Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20~ Summary of Hypothesis Testing for One Mean ~~ Summary of Hypothesis Testing for One Mean ~~ Summary of Hypothesis Testing for One Mean ~Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30Slide 31• 6.1 - One Sample Mean μ, Variance σ 2, Proportion π • 6.2 - Two Samples Means, Variances, Proportions μ1 vs. μ2 σ12 vs. σ22 π1 vs. π2• 6.3 - Multiple Samples Means, Variances, Proportions μ1, …, μk σ12, …, σk2 π1, …, πkCHAPTER 6 Statistical Inference & Hypothesis Testing• 6.1 - One Sample Mean μ, Variance σ 2, Proportion π • 6.2 - Two Samples Means, Variances, Proportions μ1 vs. μ2 σ12 vs. σ22 π1 vs. π2• 6.3 - Multiple Samples Means, Variances, Proportions μ1, …, μk σ12, …, σk2 π1, …, πkCHAPTER 6 Statistical Inference & Hypothesis Testing• 6.1 - One Sample Mean μ, Variance σ 2, Proportion π • 6.2 - Two Samples Means, Variances, Proportions μ1 vs. μ2 σ12 vs. σ22 π1 vs. π2• 6.3 - Multiple Samples Means, Variances, Proportions μ1, …, μk σ12, …, σk2 π1, …, πkCHAPTER 6 Statistical Inference & Hypothesis TestingWomen in U.S. who have given birthPOPULATION“Random Variable” X = Age (years)That is, X ~ N(μ, 1.5). Present: Assume that X follows a “normal distribution” in the population, with std dev σ = 1.5 yrs, but unknown mean μ = ? mean x = 25.6FORMULAmean μ = ???Random Sample{x1, x2, x3, x4, … , x400}standard deviation σ = 1.5This is referred to as a “point estimate” of μ from the sample.This is referred to as a “point estimate” of μ from the sample.Improve this point estimate of μ to an “interval estimate” of μ, via the… “Sampling Distribution of ” “Sampling Distribution of ” Xsize n = 400Example: One MeanObjective 1: “Parameter Estimation”Estimate the parameter value μ. Estimate the parameter value μ.X = Age of women in U.S. who have given birth Sampling Distribution ofSampling Distribution ofXPopulation Distribution of X Population Distribution of X Xμstandard deviation σ = 1.5 yrsIf X ~ N(μ, σ), then… for any sample size n.,nsm� �� �� �X ~ N ,μX“standard error”1.5 yrs.075 yrs400ns= =nms-=XZms-=XZX d X d- +Sampling Distribution ofSampling Distribution ofXμX“standard error”1.5 yrs.075 yrs400ns= =To achieve Objective 1 — obtain an “interval estimate” of μ — we first ask the following general question: Find a “margin of error” (d) so that there is a 95% probability that the interval contains μ.( )X d, X d- +� �d( )μP X - d X d< < + = 0.95( )μ μP - d X d< < + = 0.95X. .� �� �� �-d dP Z+< < = 0.95s.e s.em-=XZs.e.Suppose is any random sample mean. XX|μ( )μP X - d X d< < + = 0.95( )μ μP - d X d< < + = 0.95. .� �� �� �-d dP Z+< < = 0.95s.e s.em-=XZs.e.X|μX d X d- +Sampling Distribution ofSampling Distribution ofXμX“standard error”1.5 yrs.075 yrs400ns= =� �dX( )μP X - d X d< < + = 0.95( )μ μP - d X d< < + = 0.95. .� �� �� �-d dP Z+< < = 0.95s.e s.em-=XZs.e.standard normal distributionN(0, 1)Z0.950.025 0.025+z.025-z.025d = (z.025)(s.e.) = (1.96)(.075 yrs) = 0.147 yrs( )μP X - d X d< < + = 0.95( )μ μP - d X d< < + = 0.95. .� �� �� �-d dP Z+< < = 0.95s.e s.em-=XZs.e.X|μX d X d- +� �dX( )μP X - d X d< < + = 0.95( )μ μP - d X d< < + = 0.95. .� �� �� �-d dP Z+< < = 0.95s.e s.em-=XZs.e.standard normal distributionN(0, 1)Z0.950.025 0.025+z.025-z.025d = (z.025)(s.e.) = (1.96)(.075 yrs) = 0.147 yrs The “confidence level” is 95%.IMPORTANT DEF’NS and FACTSd is called the “95% margin of error” and is equal to the product of the “.025 critical value” (i.e., z.025 = 1.96) times the “standard error” (i.e., ). nsThe “significance level” is 5%.For any random sample mean the “95% confidence interval” isIt contains μ with probability 95%.X,- +X X ( margin of error, margin of error).In this example, the 95% CI is- +X X ( 0.147, 0.147).For instance, if a particular sample yields the 95% CI is (25.6 – 0.147, 25.6 + 0.147) = (25.543, 25.747) yrs. It contains μ with 95% “confidence.”x = 25.6 yrs,( )μP X - d X d< < + = 0.95( )μ μP - d X d< < + = 0.95. .� �� �� �-d dP Z+< < = 0.95s.e s.em-=XZs.e.X|μX d X d- +� �dX( )μP X - d X d< < + = 0.95( )μ μP - d X d< < + = 0.95. .� �� �� �-d dP Z+< < = 0.95s.e s.em-=XZs.e.standard normal distributionN(0, 1)Z0.950.025 0.025+z.025-z.025d = (z.025)(s.e.) = (1.96)(.075 yrs) = 0.147 yrs The “confidence level” is 95%.IMPORTANT DEF’NS and FACTSd is called the “95% margin of error” and is equal to the product of the “.025 critical value” (i.e., z.025 = 1.96) times the “standard error” (i.e., ). nsThe “significance level” is 5%.For any random sample mean the “95% confidence interval” is It contains μ with probability 95%.X,- +X X ( margin of error, margin of error).In this example, the 95% CI is- +X X ( 0.147, 0.147).For instance, if a particular sample yields the 95% CI is (25.6 – 0.147, 25.6 + 0.147) = (25.543, 25.747) yrs. It contains μ with 95% “confidence.”x = 25.6 yrs,1 – αα/2 α/2+zα/2-zα/21 – α“α/2“100(1 – α)% margin of error”zα/2)1 – α.α.“100(1 – α)% “confidence interval” 1 – α.d = (zα/2)(s.e.) ( )μ μP - d X dα< < + = 1 -. .� �� �� �-d dP Zα+< < = 1 -s.e s.eExample: α = .05, 1 – α = .95 Example: α = .10, 1 – α = .90 Example: α = .01, 1 – α = .99 +1.645-1.645( )μP X - d X d< < + = 0.95( )μ μP - d X d< < + = 0.95. .� �� �� �-d dP Z+< < = 0.95s.e s.em-=XZs.e.X|μX d X d- +� �dX( )μP X - d X d< < + = 0.95m-=XZs.e.standard normal distributionN(0, 1)Z0.950.025 0.025+z.025-z.025The “confidence level” is 95%.IMPORTANT DEF’NS and FACTSd is called the “95% margin of error” and is equal to the product of the “.025 critical value” (i.e., z.025 = 1.96) times the “standard error” (i.e., ). nsThe “significance level” is 5%.For any random sample mean the “95% confidence interval” is