CHAPTER 6. Estimation

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1 CHAPTER 6 Etimation Definition. Statitical inference i the procedure by which we reach a concluion about a population on the bai of information contained in a ample drawn from that population. Definition. A point etimate i a ingle numerical value ued to etimate the correponding population parameter. Definition. An interval etimate conit of two numerical value that, with a pecified degree of confidence, mot likely include the parameter being etimated. Definition. An etimator, ay T, of the parameter i aid to be an unbiaed etimator of if the mean of the ampling ditribution µ T =. Example of unbiaed etimator are the ample mean, the ample proportion, and the di erence between ample mean and proportion. Definition. The ampled population i the population from which one actually draw a ample. Definition. The target population i the population about which one wihe to make an inference. Note. Statitical method aume random ample. Ideally, we want the ampled population to be like the target population. 6

2 6. ESTIMATION 63

3 64 6. ESTIMATION

4 Proper Sampling 6. ESTIMATION 65 a) Simple Random Sampling (SRS) each et of n individual ha an equal chance of being elected. Thi i a ample where not only doe each core have an equal chance of being choen, but each group of n core ha an equal chance of being choen. Example. Suppoe a high chool ha 300 enior who have taken the SAT and their core, along with their gender, are on the previou two page. We will ue ample of ize 1 to etimate the mean of the verbal core. We aume we know the following: = , n = 1, p n = p 1 = Aume alo that the verbal core are normally ditributed and note that the core are ordered according to the Math core. (1) To etimate the mean of the verbal core, we take a imple random ample (SRS) of 1 of the 300 verbal core. We ue the TI to chooe 1 random number. Pre ndmath > 7 : Probability > 4 : rand( and complete the command to rand(300) and pre ENTER 1 time. We get the number 1, 14, 10, 99, 61, 40, 86, 67, 111, 3, 81, 33. Throw out any number that are repeat ince 1 ditinct number are neeeded. Thee number correpond to verbal core of 540, 660, 400, 410, 690, 410, 500, 40, 50, 570, 630, 510. The mean of thi ample i x = Thi i alo our etimation of the population mean µ. Jut how good i thi etimate? We will come back to thi.

5 66 6. ESTIMATION (b) Stratified Sample divide the population into trata by major characteritic to enure repreentation from each group (male-female, race, etc.). Example. () Ue two SRS to chooe 6 male and 6 female. (3) Ue ix SRS to chooe from each column (trata by SAT Math). (4) Ue two SRS to chooe 6 from each page (trata by SAT Math). (c) Multitage Sample e.g., pick 10 tate, 3 countie in each, and then two citie per county. Example. (5) Ue an SRS to pick 6 of the 50 row, then 6 SRS to pick two from each row. Bad Sampling (a) Convenience Sampling (often done in biology and medicine) Example. (6) Take the firt 1. (7) Chooe a tarting point (poibly randomly) and take it and the next 11. (b) The ampled population i not imilar to the target population. Example. (8) Take an SRS of 1 from the firt page (9) Take an SRS of 1 from the econd page

6 6. ESTIMATION 67 How good i our extimate of µ = in (1)? Conider the following. From the tandard normal curve, P ( 1.96 apple z apple 1.96) =.95. We denote the tandard error of the mean the ample, Then o after the ample i drawn, P (µ 1.96 x apple x apple µ x ) =.95. x 1.96 x apple µ apple x x x 1.96 x apple µ apple x x, P (x 1.96 x apple µ apple x x ) =.95. x = p n. Then, prior to taking We ay x 1.96 x apple µ apple x x i a 95% confidence interval for the population mean µ. Thi mean that for every 100 time we take uch a imple random ample, we have that i true an average of 95 time. x 1.96 x apple µ apple x x Similarly, a 99% confidence interval for the population mean µ i x.5758 x apple µ apple x x Now let return to our example. We have x = p n = p 1 = , o for the 95% confidence interval we have (3.1564) apple µ apple (3.1564) or apple µ apple ,

7 68 6. ESTIMATION and for the 99% confidence interval we have (3.1564) apple µ apple (3.1564) or 46.0 apple µ apple Obviouly, larger ample will give tighter confidence interval ince maller. Thee computation are eay to do on the TI-89 calculator. Pre x will be APPS > 1 : FlahApp > Stat/LitEditor > F7 : Int > 1 : ZInterval In the window that open, chooe Stat for Data Input Method, put in for, for x, 1 for n,.95 for C-Level, and pre ENTER (poibly twice) to get a window with the confidence interval. The ME tand for margin of error. For a 99% confidence interval, jut change C Level to.99. If your ample i in a lit, ay lit1, chooe Data intead of Stat, put in lit1 for Lit, and 1 for Freq. There i a good applet illutrating all of thi at The applet we are uing i Simulating Confidence Interval for Population Parameter. Put in 1 in the box for n, Normal for Ditribution, for Mean, for Std.Dev., 100 for Interval, and then pre Sample. If you have trouble entering the data, double-click on the number you wih to replace, then enter the data.

8 6. ESTIMATION 69 For 100 uch ample taken here, we ee that 6 of the green bar (changed to red) do not cover our mean (for the 95% confidence interval). Now change the confidence level to 99% and hit Recalculate. We ee there are no longer any red bar in thi intance. If you continue to click on Sample, the number for the Running Total hould approach.95 and.99, repectively. Finally, the true population mean for our 300 verbal core i µ =

9 70 6. ESTIMATION Example. Normally ditributed population, = 100. Take SRS of ize n = 400 with x = 461. x = p = p 100 = 100 n = 5. The point etimate i x = 461. For our interval etimate, we ue the 95% confidence interval. Since P (z apple 1.96) =.975 or P ( 1.96 apple z apple 1.96) =.95, the 95% confidence interval i or In general, 461 ± 1.96(5) {z } margin of error (451., 470.8). = 461 ± 9.8 {z} ME interval etimate = etimator ± (reliability coe cient)(tandard error) {z } margin of error=me = x ± z (1 ) x where z (1 ) i the value of z where 1 of it area lie to it left and to it right. Thi give a 1 confidence interval for µ. Confidence level 1 z (1 ) z.95 = z.975 = z.995 =.5758 Thu a 99% confidence interval for µ in the example above i or 461 ±.5758(5) = 461 ± ( , ).

10 6. ESTIMATION 71 Sampling from Nonnormal Population If n i large (n 30), the Central Limit Theorem applie and the ampling ditribution of x i approximately normally ditributed. Problem (6..). n = 16, x = 5.98, = 3.5, x = = % confidence interval: 95% confidence interval: 99% confidence interval: 5.98 ± (.875) = 5.98 ± 1.44 (4.59, 7.4) 5.98 ± 1.96(.875) = 5.98 ± 1.7 (4.7, 7.70) 5.98 ±.5758(.875) = 5.98 ±.5 (3.7, 8.3) Practical interpretation ince the population i approximately normal, we are 99% confident that the interval (3.7, 8.3) contain the population mean µ. Probabilitic interpretation ince the population i approximately normal, in repeated ampling, 99% of the confidence interval formed in thi way will contain the population mean µ. The t Ditribution There i a problem uing the z ditribution that we ignored in the above how likely are we to know without knowing µ? We have been uing z = x µ for the ditribution of ample mean. p n For n 30, we can ue = p P (xi x) /(n 1) to replace, and till ue z.

11 7 6. ESTIMATION However, if the population i normal, we can ue t = x µ p, n which follow Student t ditribution. Actually, there i a t ditribution for each degree of freedom df = n 1. The blue graph i the normal ditribution, and the green, red, and violet graph are t ditribution with degree of freedom (df) of 30, 5, and, repectively. Propertie of the t Ditribution (1) The mean i 0. () It i ymmetric about the mean. (3) In general, the variance = df df ize become large. (4) The variable t range from 1 to 1. for df > approache 1 a the ample

12 6. ESTIMATION 73 (5) The t ditribution i a family of ditribution, with a di erent one for each df = n 1, the divior ued in computing. (6) Compared to the normal ditribution, the t ditribution i le peaked in the center and ha thicker tail. (7) The t ditribution! the normal ditribution a df! 1. For a 100(1 )% confidence interval for µ, where ampling i from a normal ditribution with unknown tandard deviation, we ue interval etimate = etimator ± (reliability coe cient) (tandard error of the etimator) = x ± t (1 /) p. n Problem (6.3.). (a) x = (b) =.9. (c) x p n =.9 p 8 = (d) 95% CI = x ± t 7.975(1.053) = ± (.3646)(1.053) = ±.444 = (e) preciion or MOE =.444 (6.4506, ) Nonnormality Moderate departure from normality can be tolerated. However, mot reearcher require at leat a mound-haped ditribution.

13 74 6. ESTIMATION The following flow chart, which i alo on page 175, can help you decide whether to ue z, t, or a nonparametric tet (?). Confidence Interval for the Di erence of Two Population Mean Recall that µx1 x = µ1 µ and x1 x = 1 n1 + n. Cae 1: n1 and n are large, o the ampling ditribution for µ1 approximately normal. The 100(1 )% CI i (x1 If 1 and x) ± z1 1 n1 + n µ i at leat. are unknown, then ue 1 and to approximate them.

14 Problem (Page 06 #0). 6. ESTIMATION 75 Group (Treatment) x n A B Since n A and n B are large, aume x B x A i approximately normal. Alo, approximate B and A by B = 30 and A = 5. The 95% CI i r 30 (15 95) ± z = 30 ± 1.96(6.6009) = 30 ± 1.94 = (17.06, 4.94) Since 0 i not in the CI, we conclude the mean µ A and µ B are not the ame. Cae : Both population are normally ditributed with known variance. Then the 100(1 ) CI i 1 (x 1 x ) ± z 1 +. n 1 n Cae 3: Population are normally ditributed, variance are unknown, but 1 =. In thi cae, we ue the t ditribution. We aume 1 and are both etimate of the common variance. We ue a pooled etimate, a weighted average, of the ample variance, p = (n 1 1) 1 + (n 1) n 1 + n and the 100(1 )% CI for µ 1 µ i (x 1 x ) ± t n 1+n 1 =) x1 x = p n 1 + p n. p n 1 + p n

15 76 6. ESTIMATION Problem (6.4.4). We aume two independent SRS from normally ditributed population with equal variance. Gender x n F M p = 5(4.36) + 13(3.36) = x1 x = r = 1.79 The 95% CI i ( ) ± t x1 x = 3.38 ±.1009(1.79) = 3.38 ± 3.76 = ( 7.14,.38) Cae 4: The population are normally ditributed, the variance are unknown and 1 6= or their equality i unknown. A problem: (x 1 x ) (µ 1 µ ) 1 + n 1 n doe not follow a t ditribution with df = n 1 + n.

16 6. ESTIMATION 77 Note. We will follow the method ued by the TI-89 and SPSS, which di er from our text. We will ue t with degree of freedom, where apple 1 + n 1 = " 1 n 1 n n n n 1 Since i eldom an integer, we round it o to the nearet integer for table ue. Then the 100(1 )% CI i (x 1 x ) ± t n 1 n Problem (6.4.10). We aume the population are normally ditributed and the variance are unknown and whether they are equal i alo unknown. Thu the 95% CI i xn x S = = #. Catagory x n N S N n N + S n S = " r =.33 # = ( ) ± t xn x S = 5.69 ±.0555(.33) = 5.69 ± 4.79 = ( 10.48,.90)

17 78 6. ESTIMATION Confidence Interval for a Population Proportion Ue bp a the point etimate. The interval etimate i etimator ± (reliability coe cient) tandard error of the etimator. If nbp 5 and n(1 bp) 5, aume the ampling ditribution i approximately normally ditributed and ue z. Etimate the tandard error r r p(1 p) bp(1 bp) bp = by bp =. n n Then a 100(1 )% CI for p i Problem (6.5.). r bp(1 bp) bp ± z 1. n bp = =.584, n = 0, z.95 = nbp = 0(.584) = and n(1 bp) = 0(.4158) = Then a 90% CI for p i r.548(.4158).584 ± = ± (.0347) =.584 ±.0571 = (.571,.6413)

18 6. ESTIMATION 79 Confidence Interval for the Di erence of Two Population Proportion If n 1 bp 1 5, n 1 (1 bp 1 ) 5, n bp 5, n (1 bp ) 5, we aume the ampling ditribution of p 1 p i approximately normal with point etimate bp 1 bp and tandard error of the mean bp 1 (1 bp 1 ) bp 1 bp + bp (1 bp ). n 1 n Then the 100(1 )% CI for p 1 p i (bp 1 bp ) ± z 1 bp 1 (1 bp 1 ) Problem (6.6.). bp 1 = =.7899, bp = 10 r.7899(.101) bp 1 bp = Then a 95% CI for p 1 + bp (1 bp ). n 1 n 136 =.884, n 1 = 138, n = 136, z.975 = 1.96 p i (.1176) 136 ( ) ± 1.96 bp1 bp =.095 ± 1.96(.0443) =.095 ±.0868 = (.1793,.0057) =.0443

19 80 6. ESTIMATION Determining the Sample Size for Etimating Mean Suppoe we want a CI of length d, where d = (reliability coe cient) (tandard error of the etimator). {z } margin of error (MOE) (1) If ampling with replacement or from a u ciently large population (i.e., n apple.05): N d = zp =) n = z n d () If ampling without replacement from a mall population: Etimating r N n d = zp n N 1 for the Formula =) n = Nz d (N 1) + z (1) Draw a pilot or preliminary ample of ize n 1, compute, and ue it to approximate. Then compute n, and chooe n = n n 1 more ubject. () Etimate may be available from previou or imilar tudie. (3) If the approximation i approximately normal, ue range R 6, which implie that R. However, thi aume knowledge of the larget and 6 mallet value of the variable in the population. Problem (6.7.4). We aume a u ciently large population. d = 1 (10) = 5, z.975 = 1.96, = 90 n = z = (1.96) (90) = d 5 Then round up and chooe n = 14.

20 6. ESTIMATION 81 Determination of Sample Size for Etimating Proportion Auming the ampling ditribution of bp i approximately normal: (1) When ampling with replacement, or from an infinite population, or n apple.05: N n = z pq where q = 1 p d () If the finite population correction i needed: Etimating p (1) A pilot ample. n = Nz pq d (N 1) + z pq () If it i known (or aumed) that p < p 0 for ome 0 < p 0 <.5, ue p 0. (3) p =.5 yield the maximum value for n. (See graph below) Problem (6.8.4). n N apple.05, d =.03, p.0. For a 95% CI, n = (1.96) (.0)(.80) (.03) = Rounding up, ue n = 683. Maple. See t.mw or t.pdf.

Social Studies 201 Notes for March 18, 2005

Social Studies 201 Notes for March 18, 2005 1 Social Studie 201 Note for March 18, 2005 Etimation of a mean, mall ample ize Section 8.4, p. 501. When a reearcher ha only a mall ample ize available, the central limit theorem doe not apply to the