- A researcher wants a point estimate of the population mean daily screen time μ for all adults in a state. She selects a simple random sample of 100 adults from the state and computes the sample mean xˉ. She then reports the point estimate 2xˉ (twice the sample mean) to "account for multitasking." Which statement describes whether 2xˉ is biased for μ?
Answer: The estimate is biased high because its expected value is 2μ rather than μ.
This question tests identifying bias in modified estimators, like scaling the sample mean 2\bar{x} for mu. Multiplying \bar{x} by 2 scales the expected value to 2mu, creating a high bias; over many SRSs, the long-run average would be 2mu, not mu, due to this arbitrary adjustment. The sampling distribution's center is shifted proportionally. Distractor choice C incorrectly states scaling doesn't affect bias, but it does when the factor isn't 1. Mini-lesson: Bias is a systematic deviation in the estimator's expected value from the parameter, unaffected by sample size n, whereas variability measures spread and reduces with larger n—here, n=100 lowers variability but the bias remains 2mu - mu. Thus, 2\bar{x} is biased high. - A factory wants a point estimate of the population mean fill amount μ (in ounces) for all bottles produced on a given day. Inspectors take an SRS of 50 bottles and compute xˉ, but then they discard the 5 smallest fill amounts and compute the mean of the remaining 45 bottles. They report this trimmed mean as the point estimate for μ. Which statement describes whether the estimate is biased?
Answer: The estimate is biased high because removing the smallest values tends to increase the expected value of the estimator.
This question examines bias in point estimates, focusing on a trimmed mean that discards only the smallest values when estimating mu. Removing the 5 smallest from 50 shifts the mean higher, so the expected value exceeds mu, introducing high bias; long-run averages over repeated SRSs would be systematically inflated. This asymmetric trimming alters the center of the sampling distribution. Choice A is a distractor claiming trimming reduces variability without bias, but while it may lower variability, it biases unless symmetrically applied to a symmetric distribution. Bias vs. variability mini-lesson: Bias offsets the center and isn't mitigated by n=50, while variability shrinks with n but here the bias persists from the method. Hence, the trimmed mean is biased high.
- A company wants a point estimate of the population mean commute time μ (in minutes) for all its employees. To construct the estimate, the HR manager surveys the first 50 employees who arrive at work on Monday and computes their sample mean commute time xˉ, reporting xˉ as the point estimate for μ. Which statement describes whether the estimate is biased?
Answer: The estimate is biased because surveying early arrivals is not a random sample and could systematically misrepresent commute times.
This question tests recognizing bias from non-random sampling methods. The HR manager surveys only the first 50 employees to arrive, which is not a random sample of all employees. Early arrivals might have systematically different commute times (perhaps shorter commutes make it easier to arrive early), so this sampling method could produce a biased estimate. Choice B correctly identifies this bias. Choice A is wrong because being an unbiased estimator requires random sampling from the target population. Remember: the formula for xˉ doesn't create bias, but the sampling method can—if your sample systematically excludes or overrepresents certain groups, your estimate will be biased regardless of what statistic you calculate. - A teacher wants a point estimate of the population mean score μ on an end-of-unit test for all students in the school who took the test. The teacher randomly selects 40 test papers from all completed tests and computes the sample mean score xˉ, reporting xˉ as the point estimate for μ. Which statement describes whether the estimate is biased?
Answer: The estimate is unbiased because a randomly selected sample mean xˉ has expected value μ.
This question asks about bias when using a random sample to estimate a population mean. Since the teacher randomly selects 40 test papers from all completed tests, the sample mean xˉ is an unbiased estimator of the population mean μ. Choice A correctly states this. Choice B incorrectly claims small samples produce biased estimates—sample size affects precision (variability) but not bias. Choice C confuses the fact that individual estimates differ from the parameter with bias, which is about long-run average behavior. The key principle: random sampling from the target population produces unbiased estimators, regardless of sample size or whether individual estimates equal the true parameter. - To estimate the population mean height μ of all students at a university, a student selects an SRS of 30 students and measures height, but rounds every measurement to the nearest 2 inches before computing the sample mean xˉ from the rounded values. She reports this rounded-data mean as the point estimate for μ. Which statement describes whether the estimate is biased?
Answer: The estimate is unbiased because rounding introduces random error with mean 0, so the expected value remains μ.
This question assesses whether rounding measurements introduces bias in the sample mean for estimating mu. Rounding to the nearest 2 inches adds errors between -1 and +1 inch; if fractional parts are uniformly distributed, the expected error is 0, so E(rounded \bar{x}) = mu, making it unbiased—long-run averages would still hit mu. The sampling distribution remains centered at mu despite added noise. Distractor choice E confuses bias with variability, as sample-to-sample changes don't shift the center. Mini-lesson: Bias occurs if the center deviates systematically, but here it doesn't; variability increases slightly from rounding but decreases with n=30, distinguishing spread from center shift. Thus, the estimate is unbiased. - A store manager wants to estimate the population mean amount μ that customers spend per visit. The manager records the spending amounts for the first 80 customers who enter the store on a Saturday morning and computes the point estimate xˉ from those 80 purchases. Which statement describes whether the estimate is biased?
Answer: The estimate is biased because selecting the first customers on a Saturday morning may not represent all customers, causing systematic over- or underestimation of μ.
This question highlights sampling bias from non-representative selection. Saturday morning shoppers may systematically differ from the store's overall customer base - perhaps they're weekend grocery shoppers who spend more, or early-bird bargain hunters who spend less. By only sampling from this specific time period, the estimate xˉ will likely systematically over- or underestimate the true mean spending μ across all customers and all times. This is a classic example of convenience sampling creating bias. Choice A incorrectly assumes any sample mean is unbiased, ignoring the importance of how the sample was selected. Choice D confuses variability with bias - having sampling variability is normal and doesn't indicate bias; bias is about systematic deviation from the true parameter. To get an unbiased estimate, the manager would need to sample customers randomly across different days and times. - A city wants to estimate the population proportion p of households that have a working smoke detector. Inspectors visit a convenience sample of 120 households by stopping at homes on two nearby streets and record whether each home has a working detector. The point estimate is the sample proportion p^, calculated as (number of sampled homes with working detectors)/120. Which statement describes whether the estimate is biased?
Answer: The estimate is biased because using a convenience sample can systematically overestimate or underestimate p.
This question addresses bias from non-random sampling methods. A convenience sample, where inspectors only visit homes on two nearby streets, is not representative of all households in the city. Homes on these particular streets might have systematically different smoke detector rates than the city as a whole - perhaps they're in a newer neighborhood with stricter building codes, or an older area where detectors are less common. This sampling method will cause the sample proportion p^ to systematically over- or underestimate the true population proportion p in the long run. Choice A is wrong because the sampling method matters greatly for bias. Choice C incorrectly claims that large samples eliminate bias, but no sample size can fix the fundamental problem of non-representative sampling. Bias is about the sampling method's systematic tendency, not sample-to-sample variability. - To estimate the population mean commute time μ (in minutes) for employees at a large company, an SRS of 100 employees is selected and each is asked for their commute time. The point estimate is the sample mean xˉ. Suppose some employees round their commute times to the nearest 5 minutes, sometimes rounding up and sometimes rounding down. Which statement describes whether the estimate is biased?
Answer: The estimate is unbiased because with an SRS, xˉ targets μ, and random rounding up/down does not systematically push the mean in one direction.
This question tests understanding of when measurement error creates bias versus when it doesn't. When employees round to the nearest 5 minutes, some round up (27 becomes 30) and others round down (28 becomes 25). If this rounding is truly random - meaning there's no systematic tendency to round more in one direction - then these errors tend to balance out in the long run. The sample mean xˉ from an SRS remains an unbiased estimator of μ because the expected value of the rounding errors is zero. Choice A incorrectly claims that increased variability implies bias, but variability and bias are distinct concepts. Choice B wrongly states that rounding always decreases values, when the problem clearly states rounding goes both ways. The key insight is that random errors that balance out don't create bias, while systematic errors do. - A wildlife biologist wants a point estimate of the population proportion p of fish in a lake that are infected with a certain parasite. The biologist catches fish using a net placed near the shoreline for one afternoon, tests 80 fish, and uses the sample proportion p^ infected as the point estimate for p. Which statement describes whether the estimate is biased?
Answer: The estimate is biased because the sampling method (near-shore, one afternoon) may not represent all fish in the lake.
This question tests recognizing sampling bias from a non-representative collection method. The biologist catches fish using a net near the shoreline during one afternoon, which is not a random sample of all fish in the lake. Fish near shore might differ from those in deeper water, and afternoon behavior might differ from other times. This systematic sampling method could produce a biased estimate of the infection rate. Choice B correctly identifies this bias. Choice A incorrectly suggests sample size alone determines bias—but even large samples can be biased if not randomly selected. Remember: bias comes from systematic problems in how you collect data, not from random sampling variability.
- To estimate the population mean commute time μ for all employees at a large firm, an analyst takes an SRS of 40 employees and records each commute time. She reports the point estimate xˉ+5 minutes, explaining that people often underreport commute times by about 5 minutes. Which statement describes whether the estimate xˉ+5 is biased for μ?
Answer: The estimate is biased high because its expected value is μ+5.
This question evaluates understanding of biased and unbiased estimators, particularly how modifying the sample mean affects bias in estimating mu. Adding 5 to \bar{x} shifts the entire sampling distribution by 5, so the expected value becomes mu + 5, making it biased high; over many SRSs, the long-run average of (\bar{x} + 5) would be mu + 5, not mu. This intentional adjustment introduces systematic error, even though \bar{x} itself is unbiased. Choice E is a distractor that confuses bias with variability, as the variability of \bar{x} exists but doesn't cause bias—the addition does. Bias vs. variability mini-lesson: Bias is a fixed offset in the center of the sampling distribution from the true parameter, independent of sample size, while variability is the dispersion that shrinks as n increases, but here the bias persists regardless of n=40. Hence, \bar{x} + 5 is biased high. - A school district wants a point estimate of the population proportion p of all high school students in the district who get at least 8 hours of sleep on a typical school night. The district selects a simple random sample of 200 students from the full enrollment list and computes the sample proportion p^ who report at least 8 hours. The district reports p^ as the point estimate for p. Which statement describes whether the estimate is biased?
Answer: The estimate is unbiased because p^ is an unbiased estimator of p for a simple random sample.
This question tests understanding of biased versus unbiased point estimates. An estimator is unbiased if its expected value (long-run average) equals the population parameter it estimates. The sample proportion p^ from a simple random sample is an unbiased estimator of the population proportion p, meaning that if we took many random samples and computed p^ for each, the average of all these p^ values would equal p. Choice A incorrectly confuses variability with bias—having sample-to-sample variation doesn't make an estimator biased. Remember: bias is about whether the estimator is systematically off-target on average, while variability is about how spread out the estimates are. - A polling organization wants to estimate the population proportion p of registered voters who support a bond measure. They randomly select 1,000 registered voters to call, but only those who answer and agree to respond are included. The point estimate is the sample proportion p^ among respondents. Which statement describes whether the estimate is biased?
Answer: The estimate is biased because nonresponse can create systematic differences between respondents and nonrespondents, affecting p^.
This question addresses nonresponse bias, a critical issue in survey sampling. Even though the initial 1,000 voters were randomly selected, only including those who answer and agree to respond creates systematic differences. People with strong opinions (either for or against) might be more likely to participate, while those who are neutral or uninterested might not respond. This self-selection means the respondents don't represent the full population, causing p^ to systematically differ from the true proportion p. Choice A wrongly focuses only on the initial selection, ignoring the nonresponse problem. Choice C incorrectly claims large samples guarantee unbiasedness, but even surveying millions wouldn't fix the fundamental issue of who chooses to respond. Nonresponse bias is about systematic differences between respondents and non-respondents, not just having fewer responses. - A school wants a point estimate of the population mean number of hours of sleep per night, μ, for all students in the district. Researchers take a simple random sample of 60 students and compute the sample mean xˉ. They report xˉ as the point estimate for μ. Which statement describes whether the estimate is biased?
Answer: The estimate is unbiased because the mean of xˉ over many SRSs equals μ.
This question assesses the skill of identifying biased and unbiased point estimates in AP Statistics, focusing on the sample mean as an estimator for the population mean. An unbiased estimator has a sampling distribution whose mean equals the true population parameter, meaning that over many repeated simple random samples (SRSs), the average of the sample means \bar{x} would equal mu. Here, since the sample mean \bar{x} from an SRS is unbiased for mu, the long-run average of many \bar{x} values would center at mu, regardless of sample-to-sample variability. A common distractor is choice A, which confuses bias with variability; variability means \bar{x} changes across samples, but that doesn't make it biased—the center still hits mu. In a mini-lesson on bias versus variability: bias refers to a systematic shift in the center of the sampling distribution away from the parameter, while variability describes the spread around that center, which decreases with larger sample sizes but doesn't affect bias. Thus, \bar{x} is unbiased here. - A community health clinic wants a point estimate of the population mean systolic blood pressure μ for all adult patients who visited the clinic this month. The clinic uses a random number generator to select 60 patient records from all adult visits this month and computes the sample mean xˉ, reporting xˉ as the point estimate for μ. Which statement describes whether the estimate is biased?
Answer: The estimate is unbiased because a random sample mean xˉ has expected value μ for the defined population of adult visits this month.
This question asks about bias when using proper random sampling. The clinic uses a random number generator to select patient records, which gives each adult visit an equal chance of selection—this is simple random sampling. For such a sample, the sample mean xˉ is an unbiased estimator of the population mean μ. Choice B correctly identifies this. Choice A incorrectly suggests that natural variability in measurements creates bias—variability and bias are different concepts. Choice D wrongly claims that unbiasedness requires normality—the sample mean is unbiased regardless of the population distribution shape. Remember: unbiased means the estimator is on-target on average across many samples, not that it has low variability. - A company wants a point estimate of the population proportion p of all customers who would recommend the company. A simple random sample of 200 customers is surveyed, and the sample proportion p^ is computed. The company reports p^ as the point estimate for p. Which statement describes whether the estimate is biased?
Answer: The estimate is unbiased because E(p^)=p for an SRS.
This question tests the concept of biased and unbiased point estimates, specifically for the sample proportion hat{p} estimating the population proportion p. For an SRS, hat{p} is unbiased because the expected value E(hat{p}) equals p, so in the long run, across many repeated samples, the average of the hat{p} values would equal the true p. The sampling distribution of hat{p} is centered at p, confirming no systematic over- or underestimation. A frequent distractor is choice B, which mistakes the randomness and variability of hat{p} for bias; variability is the spread of the distribution, not a shift in its center. Mini-lesson: Bias occurs when the center of the sampling distribution deviates from the parameter, whereas variability measures how much estimates fluctuate around that center and can be reduced by increasing sample size n, but bias remains unchanged by n. Therefore, hat{p} is unbiased in this scenario. - A city wants a point estimate of the population mean weekly household water use μ (in gallons). The city takes a simple random sample of 100 households from its utility customer list and computes the sample mean xˉ, reporting xˉ as the point estimate for μ. Which statement describes whether the estimate is biased?
Answer: The estimate is unbiased because xˉ is centered at μ for a simple random sample from the target population.
This question tests understanding of when a sample mean is unbiased. The city takes a simple random sample from its utility customer list, and for this population (utility customers), xˉ is an unbiased estimator of μ. Choice B is correct. Choice A raises an interesting point about households not using city water, but the question asks about estimating mean water use, which implicitly refers to the population of water users (the customer list). Choice C confuses variability with bias—higher variability doesn't create bias. Remember: an estimator is unbiased if its expected value equals the parameter for the population from which you're sampling, regardless of the estimator's variability. - A researcher wants a point estimate of the population proportion p of voters in a state who support a new policy. The researcher calls 400 landline phone numbers selected from a directory and computes p^ from those who answer and agree to respond, then reports p^ as the point estimate for p. Which statement describes whether the estimate is biased?
Answer: The estimate is biased because the sample is not a true random sample of all voters and may underrepresent groups without landlines or who do not respond.
This question examines bias from non-representative sampling. Calling only landline numbers creates selection bias because it systematically excludes households without landlines (often younger people or lower-income households) and only includes those who answer and agree to respond. This sampling method doesn't give all voters an equal chance of selection, so p^ will be a biased estimator of the true population proportion. Choice B correctly identifies this bias. Choice C incorrectly suggests that large sample size eliminates bias—but even a very large biased sample remains biased. The lesson: bias comes from how you select your sample, not from how many observations you collect. - A wildlife biologist wants a point estimate of the population mean weight μ of adult fish in a lake. She takes an SRS of 25 fish and records their weights, but her scale consistently reads 0.2 kg too heavy for every fish. She calculates the sample mean from these recorded weights and reports it as the estimate of μ. Which statement describes whether the estimate is biased?
Answer: The estimate is biased high because all measurements are shifted upward by 0.2 kg.
This question addresses biased and unbiased point estimates, examining how measurement error impacts the sample mean's bias for mu. The scale's consistent 0.2 kg overreading shifts every weight upward, so the sample mean is biased high with expected value mu + 0.2; in the long run, repeated samples would average to mu + 0.2, not mu. This is a systematic measurement bias affecting the center of the sampling distribution. Distractor choice D mixes up bias and variability, noting high variability from small n=25, but variability doesn't cause bias—the overreading does. Mini-lesson on bias vs. variability: Bias systematically displaces the sampling distribution's mean from the parameter and isn't reduced by larger n, whereas variability is the spread that does decrease with n, but here the bias remains fixed at +0.2 kg. Thus, the estimate is biased high.
- A researcher wants to estimate the population mean systolic blood pressure μ for adults in a county. An SRS of 50 adults is selected. Each person's blood pressure is measured once using a cuff that is later found to read 5 mmHg too high for every measurement. The point estimate is the sample mean xˉ of the recorded measurements. Which statement describes whether the estimate is biased?
Answer: The estimate is biased because a consistent +5 mmHg measurement error will make xˉ too large on average.
This question demonstrates measurement bias from faulty equipment. When every blood pressure reading is systematically 5 mmHg too high, the sample mean xˉ will also be exactly 5 mmHg too high compared to what it would be with accurate measurements. This creates an upward bias - if we repeated this study many times with the same faulty cuff, the average of all sample means would be μ+5, not μ. The SRS ensures representative sampling, but cannot correct for systematic measurement error. Choice A wrongly assumes random sampling fixes all problems, but sampling method and measurement accuracy are separate issues. Choice C incorrectly claims large samples eliminate bias, but whether we measure 50 or 5,000 people, adding 5 to every measurement still shifts the mean by 5. Bias comes from systematic errors that push estimates consistently in one direction. - A wildlife biologist wants to estimate the population mean weight μ of adult fish in a lake. The biologist catches fish using a net with holes large enough that very small fish often escape before being weighed. The point estimate is the sample mean xˉ of the weights of the 70 fish that are successfully caught and weighed. Which statement describes whether the estimate is biased?
Answer: The estimate is biased because the net tends to miss lighter fish, so the sample mean will tend to be larger than μ.
This question illustrates selection bias through a flawed sampling mechanism. The net's design systematically excludes lighter fish that slip through the holes, meaning the sample only includes heavier fish. This creates an upward bias where xˉ will consistently overestimate the true population mean weight μ. No matter how many times the biologist repeats this sampling method, the average of all sample means will be larger than μ because lighter fish are systematically excluded. Choice A incorrectly assumes all sample means are unbiased, ignoring how the sample was obtained. Choice C wrongly claims large samples eliminate bias, but catching 700 or 7,000 fish with the same net would still exclude the lighter ones. The bias comes from the systematic exclusion of part of the population, not from random sampling variability.
