Point Estimation: Bias, Variance and Consistency · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Judging an estimator

Mathematics · Data & Statistics · ages 20-21
Name ______________________   Date ____________
  1. Each of two independent observations is 0 or 4 with equal probability. The four ordered samples are equally likely. What is the expected sample mean?

    Answer: ______________

  2. A sample of three readings is 4, 6 and 8. Compute the sample mean.

    Answer: ______________

  3. For the readings 11, 13 and 15 with mean 13, the squared deviations total 8. With the n minus 1 divisor, what is the sample variance?

    • 4
    • 8 over 3
    • 8
  4. Priya says the expected sample mean is greater than the population mean for independent observations from the same population with finite mean. Is this claim correct?

    Circle one:   True   False

  5. Let Q be the sum of squared deviations from the sample mean. A sample has only one observation. Why is the sample variance Q/(n - 1) unavailable?

    • The single observation is always an outlier, so variance has to be discarded.
    • Both Q and n - 1 are zero, so their quotient is undefined.
    • The denominator is one, so the quotient equals the observation itself.
    • A single observation makes the population variance infinite in every case.
  6. For five observations, dividing the sum of squared deviations from their sample mean by n gives 12. What value results when the same total is divided by n - 1?

    Answer: ______________

  7. If an estimator's bias and variance both tend to zero as the sample size grows, these conditions are sufficient for consistency.

    Circle one:   True   False

  8. For n >= 2 independent, identically distributed observations with finite positive variance, why does the unbiased sample variance divide by n - 1 instead of n?

    • It makes all squared deviations equal to each other
    • It corrects downward bias in the variance estimate
    • It cancels any unusually large values in the sample
  9. For every sample size n, estimator A has bias 1 and variance 1/n; estimator B has bias 0 and variance 4/n. Both estimate the same target theta. Which statement is correct?

    • At n = 1, A has lower MSE, but only B is consistent.
    • At n = 1, B has lower MSE, but only A is consistent.
    • At n = 1, A has lower MSE, and both estimators are consistent.
    • At n = 1, B has lower MSE, and neither estimator is consistent.
  10. A class takes one sample with mean 6 and declares the population mean is exactly 6 with no uncertainty. Which statement explains the problem?

    • One sample mean equals the population mean because the estimation rule is unbiased.
    • Unbiasedness fixes the expected sample mean, while one observed mean can differ.
    • The sample mean must use n - 1 as its divisor whenever the population mean is unknown.
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Answer key

For grown-ups. Fold this page away before handing over the rest.

Judging an estimator W1-mt_I374r2UU_i-s1

  1. 2 · The four ordered samples give means 0, 2, 2 and 4. Their equally weighted average is 8/4 = 2.
  2. 6 · The sample mean is (4 + 6 + 8)/3 = 6. This is an observed estimate of the population mean.
  3. 4 · There are three readings, so the divisor is 3 - 1 = 2. The sample variance is 8/2 = 4.
  4. False · The expected sample mean equals the population mean. This does not say which side contains more individual sample means.
  5. Both Q and n - 1 are zero, so their quotient is undefined. · The sample mean equals its only observation, giving Q = 0. Since n - 1 = 0, the formula produces 0/0, which is undefined.
  6. 15 · The squared-deviation total is 5 × 12 = 60. Dividing that total by 4 gives 15.
  7. True · Vanishing bias and variance are sufficient for consistency. Consistency itself is defined by the probability of an error exceeding any fixed positive tolerance tending to zero.
  8. It corrects downward bias in the variance estimate · The expected squared-deviation total is (n - 1)sigma^2. Dividing by n - 1 gives expectation sigma^2; dividing by n gives less.
  9. At n = 1, A has lower MSE, but only B is consistent. · At n = 1, A has MSE 2 and B has MSE 4. B's bias and variance tend to zero. A concentrates on theta + 1 instead of theta.
  10. Unbiasedness fixes the expected sample mean, while one observed mean can differ. · The sample mean varies across samples. Its expectation can equal the population mean without every observed estimate being exact.
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