Edexcel A-Level Mathematics A2 S3.3.1 Concepts of Standard Error QuestionsPractise unbiased estimates, sample variance and standard error, including estimator efficiency, with verified Edexcel IAL evidence.SyllabusFirst assessment 2019CourseMathematics YMA01LevelA2
Exam pointscalculate x̄ and s² from totals or coded data, using n - 1 for the unbiased varianceestimate the standard error from a combined sample variance and sample sizeshow an estimator is unbiased with E(U)=µ, then choose the smaller-variance estimator
Edexcel A-Level Mathematics A2 S3.3.1 Concepts of Standard Error Questions question 1[Maximum number: 8]A random sample of two observations X1X_{1}X1 and X2X_{2}X2 is taken from a population with unknown mean μ\muμ and unknown variance σ2\sigma^{2}σ2Question (a)(a)Explain what you understand by an unbiased estimator for μ\muμTwo estimators for μ\muμ are U1U_{1}U1 and U2U_{2}U2 whereU1=3X1−2X2 and U2=X1+3X24U_{1}=3 X_{1}-2 X_{2} \quad \text { and } \quad U_{2}=\frac{X_{1}+3 X_{2}}{4}U1=3X1−2X2 and U2=4X1+3X2[ 1 ]Mark as masteredShow AnswerAn estimator for μ\muμ is unbiased if its expected value is equal to μ\muμ(1)Question (b)(b)Show that both U1U_{1}U1 and U2U_{2}U2 are unbiased estimators for μ\muμThe most efficient estimator among a group of unbiased estimators is the one with the smallest variance.[ 3 ]Mark as masteredShow AnswerE(U1)=3E(X1)−2E(X2)\mathrm{E}\left(U_{1}\right)=3 \mathrm{E}\left(X_{1}\right)-2 \mathrm{E}\left(X_{2}\right)E(U1)=3E(X1)−2E(X2) or E(U2)=14(E(X1)+3E(X2))\mathrm{E}\left(U_{2}\right)=\frac{1}{4}\left(\mathrm{E}\left(X_{1}\right)+3 \mathrm{E}\left(X_{2}\right)\right)E(U2)=41(E(X1)+3E(X2))M1E(U1)=3μ−2μ=μ\mathrm{E}\left(U_{1}\right)=3 \mu-2 \mu=\muE(U1)=3μ−2μ=μ (therefore unbiased)A1csoE(U2)=14(μ+3μ)=μ\mathrm{E}\left(U_{2}\right)=\frac{1}{4}(\mu+3 \mu)=\muE(U2)=41(μ+3μ)=μ (therefore unbiased)A1csoQuestion (c)(c)By finding the variance of U1U_{1}U1 and the variance of U2U_{2}U2 state, giving a reason, the most efficient estimator for μ\muμ from these two estimators.[ 4 ]Mark as masteredShow AnswerVar(U1)=9Var(X1)+4Var(X2)=13σ2\operatorname{Var}(U_1)=9\operatorname{Var}(X_1)+4\operatorname{Var}(X_2)=13\sigma^2Var(U1)=9Var(X1)+4Var(X2)=13σ2Var(U2)=116Var(X1)+916Var(X2)=58σ2\operatorname{Var}(U_2)=\frac{1}{16}\operatorname{Var}(X_1)+\frac{9}{16}\operatorname{Var}(X_2)=\frac{5}{8}\sigma^2Var(U2)=161Var(X1)+169Var(X2)=85σ2As Var(U1)>Var(U2)\operatorname{Var}(U_1)>\operatorname{Var}(U_2)Var(U1)>Var(U2), U2U_2U2 is the most efficient estimator for μ\muμ.M1 for use of a2Var(X1)+b2Var(X2)a^2\operatorname{Var}(X_1)+b^2\operatorname{Var}(X_2)a2Var(X1)+b2Var(X2).A1 for 13σ213\sigma^213σ2, A1 for 5σ2/85\sigma^2/85σ2/8, A1 for U2U_2U2 with a correct reason.Add to Test
Question (a)(a)Explain what you understand by an unbiased estimator for μ\muμTwo estimators for μ\muμ are U1U_{1}U1 and U2U_{2}U2 whereU1=3X1−2X2 and U2=X1+3X24U_{1}=3 X_{1}-2 X_{2} \quad \text { and } \quad U_{2}=\frac{X_{1}+3 X_{2}}{4}U1=3X1−2X2 and U2=4X1+3X2[ 1 ]Mark as masteredShow AnswerAn estimator for μ\muμ is unbiased if its expected value is equal to μ\muμ(1)
Question (b)(b)Show that both U1U_{1}U1 and U2U_{2}U2 are unbiased estimators for μ\muμThe most efficient estimator among a group of unbiased estimators is the one with the smallest variance.[ 3 ]Mark as masteredShow AnswerE(U1)=3E(X1)−2E(X2)\mathrm{E}\left(U_{1}\right)=3 \mathrm{E}\left(X_{1}\right)-2 \mathrm{E}\left(X_{2}\right)E(U1)=3E(X1)−2E(X2) or E(U2)=14(E(X1)+3E(X2))\mathrm{E}\left(U_{2}\right)=\frac{1}{4}\left(\mathrm{E}\left(X_{1}\right)+3 \mathrm{E}\left(X_{2}\right)\right)E(U2)=41(E(X1)+3E(X2))M1E(U1)=3μ−2μ=μ\mathrm{E}\left(U_{1}\right)=3 \mu-2 \mu=\muE(U1)=3μ−2μ=μ (therefore unbiased)A1csoE(U2)=14(μ+3μ)=μ\mathrm{E}\left(U_{2}\right)=\frac{1}{4}(\mu+3 \mu)=\muE(U2)=41(μ+3μ)=μ (therefore unbiased)A1cso
Question (c)(c)By finding the variance of U1U_{1}U1 and the variance of U2U_{2}U2 state, giving a reason, the most efficient estimator for μ\muμ from these two estimators.[ 4 ]Mark as masteredShow AnswerVar(U1)=9Var(X1)+4Var(X2)=13σ2\operatorname{Var}(U_1)=9\operatorname{Var}(X_1)+4\operatorname{Var}(X_2)=13\sigma^2Var(U1)=9Var(X1)+4Var(X2)=13σ2Var(U2)=116Var(X1)+916Var(X2)=58σ2\operatorname{Var}(U_2)=\frac{1}{16}\operatorname{Var}(X_1)+\frac{9}{16}\operatorname{Var}(X_2)=\frac{5}{8}\sigma^2Var(U2)=161Var(X1)+169Var(X2)=85σ2As Var(U1)>Var(U2)\operatorname{Var}(U_1)>\operatorname{Var}(U_2)Var(U1)>Var(U2), U2U_2U2 is the most efficient estimator for μ\muμ.M1 for use of a2Var(X1)+b2Var(X2)a^2\operatorname{Var}(X_1)+b^2\operatorname{Var}(X_2)a2Var(X1)+b2Var(X2).A1 for 13σ213\sigma^213σ2, A1 for 5σ2/85\sigma^2/85σ2/8, A1 for U2U_2U2 with a correct reason.