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Systems Engineering Program Department of Engineering Management, Information and Systems EMIS 7370/5370 STAT 5340 : PROBABILITY AND STATISTICS FOR SCIENTISTS AND ENGINEERS Estimation of Means Dr. Jerrell T. Stracener, SAE Fellow Leadership in Engineering 1 Estimation • Estimation of the Mean - Normal Distribution • Estimation of the Difference Between two Means Normal Distribution • Estimation of the Mean - Infinite Population - Type Unknown • Estimation of the Mean - Finite Population • Estimation of Lognormal Distribution Parameters • Estimation of Weibull Distribution Parameters 2 Estimation of Means 3 Estimation of the Mean - Normal Distribution • X1, X2, …, Xn is a random sample of size n from N(, ), where both & are unknown. • Point Estimate of n 1 μ^ X i X n i 1 • (1 - ) 100% Confidence Interval for the mean μ L , μ U μ L X Δμ and μ U X Δμ where Δμ t α 2 , n 1 s n 4 Example X = diameter of a rod where X ~ N(, ) A random sample of 26 randomly selected rods resulted in x = 1.030 inches and s = 0.020 inches. Estimate in terms of a point estimate and 98% and 95% confidence intervals. 5 Example - Solution μ^ x 1.030 For = 0.02, t 2 , n 1 t0.01, 25 2.485 so that L x t 2 , n 1 s 0.020 1.030 2.485 n 26 1.030 0.0097 1.0202 6 Example - Solution and U x t 2 , n 1 s 0.020 1.030 2.485 n 26 1.030 0.0097 1.0397 Therefore a 98% confidence interval for is (1.020, 1.040) For = 0.05, t 2 , n 1 t0.025, 25 2.060 7 Example - Solution μL x t α 2 , n 1 s 0.020 1.030 2.060 n 26 1.030 0.0081 1.0219 s 0.020 μU x t α 1.030 2.060 , n 1 n 26 2 1.030 0.0081 1.0381 so that a 95% confidence interval for is (1.022, 1.038). 8 Estimation of Difference Between Two Means Normal Distribution • Let X11, X12, …, X1n, and X21, X22, …, X 2 n2be 1 random samples from N(1, 1) and N(2, 2), respectively, where 1, 1, 2 and 2 are all unknown • Point estimation of = 1 - 2 ^ ^ ^ Δ μ μ1 μ 2 X1 X 2 9 Estimation of Difference Between Two Means Normal Distribution • An approximate (1 - ) 100% Confidence Interval for = 1 - 2 is ΔμL , ΔμU , Where s12 s 22 ΔμL Δμ̂ t α 2 , ν n1 n 2 and s12 s 22 ΔμU Δμ̂ t α ,ν 2 n1 n 2 10 Estimation of Difference Between Two Means Normal Distribution where ν = degrees of freedom 2 s s n1 n2 2 2 2 2 s1 s2 n1 n2 n1 1 n2 1 2 1 2 2 11 Example - Estimation of 1 - 2 Records for the past 15 years have shown the average rainfall in a certain region of the country for the month of May to be 4.93 cm, with a standard deviation of 1.14 cm. A second region of the country has had an average rainfall in may of 2.64 cm, with a standard deviation of 0.66 cm during the past 10 years. Find a 95% confidence interval for the difference of the true average rainfalls in these two regions, assuming that the observations came from normal populations. 12 Example - Solution For the first region we have X1= 4.93, s1 = 1.14, and n1 = 15. For the second region X 2= 2.64, s2 = 0.66, and n2 = 10. We wish to find a 95% confidence interval for 1 - 2. Our point estimate of Δμ μ1 μ 2 is Δμ̂ X1 X 2 = 4.93 - 2.64 = 2.29. Since we have no information concerning the population variances and our sample sizes are not the same, we can only find an approximate 95% confidence interval based on the t distribution with ν degrees of freedom, where 13 Example - Solution s 2 1 s 2 1 n1 s n2 2 2 2 n1 / n1 1 s n2 / n2 1 2 1.14 2 1.14 15 0.66 10 15 / 15 1 0.66 10 / 10 1 2 2 2 2 2 2 2 2 2 22.7 23 Using = 0.05, we find that t0.025 = 2.069 for = 23 degrees of freedom. 14 Example - Solution Therefore, substituting in the formula ΔμL x1 x 2 t α 2 s12 s 22 n1 n 2 1.14 2 0.66 2 2.29 2.069 15 10 1.5434 15 Example - Solution ΔμU x1 x 2 t α 2 s12 s 22 n1 n 2 1.14 2 0.66 2 2.29 2.069 15 10 3.037 Hence, we are approximately 95% confident that the interval from 1.54 to 3.04 contains the true difference of the average rainfall for the two regions. 16 Estimation of Means Infinite Population - Type Unknown 17 Estimation of the Mean Infinite Population - Type Unknown • X1, X2, …, Xn is a random sample of size n • Point Estimate of 1 n μ̂ X i X n i 1 18 Estimation of the Mean Infinite Population - Type Unknown • An approximate (1 - ) 100% Confidence Interval for the mean μL , μU based on the Central Limit Theorem is: where μL X Δμ μU X Δμ and μ t α 2 , n 1 s n 19 Estimation of Means Finite Population 20 Estimation of Means - Finite Populations • X1, X2, ... , Xn is a random sample of size n from a population of size N with parameters and • Point Estimate of : μ̂ X • An approximate (1 - ) 100% Confidence Interval for is: μL , μU where μL x Δμ μU x Δμ and Δμ t α 2 , n 1 s n Nn N 1 21 Estimation of Means - Finite Populations where n 1 S Xi X n 1 i 1 2 2 t • ,n 1 is the value of T ~ tdf for which P T t 2 , n 1 2 2 • N n is the finite population correction factor N 1 22 Estimation - Lognormal Distribution 23 Estimation of Lognormal Distribution • Random sample of size n, X1, X2, ... , Xn from LN (, ) • Let Yi = ln Xi for i = 1, 2, ..., n • Treat Y1, Y2, ... , Yn as a random sample from N(, ) • Estimate and using the Normal Distribution Methods 24 Graphical Estimation of Lognormal Distribution Parameters • Take logarithms of sample values • Prepare Normal Probability Plot • μ̂= 50th percentile μ̂ x 0.1 • σ̂ 1.28 25 Estimation of the Mean of a Lognormal Distribution • Mean or Expected value of 2 mean E(x) e μ σ 2 • Point Estimate of mean ^ ^ mean E(x) e σ̂ 2 μ̂ 2 where μ̂ and σ̂ are point estimates of μ and σ respectively. 26 Estimation - Weibull Distribution 27 Estimation of Weibull Distribution • Random sample of size n, T1, T2, …, Tn, from W(, ), where both & are unknown. • Point estimates ^ • β is the solution of g() = 0 n where gβ β T i lnT i i 1 n β T i 1 1 n lnT i β n i 1 i 1 1 1 n ^β ^β • θ Ti n i 1 ^ 28 Graphical Estimation of Weibull Distribution Parameters • Prepare Weibull probability Plot • θ̂ 63.2th percentile ln - ln 1 F(x i ) • β̂ x ln θ̂ for a selected value of x. 29 Estimation of the Mean of a Weibull Distribution • Mean or Expected value of 1 μ E(x) θΓ 1 β • Point Estimate of mean 1 θ̂Γ 1 β̂ where β̂ and θ̂ are point estimates of β and θ respectively. ^ μ 30