1. This is an exercise to compare the unbiased estimator of population mean using two-stage sampling against the ratio estimator. Specifically we want to show that if the magnitude of the mean is large compared to its standard deviation, the ratio estimator is superior. A company produces drilling machines from 50 factories. The company believes that the weights of the machines are fairly constant within each factory but may vary considerably from factory to factory. Therefore, the company decides to sample approximately 10% of the machines within each of 10 factories using a two-stage cluster sample. The data are given in the accompanying table. (a) Estimate the average weights of the machines produced by the com- pany and place a bound on the error of estimation. (b) Assume that the company knows there are 2600 machines produced. Use this additional information to provide an unbiased estimate of the average weight and place a bound on the error of estimation. Compare this error bound against that in (a) and comment. Compare this estimate against all the weights in the dataset and comment. Number of Number of Number machines machines sampled 1 52 5 2 56 3 60 10 6 6 4 46 5 5 49 5 6 51 5 7 50 5 8 61 6 9 60 6 10 45 6 Weights of machines (in kg) 1012, 1011, 1012, 1010, 1013 1010, 1009, 1007, 1009, 1008, 1010 1006, 1005, 1007, 1005, 1006, 1004 1007, 1008, 1007, 1007, 1006 1010, 1011, 1013, 1012, 1012 1014, 1015, 1013, 1012, 1013 1006, 1007, 1006, 1008, 1007 1009, 1010, 1008, 1009, 1009, 1010 1007, 1010, 1008, 1009, 1009, 1010 1012, 1011, 1012, 1013, 1012, 1012 You may find the following R codes and outputs useful. > M=c (52,56,60,46,49,51,50,61,60,45) > m=c(5,6,6,5,5,5,5,6,6,6) > y=matrix(nrow=10, ncol=6,data=0) > y [1,1:5]=c(12,11,12,10,13)+1000 > y [2,]=c(10,9,7,9,8,10)+1000 > y [3,]=c(6,5,7,5,6,4)+1000 > y [4,1:5]=c(7,8,7,7,6)+1000 > y [5, 1:5]=c(10,11,13,12,12)+1000 > y [6,1:5]=c(14,15,13,12,13)+1000 > y [7,1:5]=c(6,7,6,8,7)+1000 > y [8,]=c(9,10,8,9,9,10)+1000 > y [9,]=c(7,10,8,9,9,10)+1000 > y [10,]=c(12,11,12,13,12,12) +1000 > > s2=ybar=rep(0, times=10) > for (i in 1:10){ +yrow=y [i, 1:m[i]] + ybar [i]=mean (yrow) + s2[i]=var (yrow) + } > > ybar [1] 1011.600 1008.833 1005.500 1007.000 1011.600 1013.400 1006.800 1009.167 [9] 1008.833 1012.000 > [1] 1.3000000 1.3666667 1.1000000 0.5000000 1.3000000 1.3000000 0.7000000 [8] 0.5666667 1.3666667 0.4000000 Zཆ

1. This is an exercise to compare the unbiased estimator of population mean using two-stage sampling against the ratio estimator. Specifically we want to show that if the magnitude of the mean is large compared to its standard deviation, the ratio estimator is superior. A company produces drilling machines from 50 factories. The company believes that the weights of the machines are fairly constant within each factory but may vary considerably from factory to factory. Therefore, the company decides to sample approximately 10% of the machines within each of 10 factories using a two-stage cluster sample. The data are given in the accompanying table. (a) Estimate the average weights of the machines produced by the com- pany and place a bound on the error of estimation. (b) Assume that the company knows there are 2600 machines produced. Use this additional information to provide an unbiased estimate of the average weight and place a bound on the error of estimation. Compare this error bound against that in (a) and comment. Compare this estimate against all the weights in the dataset and comment. Number of Number of Number machines machines sampled 1 52 5 2 56 3 60 10 6 6 4 46 5 5 49 5 6 51 5 7 50 5 8 61 6 9 60 6 10 45 6 Weights of machines (in kg) 1012, 1011, 1012, 1010, 1013 1010, 1009, 1007, 1009, 1008, 1010 1006, 1005, 1007, 1005, 1006, 1004 1007, 1008, 1007, 1007, 1006 1010, 1011, 1013, 1012, 1012 1014, 1015, 1013, 1012, 1013 1006, 1007, 1006, 1008, 1007 1009, 1010, 1008, 1009, 1009, 1010 1007, 1010, 1008, 1009, 1009, 1010 1012, 1011, 1012, 1013, 1012, 1012 You may find the following R codes and outputs useful. > M=c (52,56,60,46,49,51,50,61,60,45) > m=c(5,6,6,5,5,5,5,6,6,6) > y=matrix(nrow=10, ncol=6,data=0) > y [1,1:5]=c(12,11,12,10,13)+1000 > y [2,]=c(10,9,7,9,8,10)+1000 > y [3,]=c(6,5,7,5,6,4)+1000 > y [4,1:5]=c(7,8,7,7,6)+1000 > y [5, 1:5]=c(10,11,13,12,12)+1000 > y [6,1:5]=c(14,15,13,12,13)+1000 > y [7,1:5]=c(6,7,6,8,7)+1000 > y [8,]=c(9,10,8,9,9,10)+1000 > y [9,]=c(7,10,8,9,9,10)+1000 > y [10,]=c(12,11,12,13,12,12) +1000 > > s2=ybar=rep(0, times=10) > for (i in 1:10){ +yrow=y [i, 1:m[i]] + ybar [i]=mean (yrow) + s2[i]=var (yrow) + } > > ybar [1] 1011.600 1008.833 1005.500 1007.000 1011.600 1013.400 1006.800 1009.167 [9] 1008.833 1012.000 > [1] 1.3000000 1.3666667 1.1000000 0.5000000 1.3000000 1.3000000 0.7000000 [8] 0.5666667 1.3666667 0.4000000 Zཆ

Linear Algebra: A Modern Introduction

4th Edition

ISBN:9781285463247

Author:David Poole

Publisher:David Poole

Chapter7: Distance And Approximation

Section7.3: Least Squares Approximation

Problem 31EQ

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