University of California, Los AngelesDepartment of StatisticsStatistics 13 Instructor: Nicolas ChristouConfidence intervalsA. Confidence intervals for the population mean µ of normal population withknown population standard deviation σ:Usually (if not always) µ is unknown. Then what?What would be a good estimate for µ?Let X1, X2, ···, Xnbe a random sample from N (µ, σ). We know that¯X ∼ N(µ,σ√n).Therefore,P−zα2≤¯X − µσ√n≤ zα2= 1 − α, where −zα2and zα2are defined as follows:αα2αα2−−zαα2−−zαα2−−zαα2++zαα2++zαα2++zαα201 −− ααN(0,1)The area 1 −α is called confidence level. When we construct confidence intervals we usuallyuse the following confidence levels:1 − α zα20.90 1.6450.95 1.9600.98 2.3250.99 2.5751The expression above can be written as:P ¯x − zα2σ√n≤ µ ≤ ¯x + zα2σ√n!= 1 − α. (1)It is tempting to read this statement as “the probability . . .”. But we should not! Instead,we can say that we are 1 − α confident that µ falls in the interval ¯x ± zα2σ√n. Why?Example:Suppose that the length of iron rods from a certain factory follows the normal distributionwith known standard deviation σ = 0.2 m but unknown mean µ. Construct a 95% confi-dence interval for the population mean µ if a random sample of n = 16 of these iron rodshas sample mean ¯x = 6 m.Sample size determination for a given length of the confidence interval:Find the sample size n needed when we want the width of the confidence interval to be ±Ewith confidence level 1 − α.Answer:In the expression ¯x ±zα2σ√nthe width of the confidence interval is given by zα2σ√n(also calledmargin of error). We want this width to be equal to E. Therefore,E = zα2σ√n⇒ n =zα2σE2.Example:For the example above, suppose that we want the entire width of the confidence interval tobe equal to 0.05 m. Find the sample size n needed.Question:Is there a 100% confidence interval?2B. Confidence intervals for the population mean µ with known population stan-dard deviation σ:From the central limit theorem we know that when n ≥ 30 the distribution of the samplemean¯X approximately follows:¯X ∼ N(µ,σ√n)Therefore, the confidence interval for the population mean µ is given by the expression wefound in part (A):P ¯x − zα2σ√n≤ µ ≤ ¯x + zα2σ√n!≈ 1 − α.The mean µ falls in the interval ¯x ±zα2σ√n.Also the sample size determination is given by the same formula we found in part (A):E = zα2σ√n⇒ n =zα2σE2.Example:A sample of size n = 50 is taken from the production of lightbulbs at a certain factory. Thesample mean of the lifetime of these 50 lightbulbs is found to be ¯x = 1570 hours. Assumethat the population standard deviation is σ = 120 hours.a. Construct a 95% confidence interval for µ.b. Construct a 99% confidence interval for µ.c. What sample size is needed so that the length of the interval is 30 hours with 95%confidence?3Confidence intervals - An empirical investigationTwo dice are rolled and the sum X of the two numbers that occured is recorded. The probabilitydistribution of X is as follows:X 2 3 4 5 6 7 8 9 10 11 12P (X) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36This distribution has mean µ = 7 and standard deviation σ = 2.42. We take 100 samples of sizen = 50 each from this distribution and compute for each sample the sample mean ¯x. Pretendnow that we only know that σ = 2.42, and that µ is unknown. We are going to use these 100sample means to construct 100 confidence intervals each one with 95% confidence level for the truepopulation mean µ. Here are the results:Sample ¯x 95%C.I.f orµ: ¯x − 1.962.42√50≤ µ ≤ ¯x + 1.962.42√50Is µ = 7 included?1 6.9 6.23 ≤ µ ≤ 7.57 YES2 6.3 5.63 ≤ µ ≤ 6.97 NO3 6.58 5.91 ≤ µ ≤ 7.25 YES4 6.54 5.87 ≤ µ ≤ 7.21 YES5 6.7 6.03 ≤ µ ≤ 7.37 YES6 6.58 5.91 ≤ µ ≤ 7.25 YES7 7.2 6.53 ≤ µ ≤ 7.87 YES8 7.62 6.95 ≤ µ ≤ 8.29 YES9 6.94 6.27 ≤ µ ≤ 7.61 YES10 7.36 6.69 ≤ µ ≤ 8.03 YES11 7.06 6.39 ≤ µ ≤ 7.73 YES12 7.08 6.41 ≤ µ ≤ 7.75 YES13 7.42 6.75 ≤ µ ≤ 8.09 YES14 7.42 6.75 ≤ µ ≤ 8.09 YES15 6.8 6.13 ≤ µ ≤ 7.47 YES16 6.94 6.27 ≤ µ ≤ 7.61 YES17 7.2 6.53 ≤ µ ≤ 7.87 YES18 6.7 6.03 ≤ µ ≤ 7.37 YES19 7.1 6.43 ≤ µ ≤ 7.77 YES20 7.04 6.37 ≤ µ ≤ 7.71 YES21 6.98 6.31 ≤ µ ≤ 7.65 YES22 7.18 6.51 ≤ µ ≤ 7.85 YES23 6.8 6.13 ≤ µ ≤ 7.47 YES24 6.94 6.27 ≤ µ ≤ 7.61 YES25 8.1 7.43 ≤ µ ≤ 8.77 NO26 7 6.33 ≤ µ ≤ 7.67 YES27 7.06 6.39 ≤ µ ≤ 7.73 YES28 6.82 6.15 ≤ µ ≤ 7.49 YES29 6.96 6.29 ≤ µ ≤ 7.63 YES30 7.46 6.79 ≤ µ ≤ 8.13 YES31 7.04 6.37 ≤ µ ≤ 7.71 YES32 7.06 6.39 ≤ µ ≤ 7.73 YES33 7.06 6.39 ≤ µ ≤ 7.73 YES34 6.8 6.13 ≤ µ ≤ 7.47 YES35 7.12 6.45 ≤ µ ≤ 7.79 YES36 7.18 6.51 ≤ µ ≤ 7.85 YES37 7.08 6.41 ≤ µ ≤ 7.75 YES38 7.24 6.57 ≤ µ ≤ 7.91 YES39 6.82 6.15 ≤ µ ≤ 7.49 YES40 7.26 6.59 ≤ µ ≤ 7.93 YES41 7.34 6.67 ≤ µ ≤ 8.01 YES42 6.62 5.95 ≤ µ ≤ 7.29 YES43 7.1 6.43 ≤ µ ≤ 7.77 YES44 6.98 6.31 ≤ µ ≤ 7.65 YES45 6.98 6.31 ≤ µ ≤ 7.65 YES46 7.06 6.39 ≤ µ ≤ 7.73 YES47 7.14 6.47 ≤ µ ≤ 7.81 YES48 7.5 6.83 ≤ µ ≤ 8.17 YES49 7.08 6.41 ≤ µ ≤ 7.75 YES50 7.32 6.65 ≤ µ ≤ 7.99 YES4Sample ¯x 95%C.I.f orµ: ¯x − 1.962.42√50≤ µ ≤ ¯x + 1.962.42√50Is µ = 7 included?51 6.54 5.87 ≤ µ ≤ 7.21 YES52 7.14 6.47 ≤ µ ≤ 7.81 YES53 6.64 5.97 ≤ µ ≤ 7.31 YES54 7.46 6.79 ≤ µ ≤ 8.13 YES55 7.34 6.67 ≤ µ ≤ 8.01 YES56 7.28 6.61 ≤ µ ≤ 7.95 YES57 6.56 5.89 ≤ µ ≤ 7.23 YES58 7.72 7.05 ≤ µ ≤ 8.39 NO59 6.66 5.99 ≤ µ ≤ 7.33 YES60 6.8 6.13 ≤ µ ≤ 7.47 YES61 7.08 6.41 ≤ µ ≤ 7.75 YES62 6.58 5.91 ≤ µ ≤ 7.25 YES63 7.3 6.63 ≤ µ ≤ 7.97 YES64 7.1 6.43 ≤ µ ≤ 7.77 YES65 6.68 6.01 ≤ µ ≤ 7.35 YES66 6.98 6.31 ≤ µ ≤ 7.65 YES67 6.94 6.27 ≤ µ ≤ 7.61 YES68 6.78 6.11 ≤ µ ≤ 7.45 YES69 7.2 6.53 ≤ µ ≤ 7.87 YES70 6.9 6.23 ≤ µ ≤ 7.57 YES71 6.42 5.75 ≤ µ ≤ 7.09 YES72 6.48 5.81 ≤ µ ≤ 7.15 YES73 7.12 6.45 ≤ µ ≤ 7.79 YES74 6.9 6.23 ≤ µ ≤ 7.57 YES75 7.24 6.57 ≤ µ ≤ 7.91 YES76 6.6 5.93 ≤ µ ≤ 7.27 YES77 7.28 6.61 ≤ µ ≤ 7.95 YES78 7.18 6.51 ≤ µ ≤ 7.85 YES79 6.76 6.09 ≤ µ ≤ 7.43 YES80 7.06 6.39 ≤ µ ≤ 7.73 YES81 7 6.33 ≤ µ ≤ 7.67 YES82 7.08 6.41 ≤ µ ≤ 7.75 YES83 7.18 6.51 ≤ µ ≤ 7.85 YES84 7.26
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