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統計推論 (An Overview Of Statistical Inference) The statistical inference contains two parts • Estimation(參數估計;母數估計) 點估計 區間估計 • Hypothesis Testing(假說檢定; 假設考驗) 有母數分析(利用常態分布的特性) 無母數分析 CHAPTER 6 Estimation (參數估計) Definition _統計推論 • Statistical inference is the procedure by which we reach a conclusion about a population on the basis of the information contained in a sample drawn from that population. Definition_點估計 • A point estimate is a single numerical value used to estimate the corresponding population parameter. Definition _區間估計 • An interval estimate consists of two numerical values defining a range of values that, with a specified degree of confidence, we feel includes the parameter being estimated. 母群平均值的估計 • Estimators are usually presented as formulas. For example, x x n i Definition_不偏估計 • An estimator, say, T, of the parameter θ is said to be an unbiased estimator of θ if E (T )=θ. Definition _樣本 • The sampled population is the population from which one actually draws a sample. Definition_標的母群 • The target population is the population about which one wishes to make an inference. Point Estimation (點估計) • sample statistics population parameters X=ΣXI∕n S2=Σ(XI-X)2∕(n-1) =[ΣXi2-(ΣX)2∕n]∕(n-1) μ =ΣXI∕N σ2 =Σ(XI -μ)2∕N =[ΣXi2 -Nμ2 ]∕N • Principles of estimation – unbiasedness(不偏性): 期望值等於真值才是不偏估計 E(X)=μ – – – E(S2)=σ2 consistency(一致性) : 樣本數愈大愈趨近真值 efficiency (有效性): 估計數的變異最小 sufficiency(充分性) : 充分的樣本所估計出的 Interval Estimation (區間估計) Confidence Interval of population mean ﹦Point Estimator ± (reliability coefficient) × (standard error of point Estimator) (6.2.1) x z(1α σx 2) (6.2.2) §μ的區間估計(95%信賴區間) • 當樣本平均值(X)能座落在μ ±1.96(σ/ 圍內時,則μ就能座落在X ±1.96(σ/ n) 範 )範 n 圍內(此︰Z.975=1.96) • 估計參數有95%的機率可能座落之範圍 • 點估計 ±( 標準化值* 點估計的標準誤) • X ±Z.975 SEM = X ±1.96(σ/ n) Example 6.2.1 • Suppose a researcher, interested in obtaining an estimate of the average level of some enzyme in a certain human population, takes a sample of 10 individuals, determines the level of the enzyme in each, and computes a sample mean of x =22. Suppose further it is known that the variable of interest is approximately normally distributed with a variance of 45. We wish to estimate μ. • μ的95% C.I.等於: x 2σ x 45 22 2 10 22 2( 2.1213) 17.76, 26.24 Example 6.2.2 • A physical therapist wished to estimate, with 99 percent confidence, the mean maximal strength of a particular muscle in a certain group of individuals. He is willing to assume that strength scores are approximately normally distributed with a variance of 144. A sample of 15 subjects who participated in the experiment yielded a mean of 84.3. z 2.58 Z.995=2.58 σ x 12 / 15 3.0984 84.3 2.58(3.0984) 84.3 8.0 76.3, 92.3 Example 6.2.3 • Punctuality of patients in keeping appointments is of interest to a research team. In a study of patient flow through the offices of general practitioners, it was found that a sample of 35 patients were 17.2, minutes late for appointments, on the average. Previous research had shown the standard deviation to be about 8 minutes. The population distribution was felt to be nonnormal. What is the 90 percent confidence interval for μ, the true mean amount of time late for appointments? σ x 8/ 35 1.3522 90% C.I. 17.2 1.645(1.3522) 17.2 2.2 15.0, 19.4 Confidence interval for the difference between two population means • When the population variances are known, the 100(1-α) percent confidence interval for μ1-μ2 is given by ( x 1 x 2 ) z1α 2 σ σ n1 n2 2 1 2 2 §(μ1-μ2)的區間估計(95%信賴區間) • 估計參數有95%的機率可能座落之範圍 • 點估計 ±( 標準化值* 點估計的標準誤) • (X 1- X 2 ) ±Z.975 SE •=(X 1- X 2 ) ±1.96σ (X 1- X 2 ) (X 1- X 2 ) •= (X 1- X 2 ) ±1.96 σ12/n1+ σ22/n2 Example 6.4.1 • A research team is interested in the difference between serum uric acid levels in patients with and without Down’s syndrome. In a large hospital for the treatment of the mentally retarded, a sample of 12 individuals with Down’s syndrome yielded a mean of x1=4.5 mg/100 ml. In a general hospital a sample of 15 normal individuals of the same age and sex were found to have a mean value of x2=3.4. If it is reasonable to assume that the two populations of values are normally distributed with variances equal to 1 and 1.5, find the 95 percent confidence interval for μ1-μ2. x 1 x 2 4.5 3.4 1.1 σ x1 x 2 σ12 σ 22 n1 n2 1 1.5 .4282 12 15 95% C.I. 1.1 1.96(.4282) 1.1 .84 .26, 1.94 Example 6.4.2 • Motivated by an awareness of the existence of a body of controversial literature suggesting that stress, anxiety, and depression are harmful to the immune system, Gorman et al. conducted a study in which the subjects were homosexual men, some of whom were HIV positive and some of whom were HIV negative. Data were collected on a wide variety of medical, immunological, psychiatric, and neurological measures, one of which was the number of CD4+ cells in the blood. The mean number of CD4+ cells for the 112 men with HIV infection was 401.8 with a standard deviation of 226.4. For the 75 men without HIV infection the mean and standard deviation were 828.2 and 274.9, respectively. We wish to construct a 99 percent confidence interval for the difference between population means. 828.2 401.8 426.4 the reliability factor 2.58 274.9 2 226.4 2 the estimated standard error s x1x 2 38.2786 75 112 99% C.I. 426.4 2.58(38.2786) 327.6, 525.2 Confidence Interval of Population proportion • Estimator ± (reliability coefficient) × (standard error) • (6.5.1) § 二項分布π的區間估計(95%信賴區間) • 估計參數有95%的機率可能座落之範圍 • 點估計 ±( 標準化值* 點估計的標準誤) • p ± Z.975 SEp= p ±1.96 σp • p ± Z.975 SEp= p ±1.96 = p ±1.96 π(1-π) n p(1-p) n Example 6.5.1 • Mothers et al. (A-12) found that in a sample of 591 patients admitted to a psychiatric hospital, 204 admitted to using cannabis at least once in their lifetime. We wish to construct a 95 percent confidence interval for the proportion of lifetime cannabis users in the sampled population of psychiatric hospital admissions. ^ p 204 591 .3452 ^ ^ σ ^ p(1 p) n (.3452)(. 6548) 591 .01956 p 95% C.I. .3452 1.96(.01956) .3452 .0383 .3069, .3835 Confidence interval for the difference between two population proportions • The standard error of the estimate usually must be estimated by ^ ^ σ p1 p2 ^ ^ ^ ^ ^ p1(1 p1 ) p2 (1 p2 ) n1 n2 • 100(1-α) percent confidence interval for 1- 2 is given by §兩個二項分布成功率差異 (π1 - π2)的 區間估計(95%信賴區間) • 估計參數有95%的機率可能座落之範圍 • 點估計 ±( 標準化值* 點估計的標準誤) • (p1- p2 ) ±Z.975 SE (p11 -- pp22 )) (p •= (p1 - p2 ) ±1.96 π1 (1-π1 )/n1 +π2 (1-π2 )/n2 •= (p1 - p2 ) ±1.96 p1 (1-p1 )/n1 +p2 (1-p2 )/n2 Example 6.6.1 • Borst et al. investigated the relation of ego development, age, gender, and diagnosis to suicidality among adolescent psychiatric inpatients. Their sample consisted of 96 boys and 123 girls between the ages of 12 and 16 years selected from admissions to a child and adolescent unit of a private psychiatric hospital. Suicide attempts were reported by 18 of the boys and 60 of the girls. Let us assume that the girls behave like a simple random sample from a population of similar girls and that the boys likewise may be considered a simple random sample from a population of similar boys. For these two populations, we wish to construct a 99 percent confidence interval for the difference between the proportions of suicide attempters. ^ p G 60 123 .4878 ^ ^ p B 18 96 .1875 ^ p G p B .4878 .1875 .3003 (.4878)(. 5122) (.1875)(. 8125) .0602 123 96 99% C.I. .3003 2.58(.0602) .1450, .4556 s pG pB §布瓦松分布成功數μ的區間估計 (95%信賴區間) • 估計參數有95%的機率可能座落之範圍 • 點估計 ±( 標準化值* 點估計的標準誤) • x ±Z.975 SEx= x ±1.96 x §兩個布瓦松分布成功數差異 (μ1-μ2)的區 間估計(95%信賴區間) • 估計參數有95%的機率可能座落之範圍 • 點估計 ±( 標準化值* 點估計的標準誤) • (x -x ) 1 2 ±Z.975 SE •= (x -x ) 1 2 ±1.96 (X 1- X 2 ) x1 + x2 Interval Estimation • μ的95% Confident Interval(信賴區間): x z(1α 2 )σ x • t-Distribution: – In repeated sampling, the frequency distribution curve of sample means from normal population should be normal, but, when σ2 is estimated by S2 and the sampling size is small ( < 30), the frequency distribution curve of (X- μ)∕SX should be leptokurtic and is called: " student's tdistribution ". – Degree of Freedom(自由度;df) • μ的95% C.I.等於: X ±Z(1-α∕2) σ/ n X ±t(1-α∕2;df) S / n §z分布與 t 分布 • X~N(μ, σ2) x t(df)=(x- µ)/s Z =(x- µ)/ σσ t(df)~t(0,1) Z~N(0,1) 95% Z -1.96 +1.96 <95% t.025-1.96 +1.96 t(df) t.975 Properties of the t Distribution • It has a mean of 0. • It is symmetrical about the mean. • In general, it has a variance greater than 1, but the variance approaches 1 as the sample size becomes large. For df>2 , the variance of the t distribution is df/(df-2), where df is the degrees of freedom. Alternatively, since here df=n-1 for n>3, we may write the variance of the t distribution as (n-1)/(n-3) • The variable t ranges from –∞ to + ∞. Properties of the t Distribution • The t distribution is really a family of distributions, since there is a different distribution for each sample value of n -1, the divisor used in computing s 2. We recall that n -1 is referred to as degrees of freedom. Properties of the t Distribution • Compared to the normal distribution the t distribution is less peaked in the center and has higher tails. The t distribution approaches the normal distribution as n-1 approaches infinity. Confidence Intervals Using t • estimator ± (reliability coefficient) × (standard error) • When sampling is from a normal distribution whose standard deviation, σ is unknown, the 100 (1 -α) percent confidence interval for the population mean, μ, is given by x t (1α / 2) s n Example 6.3.1 • It is a study to evaluate the effect of on-the-job body mechanics instruction on the work performance of newly employed young workers. The experimental group received one hour of back school training provided by an occupational therapist. The control group did not receive this training. A criterion-referenced Body Mechanics Evaluation Checklist was used to evaluate each worker’s lifting, lowering, pulling, and transferring of objects in the work environment. A correctly performed task received a score of 1. The 15 control subjects, which behave as a random sample from a population, made a mean score of 11.53 on the evaluation with a standard deviation of 3.681. We wish to use these sample data to estimate the mean score for the population. x 11.53 standard error s n 3.681 15 .9504 degree freedom n 1 15 1 14 t.975 2.1448 95% C.I. 11.53 2.1448(.9504) 11.53 2.04 9.49, 13.57 假設兩母群的變異數相等 The pooled variance of the estimate ( X 1- X 2 ) S The standard error of the estimate ( X 1- X 2 ) The 100(1-α) percent confidence interval for (μ1-μ2) Example 6.4.3 • The purpose of a study by Stone et al. was to determine the effects of long-term exercise intervention on corporate executives enrolled in a supervised fitness program. Data were collected on 13 subjects (the exercise group) who voluntarily entered a supervised exercise program and remained active for an average of 13 years and 17 subjects (the sedentary group) who elected not to join the fitness program. Among the data collected on the subjects was maximum number of sit-ups completed in 30 seconds. The exercise group had a mean and standard deviation for this variable of 21.0 and 4.9, respectively. The mean and standard deviation for the sedentary group were 12.1 and 5.6, respectively. We assume that the two populations of overall muscle condition measures are approximately normally distributed and that the two population variances are equal. We wish to construct a 95 percent confidence interval for the difference between the means of the populations represented by these two samples. 2 2 ( 13 1 )( 4 . 9 ) ( 17 1 )( 5 . 6 ) s p2 28.21 13 17 2 the reliability factor 2.0484 =t.975(28) 95% C.I. (21.0 - 12.1) 2.0484 28.21 28.21 8.9 4.0085 4.9, 12.9 13 17 假設兩母群的變異數不相等 若兩母群的變異數不相等時,則 t 分布的 形狀不能遵循自由度為(n1+n2-2)的分布, 必須做調整。調整的方法有兩種︰(1)調 整t 的判定值,(2)調整t 的自由度;兩 種方法的權重均為該樣本平均數的變異數。 (1)調整t 的判定值 Example 6.4.4 • In the study by Stone et al. described in Example 6.4.3, the investigators also reported the following information on a measure of overall muscle condition scores made by the subjects: degrees of freeom 12 1 .05 2 .975 t1 2.1788 degrees of freeom 16 1 .05 2 .975 t 2 2.1199 (.3 2 13)( 2.1788) (1.0 2 17)( 2.1199) .139784 t' 2 2 .065747 (.3 13) (1.0 17) .3 2 1.0 2 95% C.I. (4.5 3.7) 2.1261 .8 2.1261(.25641101) .25, 1.34 13 17 假設兩母群的變異數不相等 (2)調整t分布的自由度 1/dfc=(c12/df1)+ (c22/df2) c1 = Sx12/(Sx12 + Sx22 ) = (S12/n1) / [(S12/n1) + (S22/n2 )] c2 = (1-c1) = Sx22/(Sx12 + Sx22 ) = (S22/n2) / [(S12/n1) + (S22/n2 ) ] {(S12/n1)/[(S12/n1)+(S22/n2)]}2 {(S12/n1)/[(S12/n1)+(S22/n2)]}2 + dfc= df df 1 dfc= df1*( n2 S12 )2+ df2*(n1 S22 )2 [(n2 S12)+(n1 S22)]2 2 Example 6.4.4 • In the study by Stone et al. described in Example 6.4.3, the investigators also reported the following information on a measure of overall muscle condition scores made by the subjects: df1=12 S1=0.3 SE12=S12/n1=0.0069 df2=16 S2=1.0 SE22=S22/n2=0.0588 c1= SE12/(SE12+ SE22) =0.0069/0.0657=0.105 95%C.I.(μ1-μ2)= c2= 1- c1= 1-0.105 =0.895 1/dfc=(c12/df1)+ (c22/df2) t.975(16.925) =0.1052/12 )+ (0.8952/16) =0.0510 =(4.5-3.7)±2.093*0.2564 dfc=16.925 t.975(16.925) =2.093 SQRT(SE12+ SE22)=0.06571/2 =0.2564 =0.8±0.5367=0.2633~1.3367 Determination of sample size--For estimation of population mean If d=unit on either side of the estimator, then: d=(reliability coefficient) ×(standard error) σ dz n σ d z n N n N 1 z 2σ 2 n 2 d Nz2σ 2 n 2 d (N 1) z 2σ 2 Example 6.7.1 • A health department nutritionist, wishing to conduct a survey among a population of teenage girls to determine their average daily protein intake (measured in grams), is seeking the advice of a biostatistician relative to the sample size that should be taken. What procedure does the biostatistician follow in providing assistance to the nutritionist? Before the statistician can be of help to the nutritionist, the latter must provide three items of information: the desired width of the confidence interval(5 grams), the level of confidence desired(95%C.I.), and the magnitude of the population variance (=20 grams). 2 2 (1.96) (20) n 61.47 2 (5) Determination of sample size--- For estimation of population proportion d=(reliability coefficient) ×(standard error) • d= Z * (p q / n) ½ 2 z pq • n d2 q 1 p Nz 2 pq • n 2 2 d (N 1) z pq Example 6.8.1 • A survey is being planned to determine what proportion of families in a certain area are medically indigent. It is believed that the proportion cannot be greater than .35. A 95 percent confidence interval is desired with d=.05. What size sample of families should be selected? 2 (1.96) (.35)(.65) n 349 . 6 2 (.05) • Sampling with Replacement 2 s i 0 2 2 0 200 E (s ) n 8 25 25 N E(s 2 ) σ 2 s 2 ( xi x)2 (n 1) σ 2 ( xi μ )2 N 2 • Sampling Without Replacement E (s ) 2 2 s i N S2 Cn 2 8 2 100 10 10 10 2 ( x μ ) (N 1) i • E(s 2)=σ2, when sampling is with replacement • E(s 2)=s 2, when sampling is without replacement Inference of population Variance n • S 2= (X-X) 2/(n-1)= 2 • S 2 =[ (X-X) 2 / 2 ] / [(n-1)/ 2 ] • S 2 =[ Z 2 ] / [(n-1) / 2 ] • S 2 =(n-1) 2 / [(n-1) / 2 ] • 2 =S 2 (n-1) / (n-1) 2 • 2 =SS / (n-1) 2 Confidence Interval of Population Variance( 2) • • • X X 1 (n 1)s 2 2 X 1(α 2) 2 2 2 (n 1)s σ (n 1)s σ 2 α2 2 X 2 α2 (n 1)s (n 1)s 2 σ 2 2 Xα 2 X 1(α 2) 2 (n 1)s 2 σ 2 X 1(α 2) 2 2 1(α 2) (n 1)s (n 1)s 2 σ 2 2 X 1(α 2) Xα 2 2 (n 1)s 2 2 Xα 2 2 Example 6.9.1 • In a study of the effect of diet on low-density lipoprotein cholesterol, Rassias et al. (A-21) used as subjects 12 mildly hypercholesterolemic men and women. The plasma cholesterol levels (mmol/L) of the subjects were as follows: 6.0, 6.4, 7.0, 5.8, 6.0, 5.8, 5.9, 6.7, 6.1, 6.5, 6.3, 5.8. Let us assume that these 12 subjects behave as a simple random sample of subjects from a normally distributed population of similar subjects. We wish to estimate, from the data of this sample, the variance of the plasma cholesterol levels in the population with a 95 percent confidence interval. s 2 .391868 df n 1 11 X 12(α 2) 21.920 Xα2 2 3.1816 11(.391868) 11(.391868) σ2 .196649087 σ 2 1.35483656 21.920 3.1816 95% C.I. for σ .4434 σ 1.1640 95% C.I. for σ 2 Confidence Interval for [σ12/σ22] • F=S12/S22 • ={(n1-1)2/[(n1-1)/2]}/{(n2-1)2/[(n2-1)/2]} • ={(n1-1)2/(n1-1)}/{(n2-1)2/(n2-1)} Fα 2 s12 σ12 2 2 F1(α 2) s2 σ 2 • Fα 2 s12 σ 22 2 2 F1(α 2) s2 σ1 σ 22 F1(α 2) 2 2 2 2 2 s1 s2 σ1 s1 s2 • s12 s 22 σ12 s12 s 22 2 Fα 2 σ 2 F1(α 2) s12 s 22 σ12 s12 s 22 2 F1(α 2) σ 2 Fα 2 • Fα 2 Confidence Interval for [σ12/σ22] s s σ s s • F1(α 2) σ Fα 2 2 1 2 2 2 1 2 2 2 1 2 1 2 2 s LCL s • 2 1 2 2 s UCL s 2 2 F1α ,df1,df2 1 Fα ,df1,df2 1 Fα 2,df2 ,df1 1 1 F1(α 2 ),df1,df 2 Example 6.10.1 • Goldberg et al. conducted a study to determine if an acute dose of dextroamphetamine might have positive effects on affect and cognition in schizophrenic patients amintained on a regimen of haloperidol. Among the variables measured was the change in patients’ tension-anxiety states. For n2=4 patients who responded to amphetamine, the standard deviation for this measurement was 3.4. For n1=11 patients who did not respond, the standard deviation was 5.8. Let us assume that these patients constitute independent simple random samples from populations of a normally distributed variable in both populations. We wish to construct a 95 percent confidence interval for the ratio of the variances of these two populations. n1 11 n2 4 s12 (5.8) 2 33.64 s 22 (3.4)2 11.56 df1 10 df2 3 α .05 F.025 .20704 F.975 14.42 33.64 11.56 σ12 33.64 11.56 2 14.42 .20704 σ2 σ12 .2018 2 14.0554 σ2 To be continued….. Thanks for your attention