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Sampling Distribution and Confidence Interval Sampling Distr ibution T he ore tical Probability Dis tribution of the Sa mple Statistic. Sa mpling D istribution of Mea ns If a random sample is taken from a normally distributed population that has a mean µ and a standard deviation σ , the sampling distribution of the sample means, x , is normal with µ µxx == µ µ σ σ σ σxx == nn W aiting tim es at a c ertain type of c linic are norm ally dis tribution wit h µµ = 8 m in. & σσ = 2 m in. If y ou s elec t random s am ples of 25 c as es , what is the s am pling dis tribution of t he m ean? If you select random samples of 25 cases from a normal distribution (or population) with µ = 8 min. & σ = 2 min. The sampling distribution of the mean is normal distribution with mean = 8 min., and standard deviation = 2/5 = .4 min. Example : W aiting tim es at a c ertain type of c lin ic are norm ally dis tribut ion with µµ = 8 m in. & σσ = 2 m in. If y ou s elec t random s am ples of 25 c as es , what is the probability th at the sa mple me an of a rand om s am ple of 25 c as es is between 7.8 & 8.2 m inutes ? xx −− µ µ 77..88 −−88 == == −−..50 50 σ σ nn 22 25 25 x − µ 8. 2 − 8 zz == x − µ == 8. 2 − 8 == ..50 50 σ σ nn 22 25 25 zz == The probability that the sample mean would be between 7.8 & 8.2 minutes is .383. Central Limit Theorem If a relative large random sample is taken from a population that has a mean µ and a standard deviation σ , regardless of the distribution of the population, the distribution of the sample means is approximately normal with µ µxx == µ µ σ σ σ σxx == nn Example: Consider the distribution of serum cholesterol levels for all 20- to 74-year-old males living in United States has a mean of 211 mg/100 ml, and the standard deviation of 46 mg/100 ml. If a random sample of 100 individuals from the population, what is the probability that the average serum cholesterol level of these 100 individuals is higher than 225? Solution: The sampling distribution of the mean is normal with mean = 211 and standard error = 4.6 (= 46/10). P( X > 225) = P(Z > [225 - 211]/4.6) = P(Z > 3.04) = 0.001 1 Sampling Distribution and Confidence Interval A special notation: z α = the z score that the proportion of the standard normal distribution to the right of it is α. z .025 = 1.96 z .010 = ? Confide nce Inte rva l Mea n (σσ Know n) σ σ σ xx −− zzαα // 22 ⋅⋅ σ ≤≤ µ µ ≤≤ xx ++ zzαα // 22 ⋅⋅ nn nn or σ xx ±± zzαα // 22 ⋅⋅ σ nn As s um ptions • Population Sta ndard Devi ation Is Known ( It m ay c om e from prior s tudy.) • Population Is Norm ally Dis tributed • If Not Norm al, Can Be Approxim ated b y Norm al Dis tribution (n ≥ 30) Example : The m ean of a ran dom s am ple of n = 25 is X = 50. Set up a 95% c onfid enc e interval es tim ate for µ if σ = 10. W e c an be 95% c onfident t hat the p opulat ion m ean is in (46. 08, 53.9 2). W e c an be 95% c onfident t hat the m axim um s am pling error us ing this interval es tim ate for es tim ating m ean is within 3.92. σ σ σ xx −− zzαα // 22 ⋅⋅ σ ≤≤ µ µ ≤≤ xx ++ zzαα // 22 ⋅⋅ nn nn 10 10 ≤ µ ≤ 50 + 1. 96 ⋅ 10 10 50 50 −−11..96 96⋅⋅ ≤ µ ≤ 50 + 1. 96 ⋅ 25 25 25 25 46 46..08 08 ≤≤ µ µ ≤≤ 53 53..92 92 or 50 ± 3.92 Example : Yo u’re a Q/C ins pec tor for Coke. The σσ for 2-liter bottles is .05 liters . A random s am ple of 100 bottles s howedx = 1.99 liters . W hat is the 90% c onfidenc e interval es tim ate of the true me an am ount in 2-l iter bottl es ? σ σ σ xx −− ZZαα // 22 ⋅⋅ σ ≤≤ µ µ ≤≤ xx ++ ZZαα // 22 ⋅⋅ nn nn ..05 05 ≤ µ ≤ 1.99 + 1.645 ⋅ ..05 05 11..99 99 −−11..645 645⋅⋅ ≤ µ ≤ 1.99 + 1.645 ⋅ 100 100 100 100 11..982 ≤ µ ≤ 1 . 998 982 ≤ µ ≤ 1.998 or 1.99 ± .008 2 Sampling Distribution and Confidence Interval Confidence Interval Mean (σσ Unknown) 1. As s um ptions • Population Sta ndard Devi ation Is Unknown • Population Mus t Be Normall y Distribute d 2. Us e Student’s t Dis tribution 3. Confidenc e Interv al Es tim ate s ss xx −− ttαα // 22,,nn−−11 ⋅⋅ s ≤≤ µ µ≤≤ xx ++ ttαα // 22,,nn−−11 ⋅⋅ nn nn or σ xx ±± zzαα // 22 ⋅⋅ σ nn Example: A ran dom s am ple of n = 25 has x = 50 & s = 8. Set up a 95% c onfidenc e interval es tim ate for µ. ss ss ≤≤ µ µ ≤≤ xx ++ ttαα // 22,,nn−−11 ⋅⋅ nn nn 88 88 50 ≤≤ µ 50 −− 22..0639 0639 ⋅⋅ µ ≤≤ 50 50 ++ 22..0639 0639⋅⋅ 25 25 25 25 46 . 69 ≤ µ ≤ 53 . 30 46.69 ≤ µ ≤ 53. 30 or 50 ± 3.30 xx −−ttαα // 22,,nn−−11 ⋅⋅ Example : You’re a t im e s tudy analys t in m anufac turing. You’ve rec orded the fo llowing tas k tim es (m in.): 3.6, 4.2, 4.0, 3.5, 3.8, 3.1. W hat is the 90% c onfidenc e interval es tim ate of the popu lation me an tas k tim e? x = 3.7 s .d. = 0.38987 n = 6, df = n - 1 = 6 - 1 = 5 s / √ n = 0.38987 / √ 6 = 0.1592 t. 05, 5 = 2.015 ( 3.7 - (2.015)(0.15 92) , 3.7 + (2.015)(0.1592) ) ( 3.379, 4.0 21 ) or 3.7 ± .321 3 Sampling Distribution and Confidence Interval One-side d confide nce inte rv al Z C. I. : Lower i nter va l : t C. I.: − ∞ ≤ µ ≤ x + zα ⋅ σ n σ ≤ µ≤ ∞ n Upper i nter va l : x − zα ⋅ Lower i nter va l : − ∞ ≤ µ ≤ x + tα ⋅ Upper i nter va l: x − tα ⋅ s n s ≤ µ≤ ∞ n Example : The m ean of a ran dom s am ple of n = 25 is x = 50. Set up a upper 95% c onfidenc e interval es tim ate for µ if σ = 10. 1 − α = .95, α = .05, zα = 1.64. σ , − ∞) n 10 ( 50 − 1.64 ⋅ , − ∞) 25 ( 46.72 , − ∞ ) ( x − zα ⋅ M argin of Error: C.I. : x ± zα ⋅ 2 σ n Margin of Error = B = zα ⋅ 2 n= σ n zα2 ⋅ σ2 2 B2 Example : W hat s am ple s iz e is needed to be 90% c onfid ent of b eing c orrec t within ± 5? A pi lot s tudy s ugges ted that t he s tandard de viati on is 45. 22 22 zz..0505 2σσ 22 (11..645 645) (45 45) nn == == == 219 219..22 ≅≅ 220 220 BB 22 (55) 22 2 Example : Yo u plan to s urvey em ploy ees to find th eir aver age aut o ins uranc e. Yo u want to be 95% c onfide nt that t he s am ple me an is within ± $50 . A pi lot s tudy s howed that σσ was about $400. W hat sa mple siz e do you us e? nn == 22 22 2 2 zz..025 σ (11..96 96) 2 (400 400) 2 025 σ = == 245 = 245..86 86 ≅≅ 246 246 22 BB22 (50 50) 4