking scheme

Scored in four parts: (a), (b), (c), and (d). Each part is scored as Essentially Correct (E), Partially Correct (P), or Incorrect (I).

Part (a) is scored as:
- E if the response correctly calculates both (i) $\text{IQR} = 9.00$ and (ii) $\widehat{\sigma}_{\text{IQR}} \approx 6.67$ with work shown.
- P if only one of the two calculations is correct with work shown.
- I if neither calculation is correct.

Part (b) is scored as:
- E if the response states that $\widehat{\sigma}_{\text{IQR}}$ is more resistant AND provides a correct explanation contrasting how outliers affect quartiles versus the sum of squared deviations from the mean.
- P if the response correctly identifies $\widehat{\sigma}_{\text{IQR}}$ as more resistant but provides an incomplete or weak explanation.
- I if the response identifies $s$ or gives an incorrect justification.

Part (c) is scored as:
- E if the response correctly states that (i) $s$ will increase substantially with justification AND (ii) $\widehat{\sigma}_{\text{IQR}}$ will experience little/no change with justification based on ranking/position.
- P if the response correctly describes and justifies the effect for only one of the two estimators, OR correctly identifies the effect for both without complete justification.
- I otherwise.

Part (d) is scored as:
- E if the response (i) selects $s$ and justifies it using the smaller standard deviation from the simulation table ($1.10 < 1.65$) AND (ii) explains that $\widehat{\sigma}_{\text{IQR}}$ is preferred when there is potential for outliers/recording errors/contamination.
- P if only component (i) or component (ii) is fully satisfied.
- I otherwise.

Holistic Score:
- 4: 4 E's
- 3: 3 E's and 1 P, or 3 E's and 1 I
- 2: 2 E's, or 1 E and 2/3 P's, or 4 P's
- 1: 1 E and 0/1 P, or 2/3 P's
- 0: 0/1 P and 3/4 I's

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