## MS Mod 1 Background and data visualization – practice problems

 Author Message NEAS Supreme Being         Group: Administrators Posts: 4.2K, Visits: 1.2K MS Module 1 Background and data visualization – practice problems(The attached PDF file has better formatting.)Exercise 1.1: BiasAn estimator has a standard error of 2 and a mean squared error of 6. A.    What is the variance of the estimator?B.    What is the bias of the estimator?Part A: The variance is the square of the standard error = 22 = 4.Part B: The mean squared error = the variance + the square of the bias ➾ the bias = the square root of (the mean squared error – the variance) = √(6 – 22) = √2, Exercise 1.2: Bias, variance, and mean squared error ●    Let D be the number of deaths in a sample of N = 1,000 men age 40. ●    An actuary estimates the mortality rate as (D + 2) / (N + 4). ●    The true mortality rate (p) is 5%.A.    What is the bias of the estimator? B.    What is the variance of the estimator? C.    What is the mean squared error? Part A: The bias of the estimator is E(x+2 / n+4) – p = (1/ n+4)E(x+2) – p = (np+2) / (n+4) – p = (2/n – 4p/n) / (1 + 4/n) =(2/1,000 – 4 × 5% / 1,000) / (1 + 4/1,000) = 0.001793Part B: The variance of the estimator is var ( (x+2) / (n+4) ) = 1/(n+4)2 × var(x+2) = var(x) / (n+4)2 = np(1–p)/(n+4)2 = p(1–p)/(n+8 + 16/n) =(5% × (1 – 5%) ) / (1,000 + 8 + 16/1,000) = 0.000047Part C: The mean squared error isMSE = p(1–p)/(n+8 + 16/n) + [ ( 2/n – 4p/n) / (1 + 4/n) ]2 =(5% × (1 – 5%) ) / (1,000 + 8 + 16/1,000) + ( (2/1,000 – 4 × 5% / 1,000) / (1 + 4/1,000) )2 = 0.000050 Exercise 1.3: Bias, variance, and mean squared error ●    Let D be the number of deaths in a sample of N = 1,000 men age 40. ●    An actuary estimates the mortality rate as (D + 2) / (N + 4). ●    The standard deviation of the estimator is 0.01.A.    What is the variance of the estimator? B.    What is the mortality rate (p)?C.    What is the bias of the estimator? D.    What is the mean squared error? Part A: The variance is the square of the standard deviation = 0.012 = 0.0001.Part B: The variance of the estimator is var ( (x+2) / (n+4) ) = 1/(n+4)2 × var(x+2) = var(x) / (n+4)2 = np(1–p)/(n+4)2 = p(1–p)/(n+8 + 16/n) ➾ p(1–p) = variance × (n+8 + 16/n) = 0.0001 × (1000 + 8 + 16/1000) = 0.1008016The solutions are (+1 ± ( (–1)2 – 4 × 1 × 0.1008016)0.5 ) / (2 × 1):●    (1 + ( (–1)2 – 4 × 1 × 0.1008016)0.5 ) / (2 × 1) = 0.886262●    (1 – ( (–1)2 – 4 × 1 × 0.1008016)0.5 ) / (2 × 1) = 0.113738The mortality rate is 88.63% or 11.37%.Part C: The bias of the estimator is E(x+2 / n+4) – p = (1/ n+4)E(x+2) – p = (np+2) / (n+4) – p = (2/n – 4p/n) / (1 + 4/n) =(2/1,000 – 4 × 11.3738% / 1,000) / (1 + 4/1,000) = 0.001539Part D: The mean squared error = the variance + the square of the bias =MSE = p(1–p)/(n+8 + 16/n) + [ ( 2/n – 4p/n) / (1 + 4/n) ]2 = 0.0001 + 0.0015392 = 0.000102 Exercise 1.4: Bias, variance, and mean squared error ●    Let D be the number of deaths in a sample of N = 1,000 men age 40. ●    An actuary estimates the mortality rate as (D + 2) / (N + 4). ●    The bias of the estimator is 0.0015.A.    What is the mortality rate (p)?B.    What is the variance of the estimator? C.    What is the mean squared error? Part A: The bias of the estimator is E(x+2 / n+4) – p = (1/ n+4)E(x+2) – p = (np+2) / (n+4) – p = (2/n – 4p/n) / (1 + 4/n).Solving for the mortality rate p givesp = (2/n – bias × (1 + 4/n) ) / (4/n) = (2 / 1000 – 0.0015 × (1 + 4/1000) ) / (4 / 1000) = 0.123500Part B: The variance of the estimator is var ( (x+2) / (n+4) ) = 1/(n+4)2 × var(x+2) = var(x) / (n+4)2 = np(1–p)/(n+4)2 = p(1–p)/(n+8 + 16/n) =(12.35% × (1 – 12.35%) ) / (1,000 + 8 + 16/1,000) = 0.000107.Part C: The mean squared error isMSE = p(1–p)/(n+8 + 16/n) + [ ( 2/n – 4p/n) / (1 + 4/n) ]2 =(12.35% × (1–12.35%) ) / (1,000 + 8 + 16/1,000) + ( (2/1,000 – 4 × 12.35%/1,000) / (1 + 4/1,000) )2 = 0.000110 Exercise 1.5: Normal distributionA statistician draws a sample of 8 values from a normal distribution. The summary statistics are●    xi = 7,113●    xi2 = 6,991,551A.    What is the estimate of μ?B.    What is the estimated variance of the population?C.    What is the estimated standard deviation of the population?D.    What is the standard error of the sample mean?E.    What is the maximum likelihood estimator of σ, the standard deviation of the population? Part A: The estimate of μ is 7,113 / 8 = 889.125Part B: The estimated variance of the population is(6,991,551 – (7,113)2 / 8 ) / (8 – 1) = 95,314.98Part C: The estimated standard deviation of the population is 95,314.980.5 = 308.73Part D: The standard error of the sample mean = 308.73 / 80.5 = 109.152538Part E: The maximum likelihood estimator of σ, the standard deviation of the population, is((6,991,551 – (7,113)2 / 8 ) / 8)0.5 = 288.79The maximum likelihood estimator uses N instead of N-1 as the denominator, so it is biased for both the variance and the standard deviation. Exercise 1.6: Mortality rateAn actuary studying mortality rates in Country W finds 20 deaths among 40,000 men age 40. The estimated mortality rate is the number of deaths / the sample size.A.    What is the estimated variance of the mortality rate for men age 40?B.    What is the standard error of the mortality rate for men age 40?Part A: The mortality rate is a percentage, and the variance of a sample percentage p is np(1–p).The estimated mortality rate is 20 / 40,000 = 0.00050, so the estimated variance is 0.0005 × (1 – 0.0005) / 40,000 = 0.000000012494Part B: The standard error of the mortality rate for men age 40 is (0.0005 × (1 – 0.0005) / 40,000)0.5 = 0.000112 Attachments MS Module 1 Backgrounddata visualization – practice problems df.pdf (241 views, 63.00 KB)
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