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Chapter 6

Chapter 6. Ratio, Regression and Difference Estimation. 6.2 Ratio Estimators. Looking at two measurements of a population, call these x and y (we will only focus on the SRS sampling strategy) Note that the ratio of population means is the same as the ratio of population totals!!

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Chapter 6

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  1. Chapter 6 Ratio, Regression and Difference Estimation

  2. 6.2 Ratio Estimators • Looking at two measurements of a population, call these x and y (we will only focus on the SRS sampling strategy) • Note that the ratio of population means is the same as the ratio of population totals!! • mx/my = Nmx/Nmy = tx/ty • Therefore, we will use R = ratio of the means

  3. 6.3 Ratio estimator for SRS • Calculating an estimate and bound on r: • r = ybar/xbar • Estimated variance of r: • (1-n/N)*(1/mx2)*(sr2/n) • where sr2 = Si(yi-rxi)2/(n-1) • Use xbar to estimate mx when necessary

  4. Ratio estimator of total and mean • For population total: • thaty = rtx • Estimated variance of thaty • N2(1-n/N) sr2 /n) (where sr2 is defined previously) • For population mean: • mhaty = rmx • Estimated variance of mhaty • (1-n/N) sr2 /n (where sr2 is defined previously)

  5. Examples • 6.1: mean of x (square-foot) = 0.558333 mean of y ( volume) = 11.83333 thaty = (11.8333/.558333)*75=1589.552204 sr2 = 1.750299 Estimated variance = 250^2*(1-12/250)* 1.750299/12 = 8678.566 Bound=2*sqrt(8678.566) = 186.3176 Without fpc (don’t really need it since 1-n/N = 0.952), B = 190.957

  6. More examples Rcode: x=dat$Basal.ar y=dat$Volume ybar=mean(y) xbar = mean(x) rhat =(ybar/xbar) t.hat.y = rhat*75 n=length(dat$Volume) s.r.2 = sum((y-rhat*x)^2)/(n-1) var.that = (250^2)*(1-12/250)*s.r.2/12 ___________________________________________________ • 6.2: that =250* 11.83333 = 2958.333 • Variance for that=250^2*(1-12/250)*26.87879/12 = 133274 • Bound = 730.1342 • Without fpc, bound = 748.3146

  7. More examples • Do problem 6.6 mean(x)=16.47273,mean(y) = 16.84545, so rhat = 16.84545/16.47273=1.022627; sr2=0.2049424 • Coefficient of variation: CV(x) = sx/xbar, where sx is the standard deviation of x. It is desired that the CV is lower than 1. Find the CV for x and the CV for y in problem 6.6.

  8. 6.4 Sample size estimation • n = Ns2/(ND + s2) where we can estimate s2 as Si(yi-rxi)2/(n’-1) with n’ a small preliminary sample With D=B2mx2/4 for estimating R D = B2/4 for estimating my D = B2/(4N2) for estimating ty Do problem 6.13 (x=dat$Precamp; y=dat$Present) sr2 = 241.9403

  9. 6.6 Regression Estimation • Need to perform a simple linear regression and get a and b, where y-hati= a + bxi (a is y-intercept and b is slope of line)…Note that the formula to find a is ybar – b(xbar) • Estimate of mhatyL = ybar + b(mx – xbar) • Estimated variance of mhatyL = (1-n/N)(MSE/n)….MSE is the estimate variance in a linear regression problem!!! • Refer back to 6.1 and find a regression estimator for mY

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