Statistical Bias and Sampling

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Statistical Bias and Sampling · Grade 10 · Worksheet 2

1. Mason wants to estimate the average number of hours students at his school spend on homework each week. He surveys 72 students who are members of the chess club. Identify the type of sampling bias present and explain why it is a problem. Answer: Selection bias (undercoverage bias) because the sample is not representative of the entire student population; chess club members may have different homework habits than non-members. Solution: Identify the sampling method. Mason used a convenience sample by only surveying chess club members. Step 2: Determine the bias.
   Full step-by-step solution

   Step 1: Identify the sampling method. Mason used a convenience sample by only surveying chess club members. Step 2: Determine the bias. This is selection bias, specifically undercoverage bias, because the sample excludes students who are not in the chess club. Step 3: Explain the issue. Chess club members may be more academically inclined or have different schedules, leading to a systematic overestimate or underestimate of the true average homework hours for all students. The sample is not representative, so the estimate is likely biased.

2. Aroha wants to estimate the average height of trees in a large forest. She stands at the entrance and measures the first 75 trees she sees. Identify the sampling method used, the type of bias present, and explain why this sample may not represent the entire forest. Answer: Convenience sampling; selection bias; the sample is not random and may overrepresent trees near the entrance, which could be different in height from trees deeper in the forest. Solution: Identify the sampling method. Aroha measured the first 75 trees she saw at the entrance. This is convenience sampling because she selected trees that were easiest to access.
   Full step-by-step solution

   Step 1: Identify the sampling method. Aroha measured the first 75 trees she saw at the entrance. This is convenience sampling because she selected trees that were easiest to access. Step 2: Identify the bias. This introduces selection bias because the sample is not random. Step 3: Explain the issue. Trees near the entrance may be younger, shorter, or a different species than trees deeper in the forest, so the sample may not represent the entire forest's average height. The answer is: convenience sampling; selection bias; the sample may overrepresent trees near the entrance.

3. Ava is a school board member investigating whether students at her high school support a new later start time of 9:00 AM. She posts a question on the school's main social media page and asks students to comment 'yes' or 'no' with their reasons. Within 24 hours, she receives 450 responses, with 76% of them in favor of the later start time. Identify the sampling method used, name the type of bias present, and explain why this method leads to biased results. Answer: Voluntary response sample; non-response bias and voluntary response bias; the sample only includes students who choose to respond, likely overrepresenting those with strong opinions (especially those in favor of a later start), and excludes students who do not use social media or choose not to participate. Solution: Identify the sampling method. Ava posts a question on social media and waits for students to voluntarily comment. Additionally, students who do not use social media or are not online within the 24-hour window are excluded entirely.
   Full step-by-step solution

   Step 1: Identify the sampling method. Ava posts a question on social media and waits for students to voluntarily comment. There is no random selection; students self-select to participate. This is a voluntary response sample. Step 2: Name the bias type. The main biases are voluntary response bias (only people with strong feelings respond) and non-response bias (many students do not participate). Step 3: Explain why the method is biased. In a voluntary response sample, individuals who feel strongly about the issue are more likely to respond. Students who favor a later start time may be more motivated to comment, while those who are indifferent or opposed might not bother. Additionally, students who do not use social media or are not online within the 24-hour window are excluded entirely. Therefore, the 76% support figure likely overestimates true support among all students. The sample is not representative of the entire student population. Step 4: Conclusion. The sampling method is a voluntary response sample, which introduces voluntary response bias and non-response bias, making the results unreliable for generalizing to all students. The answer is: Voluntary response sample; non-response bias and voluntary response bias; the sample only includes students who choose to respond, likely overrepresenting those with strong opinions (especially those in favor of a later start), and excludes students who do not use social media or choose not to participate.

4. Ava wants to estimate the average number of hours per week that students at her high school spend on homework. She posts a survey link on the school's social media page and receives responses from 150 students. Identify the sampling method used, describe the type of bias present, and explain why this bias could affect her estimate. Answer: Voluntary response sampling; nonresponse bias and selection bias; the sample overrepresents students who are active on social media and have strong opinions about homework, leading to an overestimate or underestimate of the true average. Solution: Identify the sampling method. Ava posts a link on social media and waits for students to respond. Step 2: Identify the bias.
   Full step-by-step solution

   Step 1: Identify the sampling method. Ava posts a link on social media and waits for students to respond. This is voluntary response sampling because participants self-select. Step 2: Identify the bias. Voluntary response samples suffer from nonresponse bias (only those who see the post and care enough respond) and selection bias (students not on social media have no chance to participate). Step 3: Explain the effect on the estimate. Students who spend many hours on homework may be more motivated to respond, skewing the average upward. Conversely, those who spend little time might ignore the survey. Thus, the estimate is unlikely to reflect the true average for all students. The answer is voluntary response sampling with nonresponse and selection bias.

5. Sophia is a school board member investigating whether the new flexible scheduling policy has improved student attendance at Westwood High School, which has 1,200 students. She obtains a list of all students and randomly selects 160 of them to survey. However, she distributes the survey only during the first period class, which begins at 8:00 AM. Of the 160 selected students, only 96 are present in first period to receive the survey, and all 96 complete it. Among the respondents, 86 report that the flexible schedule has improved their attendance. Identify and explain two distinct types of bias present in Sophia's sampling method, and discuss how each bias could affect her conclusion about the entire student body. Answer: Selection bias (specifically, exclusion of late or absent students) and non-response bias (students who are not in first period are systematically excluded). The sample only includes students who are present at 8:00 AM, so it underrepresents students who arrive late or skip first period—who likely have worse attendance. This overestimates the perceived improvement. Also, the 64 students not present are a form of non-response, and their absence is related to the topic, further skewing results. Solution: Identify the first type of bias. Sophia randomly selected 160 students, but she only distributed surveys to those present in first period at 8:00 AM.
   Full step-by-step solution

   Step 1: Identify the first type of bias. Sophia randomly selected 160 students, but she only distributed surveys to those present in first period at 8:00 AM. Students who are late or skip first period are excluded entirely from the sampling frame. This is selection bias (specifically, exclusion bias). These excluded students likely have poorer attendance, so their views on the flexible schedule are missing. The sample overrepresents students who are punctual, leading to an overestimate of positive opinions about attendance improvement. Step 2: Identify the second type of bias. Of the 160 selected students, only 96 responded (the 64 absent students are non-respondents). This is non-response bias. The non-respondents are systematically different from respondents because their absence is directly related to the topic of attendance. They are likely less satisfied with the flexible schedule or have attendance issues, so their omission inflates the proportion of positive responses. Step 3: Explain the effect on generalizability. Both biases work together to make the sample unrepresentative of the entire 1,200-student population. The 86 out of 96 (approximately 89.6%) who report improvement is likely an overestimate. The true proportion for all students is probably lower. Therefore, Sophia cannot validly conclude that the flexible schedule improved attendance for the whole school based on this biased sample. Final answer: Selection bias (exclusion of late/absent students) and non-response bias (absent students are non-respondents). Both biases overestimate positive opinions and prevent generalizing to all students.

6. A triangular prism has a right triangle base with legs measuring 6 cm and 8 cm, and a height of 15 cm. The prism is sliced by a plane that passes through the midpoint of the prism's height and is parallel to the triangular bases. What is the area of the cross-section formed by this slice? Answer: 24 cm² Solution: We have a triangular prism with a right triangle base (legs 6 cm and 8 cm) and prism height 15 cm.
   Full step-by-step solution

   Let's go step-by-step. --- **Step 1: Understand the problem** We have a triangular prism with a right triangle base (legs 6 cm and 8 cm) and prism height 15 cm. A plane cuts the prism parallel to the triangular base, at the midpoint of the prism's height (so 15/2 = 7.5 cm from one base). Because the plane is parallel to the triangular base, the cross-section is a triangle similar to the base triangle. --- **Step 2: Find the area of the base triangle** Base triangle: right triangle with legs 6 cm and 8 cm. Area of base triangle = (1/2) * (6 cm) * (8 cm) = (1/2) * 48 = 24 cm². --- **Step 3: Determine the scaling factor for the cross-section** The cross-section is parallel to the base and at the midpoint along the prism's height. But wait — in a prism, cross-sections parallel to the base are congruent to the base, not scaled, unless it's a pyramid. Let's check carefully. Actually, in a prism, all cross-sections parallel to the base are identical in shape and size to the base. So the cross-section here should be exactly the same triangle as the base triangle, regardless of where you slice parallel to the base. That means the cross-section area = area of base triangle = 24 cm². --- **Step 4: Verify the "midpoint of prism's height" detail** The problem says: "passes through the midpoint of the prism's height and is parallel to the triangular bases." Yes — in a prism, any plane parallel to the base yields a cross-section congruent to the base. So the midpoint information is irrelevant for the area — it's still the same size triangle. --- **Step 5: Conclusion** Area of cross-section = area of base triangle = 24 cm². --- **Final answer:** 24 cm²