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Estimators

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Difficulty: 1 – 5

54758912

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: \(\$1450\).

54759012

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 \(\hat p\).

54759112

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 \(\theta\).

54759812

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 \(n=100\).

54761112

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) \(14.8\).

54761712

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) \(31\).

54763412

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. c) Which estimator is more efficient?

Hints

- Compare the expected values before considering spread. - Account for the factor outside the sum when finding the second estimator's variance. - Among unbiased estimators, smaller variance means greater efficiency.

Solution

1. Both estimators have expected value \(\mu\), so both are unbiased. 2. Estimator \(A\) has variance \(36\) and standard deviation \(6\). 3. Independence gives \(\operatorname{Var}(B)=\frac14(36+36)=18\), so its standard deviation is \(\sqrt{18}=3\sqrt2\approx4.243\). 4. Estimator \(B\) is more efficient because it has smaller variance.

Answer

a) Both are unbiased. b) \(A\): variance \(36\), standard deviation \(6\). \(B\): variance \(18\), standard deviation \(3\sqrt2\approx4.243\). c) \(B\) is more efficient.

54759212

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) \(20.5\).

54759312

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.

54759612

Three independent measurements \(X_1,X_2,X_3\) each have mean \(\mu\) and 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

- Examine the sum of the coefficients for each weighted average. - Square coefficients when combining independent variance contributions. - Precision is determined by sampling variability, not by how complicated the estimator looks.

Solution

1. The coefficients in each estimator sum 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. Since \(\frac{\sigma^2}{3}<\frac{3\sigma^2}{8}\), estimator \(A\) is more precise.

Answer

a) Both are unbiased for \(\mu\). b) \(\operatorname{Var}(A)=\frac{\sigma^2}{3}\) and \(\operatorname{Var}(B)=\frac{3\sigma^2}{8}\). c) \(A\) is more precise because it has the smaller variance.

54759912

Independent random samples from two populations produce sample means \(\bar x_A=74.2\) and \(\bar x_B=69.8\). Their sampling standard deviations are \(1.5\) and \(1.2\), respectively. a) Give the point estimate of \(\mu_A-\mu_B\). b) Find the standard deviation of the estimator \(\bar X_A-\bar X_B\). c) State one condition needed for the calculation in part b).

Hints

- Use the same subtraction order for the estimate as for the target parameter. - Combine independent uncertainty through variances rather than standard deviations. - Identify the assumption that removes a covariance term.

Solution

1. The point estimate is \(74.2-69.8=4.4\). 2. For independent samples, the variance of the difference is \(1.5^2+1.2^2=3.69\). 3. The standard deviation is \(\sqrt{3.69}\approx1.921\). 4. The variance calculation requires the two sample statistics to be independent.

Answer

a) \(4.4\). b) Approximately \(1.921\). c) The two samples, and therefore the two sample means, must be independent.

54760312

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) \(6.5\).

54760412

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 \(9\).

54760512

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 data \(10,13,16\), calculate the corrected estimate.

Hints

- Add the weights before judging the estimator's expected value. - Determine the multiplicative factor relating its center to the target. - Correct the scale before substituting the observed data.

Solution

1. The coefficients sum to \(0.9\), so \(E(T)=0.9\mu\). 2. The bias is \(0.9\mu-\mu=-0.1\mu\). 3. Dividing by \(0.9\) gives the unbiased estimator \(U=\frac{T}{0.9}\). 4. The observed value is \(T=0.2(10)+0.3(13)+0.4(16)=12.3\). 5. The corrected estimate is \(\frac{12.3}{0.9}\approx13.667\).

Answer

a) The bias is \(-0.1\mu\). b) \(U=\frac{T}{0.9}\). c) Approximately \(13.667\).

54760712

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 \(99.36\).

54761212

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

- Identify the multiplicative shrinkage in the stated expected value. - Reverse that shrinkage without adding a constant. - Substitute the sample size before applying the correction to the observed maximum.

Solution

1. The expected sample maximum is too small by the factor \(\frac{n}{n+1}\). 2. Multiply by its reciprocal to form \(U=\frac{n+1}{n}M\). 3. Then \(E(U)=\frac{n+1}{n}E(M)=\theta\). 4. For \(n=5\), \(U=\frac65(18)=21.6\).

Answer

a) \(U=\frac{n+1}{n}M\). b) \(21.6\).

54761412

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.

54761512

For a random sample of size \(n\), 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 stated expected value with the target variance. - Undo the multiplicative factor causing the downward bias. - Recognize how changing the divisor changes the estimator.

Solution

1. The estimator \(V\) is too small on average by the factor \(\frac{n-1}{n}\). 2. Multiply by \(\frac{n}{n-1}\) to obtain \(S^2=\frac{n}{n-1}V\). 3. This is equivalent to dividing the sum of squared deviations by \(n-1\). 4. For \(n=10\), the estimate is \(\frac{10}{9}(18)=20\).

Answer

a) \(S^2=\frac{n}{n-1}V=\frac{1}{n-1}\sum_{i=1}^n(X_i-\bar X)^2\). b) \(20\).

54761612

A study measures the same participants before and after a program. Let \(\bar X_{\text{before}}\) and \(\bar X_{\text{after}}\) have variances \(25\) and \(16\), with covariance \(18\). The observed means are \(68\) and \(72\). a) Give the point estimate of the population mean change, defined as after minus before. b) Find the variance and standard deviation of this estimator. c) Compare its variance with the value obtained by incorrectly treating the two means as independent.

Hints

- Preserve the requested subtraction order. - Include the dependence term for two measurements on the same people. - Compare what positive association does to the uncertainty of a difference.

Solution

1. The point estimate is \(72-68=4\). 2. The variance of the paired difference is \(16+25-2(18)=5\). 3. Its standard deviation is \(\sqrt5\approx2.236\). 4. Ignoring the pairing would give variance \(16+25=41\), greatly overstating the uncertainty.

Answer

a) \(4\). b) Variance \(5\); standard deviation \(\sqrt5\approx2.236\). c) Treating the means as independent gives variance \(41\), which is much larger.

54761812

For a sample proportion \(\hat p\) based on \(n\) independent trials, it is known that \(E[\hat p(1-\hat p)]=\frac{n-1}{n}p(1-p)\). a) Construct an unbiased estimator of the population quantity \(p(1-p)\). b) For \(n=50\) and \(\hat p=0.30\), calculate the estimate.

Hints

- Identify the multiplicative factor between the estimator's expected value and the target. - Reverse that factor to remove the bias. - Apply the adjustment to the entire plug-in expression.

Solution

1. The plug-in quantity is too small on average by the factor \(\frac{n-1}{n}\). 2. Multiply by \(\frac{n}{n-1}\) to obtain \(U=\frac{n}{n-1}\hat p(1-\hat p)\). 3. For \(n=50\) and \(\hat p=0.30\), \(U=\frac{50}{49}(0.30)(0.70)=\frac{3}{14}\approx0.2143\).

Answer

a) \(U=\frac{n}{n-1}\hat p(1-\hat p)\). b) \(\frac{3}{14}\approx0.2143\).

54761912

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.

54762312

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.

54762512

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 \(1/T\) to estimate \(1/\mu\). a) Verify that \(T\) is unbiased for \(\mu\). b) Find \(E(1/T)\). c) Find the bias of \(1/T\) as an estimator of \(1/\mu\).

Hints

- Check the original estimator's center first. - Transform each possible estimator value before averaging. - A nonlinear transformation need not preserve unbiasedness.

Solution

1. \(E(T)=0.5(1)+0.5(3)=2=\mu\). 2. The reciprocal estimator takes values \(1\) and \(\frac13\), so \(E(1/T)=0.5(1)+0.5(\frac13)=\frac23\). 3. The target reciprocal is \(\frac1\mu=\frac12\). 4. The bias is \(\frac23-\frac12=\frac16\).

Answer

a) \(E(T)=2\), so \(T\) is unbiased. b) \(E(1/T)=\frac23\). c) The bias is \(\frac16\).

54762612

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 \(4\).

54762812

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) \(28\).

54762912

Two estimators target a population center \(\theta=100\): the sample mean and a trimmed mean. Simulation gives the following sampling-distribution summaries. <table> <tr><th>Population condition</th><th>Estimator</th><th>Sampling mean</th><th>Sampling standard deviation</th></tr> <tr><td>Clean</td><td>Sample mean</td><td>\(100\)</td><td>\(2\)</td></tr> <tr><td>Clean</td><td>Trimmed mean</td><td>\(99.5\)</td><td>\(2.4\)</td></tr> <tr><td>Occasional extreme values</td><td>Sample mean</td><td>\(107\)</td><td>\(5\)</td></tr> <tr><td>Occasional extreme values</td><td>Trimmed mean</td><td>\(101\)</td><td>\(3\)</td></tr> </table> Compute the mean squared error of each estimator under each condition. Which estimator is preferred under each condition by this criterion?

Hints

- For each table row, compare the sampling center with the target. - Add squared bias to variance. - Allow the preferred estimator to depend on the population condition.

Solution

1. Under clean conditions, the sample mean has MSE \(2^2+0^2=4\). 2. The trimmed mean has MSE \(2.4^2+(-0.5)^2=5.76+0.25=6.01\). 3. With extreme values, the sample mean has MSE \(5^2+7^2=74\). 4. The trimmed mean has MSE \(3^2+1^2=10\). 5. The sample mean is preferred for the clean population, while the trimmed mean is preferred under contamination.

Answer

Clean population: sample mean MSE \(4\); trimmed mean MSE \(6.01\). Prefer the sample mean. Occasional extreme values: sample mean MSE \(74\); trimmed mean MSE \(10\). Prefer the trimmed mean.

54763312

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. c) Is \(T_n\) consistent for \(\mu\)?

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. - Examine both expressions as the sample size increases without bound.

Solution

1. The expected value is \(\frac{n}{n+1}\mu\), so the bias is \(-\frac{\mu}{n+1}\). 2. Its standard deviation is \(\frac{n}{n+1}\frac{\sigma}{\sqrt n}\). 3. At \(n=99\), the bias is \(-\frac{20}{100}=-0.2\). 4. The standard deviation is \(\frac{99}{100}\frac{8}{\sqrt{99}}\approx0.796\). 5. As \(n\) grows, both the bias and standard deviation approach \(0\), so the estimator 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 approximately \(0.796\). c) Yes, it is consistent.

54793712

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.

54793812

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.

54849612

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.

54759412

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 \(\frac{36}{13}\approx2.769\).

54759512

A box is known to contain 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

- It may be easier to first find the probability that both draws are at most a given value. - Convert cumulative probabilities into point probabilities. - Compare the estimator's long-run center with the true largest label.

Solution

1. For \(m=1,2,3,4\), \(P(M\le m)=(\frac{m}{4})^2\). 2. Therefore the point probabilities are \(\frac1{16},\frac3{16},\frac5{16},\frac7{16}\). 3. The expected value 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>\(\frac1{16}\)</td><td>\(\frac3{16}\)</td><td>\(\frac5{16}\)</td><td>\(\frac7{16}\)</td></tr> </table> b) The bias is \(-0.875\), so the sample maximum underestimates \(N\) on average.

54759712

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.

54760012

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) \(0.30\).

54760112

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 \(n\ge25\).

54760212

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 \(4\).

54760612

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 \(0.5\).

54760812

Two remote stations produce \(40\) and \(100\) data files in a day. Only one station can be inspected: station A is selected with probability \(0.25\), and station B with probability \(0.75\). To estimate the total number of files, use the observed station's count divided by its selection probability. a) Give the two possible estimates and their probabilities. b) Show that this estimator is unbiased for the total. c) What estimate is reported if station B is selected?

Hints

- Adjust each observed value by how likely that unit was to be selected. - Weight each possible estimate by its selection probability. - Compare the resulting expected value with the sum of both station counts.

Solution

1. If A is selected, the estimate is \(\frac{40}{0.25}=160\); if B is selected, it is \(\frac{100}{0.75}=\frac{400}{3}\approx133.333\). 2. The expected estimate is \(0.25(160)+0.75(\frac{400}{3})=40+100=140\). 3. The true total is \(40+100=140\), so the estimator is unbiased. 4. Selecting B produces the estimate \(\frac{400}{3}\approx133.333\).

Answer

a) \(160\) with probability \(0.25\), and \(\frac{400}{3}\approx133.333\) with probability \(0.75\). b) Its expected value is \(140\), equal to the true total. c) Approximately \(133.333\) files.

54760912

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 \(71\).

54761012

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.

54761312

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) \(0.75\).

54762012

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 \(0.55\).

54762112

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 \(\frac16\).

54762212

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.

54762412

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, 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

- Compute the ratio of totals before examining the sample rule. - List the estimator value from each possible selected unit. - Do not assume that a ratio of sample quantities inherits unbiasedness.

Solution

1. The true ratio is \(R=\frac{1+6}{1+3}=\frac74=1.75\). 2. Selecting the first unit gives \(T=1\), and selecting the second gives \(T=2\); each has probability \(0.5\). 3. The expected value is \(E(T)=0.5(1)+0.5(2)=1.5\). 4. The bias is \(1.5-1.75=-0.25\).

Answer

a) \(R=1.75\). b) \(T=1\) or \(2\), each with probability \(0.5\), and \(E(T)=1.5\). c) The bias is \(-0.25\).

54762712

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) \(10\).

54763012

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.

54763112

A sample is \(3,7,8,14\). Form four leave-one-out means by omitting one observation at a time. a) Calculate the four leave-one-out means. b) Average them. c) Show algebraically that for any sample of size \(n\), the average of all \(n\) leave-one-out means equals the full sample mean \(\bar X\).

Hints

- Use the full sample total to calculate each omitted-case mean efficiently. - Track how many times each observation appears across all reduced samples. - Simplify the general average in terms of the original sample total.

Solution

1. The leave-one-out means are \(\frac{29}{3},\frac{25}{3},8,6\). 2. Their average is \(\frac14(\frac{29}{3}+\frac{25}{3}+8+6)=8\), equal to the full sample mean. 3. If the sample total is \(S\), the mean omitting \(X_i\) is \(\frac{S-X_i}{n-1}\). 4. Averaging these gives \(\frac1n\sum_{i=1}^n\frac{S-X_i}{n-1}=\frac{nS-S}{n(n-1)}=\frac{S}{n}=\bar X\).

Answer

a) \(\frac{29}{3},\frac{25}{3},8,6\). b) \(8\). c) The average is always \(\bar X\) because each observation is included in exactly \(n-1\) of the leave-one-out totals.

54763212

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 \(\frac{150}{13}\approx11.538\).

54763512

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>\(\frac13\)</td><td>\(\frac7{15}\)</td><td>\(\frac15\)</td></tr> </table> An analyst uses \(S=\sqrt{S^2}\) to estimate the population standard deviation \(\sigma=2\). a) Verify that \(S^2\) is unbiased for \(\sigma^2\). b) Find the expected value and bias of \(S\). c) What does this show about applying a square root to an unbiased variance estimator?

Hints

- Check the probability-weighted mean before transforming the estimator. - Apply the square root to each possible value, then average again. - Do not interchange a nonlinear transformation and expected value.

Solution

1. \(E(S^2)=1(\frac13)+4(\frac7{15})+9(\frac15)=4\), so \(S^2\) is unbiased. 2. The estimator \(S\) takes values \(1,2,3\) with the same probabilities. 3. Its expected value is \(\frac13+2(\frac7{15})+3(\frac15)=\frac{28}{15}\approx1.867\). 4. The bias is \(\frac{28}{15}-2=-\frac{2}{15}\approx-0.133\). 5. A nonlinear square-root transformation does not preserve unbiasedness.

Answer

a) \(E(S^2)=4\). b) \(E(S)=\frac{28}{15}\approx1.867\), with bias \(-\frac{2}{15}\approx-0.133\). c) Taking the square root of an unbiased variance estimator can produce a biased standard-deviation estimator.

54763612

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 parameter term and a constant term. - Match both parts to the target for every possible parameter value. - After finding the coefficients, use independence to combine the variance contributions.

Solution

1. The expected value is \(E(T)=a(\theta+2)+b(2\theta-1)=(a+2b)\theta+(2a-b)\). 2. Unbiasedness for every \(\theta\) requires \(a+2b=1\) and \(2a-b=0\). 3. Solving gives \(a=\frac15\) and \(b=\frac25\). 4. Independence gives \(\operatorname{Var}(T)=(\frac15)^2(9)+(\frac25)^2(16)=\frac{73}{25}=2.92\). 5. The observed estimate is \(\frac15(14)+\frac25(21)=\frac{56}{5}=11.2\).

Answer

a) \(a=\frac15\) and \(b=\frac25\). b) \(\operatorname{Var}(T)=\frac{73}{25}=2.92\). c) \(11.2\).

54793912

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. c) Explain why simply averaging \(18\) and \(27\) is not the appropriate estimator.

Hints

- Match each sample statistic to its group's share of the population. - Square both weights when combining independent uncertainty. - Distinguish equal sample treatment from correct population weighting.

Solution

1. Weight each group mean by its population share: \(\widehat{\mu}=0.4\bar{X}_U+0.6\bar{X}_R\). 2. The estimate is \(0.4(18)+0.6(27)=23.4\) miles. 3. Independence gives variance \(0.4^2(1.2)^2+0.6^2(0.8)^2=0.4608\). 4. The standard deviation is \(\sqrt{0.4608}\approx0.679\) mile. 5. An unweighted average would incorrectly treat the two groups as equally large.

Answer

a) \(0.4\bar{X}_U+0.6\bar{X}_R\), giving \(23.4\) miles. b) Approximately \(0.679\) mile. c) The groups are not equally represented in the population, so equal weights target the wrong overall mean.

54845512

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 \(0.20\hat p_G+0.80\hat p_U\).

54855812

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 \(\hat p_{\text{recorded}}\). c) Find the bias of \(\hat p_{\text{recorded}}\) as an estimator of the true condition proportion.

Hints

- Break the recorded-positive event into the two ways it can occur. - Use the recorded outcome probability, not the true condition proportion, for the sampling distribution. - Compare the estimator’s expected value with the parameter it is intended to estimate.

Solution

1. The recorded-positive probability is \((0.20)(0.90)+(0.80)(0.05)=0.22\). 2. Therefore, \(\mu_{\hat p_{\text{recorded}}}=0.22\). 3. The expected recorded-positive and recorded-negative counts are \(400(0.22)=88\) and \(400(0.78)=312\), so the sampling distribution is approximately normal. 4. The standard deviation is \(\sqrt{\frac{(0.22)(0.78)}{400}}\approx 0.02071\). 5. The bias relative to the true condition proportion is \(0.22-0.20=0.02\).

Answer

a) \(0.22\). b) The mean is \(0.22\), and the standard deviation is approximately \(0.02071\). c) The estimator has an upward bias of \(0.02\), or \(2\) percentage points.

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