Aimathic
Login | English | Deutsch

Statistics math worksheets

Build your own math worksheets from 35,000+ problems for grades 3 to 12, from fractions to AP Calculus. Every problem comes with step-by-step solutions.

Estimators

Learning goals

  • Distinguishes a sample statistic from its population parameter.
  • Evaluates estimators in terms of bias and variability.
  • Explains how sample size affects estimator variability.
  • Compares competing estimators using sampling-distribution behavior.

Click problems to add them to your worksheet.

Estimators 1
A random sample of \(80\) apartments has a mean monthly rent of \(\$1450\). The goal is to estimate the mean monthly rent \(\mu\) of all apartments in the city. Identify the parameter, the estimator, and the numerical estimate.

Hints

- Separate the fixed population quantity from the random sample-based rule. - Distinguish the name of a statistic from the value it produced in this sample. - Identify which quantity is known only after the data are collected.

Solution

1. The parameter is the unknown population mean monthly rent \(\mu\). 2. The estimator is the rule that uses the sample mean \(\bar X\). 3. For this sample, the realized estimate is \(\bar x=\$1450\).

Answer

Parameter: the population mean \(\mu\). Estimator: the sample mean \(\bar X\). Estimate: \(\$1450\).
Estimators 2
A school district wants to know the true mean commute time \(\mu\) for all of its employees. A random sample of employees has sample mean \(\bar X\), and the observed sample gives \(\bar x=24.6\) minutes. Match each item with the correct term: parameter, estimator/statistic, or estimate. 1. \(\mu\) 2. \(\bar X\) 3. \(24.6\) minutes

Hints

- A parameter describes the population. - A statistic is calculated from a random sample. - An estimate is the particular numerical value produced by the statistic for one observed sample.

Solution

1. \(\mu\) describes the entire population, so it is the parameter. 2. \(\bar X\) is the sample-based rule or statistic used to estimate \(\mu\), so it is the estimator/statistic. 3. \(24.6\) minutes is the realized numerical value from the observed sample, so it is the estimate.

Answer

1. Parameter. 2. Estimator/statistic. 3. Estimate.
Estimators 3
Two estimators, \(A\) and \(B\), are used to estimate a parameter whose true value is \(50\). Their sampling distributions have the following summaries. <table> <tr><th>Estimator</th><th>Sampling mean</th><th>Sampling standard deviation</th></tr> <tr><td>\(A\)</td><td>\(50\)</td><td>\(5\)</td></tr> <tr><td>\(B\)</td><td>\(50\)</td><td>\(2\)</td></tr> </table> a) Which estimators are unbiased? b) Which estimator is more precise? Explain.

Hints

- Compare each sampling mean with the target value \(50\). - Precision concerns the spread of the sampling distribution. - Smaller sampling standard deviation means estimates vary less from sample to sample.

Solution

1. An estimator is unbiased when its sampling mean equals the parameter. 2. Both sampling means equal \(50\), so both \(A\) and \(B\) are unbiased. 3. Estimator \(B\) has the smaller sampling standard deviation, \(2\) instead of \(5\). 4. Therefore, \(B\) is more precise.

Answer

a) Both \(A\) and \(B\) are unbiased. b) \(B\) is more precise because its sampling standard deviation is smaller.
Estimators 4
A city wants to know the average number of library books borrowed per resident each year. It surveys \(80\) residents, and their sample mean is \(6.2\) books. What population quantity is being estimated, and which number is the estimate?

Hints

- Separate what the city wants to know about everyone from what it observed in its survey. - Identify the number calculated from the selected residents.

Solution

1. The target parameter is the mean annual number of books borrowed by all city residents. 2. The sample mean is the estimator used for that parameter. 3. The observed sample mean, \(6.2\) books, is the estimate from this particular sample.

Answer

The target is the population mean number of books per resident per year. The estimate is \(6.2\) books.
Estimators 5
A vertical reference line marks the true population proportion \(p\). The histogram shows estimates from many random samples. Do the estimates tend to be too high, too low, or centered correctly? Explain.
Figure for problem 563503

Hints

- Find the vertical line marking the true target. - Compare it with the center of the histogram, not one unusual estimate. - Decide whether repeated estimates tend to fall above or below the target.

Solution

1. The vertical reference line is at \(p=0.50\). 2. Most estimates and the center of the histogram are to the right of that line. 3. The estimator tends to produce values above the true proportion, so it has upward bias.

Answer

Too high on average. The distribution of estimates is centered to the right of the true proportion, showing upward bias.
Estimators 6
Two unbiased estimators of the same population mean produce the sampling distributions shown. Which estimator is more precise, a) or b)? Explain.
Figure for problem 563504

Hints

- Precision is about variation among repeated estimates. - Compare the spread of the two histograms, not the height of one bar alone. - Both centers already agree with the population mean.

Solution

1. Both distributions are centered at the same target, so neither has a systematic high or low tendency. 2. Panel b) is more tightly concentrated around that center. 3. The estimator in panel b) varies less between samples and is therefore more precise.

Answer

Estimator b). Its sampling distribution is narrower, so it gives more consistent estimates.
Estimators 7
A population has true mean \(50\). Two estimators are used many times: <table> <tr><th>Estimator</th><th>Mean of estimates</th><th>SD of estimates</th></tr> <tr><td>A</td><td>\(50\)</td><td>\(4\)</td></tr> <tr><td>B</td><td>\(53\)</td><td>\(1\)</td></tr> </table> Which estimator is unbiased? Which is more precise?

Hints

- Compare each mean of estimates with the target \(50\). - For precision, compare the standard deviations rather than the means.

Solution

1. Unbiasedness means the mean of repeated estimates equals the population mean, so A is unbiased. 2. Precision refers to small variation among estimates. B has smaller standard deviation, so B is more precise.

Answer

Estimator A is unbiased; estimator B is more precise.
Estimators 8
A researcher estimates a city's average daily water use by taking a random sample of households. The researcher could use either \(20\) or \(200\) households. Which sample size would generally give a more precise sample-mean estimate, assuming the same sampling method? Explain.

Hints

- Compare the amount of independent information in each sample. - Precision describes how much an estimate would vary if sampling were repeated.

Solution

1. A sample mean based on more independent observations generally has less sample-to-sample variability. 2. A sample of \(200\) households therefore gives a more precise estimate than a sample of \(20\).

Answer

\(200\) households, because estimates from larger random samples generally vary less between samples.
Estimators 9
In a simple random sample of \(240\) registered voters, \(78\) support a proposed transit plan. The city has \(12{,}000\) registered voters. a) Give the point estimate of the population proportion who support the plan. b) Use that estimate to estimate the number of registered voters who support the plan. c) Name the estimator used in part a).

Hints

- Use the sample's fraction with the characteristic as the estimate of the population fraction. - Apply the estimated fraction to the known population size. - Distinguish the estimator's symbolic name from its observed numerical value.

Solution

1. The sample proportion is \(\hat p=\frac{78}{240}=0.325\). 2. Multiplying by the population size gives \(12{,}000(0.325)=3900\). 3. The estimator of the population proportion is the sample proportion \(\hat p\).

Answer

a) \(0.325\). b) Approximately \(3900\) registered voters. c) The sample proportion \(\hat p\).
Estimators 10
An estimator \(T\) is used for a parameter whose true value is \(\theta=5.4\). Its sampling distribution is shown below. <table> <tr><th>\(T\)</th><th>\(4\)</th><th>\(5\)</th><th>\(6\)</th></tr> <tr><th>Probability</th><td>\(0.20\)</td><td>\(0.50\)</td><td>\(0.30\)</td></tr> </table> Find the expected value and bias of \(T\). State whether it tends to overestimate or underestimate \(\theta\).

Hints

- Find the center of the estimator's sampling distribution. - Compare that center with the target parameter. - Use the sign of the difference to interpret the direction of the error.

Solution

1. The expected value is \(E(T)=4(0.20)+5(0.50)+6(0.30)=5.1\). 2. The bias is \(E(T)-\theta=5.1-5.4=-0.3\). 3. The negative bias means that \(T\) tends to underestimate the parameter by \(0.3\).

Answer

The expected value is \(5.1\), and the bias is \(-0.3\). The estimator tends to underestimate \(\theta\).
Estimators 11
An estimator \(T_n\) based on a sample of size \(n\) satisfies \(E(T_n)=\theta+\frac{10}{n}\). a) Write an unbiased estimator of \(\theta\). b) For \(n=25\), the observed value is \(T_{25}=18.7\). Find the corrected estimate. c) Compare the size of the correction at \(n=25\) and \(n=100\).

Hints

- Identify the amount by which the estimator's expected value misses the target. - Reverse that systematic shift in the estimator itself. - Examine how the adjustment changes when the sample size is multiplied by four.

Solution

1. Subtract the known bias to obtain \(U_n=T_n-\frac{10}{n}\). 2. Then \(E(U_n)=\theta\), so \(U_n\) is unbiased. 3. At \(n=25\), the correction is \(\frac{10}{25}=0.4\), giving \(18.7-0.4=18.3\). 4. At \(n=100\), the correction is only \(0.1\), showing that the original bias shrinks with sample size.

Answer

a) \(U_n=T_n-\frac{10}{n}\). b) \(18.3\). c) The correction is \(0.4\) at \(n=25\) and \(0.1\) at \(n=100\).
Estimators 12
A measurement is uniformly distributed from \(0\) to an unknown upper limit \(\theta\). For a random sample, the sample mean is \(\bar X\). It is known that \(E(X)=\frac{\theta}{2}\). a) Construct an unbiased estimator of \(\theta\) using \(\bar X\). b) Calculate the estimate when \(\bar x=7.4\).

Hints

- Relate the expected sample mean to the unknown endpoint. - Undo the factor separating the estimator's center from the target. - Verify the new rule by taking its expected value.

Solution

1. Since \(E(\bar X)=\frac{\theta}{2}\), multiply the sample mean by \(2\). 2. Let \(T=2\bar X\). Then \(E(T)=2E(\bar X)=\theta\), so \(T\) is unbiased. 3. For \(\bar x=7.4\), the estimate is \(2(7.4)=14.8\).

Answer

a) \(T=2\bar X\). b) \(14.8\).
Estimators 13
A parameter of interest is \(\psi=3\mu-2\), where \(\mu\) is a population mean. A sample mean \(\bar X\) is unbiased for \(\mu\), and its standard deviation is \(1.4\). a) Construct an unbiased estimator of \(\psi\). b) Find its standard deviation. c) Calculate the estimate when \(\bar x=11\).

Hints

- Transform the estimator in the same way the target parameter is defined. - Separate the effects of scaling and shifting on variability. - Verify the new estimator's center before using the observed value.

Solution

1. Apply the same linear transformation to the unbiased estimator: \(T=3\bar X-2\). 2. Its expected value is \(3\mu-2=\psi\), so it is unbiased. 3. Subtracting a constant does not affect spread, while multiplying by \(3\) gives standard deviation \(3(1.4)=4.2\). 4. The observed estimate is \(3(11)-2=31\).

Answer

a) \(T=3\bar X-2\). b) \(4.2\). c) \(31\).
Estimators 14
Two estimators are used to estimate the population median \(\theta=100\): the sample mean and a trimmed mean. Simulations give these sampling-distribution summaries for two populations that both have median \(100\). <table> <tr><th>Population</th><th>Estimator</th><th>Sampling mean</th><th>Sampling standard deviation</th></tr> <tr><td>Clean, nearly symmetric</td><td>Sample mean</td><td>\(100\)</td><td>\(2\)</td></tr> <tr><td>Clean, nearly symmetric</td><td>Trimmed mean</td><td>\(99.5\)</td><td>\(2.4\)</td></tr> <tr><td>Right-skewed with extreme values</td><td>Sample mean</td><td>\(107\)</td><td>\(5\)</td></tr> <tr><td>Right-skewed with extreme values</td><td>Trimmed mean</td><td>\(101\)</td><td>\(3\)</td></tr> </table> a) Find the bias of each estimator in each population. b) For the clean population, which estimator has the better long-run center and which is more precise? c) For the right-skewed population, which estimator has the better long-run center and which is more precise? d) Explain why the preferred estimator can change.

Hints

- Compare each sampling-distribution mean with the stated population median. - Use sampling standard deviation, not bias, to compare precision. - Compare the two estimators separately for the two population shapes.

Solution

1. Bias equals an estimator's sampling mean minus the population median \(100\). 2. In the clean population, the biases are \(100-100=0\) for the sample mean and \(99.5-100=-0.5\) for the trimmed mean. 3. The sample mean is also more precise in the clean population because its sampling standard deviation is \(2\), versus \(2.4\). 4. In the right-skewed population, the biases are \(107-100=7\) and \(101-100=1\). 5. The trimmed mean is more precise there because its sampling standard deviation is \(3\), versus \(5\) for the sample mean. 6. The sample mean works better for the nearly symmetric population; the trimmed mean works better when extreme values pull the sample mean away from the median.

Answer

a) Clean: sample mean bias \(0\); trimmed mean bias \(-0.5\). Right-skewed: sample mean bias \(7\); trimmed mean bias \(1\). b) The sample mean has the smaller bias and standard deviation. c) The trimmed mean has the smaller bias and standard deviation. d) Extreme values affect the sample mean more strongly, so the population's shape can change which estimator works better for the median.
Estimators 15
Independent observations \(X_1\) and \(X_2\) have mean \(\mu\) and standard deviation \(6\). Compare the estimators \(A=X_1\) and \(B=\frac{X_1+X_2}{2}\) for \(\mu\). a) Determine whether each estimator is unbiased. b) Find each variance and standard deviation. Give the standard deviation of \(B\) exactly and to three decimal places. c) Which estimator is more precise? Explain.

Hints

- Compare the expected values of the two proposed estimators. - How does independence affect the variance of an average? - Compare their sampling variances, not just their centers.

Solution

1. \(E(A)=\mu\) and \(E(B)=\frac{E(X_1)+E(X_2)}{2}=\mu\), so both are unbiased. 2. Estimator \(A\) has variance \(36\) and standard deviation \(6\). 3. By independence, \(\operatorname{Var}(B)=\frac{1}{4}(36+36)=18\). 4. Hence \(\operatorname{SD}(B)=\sqrt{18}=3\sqrt{2}\approx4.243\). 5. Estimator \(B\) has smaller variance, so it is more precise.

Answer

a) Both are unbiased for \(\mu\). b) \(A\): variance \(36\), standard deviation \(6\). \(B\): variance \(18\), standard deviation \(3\sqrt{2}\approx4.243\). c) \(B\) is more precise because its variance is smaller.
Estimators 16
In random samples of size \(20\) from a population with proportion \(p=0.30\), let \(X\) be the number of successes. Two estimators are proposed: \(A=\frac{X}{20}\) and \(B=\frac{X+1}{22}\). Find the mean of the sampling distribution of each estimator. Which estimator is unbiased for \(p\)?

Hints

- Use the long-run average number of successes in samples of the stated size. - Apply each estimator’s transformation to that long-run center. - Compare each resulting mean with the population proportion being estimated.

Solution

1. Since \(E(X)=np=20(0.30)=6\), \(E(A)=\frac{E(X)}{20}=\frac{6}{20}=0.30\). 2. For the second estimator, \(E(B)=\frac{E(X)+1}{22}=\frac{7}{22}\approx 0.3182\). 3. Estimator \(A\) is unbiased because its sampling-distribution mean equals \(p=0.30\). Estimator \(B\) is biased upward for this population proportion.

Answer

\(E(A)=0.30\) and \(E(B)=\frac{7}{22}\approx 0.3182\). Estimator \(A\) is unbiased for \(p\); estimator \(B\) is not.
Estimators 17
The true parameter is \(\theta=20\). Across repeated samples, estimator A has mean \(20\) and standard deviation \(3\), while estimator B has mean \(22\) and standard deviation \(1\). Mean squared error is variance plus the square of bias. Find the mean squared error of each estimator and decide which is smaller.

Hints

- Find each estimator's bias by comparing its long-run mean with the target. - Convert the stated standard deviations to variances. - The two parts of mean squared error measure different kinds of estimation error.

Solution

1. Estimator A has bias \(20-20=0\) and variance \(3^2=9\), giving mean squared error \(9\). 2. Estimator B has bias \(22-20=2\) and variance \(1^2=1\), giving mean squared error \(1+2^2=5\). 3. Estimator B has the smaller mean squared error despite being biased.

Answer

Estimator A: \(9\). Estimator B: \(5\). B has the smaller mean squared error.
Estimators 18
A survey method is known to be unbiased for a population mean of \(50\). One random sample gives an estimate of \(53\). A student says, “The method must be biased because this estimate is too high.” Explain the mistake.

Hints

- Ask whether the student is discussing a whole sampling method or one sample result. - Think about what can happen to an unbiased estimate in a single random sample.

Solution

1. Bias concerns the mean of estimates over repeated samples, not whether one estimate differs from the target. 2. Even an unbiased method sometimes gives estimates above or below \(50\) because of random sampling variation. 3. The observed \(53\) alone does not contradict the method's unbiasedness.

Answer

One high estimate is a sampling error, not proof of bias. An unbiased method is centered at \(50\) across repeated samples.
Estimators 19
In a town, \(80\%\) of households live in apartments and \(20\%\) in houses. Representative surveys estimate the average monthly electricity bill at \(\$90\) for apartments and \(\$150\) for houses. A student averages the two estimates and gets \(\$120\). Find the correctly weighted estimate for the town and explain the student's error.

Hints

- The two group averages describe different proportions of households. - Think about how much each type contributes to the overall average. - Check whether an unweighted average treats the groups as equally common.

Solution

1. The two household types do not occur equally often in the town. 2. Weight the group means by their population shares: \(0.80(90)+0.20(150)=72+30=102\). 3. The student's simple average wrongly gives equal weight to a group containing \(80\%\) of households and one containing \(20\%\).

Answer

\(\$102\) per month. The student used equal group weights instead of the town's \(80\%\) and \(20\%\) shares.
Estimators 20
The population mean is \(100\). An estimator has mean \(102\) for every sample size. Its standard deviation decreases toward \(0\) as the sample size grows. Does the estimator become more precise? Will its estimates approach the true mean as the sample size grows? Explain both answers.

Hints

- Precision asks how tightly repeated estimates are grouped. - Separately, identify where the center of those estimates stays as sample size grows. - A shrinking spread does not guarantee that the center is correct.

Solution

1. The decreasing standard deviation means repeated estimates are becoming less spread out, so precision improves. 2. The estimator's mean remains \(102\), two units above the true value of \(100\). 3. As the spread tends to zero, estimates cluster near \(102\), not \(100\). The estimator is not consistent for the true mean.

Answer

Precision improves, but the estimates approach \(102\), not the true mean \(100\). The remaining bias prevents consistency.
Estimators 21
Two methods estimate the same target value, \(20\). Their probability distributions are shown. a) Find the mean estimate for each method from the bars. b) Which method is more likely to miss the target by more than \(8\) units? Find that probability for each.
Figure for problem 563511

Hints

- Translate “more than \(8\) units” into possible estimate values outside a central interval. - Read the probabilities of those values from each graph. - First use each bar's value and height to work out the long-run mean.

Solution

1. For panel a), the mean is \(10(0.10)+20(0.80)+30(0.10)=20\). 2. For panel b), the mean is \(16(0.50)+24(0.50)=20\). 3. A miss greater than \(8\) units occurs for an estimate below \(12\) or above \(28\). 4. Panel a) gives \(10\) with probability \(0.10\) and \(30\) with probability \(0.10\), so the total probability is \(0.20\). 5. Panel b) gives only \(16\) or \(24\), both within \(8\) units of the target, so the probability is \(0\).

Answer

a) Both means equal \(20\). b) Panel a): \(0.20\). Panel b): \(0\). Panel a) is more likely to miss by more than \(8\) units.
Estimators 22
A school wants to estimate its average number of absences per student, which is \(5\). It compares two possible estimates: - Method A gives \(4\) with probability \(0.75\) and \(9\) with probability \(0.25\). - Method B gives \(2\) or \(8\), each with probability \(0.5\). Mean squared error (MSE) is the average of the squared differences between an estimate and the true value. Find the MSE for each method and decide which is smaller.

Hints

- Compare each possible estimate with the true value before averaging anything. - Why does the definition say to square each error? - Weight each squared error by how likely that estimate is.

Solution

1. For method A, the squared errors are \((4-5)^2=1\) and \((9-5)^2=16\). 2. Weighting these by their probabilities gives \(MSE_A=0.75(1)+0.25(16)=4.75\). 3. For method B, both estimates are \(3\) from the target, so \(MSE_B=0.5(3^2)+0.5(3^2)=9\). 4. Method A has the smaller mean squared error, even though it can make a large error sometimes.

Answer

Method A: \(MSE=4.75\). Method B: \(MSE=9\). Method A has the smaller MSE.
Estimators 23
A population consists of \(-1\) and \(1\), each equally likely. Two observations are sampled independently with replacement. Let \(S^2\) be the sample variance calculated with denominator \(n-1\). a) Construct the sampling distribution of \(S^2\). b) Find \(E(S^2)\) and compare it with the population variance.

Hints

- List all ordered samples and compute the statistic for each one. - Separate samples with matching values from samples with different values. - Use the sampling probabilities to find the statistic's expected value.

Solution

1. The four ordered samples are equally likely. The samples \((-1,-1)\) and \((1,1)\) have sample variance \(0\). 2. The samples \((-1,1)\) and \((1,-1)\) have mean \(0\) and sample variance \(2\). 3. Thus, \(P(S^2=0)=1/2\) and \(P(S^2=2)=1/2\). 4. The expected sample variance is \(0(1/2)+2(1/2)=1\), equal to the population variance \(1\).

Answer

a) <table> <tr><th>\(s^2\)</th><th>\(0\)</th><th>\(2\)</th></tr> <tr><th>\(P(S^2=s^2)\)</th><td>\(\frac{1}{2}\)</td><td>\(\frac{1}{2}\)</td></tr> </table> b) \(E(S^2)=1\), which equals the population variance.
Estimators 24
For every possible value of a parameter \(\theta\), an estimator \(A\) satisfies \(E(A)=0.8\theta+2\). a) Create an unbiased estimator of \(\theta\) using \(A\). b) A sample gives \(A=18.4\). What estimate does your corrected estimator produce?

Hints

- Work backward through the transformation affecting the estimator's expected value. - Undo the shift before undoing the scale. - Check the correction by taking its expected value symbolically.

Solution

1. Remove the additive bias by subtracting \(2\), giving expectation \(0.8\theta\). 2. Divide by \(0.8\), so \(U=\frac{A-2}{0.8}\). 3. Its expectation is \(E(U)=\frac{E(A)-2}{0.8}=\theta\), so it is unbiased. 4. For \(A=18.4\), \(U=\frac{18.4-2}{0.8}=20.5\).

Answer

a) \(U=\frac{A-2}{0.8}\). b) \(20.5\).
Estimators 25
Two estimators are used for a parameter with true value \(\theta=50\). Estimator \(T_1\) has mean \(50\) and standard deviation \(4\). Estimator \(T_2\) has mean \(52\) and standard deviation \(2\). a) Find the bias of each estimator. b) Compute each mean squared error using \(\operatorname{MSE}=\operatorname{variance}+(\operatorname{bias})^2\). c) Which estimator has the smaller mean squared error?

Hints

- Compare each sampling-distribution center with the parameter. - Convert each standard deviation to a variance before combining error components. - An unbiased estimator is not automatically the best under every error criterion.

Solution

1. The biases are \(50-50=0\) for \(T_1\) and \(52-50=2\) for \(T_2\). 2. \(\operatorname{MSE}(T_1)=4^2+0^2=16\). 3. \(\operatorname{MSE}(T_2)=2^2+2^2=8\). 4. Although \(T_2\) is biased, its smaller variability gives it the lower mean squared error.

Answer

a) Bias of \(T_1\): \(0\). Bias of \(T_2\): \(2\). b) \(\operatorname{MSE}(T_1)=16\); \(\operatorname{MSE}(T_2)=8\). c) \(T_2\) has the smaller mean squared error.
Estimators 26
Three independent measurements \(X_1,X_2,X_3\) each have mean \(\mu\) and positive variance \(\sigma^2\). Two estimators of \(\mu\) are \(A=\frac{X_1+X_2+X_3}{3}\) and \(B=\frac{X_1+2X_2+X_3}{4}\). a) Show that both estimators are unbiased. b) Find each variance. c) Which estimator is more precise?

Hints

- Compare the sums of the coefficients in the two averages. - When combining independent measurements, how does each coefficient affect variance? - Compare the two expressions for variance, using the positive-variance condition.

Solution

1. The coefficients in both estimators add to \(1\), so \(E(A)=E(B)=\mu\). 2. Independence gives \(\operatorname{Var}(A)=\frac{3\sigma^2}{9}=\frac{\sigma^2}{3}\). 3. For \(B\), \(\operatorname{Var}(B)=\frac{(1^2+2^2+1^2)\sigma^2}{16}=\frac{3\sigma^2}{8}\). 4. Because \(\sigma^2>0\) and \(\frac{1}{3}<\frac{3}{8}\), estimator \(A\) has the smaller variance.

Answer

a) Both estimators are unbiased for \(\mu\). b) \(\operatorname{Var}(A)=\frac{\sigma^2}{3}\); \(\operatorname{Var}(B)=\frac{3\sigma^2}{8}\). c) \(A\) is more precise because its variance is smaller.
Estimators 27
The mean of a population is known to be \(\mu=11\), but its variance \(\sigma^2\) is unknown. For independent observations \(X_1,\ldots,X_n\), consider the estimator \(U=\frac{1}{n}\sum_{i=1}^n(X_i-11)^2\). a) Explain why \(U\) is unbiased for \(\sigma^2\). b) Calculate the estimate from the sample \(8,10,11,15\).

Hints

- Use the fact that the center in the estimator is the true population mean. - Examine the expected value of one squared deviation before averaging. - For the data, calculate deviations first and square only afterward.

Solution

1. For each observation, \(E[(X_i-11)^2]=\sigma^2\) because \(11\) is the true population mean. 2. Averaging \(n\) quantities each with expectation \(\sigma^2\) gives \(E(U)=\sigma^2\). 3. The squared deviations are \(9,1,0,16\), with total \(26\). 4. The estimate is \(U=\frac{26}{4}=6.5\).

Answer

a) \(U\) is unbiased because every squared deviation from the known population mean has expected value \(\sigma^2\). b) \(6.5\).
Estimators 28
Estimators \(A\) and \(B\) are both unbiased for \(\theta\). Their variances are \(9\) and \(16\), and their covariance is \(6\). Consider the average estimator \(C=\frac{A+B}{2}\). a) Show that \(C\) is unbiased. b) Find \(\operatorname{Var}(C)\). c) Is \(C\) more precise than \(A\)?

Hints

- Average the expected values to assess center. - Include the dependence term when finding the variance of a sum. - Compare variances directly to judge precision.

Solution

1. \(E(C)=\frac{E(A)+E(B)}{2}=\frac{\theta+\theta}{2}=\theta\). 2. The variance is \(\operatorname{Var}(C)=\frac14[9+16+2(6)]=\frac{37}{4}=9.25\). 3. Estimator \(A\) has variance \(9\), so \(C\) is slightly less precise. 4. The positive covariance reduces the benefit that averaging might otherwise provide.

Answer

a) \(C\) is unbiased. b) \(\operatorname{Var}(C)=9.25\). c) No. \(A\) is slightly more precise because its variance is \(9\).
Estimators 29
Independent observations \(X_1,X_2,X_3\) each have mean \(\mu\). A proposed estimator is \(T=0.2X_1+0.3X_2+0.4X_3\). a) Find the bias of \(T\) as an estimator of \(\mu\). b) Modify \(T\) to make it unbiased. c) For observed data \(10,13,16\), calculate the corrected estimate exactly.

Hints

- Add the weights to compare the expected estimator with the population mean. - How can you undo a multiplicative factor in an expected value? - Keep the final estimate exact rather than rounding the division.

Solution

1. The coefficients total \(0.9\), so \(E(T)=0.9\mu\). 2. The bias is \(E(T)-\mu=-0.1\mu\). 3. The corrected estimator is \(U=\frac{T}{0.9}\), which has expected value \(\mu\). 4. For the data, \(T=0.2(10)+0.3(13)+0.4(16)=12.3\). 5. Therefore, \(U=\frac{12.3}{0.9}=\frac{41}{3}\).

Answer

a) Bias: \(-0.1\mu\). b) \(U=\frac{T}{0.9}\). c) \(\frac{41}{3}\).
Estimators 30
A population has unknown mean \(\mu\) and known variance \(16\). A random sample of size \(25\) has sample mean \(\bar X\). Because \(E(\bar X^2)=\mu^2+\frac{16}{25}\), construct an unbiased estimator of \(\mu^2\). Then calculate its value when \(\bar x=10\).

Hints

- Compare the stated expected value with the exact target. - Remove only the extra term that causes the systematic error. - Apply the correction after squaring the observed sample mean.

Solution

1. The excess expected value of \(\bar X^2\) above \(\mu^2\) is \(\frac{16}{25}\). 2. Subtract that amount to form \(U=\bar X^2-\frac{16}{25}\). 3. Its expected value is \(E(U)=\mu^2\). 4. For \(\bar x=10\), the estimate is \(100-\frac{16}{25}=99.36\).

Answer

The unbiased estimator is \(\bar X^2-\frac{16}{25}\), and the observed estimate is \(99.36\).
Estimators 31
A random sample of size \(n\) is taken from a uniform distribution on \([0,\theta]\). Let \(M\) be the sample maximum. It is known that \(E(M)=\frac{n}{n+1}\theta\). a) Modify \(M\) to obtain an unbiased estimator of \(\theta\). b) For \(n=5\) and observed maximum \(M=18\), calculate the estimate.

Hints

- Compare the maximum's expected value with the unknown population endpoint. - What multiplication would undo the factor on that endpoint? - Apply your corrected formula to the observed maximum.

Solution

1. The expected maximum is the target multiplied by \(\frac{n}{n+1}\). 2. Correct this factor by setting \(U=\frac{n+1}{n}M\). 3. Its expectation is \(E(U)=\frac{n+1}{n}E(M)=\theta\), so it is unbiased. 4. For \(n=5\) and \(M=18\), the estimate is \(U=\frac{6}{5}\cdot18=21.6\).

Answer

a) \(U=\frac{n+1}{n}M\). b) \(21.6\).
Estimators 32
An estimator \(R\) reports only whole-number values. For a parameter with true value \(\theta=10.2\), its sampling distribution is <table> <tr><th>\(R\)</th><th>\(9\)</th><th>\(10\)</th><th>\(11\)</th></tr> <tr><th>Probability</th><td>\(0.10\)</td><td>\(0.60\)</td><td>\(0.30\)</td></tr> </table> a) Determine whether \(R\) is unbiased. b) Find its mean squared error. c) Explain how an estimator restricted to integers can still be unbiased for a noninteger parameter.

Hints

- Find the probability-weighted center of the estimator's distribution. - For squared error, measure each possible estimate from the actual parameter. - Distinguish the support of an estimator from its expected value.

Solution

1. The expected value is \(9(0.10)+10(0.60)+11(0.30)=10.2\), equal to \(\theta\), so the estimator is unbiased. 2. The mean squared error is \(0.10(9-10.2)^2+0.60(10-10.2)^2+0.30(11-10.2)^2=0.36\). 3. Unbiasedness concerns the probability-weighted long-run average, not whether every possible estimate can equal the parameter.

Answer

a) Yes. Its expected value is \(10.2\). b) \(0.36\). c) Its possible values can average to \(10.2\) across repeated samples even though each individual estimate is an integer.
Estimators 33
For a random sample of size \(n\ge2\), an analyst computes \(V=\frac{1}{n}\sum_{i=1}^{n}(X_i-\bar X)^2\). It is known that \(E(V)=\frac{n-1}{n}\sigma^2\). a) Construct an unbiased estimator of the population variance \(\sigma^2\). b) When \(n=10\) and \(V=18\), find the unbiased estimate.

Hints

- Compare the given expected value with the target population variance. - Which factor would undo the difference between them? - Connect the corrected factor with the divisor of the sum of squared deviations.

Solution

1. The expected value of \(V\) is the target variance multiplied by \(\frac{n-1}{n}\). 2. Since \(n\ge2\), multiply by \(\frac{n}{n-1}\) to obtain \(S^2=\frac{n}{n-1}V\). 3. Equivalently, divide the sum of squared deviations from the sample mean by \(n-1\). 4. For \(n=10\), \(S^2=\frac{10}{9}\cdot18=20\).

Answer

a) \(S^2=\frac{n}{n-1}V=\frac{1}{n-1}\sum_{i=1}^{n}(X_i-\bar X)^2\). b) \(20\).
Estimators 34
A sample proportion \(\hat p\) is based on \(n\ge2\) independent trials with success probability \(p\). It is known that \(E[\hat p(1-\hat p)]=\frac{n-1}{n}p(1-p)\). a) Construct an unbiased estimator of \(p(1-p)\). b) For \(n=50\) and \(\hat p=0.30\), calculate the estimate. Round to four decimal places.

Hints

- Find the factor by which the stated expected value differs from the target. - Correct that factor without changing the target quantity. - Apply the correction to the entire sample expression and round at the end.

Solution

1. The expected value of \(\hat p(1-\hat p)\) is the target multiplied by \(\frac{n-1}{n}\). 2. Because \(n\ge2\), correct the factor with \(U=\frac{n}{n-1}\hat p(1-\hat p)\). 3. For \(n=50\) and \(\hat p=0.30\), \(U=\frac{50}{49}\cdot0.30\cdot0.70=\frac{3}{14}\approx0.2143\).

Answer

a) \(U=\frac{n}{n-1}\hat p(1-\hat p)\). b) \(\frac{3}{14}\approx0.2143\).
Estimators 35
A population has two groups. Group L is \(60\%\) of the population with mean value \(20\); Group H is \(40\%\) with mean value \(50\). In a very large survey, members of Group L respond independently with probability \(0.80\), while members of Group H respond independently with probability \(0.40\). Treat the respondent group proportions as their large-sample response-weighted proportions. The analyst estimates the population mean using the mean among respondents without adjustment. a) Find the true population mean. b) Find the large-sample center of the respondent-mean estimator. c) Find and interpret its bias.

Hints

- First weight group means by their full-population shares. - For respondents, combine population share with response probability. - Renormalize the response weights before comparing centers.

Solution

1. The population mean is \(0.60(20)+0.40(50)=32\). 2. In a very large survey, the respondent shares are proportional to \(0.60(0.80)=0.48\) and \(0.40(0.40)=0.16\), for total response weight \(0.64\). 3. The large-sample respondent mean is centered at \(\frac{0.48(20)+0.16(50)}{0.64}=27.5\). 4. The bias is \(27.5-32=-4.5\), so the unadjusted respondent mean underestimates the population mean.

Answer

a) \(32\). b) The large-sample center is \(27.5\). c) The bias is \(-4.5\); lower-valued Group L is overrepresented among respondents.
Estimators 36
An estimator \(T\) is used for a parameter with true value \(\theta=10\). Its sampling distribution is <table> <tr><th>\(T\)</th><th>\(8\)</th><th>\(10\)</th><th>\(20\)</th></tr> <tr><th>Probability</th><td>\(0.45\)</td><td>\(0.10\)</td><td>\(0.45\)</td></tr> </table> An estimator is called median-unbiased when at least half its probability is at or below the parameter and at least half is at or above it. Determine whether \(T\) is mean-unbiased and whether it is median-unbiased.

Hints

- Use the probability-weighted average for one definition of unbiasedness. - For the other definition, accumulate probability on each side of the target. - A skewed sampling distribution can satisfy one center condition but not the other.

Solution

1. The expected value is \(8(0.45)+10(0.10)+20(0.45)=13.6\), so the mean bias is \(13.6-10=3.6\). 2. Thus \(T\) is not mean-unbiased. 3. The probability at or below \(10\) is \(0.45+0.10=0.55\). 4. The probability at or above \(10\) is \(0.10+0.45=0.55\). 5. Therefore \(T\) is median-unbiased.

Answer

The estimator is not mean-unbiased; its mean is \(13.6\), giving bias \(3.6\). It is median-unbiased because both \(P(T\le10)\) and \(P(T\ge10)\) equal \(0.55\).
Estimators 37
Recorded measurements satisfy \(Y=X+E\), where the true values \(X\) and measurement errors \(E\) are independent. The error variance is known to be \(4\). Suppose \(S_Y^2\) is an unbiased estimator of \(\operatorname{Var}(Y)\). a) Construct an unbiased estimator of \(\operatorname{Var}(X)\). b) Find the estimate when \(S_Y^2=9\). c) The same rule gives \(-1\) when another sample has \(S_Y^2=3\). Explain how an unbiased estimator of a nonnegative parameter can produce a negative estimate.

Hints

- Decompose the recorded variance into true and error components. - Remove the known component from the sample-based estimate. - Distinguish a long-run property from restrictions on individual realized values.

Solution

1. Independence gives \(\operatorname{Var}(Y)=\operatorname{Var}(X)+4\). 2. Therefore use \(U=S_Y^2-4\). Its expected value is \(\operatorname{Var}(X)\). 3. For \(S_Y^2=9\), the estimate is \(5\). 4. For \(S_Y^2=3\), the estimate is \(-1\). Unbiasedness concerns the long-run average, not whether every sample estimate lies in the parameter's allowed range.

Answer

a) \(U=S_Y^2-4\). b) \(5\). c) Sampling variation can make the unbiased rule negative in an individual sample even though its expected value is the nonnegative true variance.
Estimators 38
The ordered finite population is shown. A systematic sample of size \(2\) selects positions three apart. The random start is equally likely to be position \(1,2,\) or \(3\). The sample mean estimates the population mean. a) Find the three possible estimates. b) Determine whether the estimator is unbiased over the random starts. c) Does unbiasedness imply that each possible estimate is close to the population mean?
Figure for problem 547623

Hints

- Match each possible random start with the position three places later. - Read the two population values for each sample from the chart and average them. - Separate a statement about long-run center from a statement about precision.

Solution

1. Starting at positions \(1,2,\) and \(3\) produces position pairs \((1,4)\), \((2,5)\), and \((3,6)\). 2. Reading the population values from the chart, the three sample means are \(\frac{2+11}{2}=6.5\), \(\frac{8+7}{2}=7.5\), and \(\frac{5+15}{2}=10\). 3. Their expected value is \(\frac{6.5+7.5+10}{3}=8\). 4. The population mean is \(\frac{2+8+5+11+7+15}{6}=8\), so the estimator is unbiased. 5. Individual estimates still range from \(6.5\) to \(10\), so unbiasedness alone does not guarantee small sampling error.

Answer

a) \(6.5,7.5,10\). b) Yes. Their mean is \(8\), equal to the population mean. c) No. Unbiasedness is a long-run center property, and individual estimates can still vary substantially.
Estimators 39
A population has two units with paired values \((x, y)=(1, 1)\) and \((3, 6)\). The parameter of interest is the ratio of population totals, \(R=\frac{\sum y}{\sum x}\). One unit is selected at random with equal probability, and the analyst estimates \(R\) using the sample ratio \(T=\frac{y}{x}\). a) Find the true value of \(R\). b) Find the sampling distribution and expected value of \(T\). c) Determine the bias.

Hints

- Find the ratio of the two population totals before using the sample. - What estimate comes from selecting each of the two units? - Compare the average of those estimates with the actual population ratio.

Solution

1. The population ratio is \(R=\frac{1+6}{1+3}=\frac{7}{4}=1.75\). 2. Selecting the first unit gives \(T=1\), and selecting the second gives \(T=2\), each with probability \(0.5\). 3. The expected sample ratio is \(E(T)=0.5(1)+0.5(2)=1.5\). 4. The bias is \(E(T)-R=1.5-1.75=-0.25\).

Answer

a) \(R=1.75\). b) \(P(T=1)=P(T=2)=0.5\), and \(E(T)=1.5\). c) Bias: \(-0.25\).
Estimators 40
An estimator \(T\) is unbiased for a positive parameter \(\mu=2\). Its sampling distribution places probability \(0.5\) at \(T=1\) and probability \(0.5\) at \(T=3\). An analyst uses \(\frac{1}{T}\) to estimate \(\frac{1}{\mu}\). a) Verify that \(T\) is unbiased for \(\mu\). b) Find \(E\left(\frac{1}{T}\right)\). c) Find the bias of \(\frac{1}{T}\) as an estimator of \(\frac{1}{\mu}\).

Hints

- Check the original estimator's expected value first. - Transform each of its possible values before finding the new expected value. - Compare the expected reciprocal with the reciprocal of the target.

Solution

1. \(E(T)=0.5(1)+0.5(3)=2=\mu\), so \(T\) is unbiased. 2. The reciprocal takes values \(1\) and \(\frac{1}{3}\), so \(E\left(\frac{1}{T}\right)=0.5(1)+0.5\left(\frac{1}{3}\right)=\frac{2}{3}\). 3. The target is \(\frac{1}{\mu}=\frac{1}{2}\). 4. The bias is \(\frac{2}{3}-\frac{1}{2}=\frac{1}{6}\).

Answer

a) \(E(T)=2\); \(T\) is unbiased. b) \(E\left(\frac{1}{T}\right)=\frac{2}{3}\). c) Bias: \(\frac{1}{6}\).
Estimators 41
An unbiased estimator \(X\) of \(\theta\) has variance \(4\). An analyst adds an independent random value \(C\), equal to \(3\) or \(-3\) with equal probability, and reports \(T=X+C\). a) Show that \(T\) is unbiased. b) Find its variance. c) Explain why replacing \(T\) by the conditional average over the added randomization returns the more precise estimator \(X\).

Hints

- Separate the data-based estimator from the extra random component. - Combine independent variance contributions. - Ask what remains after averaging over only the unnecessary random choice.

Solution

1. The added random value has mean \(0\), so \(E(T)=E(X)+E(C)=\theta\). 2. Its variance is \(4+9=13\) because \(X\) and \(C\) are independent. 3. For fixed observed \(X\), averaging \(X+3\) and \(X-3\) gives \(X\). 4. This removes the irrelevant randomization and reduces variance from \(13\) to \(4\) without changing the expected value.

Answer

a) \(T\) is unbiased. b) \(\operatorname{Var}(T)=13\). c) Averaging out the added coin-flip noise gives \(X\), which remains unbiased and has smaller variance \(4\).
Estimators 42
Independent observations \(X_1\) and \(X_2\) come from a population with unknown mean \(\mu\). a) Show that \(X_1X_2\) is an unbiased estimator of \(\mu^2\). b) Explain why independence is essential. c) Find the estimate when \(X_1=4\) and \(X_2=7\).

Hints

- Take the expected value of the product. - Identify the condition that allows a product of expectations. - Keep the symbolic verification separate from the numerical estimate.

Solution

1. Independence gives \(E(X_1X_2)=E(X_1)E(X_2)=\mu\cdot\mu=\mu^2\). 2. Without independence, \(E(X_1X_2)=\mu^2+\operatorname{Cov}(X_1,X_2)\), so the estimator can be biased. 3. The observed estimate is \(4(7)=28\).

Answer

a) \(E(X_1X_2)=\mu^2\), so it is unbiased. b) Dependence can add a covariance term to the expected product. c) \(28\).
Estimators 43
An unbiased estimator \(T\) of a population proportion \(p=0.10\) has sampling distribution <table> <tr><th>\(T\)</th><th>\(-0.20\)</th><th>\(0.10\)</th><th>\(0.40\)</th></tr> <tr><th>Probability</th><td>\(0.20\)</td><td>\(0.60\)</td><td>\(0.20\)</td></tr> </table> To avoid impossible reported proportions, an analyst clips negative estimates to \(0\), producing \(C=\max(0,T)\). a) Verify that \(T\) is unbiased. b) Find the bias of \(C\). c) Explain why enforcing the parameter boundary changes the long-run center.

Hints

- Compute the center before and after applying the boundary rule. - Transform each possible estimator value, not just the mean. - Check whether the adjustment is symmetric around the target.

Solution

1. \(E(T)=(-0.20)(0.20)+(0.10)(0.60)+(0.40)(0.20)=0.10\), so \(T\) is unbiased. 2. The clipped estimator takes values \(0,0.10,0.40\) with the same probabilities. 3. Its expected value is \(0(0.20)+0.10(0.60)+0.40(0.20)=0.14\). 4. Its bias is \(0.14-0.10=0.04\). The clipping raises low estimates but leaves high estimates unchanged.

Answer

a) \(E(T)=0.10\), so \(T\) is unbiased. b) The clipped estimator has bias \(0.04\). c) Clipping moves negative outcomes upward without an offsetting downward change to positive outcomes.
Estimators 44
Independent observations \(X_1,\dots,X_n\), with \(n\ge2\), have common mean \(\mu\). Let \(L_i\) be the mean after leaving out \(X_i\), and define \(T=\frac{1}{n}\sum_{i=1}^{n}L_i\). For the sample \(3,7,8,14\): a) Find the four leave-one-out means and \(T\). b) Show that \(T=\bar X\) for any sample with at least two observations. c) Is \(T\) an unbiased estimator of \(\mu\)? d) Does averaging all leave-one-out means give a different estimator from the ordinary sample mean?

Hints

- Use the full sample sum to find a mean with one observation left out. - Count how many times each observation contributes when all leave-one-out means are averaged. - Once you have an identity for the estimator, what does it tell you about its expected value?

Solution

1. The full sample sum is \(32\). The four leave-one-out means are \(\frac{29}{3},\frac{25}{3},8,6\). 2. Their mean is \(T=\frac{1}{4}\left(\frac{29}{3}+\frac{25}{3}+8+6\right)=8\). 3. In general, let \(S=\sum_{i=1}^{n}X_i\). Then \(L_i=\frac{S-X_i}{n-1}\) and \(T=\frac{1}{n}\sum_{i=1}^{n}\frac{S-X_i}{n-1}=\frac{nS-S}{n(n-1)}=\frac{S}{n}=\bar X\). 4. Therefore, \(E(T)=E(\bar X)=\mu\), so \(T\) is unbiased. 5. The leave-one-out average is exactly the ordinary sample mean for every such sample, not a new point estimator.

Answer

a) \(\frac{29}{3},\frac{25}{3},8,6\); \(T=8\). b) \(T=\bar X\) for all \(n\ge2\). c) Yes, \(E(T)=\mu\). d) No. The two estimators are identical.
Estimators 45
For a population with mean \(\mu\) and standard deviation \(\sigma\), consider \(T_n=\frac{n}{n+1}\bar X.\) a) Find the bias and standard deviation of \(T_n\). b) For \(\mu=20\), \(\sigma=8\), and \(n=99\), calculate both quantities. Give the standard deviation exactly and to three decimal places. c) Is \(T_n\) consistent for \(\mu\)? Justify your answer using both the bias and standard deviation.

Hints

- Apply the scale factor to both the center and spread of the sample mean. - Compare the scaled center with the target to obtain bias. - For consistency, examine what happens to both systematic error and sampling variability as \(n\) grows.

Solution

1. The expected value is \(E(T_n)=\frac{n}{n+1}\mu,\) so the bias is \(-\frac{\mu}{n+1}.\) 2. Scaling a random variable by \(n/(n+1)\) scales its standard deviation by the same positive factor. Since \(\operatorname{SD}(\bar X)=\frac{\sigma}{\sqrt n},\) we have \(\operatorname{SD}(T_n)=\frac{n}{n+1}\frac{\sigma}{\sqrt n}.\) 3. At \(n=99\), the bias is \(-\frac{20}{100}=-0.2.\) 4. The standard deviation is \(\frac{99}{100}\frac{8}{\sqrt{99}} =\frac{198}{25\sqrt{99}} \approx0.796.\) 5. As \(n\to\infty\), the bias tends to \(0\) and the standard deviation also tends to \(0\). Thus, the estimator becomes concentrated around \(\mu\), so \(T_n\) is consistent.

Answer

a) Bias \(-\frac{\mu}{n+1}\); standard deviation \(\frac{n}{n+1}\frac{\sigma}{\sqrt n}\). b) Bias \(-0.2\); standard deviation \(\frac{198}{25\sqrt{99}}\approx0.796.\) c) Yes. Both its bias and sampling spread tend to \(0\) as \(n\) increases.
Estimators 46
An estimator \(S^2\) is unbiased for a population variance \(\sigma^2=4\). Its sampling distribution is <table> <tr><th>\(S^2\)</th><th>\(1\)</th><th>\(4\)</th><th>\(9\)</th></tr> <tr><th>Probability</th><td>\(\frac{1}{3}\)</td><td>\(\frac{7}{15}\)</td><td>\(\frac{1}{5}\)</td></tr> </table> An analyst uses \(S=\sqrt{S^2}\) to estimate \(\sigma=2\). a) Verify that \(S^2\) is unbiased for \(\sigma^2\). b) Find the expected value and bias of \(S\). Give exact answers and approximations to three decimal places. c) What does this show about taking the square root of an unbiased variance estimator?

Hints

- First calculate the mean of the original estimator's sampling distribution. - Apply the square root to each possible value before calculating a new mean. - Compare the transformed mean with the population standard deviation.

Solution

1. \(E(S^2)=1\left(\frac{1}{3}\right)+4\left(\frac{7}{15}\right)+9\left(\frac{1}{5}\right)=4\), so \(S^2\) is unbiased. 2. The standard-deviation estimator \(S\) takes values \(1,2,3\) with the same probabilities. 3. Its expected value is \(E(S)=\frac{1}{3}+2\left(\frac{7}{15}\right)+3\left(\frac{1}{5}\right)=\frac{28}{15}\approx1.867\). 4. Its bias is \(\frac{28}{15}-2=-\frac{2}{15}\approx-0.133\). 5. The nonlinear square root does not generally preserve unbiasedness.

Answer

a) \(E(S^2)=4\); it is unbiased. b) \(E(S)=\frac{28}{15}\approx1.867\); bias \(-\frac{2}{15}\approx-0.133\). c) Taking the square root of an unbiased variance estimator may produce a biased standard-deviation estimator.
Estimators 47
A preserve is divided into \(600\) equal-area plots. A simple random sample of \(30\) plots has a mean of \(18.4\) mature trees per plot. Let \(T\) be the total number of mature trees in the preserve. a) Give a point estimator for \(T\) and calculate the estimate. b) If the standard deviation of the sampling distribution of \(\bar{X}\) is \(1.2\) trees per plot, find the standard deviation of the total estimator. c) Explain why the total estimator is unbiased when \(\bar{X}\) is unbiased for the population mean per plot.

Hints

- Express the population total in terms of the number of plots and a per-plot mean. - Apply the same scale factor to the estimator's spread. - Verify unbiasedness by comparing expected values symbolically.

Solution

1. The total equals the number of plots times the population mean per plot, so use \(\widehat{T}=600\bar{X}\). 2. The estimate is \(600(18.4)=11{,}040\) trees. 3. Scaling by \(600\) scales the standard deviation to \(600(1.2)=720\) trees. 4. If \(E(\bar{X})=\mu\), then \(E(\widehat{T})=600\mu=T\).

Answer

a) \(\widehat{T}=600\bar{X}\), giving \(11{,}040\) trees. b) \(720\) trees. c) Its expected value is \(600\mu\), which equals the population total.
Estimators 48
Wildlife researchers mark \(60\) turtles and release them. Later, they capture \(50\) turtles, of which \(12\) are marked. A common capture-recapture estimator of the population size is \(\widehat{N}=\frac{n_1n_2}{m}\), where \(n_1\) is the number initially marked, \(n_2\) is the second sample size, and \(m\) is the number of marked turtles recaptured. a) Estimate the turtle population. b) Suppose marked turtles become easier to catch than unmarked turtles. Would this tend to make the estimate too high or too low? Explain.

Hints

- Substitute each observed count into the stated estimator. - Identify which observed quantity would be distorted by unequal catchability. - Track how increasing that quantity changes the estimate.

Solution

1. The estimate is \(\widehat{N}=\frac{60\cdot50}{12}=250\). 2. If marked turtles are easier to catch, the recapture count \(m\) tends to be larger than it would be under equal catchability. 3. Because \(m\) is in the denominator, an inflated recapture count makes \(\widehat{N}\) too small.

Answer

a) Approximately \(250\) turtles. b) Too low. Easier recapture increases the denominator and lowers the population estimate.
Estimators 49
A population is \(40\%\) urban and \(60\%\) rural. Independent samples from the two groups give mean weekly travel distances of \(18\) miles and \(27\) miles. The sampling standard deviations of the two sample means are \(1.2\) miles and \(0.8\) mile. a) Give the correctly weighted estimator of the overall population mean and calculate the estimate. b) Find the estimator's standard deviation. Give the exact value and a decimal approximation to three decimal places. c) Explain why simply averaging \(18\) and \(27\) is not the appropriate estimator.

Hints

- Match each sample mean to its group's share of the population. - Square the weights when combining independent variance contributions. - Keep the exact square-root form until the final approximation. - Distinguish population weighting from equal weighting of the two sample statistics.

Solution

1. Weight each group mean by its population share: \(\widehat{\mu}=0.4\bar X_U+0.6\bar X_R.\) 2. The observed estimate is \(0.4(18)+0.6(27)=23.4\) miles. 3. Independence gives \(\operatorname{Var}(\widehat{\mu}) =0.4^2(1.2)^2+0.6^2(0.8)^2 =0.4608.\) 4. Therefore, \(\operatorname{SD}(\widehat{\mu}) =\sqrt{0.4608} =\frac{12\sqrt2}{25} \approx0.679\) mile. 5. An unweighted average would incorrectly treat the two groups as if they were equally large in the population.

Answer

a) \(\widehat{\mu}=0.4\bar X_U+0.6\bar X_R\), giving \(23.4\) miles. b) \(\frac{12\sqrt2}{25}\approx0.679\) mile. c) The groups have different population shares, so equal weights target the wrong overall population mean.
Estimators 50
A college population is \(20\%\) graduate students and \(80\%\) undergraduate students. A researcher repeatedly samples \(100\) graduate students and \(100\) undergraduate students, then computes the unweighted proportion of sampled students who use a particular service. a) Explain why this unweighted sample proportion need not be centered at the service-use proportion for the full college population. b) Under what condition would the unweighted sample proportion happen to be unbiased for the full-population proportion? c) Describe how the researcher should combine the two group sample proportions to estimate the full-population proportion.

Hints

- Compare each group's share of the sample with its share of the population. - Consider when unequal representation would make no difference to the overall estimate. - Combine group estimates using population shares rather than sample shares.

Solution

1. The sampling design gives graduate students half of the sample even though they are only \(20\%\) of the population. An unweighted average therefore overrepresents graduate students and underrepresents undergraduates. 2. The unweighted sample proportion would be unbiased for the full population only if the two groups had the same true service-use proportion. 3. If \(\hat p_G\) and \(\hat p_U\) are the group sample proportions, the appropriate weighted estimator is \(0.20\hat p_G+0.80\hat p_U\).

Answer

a) The sample composition does not match the population composition, so an unweighted proportion overweights graduate students. b) It is unbiased only if graduate and undergraduate students have the same population service-use proportion. c) Use \(0.20\hat p_G+0.80\hat p_U\).
Estimators 51
In a population, the true proportion of people with a certain condition is \(0.20\). A screening system correctly records a person with the condition as positive with probability \(0.90\), but it also records a person without the condition as positive with probability \(0.05\). A random sample of \(400\) people is screened, with screening outcomes treated as independent, and \(\hat p_{\text{recorded}}\) is the sample proportion recorded as positive. a) Find the probability that a randomly selected person is recorded as positive. b) Find the mean and standard deviation of the sampling distribution of \(\hat p_{\text{recorded}}\). Give the standard deviation exactly and to five decimal places. c) Find the bias of \(\hat p_{\text{recorded}}\) as an estimator of the true condition proportion, and interpret its sign.

Hints

- Split the recorded-positive event into true positives and false positives. - Use the recorded-positive probability, not the true condition proportion, for the sampling distribution of the recorded sample proportion. - Bias is the estimator's sampling mean minus the parameter it is intended to estimate.

Solution

1. A recorded positive can occur in two ways: \(P(\text{recorded positive}) =(0.20)(0.90)+(0.80)(0.05) =0.22.\) 2. Therefore, \(\mu_{\hat p_{\text{recorded}}}=0.22.\) 3. The standard deviation of a sample proportion with recorded-positive probability \(0.22\) is \(\sigma_{\hat p_{\text{recorded}}} =\sqrt{\frac{(0.22)(0.78)}{400}} \approx0.02071.\) 4. The bias relative to the true condition proportion is \(0.22-0.20=0.02.\) The positive sign means the recorded-positive proportion is centered \(0.02\), or \(2\) percentage points, above the true condition proportion.

Answer

a) \(0.22\). b) Mean \(0.22\); standard deviation \(\sqrt{\frac{(0.22)(0.78)}{400}}\approx0.02071.\) c) Bias \(+0.02\), or \(+2\) percentage points; the estimator is biased upward.
Estimators 52
Independent estimators \(A\) and \(B\) are both unbiased for the same parameter \(\theta\). Their variances are \(9\) and \(4\), respectively. Consider \(W=wA+(1-w)B\). a) Explain why \(W\) is unbiased for every real value of \(w\). b) Find the value of \(w\) that gives \(W\) the smallest variance. c) Find that minimum variance.

Hints

- Take the expected value of the weighted combination first. - Use independence when expressing its variance. - Rewrite the variance as a quadratic and locate its minimum.

Solution

1. Since \(E(A)=E(B)=\theta\), \(E(W)=w\theta+(1-w)\theta=\theta\). 2. Independence gives \(\operatorname{Var}(W)=9w^2+4(1-w)^2\). 3. Expanding gives \(13w^2-8w+4\), whose minimum occurs at \(w=\frac{8}{26}=\frac{4}{13}\). 4. The minimum variance is \(9(\frac{4}{13})^2+4(\frac{9}{13})^2=\frac{36}{13}\approx2.769\).

Answer

a) \(W\) is unbiased because its coefficients sum to \(1\). b) \(w=\frac{4}{13}\). c) The minimum variance is \(\frac{36}{13}\approx2.769\).
Estimators 53
A box contains tickets numbered consecutively from \(1\) through an unknown largest number \(N\). To study the estimator that uses the largest observed ticket, suppose the true value is \(N=4\) and two tickets are drawn independently with replacement. Let \(M\) be the larger ticket number drawn. a) Construct the sampling distribution of \(M\). b) Find the bias of \(M\) as an estimator of \(N\).

Hints

- Consider first the event that both ticket numbers are at most a chosen value. - How can you turn cumulative probabilities into probabilities for each possible maximum? - Compare the mean of the maximum's distribution with the true largest label.

Solution

1. For \(m=1,2,3,4\), \(P(M\le m)=\left(\frac{m}{4}\right)^2\). 2. Subtract consecutive cumulative probabilities to obtain point probabilities \(\frac{1}{16},\frac{3}{16},\frac{5}{16},\frac{7}{16}\). 3. The expected maximum is \(E(M)=\frac{1+6+15+28}{16}=\frac{25}{8}=3.125\). 4. The bias is \(E(M)-N=3.125-4=-0.875\).

Answer

a) <table> <tr><th>\(M\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th><th>\(4\)</th></tr> <tr><th>Probability</th><td>\(\frac{1}{16}\)</td><td>\(\frac{3}{16}\)</td><td>\(\frac{5}{16}\)</td><td>\(\frac{7}{16}\)</td></tr> </table> b) The bias is \(-0.875\); the sample maximum underestimates \(N\) on average.
Estimators 54
A population variable equals \(0\) with probability \(0.75\) and \(8\) with probability \(0.25\), so the population mean is \(\mu=2\). Compare two estimators of \(\mu\): \(X\), one randomly selected observation, and \(M\), the median of three independent observations. a) Find the bias of each estimator. b) Find the mean squared error of each estimator. c) Which estimator has the smaller mean squared error?

Hints

- Determine when the middle of three observations is the larger population value. - Separate systematic error from random variation. - Compare total squared error rather than bias alone.

Solution

1. Since \(E(X)=2\), estimator \(X\) is unbiased. Its variance and mean squared error are \(0.75(0-2)^2+0.25(8-2)^2=12\). 2. The median is \(8\) only when at least two observations equal \(8\), with probability \(3(0.25)^2(0.75)+(0.25)^3=\frac{5}{32}\). 3. Thus \(E(M)=8\cdot\frac{5}{32}=1.25\), so its bias is \(-0.75\). 4. The variance of \(M\) is \(64\cdot\frac{5}{32}\cdot\frac{27}{32}=8.4375\). 5. Its mean squared error is \(8.4375+(-0.75)^2=9\), which is smaller than \(12\).

Answer

a) Bias of \(X\): \(0\). Bias of \(M\): \(-0.75\). b) \(\operatorname{MSE}(X)=12\); \(\operatorname{MSE}(M)=9\). c) The biased median estimator \(M\) has the smaller mean squared error.
Estimators 55
A screening test is positive for \(90\%\) of people who have a condition and for \(10\%\) of people who do not. Let \(p\) be the population proportion with the condition, and let \(\hat r\) be the positive-test proportion in a random sample. a) Show that the population positive-test rate is \(r=0.10+0.80p\). b) Construct an unbiased estimator of \(p\) from \(\hat r\). c) If \(\hat r=0.34\), find the estimate of \(p\).

Hints

- Break positive results into contributions from the two underlying groups. - Solve the population relationship for the target proportion. - Apply the same inverse transformation to the sample statistic.

Solution

1. The positive rate is \(r=0.90p+0.10(1-p)=0.10+0.80p\). 2. Solving for \(p\) gives \(p=\frac{r-0.10}{0.80}\), so use \(\hat p_c=\frac{\hat r-0.10}{0.80}\). 3. Since \(E(\hat r)=r\), the corrected estimator has expectation \(p\). 4. For \(\hat r=0.34\), \(\hat p_c=\frac{0.34-0.10}{0.80}=0.30\).

Answer

a) \(r=0.10+0.80p\). b) \(\hat p_c=\frac{\hat r-0.10}{0.80}\). c) \(0.30\).
Estimators 56
Estimator \(A_n\) has bias \(0\) and standard deviation \(\frac{5}{\sqrt n}\). Estimator \(B_n\) has bias \(1\) and standard deviation \(\frac{1}{\sqrt n}\). a) Compare their mean squared errors at \(n=16\). b) Compare their mean squared errors at \(n=100\). c) For which integer sample sizes does \(A_n\) have the smaller mean squared error?

Hints

- Write each total squared error as variance plus squared bias. - Evaluate the two expressions before making a general comparison. - Solve the inequality only after simplifying both sides.

Solution

1. \(\operatorname{MSE}(A_n)=\frac{25}{n}\) and \(\operatorname{MSE}(B_n)=1+\frac{1}{n}\). 2. At \(n=16\), the values are \(1.5625\) and \(1.0625\), so \(B_n\) is better by this criterion. 3. At \(n=100\), the values are \(0.25\) and \(1.01\), so \(A_n\) is better. 4. Solving \(\frac{25}{n}<1+\frac{1}{n}\) gives \(n>24\).

Answer

a) At \(n=16\), \(B_n\) has the smaller mean squared error: \(1.0625\) versus \(1.5625\). b) At \(n=100\), \(A_n\) has the smaller mean squared error: \(0.25\) versus \(1.01\). c) \(A_n\) has the smaller mean squared error for integer \(n\ge25\).
Estimators 57
Two unbiased estimators of \(\theta\) are available. Estimator \(A\) has standard deviation \(2\), and estimator \(B\) has standard deviation \(5\). A procedure randomly uses \(A\) with probability \(0.70\) and \(B\) with probability \(0.30\), independently of the data. Let \(T\) be the selected estimate. a) Is \(T\) unbiased? b) Find its variance. c) Compare its precision with always using \(A\).

Hints

- Condition on which estimator the random procedure selects. - Compare the conditional centers before combining the conditional spreads. - Randomly mixing procedures does not guarantee improved precision.

Solution

1. Both conditional expected values equal \(\theta\), so \(E(T)=0.70\theta+0.30\theta=\theta\). 2. Because the conditional means are identical, the variance is the weighted average of the conditional variances: \(0.70(2^2)+0.30(5^2)=10.3\). 3. Always using \(A\) gives variance \(4\), which is smaller than \(10.3\).

Answer

a) Yes, \(T\) is unbiased. b) \(\operatorname{Var}(T)=10.3\). c) Always using \(A\) is more precise because its variance is only \(4\).
Estimators 58
Two independent Bernoulli observations have success probability \(p\). Let \(D=1\) if the two outcomes are different and \(D=0\) if they are the same. a) Show that \(\frac{D}{2}\) is an unbiased estimator of \(p(1-p)\). b) What estimate is produced when the two observed outcomes are different?

Hints

- List the two ordered ways the observations can disagree. - Find the expected value of the indicator before scaling it. - Use the observed indicator value only after establishing the estimator.

Solution

1. The outcomes differ through success-failure or failure-success, so \(P(D=1)=p(1-p)+(1-p)p=2p(1-p)\). 2. Therefore \(E(D)=2p(1-p)\). 3. It follows that \(E(\frac{D}{2})=p(1-p)\), so the estimator is unbiased. 4. Different observed outcomes give \(D=1\), so the estimate is \(\frac12\).

Answer

a) \(E(\frac{D}{2})=p(1-p)\), so it is unbiased. b) The estimate is \(0.5\).
Estimators 59
Two weather stations send daily data files: Pine Creek sends \(40\) files, and Lake Point sends \(100\). Only one station can be checked. Pine Creek is chosen with probability \(0.25\), and Lake Point with probability \(0.75\). The estimate of the total number of files is the chosen station's count divided by its selection probability. a) Give the two possible estimates and their probabilities. Give exact values. b) Show that this estimator is unbiased for the total. c) What exact estimate is reported if Lake Point is selected?

Hints

- Adjust the chosen station's count for the chance of selecting that station. - Find the expected value of the two possible estimates. - Compare that expected value with the actual combined count.

Solution

1. If Pine Creek is selected, the estimate is \(\frac{40}{0.25}=160\). If Lake Point is selected, it is \(\frac{100}{0.75}=\frac{400}{3}\). 2. The expected estimate is \(0.25(160)+0.75\left(\frac{400}{3}\right)=40+100=140\). 3. Since the actual total is \(140\), the estimator is unbiased. 4. When Lake Point is selected, the estimate is \(\frac{400}{3}\) files. An estimator need not produce a whole-number estimate.

Answer

a) \(160\) with probability \(0.25\); \(\frac{400}{3}\) with probability \(0.75\). b) Its expected value is \(140\), equal to the true total. c) \(\frac{400}{3}\) files.
Estimators 60
A study estimates an unknown population mean \(\mu_Y\) using a sample mean \(\bar Y\). The same sample also gives \(\bar X\) for an auxiliary variable whose population mean is known to be \(50\). Suppose \(\operatorname{Var}(\bar Y)=9\), \(\operatorname{Var}(\bar X)=4\), and \(\operatorname{Cov}(\bar Y,\bar X)=5\). Consider \(T=\bar Y+(50-\bar X)\). a) Show that \(T\) is unbiased for \(\mu_Y\). b) Find its variance and compare it with using \(\bar Y\) alone. c) Calculate the estimate when \(\bar y=68\) and \(\bar x=47\).

Hints

- Use the known population mean to evaluate the adjustment's expected value. - Keep the subtraction sign when including covariance in the variance. - Interpret whether the auxiliary information removes or adds sampling noise.

Solution

1. Since \(E(\bar X)=50\), \(E(T)=\mu_Y+50-50=\mu_Y\). 2. The variance is \(9+4-2(5)=3\). 3. This is smaller than the variance \(9\) of \(\bar Y\), so the adjustment improves precision. 4. The observed estimate is \(68+(50-47)=71\).

Answer

a) \(T\) is unbiased for \(\mu_Y\). b) Its variance is \(3\), compared with \(9\) for \(\bar Y\). c) The estimate is \(71\).
Estimators 61
A population has mean \(\mu\) and standard deviation \(6\). For a sample of size \(n\), let \(\bar X_n\) be the sample mean. An analyst reports \(T_n=\bar X_n+Z\), where \(Z\) is independent of the sample and equals \(2\) or \(-2\) with equal probability. a) Is \(T_n\) unbiased for \(\mu\)? b) Find its standard deviation when \(n=36\). c) Is \(T_n\) consistent for \(\mu\)? Explain.

Hints

- Separate the center contributed by the sample from the center of the added noise. - Combine the independent variance components. - For consistency, examine what remains as the sample size becomes very large.

Solution

1. Since \(E(Z)=0\), \(E(T_n)=E(\bar X_n)+E(Z)=\mu\), so the estimator is unbiased. 2. At \(n=36\), \(\operatorname{Var}(\bar X_n)=\frac{36}{36}=1\), while \(\operatorname{Var}(Z)=4\). 3. Independence gives \(\operatorname{SD}(T_{36})=\sqrt{1+4}=\sqrt5\approx2.236\). 4. As \(n\) grows, the sample-mean variance approaches \(0\), but the added-noise variance remains \(4\). Therefore \(T_n\) does not concentrate at \(\mu\) and is not consistent.

Answer

a) Yes, it is unbiased. b) \(\sqrt5\approx2.236\). c) No. Its variance approaches \(4\), not \(0\), because the added noise never shrinks.
Estimators 62
For \(X\) successes in \(n\) independent trials with success probability \(p\), an analyst uses the smoothed estimator \(S=\frac{X+1}{n+2}\) instead of \(\hat p=\frac{X}{n}\). a) Show that the bias of \(S\) is \(\frac{1-2p}{n+2}\). b) Find its expected value and bias when \(n=18\) and \(p=0.70\). c) Find the observed estimate when \(X=14\).

Hints

- Start from the expected number of successes. - Compare the estimator's expected value with the target using a common denominator. - Keep the long-run bias calculation separate from the estimate produced by one sample.

Solution

1. Since \(E(X)=np\), \(E(S)=\frac{np+1}{n+2}\). 2. Subtracting \(p\) gives \(\frac{np+1-p(n+2)}{n+2}=\frac{1-2p}{n+2}\). 3. For \(n=18\) and \(p=0.70\), the bias is \(\frac{1-1.4}{20}=-0.02\), so the expected value is \(0.68\). 4. With \(X=14\), the estimate is \(\frac{15}{20}=0.75\).

Answer

a) The bias is \(\frac{1-2p}{n+2}\). b) Expected value \(0.68\); bias \(-0.02\). c) \(0.75\).
Estimators 63
A fair coin is flipped until the first head appears or until three flips have been made, whichever comes first. An analyst estimates the head probability \(p\) by the observed proportion of heads among the flips actually made. a) List the possible stopped sequences, their probabilities, and the resulting estimates. b) Find the expected estimate. c) Determine the bias when the true value is \(p=0.5\).

Hints

- Stop each possible sequence exactly when the stated rule requires. - The denominator of the observed proportion is not fixed across outcomes. - Compare the probability-weighted estimate with the true coin probability.

Solution

1. The stopped sequences are \(\mathrm{H}\), \(\mathrm{TH}\), \(\mathrm{TTH}\), and \(\mathrm{TTT}\) with probabilities \(\frac12,\frac14,\frac18,\frac18\). 2. Their estimates are \(1,\frac12,\frac13,0\), respectively. 3. The expected estimate is \(\frac12(1)+\frac14(\frac12)+\frac18(\frac13)=\frac23\). 4. The bias is \(\frac23-\frac12=\frac16\), so the stopping rule causes upward bias.

Answer

a) \(\mathrm{H}\): probability \(\frac12\), estimate \(1\); \(\mathrm{TH}\): \(\frac14\), estimate \(\frac12\); \(\mathrm{TTH}\): \(\frac18\), estimate \(\frac13\); \(\mathrm{TTT}\): \(\frac18\), estimate \(0\). b) \(\frac23\). c) Bias \(\frac16\).
Estimators 64
Measurements taken at times \(t=1,2,3,4,5\) satisfy \(E(X_t)=\mu+\beta t\), where \(\mu\) is the baseline value and \(\beta\) is an unknown linear trend. a) Show that the simple average of the five measurements is not generally unbiased for \(\mu\). b) Construct an unbiased estimator of \(\mu\) using only \(X_1\) and \(X_5\). c) Calculate it when \(X_1=12\) and \(X_5=20\).

Hints

- Write the expected value at each measurement time. - Choose a combination whose trend contributions cancel while the baseline coefficients sum to one. - Verify the proposed rule symbolically before using the observed values.

Solution

1. The expected simple average is \(\mu+\beta\frac{1+2+3+4+5}{5}=\mu+3\beta\), so it is biased for the baseline unless \(\beta=0\). 2. Seek coefficients that preserve \(\mu\) and cancel \(\beta\). The combination \(T=\frac{5X_1-X_5}{4}\) has expectation \(\frac{5(\mu+\beta)-(\mu+5\beta)}{4}=\mu\). 3. The observed estimate is \(\frac{5(12)-20}{4}=10\).

Answer

a) Its expected value is \(\mu+3\beta\), so it is biased for \(\mu\) when a trend is present. b) \(T=\frac{5X_1-X_5}{4}\). c) \(10\).
Estimators 65
A sample mean \(\bar X\) estimates a population mean \(\mu\). For the population being studied, \(\mu=12\) and \(\operatorname{Var}(\bar X)=9\). An analyst considers the shrinkage estimator \(T_a=a\bar X+(1-a)10\), where \(0\le a\le1\). a) Write the bias, variance, and mean squared error of \(T_a\). b) Find the value of \(a\) that minimizes mean squared error. c) Find the minimum mean squared error and the estimate produced when \(\bar x=15\).

Hints

- Find how far the estimator's expected value is from the target. - Only the random sample component contributes to variance. - Minimize the resulting quadratic after combining squared bias and variance.

Solution

1. The expected value is \(12a+10(1-a)=10+2a\), so the bias is \(2a-2=-2(1-a)\). 2. The variance is \(9a^2\). 3. The mean squared error is \(9a^2+4(1-a)^2\). 4. This quadratic is minimized at \(a=\frac{4}{13}\). 5. The minimum mean squared error is \(\frac{36}{13}\approx2.769\). 6. With \(\bar x=15\), the estimate is \(\frac{4}{13}(15)+\frac{9}{13}(10)=\frac{150}{13}\approx11.538\).

Answer

a) Bias \(-2(1-a)\); variance \(9a^2\); MSE \(9a^2+4(1-a)^2\). b) \(a=\frac{4}{13}\). c) Minimum MSE \(\frac{36}{13}\approx2.769\); estimate \(\frac{150}{13}\approx11.538\).
Estimators 66
Independent estimators \(A\) and \(B\) have \(E(A)=\theta+2\), \(E(B)=2\theta-1\), \(\operatorname{Var}(A)=9\), and \(\operatorname{Var}(B)=16\). Consider an estimator with no added constant, \(T=aA+bB\). a) Find \(a\) and \(b\) so that \(T\) is unbiased for every value of \(\theta\). b) Find the variance of \(T\). c) Calculate the estimate when \(A=14\) and \(B=21\).

Hints

- Expand the expected value into a term containing the parameter and a constant term. - Compare each part with what an unbiased estimator requires. - Use independence when finding the variance of the resulting combination.

Solution

1. The expected value is \(E(T)=(a+2b)\theta+(2a-b)\). 2. For this to equal \(\theta\) for all \(\theta\), require \(a+2b=1\) and \(2a-b=0\). 3. Solving gives \(a=\frac{1}{5}\) and \(b=\frac{2}{5}\). 4. Independence gives \(\operatorname{Var}(T)=\left(\frac{1}{5}\right)^2\cdot9+\left(\frac{2}{5}\right)^2\cdot16=\frac{73}{25}=2.92\). 5. For \(A=14\) and \(B=21\), the estimate is \(\frac{1}{5}\cdot14+\frac{2}{5}\cdot21=\frac{56}{5}=11.2\).

Answer

a) \(a=\frac{1}{5}\); \(b=\frac{2}{5}\). b) \(\operatorname{Var}(T)=\frac{73}{25}=2.92\). c) \(11.2\).

All problems may be used, copied and printed free of charge for school and tutoring, including paid tutoring. Commercial adaptations as well as publication or redistribution on the internet are not permitted.