CHAPTER ESTIMATION INTRODUCTION In almost all realistic situations parameters are unknown We will use the sampling distribution to draw inferences about the unknown population parameters Statistical inference is the process by which we acquire information and draw conclusions about populations from samples Statistics Data Population Information Sample Inference Statistic Parameter There are two procedures for making inferences: • Estimation • Hypotheses testing The objective of estimation is to determine the approximate value of a population parameter on the basis of a sample statistic E.g., the sample mean ( ) is employed to estimate the population mean ( ) There are two types of estimators: Point Estimator Interval Estimator A point estimator draws inferences about a population by estimating the value of an unknown parameter using a single value or point We saw earlier that point probabilities in continuous distributions were virtually zero The probability of the point estimator being correct is zero An interval estimator draws inferences about a population by estimating the value of an unknown parameter using an interval That is we say (with some _% certainty) that the population parameter of interest is between some lower and upper bounds For example, suppose we want to estimate the mean summer income of a class of business students For n = 25 students, is calculated to be 400 $/week point estimate interval estimate An alternative statement is: The mean income is between 380 and 420 $/week ESTIMATING WHEN IS KNOWN… From Chapter 9, the sampling distribution of X is approximately normal with mean µ and standard deviation σ / n Thus X −µ Z= σ/ n is (approximately) standard normally distributed From Chapter 8, P(− Z α / < Z < Z α / ) = − α Thus, substituting Z produces x −µ P(− z α / < < zα / ) = − α σ/ n With a little bit of algebra, σ σ   P µ − z α / < x < µ + zα /  =1− α n n  With a little bit of different algebra we have σ σ   P x − z α / < µ < x + zα /  =1− α n n  The confidence interval Lower confidence limit (LCL) =  σ   x − zα /  n   Upper confidence limit (UCL) =   x + zα /  σ   n  The probability – α is the confidence level, which is a measure of how frequently the interval will actually include µ Four commonly used confidence levels α 0.10 0.05 0.02 0.01 α/ 0.05 0.025 0.01 0.005 zα/ 1.645 1.96 2.33 2.575 11 Confidence level 0.90 0.95 0.98 0.99 Example 2: Doll Computer Comp found that the demand over the lead time is normally distributed with a standard deviation of 75 Estimate the expected demand over x = 370 16 the lead time at 95% confidence level Assume N=25 and x ± zα σ n = 370.16 ± z 025 = 370.16 ± 1.96 75 25 75 25 = 370.16 ± 29.40 = [ 340.76,399.56] Comparing two confidence intervals with the same level of confidence, the narrower interval provides more information than the wider interval The width of the confidence interval is calculated by and therefore is affected by • the population standard deviation (s) • the confidence level (1-a) • the sample size (n) 13  σ   σ  σ  x + z α  −  x − z α  = 2Z α / n  n n  If the standard deviation grows larger, a longer confidence interval is needed to maintain the confidence level Note what happens when σ increases to 1.5 σ α/2 α/2 1-α Confidence level 2z α / n 1.5σ n 14 2z α / σ Example 1: Estimate the mean value of the distribution resulting from the 100 repeated throws of the die It is known that σ = 1.71 Use 90% confidence level: 1.71 σ = x ± 28 x ± zα = x ± 1.645 100 n Use 95% confidence level: 15 x ± zα σ 1.71 = x ± 1.96 = x ± 34 n 100 Larger confidence level requires longer confidence interval α/2 = 2.5% α/2 = 5% 90% 95% Confidence level 2z.05 2z 025 σ n σ n = 2(1.645) = 2(1.96) α/2 = 2.5% σ n σ n 16 α/2 = 5% There is an inverse relationship between the width of the interval and the sample size Interval width = 2z α / σ n 17 By increasing the sample size we can decrease the width of the confidence interval while the confidence level can remain unchanged The phrase “estimate the mean to within W units”, translates to an interval estimate of x ±the W form W zα / σ n = 18 where W is the margin of error The required sample size to estimate the mean is 22 19 zzαα 22σσ nn ==    W W    Example 4: To estimate the amount of lumber that can be harvested in a tract of land, the mean diameter of trees in the tract must be estimated to within one inch with 99% confidence What sample size should be taken for the margin of error +/-1 inch? (assume diameters are normally distributed with σ = inches) The confidence level 99% leads to α = 01, thus zα/2 = z.005 = 2.575 2 20  z α 2σ   2.575(6)  n= = 239  =  W     ... confidence levels α 0.10 0.05 0.02 0.01 α/ 0.05 0.025 0.01 0.005 zα/ 1.645 1 .96 2.33 2.575 11 Confidence level 0 .90 0 .95 0 .98 0 .99 Example 2: Doll Computer Comp found that the demand over the lead time... demand over x = 370 16 the lead time at 95 % confidence level Assume N=25 and x ± zα σ n = 370.16 ± z 025 = 370.16 ± 1 .96 75 25 75 25 = 370.16 ± 29. 40 = [ 340.76, 399 .56] Comparing two confidence intervals... one inch with 99 % confidence What sample size should be taken for the margin of error +/-1 inch? (assume diameters are normally distributed with σ = inches) The confidence level 99 % leads to α