This question tests understanding of unbiased estimators for population means. When we have a simple random sample (SRS), the sample mean $\bar{x}$ is an unbiased estimator of the population mean $\mu$, meaning that if we repeated the sampling process many times, the average of all those sample means would equal $\mu$. The fact that data is self-reported doesn't create bias in the estimator itself - it might affect the accuracy of individual values, but the sample mean from an SRS still targets the true population mean. Choice A incorrectly confuses variability with bias; increased variability affects precision but not whether the estimator is unbiased. Remember: bias is about whether an estimator systematically over- or underestimates the parameter in the long run, while variability is about how spread out the estimates are.

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 $\bar{x}$ will consistently overestimate the true population mean weight $\mu$. No matter how many times the biologist repeats this sampling method, the average of all sample means will be larger than $\mu$ 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.

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 $\hat{p}$ from a simple random sample is an unbiased estimator of the population proportion $p$, meaning that if we took many random samples and computed $\hat{p}$ for each, the average of all these $\hat{p}$ 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.

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 $\bar{x}$ 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.

This question evaluates understanding of biased and unbiased estimators, particularly how modifying the sample mean affects bias in estimating mu. Adding 5 to ar{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 (ar{x} + 5) would be mu + 5, not mu. This intentional adjustment introduces systematic error, even though ar{x} itself is unbiased. Choice E is a distractor that confuses bias with variability, as the variability of ar{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, ar{x} + 5 is biased high.

This question evaluates bias from nonresponse in an SRS when estimating mu. With 25% nonresponse, the measured group's mean may differ systematically from the full sample, potentially biasing the estimate if non-respondents have different blood pressures; long-run repetitions could center away from mu due to this self-selection. Nonresponse bias affects the sampling distribution's center. Choice C is a distractor stating nonresponse only increases variability, but it can also introduce bias. Bias vs. variability mini-lesson: Bias systematically shifts the center and isn't fixed by remaining sample size, while variability grows with nonresponse but is separate—here, bias may arise from who responds. Therefore, the estimate may be biased.

This question tests recognizing potential bias from excluding absent students when estimating mu via selected quizzes. Excluding absentees may systematically alter the mean if their scores differ (e.g., lower), so the expected value might not equal mu; over many selections, the long-run average could deviate due to this undercoverage. The method doesn't ensure the sampling distribution centers at mu. Distractor choice E mistakes variability for bias, as variation exists but doesn't cause the systematic shift. Mini-lesson: Bias displaces the center from the parameter due to flawed sampling and persists regardless of size (here 10), whereas variability is spread that reduces with larger n but can't eliminate bias. Thus, the estimate may be biased.

This question tests identifying bias in modified estimators, like scaling the sample mean 2ar{x} for mu. Multiplying ar{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, 2ar{x} is biased high.

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 $\hat{p}$ 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.

This question asks about bias in estimating a population mean. The sample mean $\bar{x}$ from a simple random sample is an unbiased estimator of the population mean $\mu$, regardless of sample size or population distribution shape. This means that over many random samples, the average of all sample means equals the population mean. Choice C incorrectly suggests that because $\bar{x}$ won't exactly equal $\mu$ for most samples, the estimator is biased—but bias is about long-run average behavior, not individual sample results. The key distinction: an unbiased estimator can still produce estimates that differ from the true parameter value; it just doesn't systematically over- or underestimate on average.