StatAppHW3

StatAppHW3 - Homework 3 for STA 5166 (Assigned, Oct. 8)...

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Unformatted text preview: Homework 3 for STA 5166 (Assigned, Oct. 8) Statistics in Applications I Due: Oct. 17, 2007 (Wednesday) s 1: BHH Ch.2; Problems 10, 12, 13(a,b,c,d); Pages 62—65. (40) 2: BHH Ch.3; Problem 2 (Pages 124-125). Submit both your summary results and R/Splus program for the problem. (20) 3: BHH Ch.3; Problems 4, 7, and 13 (Pages 125—128). For each of the three problems, perform a t—test on the difference of the two means and perform a test based on a ran- domization distribution (use R/Splus to generate 10000 samples and plot the histogram of the differences). Submit both your summary results and R/Splus program for each of the problems. (40) \o) Q (\ X 7f '3.“ 3 Q L1 ‘3‘ - "M ~41 0.1»0903 1 33'7"» fiSfiNfi-k 3m: {‘mcfiw‘g ‘QAW 0.1.053 on; M g gyfl % Q'Wam w mama Mr: w» w 5”? 0‘ “(if Mr:— ;2.'\ 1“ o? oaswwsfi) \m W33 QM.- ”le {‘mmfiga AA 1 ‘92." dxg\‘-€W\ag¥§~3’k E . 3 (5)) x“ tha’m) a),?ka<Q%§)< VLZK ) M .'§ fl.9%%wm»f \3) «7 \L‘KVU‘ZO) « \{ka w? 2:33am» m E m f 3‘ Q\ n \ k Q’ X (f 3313/13} ? QWDHQJ v ( E- < ‘\ (Lia \W a): 9 31M \ c9 0 E 2% Eq/%” 1 5 WW 3 kg 0' [got ( 3?;36‘4 3 3 a? w i: 3 t a; if [fig/(fig '''' ' W ?3 :3 0% 21¢: ) "3” "T is w “x “m; :3; WM“) “My '? f: Q 3‘32“ M L mm}? Y ‘7}, mm \ICF‘ICHUL. Ami} W»! k, WQQWHM v0 \ X gmf \Ew {X 9&2 a : web-m J {a x“ f \ «\5‘ E (’"X \W L». E J 45> . k a, 2x5” Tam—«ow WM W “ W WM «MW {3 O M Chapter 3.2 (:0 :1) Summary: Will try to test the hypothesis to see if there exists a significant difference between the mean values of levels of asbestos fiber in the air of the industrial plant With and Without S—l42 chemical. From the comparative trail in the plant, the four consecutive readings had a mean difference of —3.5. The null hypothesis is that with or Without S-l42, the asbestos levels will not change, the alternative is that with S-142, the level will decrease since the mean difference is negative. To test this, used as a reference the past observations of asbestos levels Without S~142. From the dataset, obtained a probability that 1/109 (=0.009l743119) that there exists a mean difference less that the comparative trail. Since this probability is less that 5%, we reject the null hypothesis and accept the salesman claim that S—142 is beneficial to reduce the level of asbestos levels in the air of the industrial plant. WWWMMWWWWWMWMWMMW u .- CODE MKMquWmWWthMiW « WWWWWWmWWMWmW data=scan("C:/Documents and Settings/Jaime/Desktop/FALL07/STA5166/BHH2— Data/datahw3.dat") data nl=0 Meanlwout = mean(c(8,6)) MeanZwith = mean(c(3,4)) diff_means = Mean2with—Meanlwout y = c(rep(NA, (109))) X = c(rep(NA, (109))) for(i in l:lO9){ y[i] (data[i]+data[i+l])/2} f0r(j in 11109){ X[j] Y[j+2] - Y[j]; if(x[j]<= difffimeans) n1=nl+l} x sort(x) nl diff_means nl/lO9 mmmmwwmmmwmwwmmmrwwwwmwww ,, _ .. .. , WWWWVAWWLW/memfl“WWWWLWWWWM WW-flsmflwgvmmwflxammwmimawmemwwlwfiw“mewwwm, _ anm mm > data=scan("C:/Documents and Settings/Jaime/Desktop/FALL07/STAS166/BHH2— Data/datahw3.dat") Read 112 items >data [1]9108988876910119101111111110111213121312 [26]14151412131312131313131310898677656564 [51]544245456556567888791091098 [76]9877877788887656567665665 [101]434455656765 >n1=0 > Meanlwout = mean(c(8,6)) > MeanZWith = mean(c(3,4)) > diff__means = Mean2with—Mean1wout > y = C(rep(NA, (109)» > x = c(rep(NA, (109))) > for(i in 1:109){ y[i] = (data[i]+data[i+1])/2} > f0r0 in 11109){ KB] = YU+21 — y[i]; + if(x[j]<= diff_means) n1=—n1+1} Error in if (x[j] <= diff_means) n1 = n1 + 1 : missing value where TRUE/FALSE needed >’X [1] —1.0 —0.5 —0.5 —0.5 —0.5 —1.5 0.0 3.0 3.0 0.5 -1.0 0.5 1.5 0.5 0.0 [16] —0.5 05 1.0 2.0 1.0 0.0 0.0 0.5 2.0 1.5 —1.5 -2.0 0.0 0.0 -0.5 [31] 0.5 0.5 0.0 0.0 —1.5 -4.0 -3.0 —0.5 —1.5 —2.0 0.0 0.0 —1.5 ~1.0 0.0 [46] 0.0 —0.5 -1.0 -0.5 —0.5 -1.5 —1.0 1.5 1.5 0.0 1.0 1.0 -0.5 0.0 0.5 [61] 0.0 1.0 2.0 1.5 0.5 —0.5 0.0 2.0 1.5 0.0 0.0 —1.0 -1.0 0.0 —1.0 [76] —1.5 0.0 0.5 ~05 —O.5 0.5 1.0 0.5 0.0 -0.5 —1.5 -2.0-1.0 0.0 0.0 [91] 1.0 1.0-0.5 —1.0—0.5 0.5 0.0 —1.5 —2.0 ~1.0 0.5 1.0 1.0 1.0 0.5 [106] 0.0 1.0 NA NA >sort(x) [1] —4.0 —3.0 ~2.0 —2.0 -2.0 -2.0 —1.5 -1.5 -1.5 -1.5 -1.5 —1.5 -1.5 -1.5 -1.5 [16] —1.0 -1.0 —1.0 —1.0 ~1.0 ~1.0 ~1.0 —1.0 —1.0 —1.0 —1.0 -0.5 -0.5 -0.5 -0.5 [31] —0.5 -0.5 —0.5 -0.5 -0.5 -0.5 —0.5 -0.5 -0.5 -0.5 -0.5 —0.5 -0.5 -0.5 0.0 [46] 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 [61] 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.5 0.5 0.5 0.5 0.5 0.5 [76] 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 1.0 1.0 1.0 1.0 1.0 1.0 1.0 [91] 1.0 1.0 1.0 1.0 1.0 1.5 1.5 1.5 1.5 1.5 1.5 2.0 2.0 2.0 2.0 [106] 3.0 3.0 >n1 [1] 1 >diff_means [1] -3.5 >n1/109 [1] 0009174312 Chapter 3.4 (:0 Ill) Assumptions: Ratings are both approximately normal distributed. Two samples, A and B, are independent. Ratings in each brand are i.i.d. Summary: To test the hypothesis that 77A = 773 , against the 77A 7: 773 . I used a t—test to check if the difference in means is not equal to zero. The p-value obtained 0.3316. Therefore there is not sufficient evidence to reject the null hypothesis. The assumptions were needed to conduct test. Using a randomization distribution, I also tested the above hypothesis. Here, I did not make assumptions about the distributions of the ratings. The mean of brand A: (3.875) and brand B: (5.285714), to obtain a difference of 1.4107. There exist 6435 possible permutations of 8 ratings of brand A and 7 ratings of brand B. Assuming that the null hypothesis, then there exist no difference in the ratings of brand A and brand B. Can arrange and for each calculate the differences that are less than 1.4107. Count the number of occurrences and this will lead to a calculation of the p—value. The p—value obtained after a large number of observations should be approximately equal to the p—value obtained from the t-test above. I obtained the p—value: 0.3551. This also leads to the conclusion that one cannot reject the null hypothesis. unwavwaFWAWMWWWWWWMKWWWWWWWWW . v , CODE y = t.test(brandA, brandB) y nl=0 hl=0 yl= c(2,4,2,l,9,9,2,2,8,3,5,3,7,7,4) Cl=c(rep("A“, 8), rep("B", 7)) dl = c(rep(0,10000)) diff: 5.285714—3.875 for(i in 1:10000){ C2=sample(cl)i xl=yl[c2=="A"]; X2=yl[C2=="B" ; ml = mean(x;); m2 = mean(x2); dl[i] : m2—m1; hl=c(h1,dl[il); if(abs(dl[i]) >= 1.4107)nl=nl+l } n1 hist(hl, main=“Randomization Distribution“) pvalue= nl/lOOOO pvalue »»»»»»»»»»»»»»»»»»»»»»» v. , , I OUTPUT r WWWWWWMWWWIWWWIWNWWWJWMO > brandA = c(2,4,2,1,9,9,2,2) > brandB = c(8,3,5,3,7,7,4) > > y = t.test(brandA, brandB) > y Welch Two Sample t—test data: brandA and brandB t = ~1.0122, df= 11.923, p-Value = 0.3316 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -4.449587 1.628159 sample estimates: mean of x mean of y 3.875000 5.285714 >n1=0 > h1=0 > y1= c(2,4,2,1,9,9,2,2,8,3,5,3,7,7,4) > cl=c(rep("A", 8), rep("B", 7)) > d1 = c(rep(0,10000)) > diff: 5.285714—3.875 > for(i in 1:10000){ + 02=sample(cl); + x1=y1[02=="A"]; + x2=y1[c2=="B"]; + m1= mean(x1); + m2 = mean(x2); + d1[i] = m2—m1; + h1=c(h1,d1[i]); + if(abs(d1[i]) >= 1.4107)n1=n1+1 + } > n1 [1] 3551 > hist(h1, main="Randomizati0n Distribution") > pvalue= n1/ 10000 > pvalue [1] 0.3551 Randomization Distribution 1500 1000 Frequency 500 Chapter 3.7 (3.0 ’b) Assumptions: Results are both approximately normal distributed. Two samples, designs A and B, are independent. Results in each design are i.i.d. Summary: ~ Will try to test the hypothesis to see if there exists a significant difference between the mean values for the power attainable for the two designs. The null hypothesis assumes there is no difference in the mean values. I used a t-test to check if the difference in means is not equal to zero. The p-value obtained 0.4343. Therefore there is not sufficient evidence to reject the null hypothesis. The assumptions were needed to conduct test. Using a randomization distribution, I also tested the above hypothesis. Here, I did not make assumptions about the distributions of the ratings. The mean of design A: (1.55) and brand B: (1.75), to obtain a difference of 0.2 The p—value obtained after a large number of observations should be approximately equal to the p—value obtained from the t-test above. I obtained the p-value: 0.4454. This also leads to the conclusion that one cannot reject the null hypothesis. W’Ww »c,_.MWW ,.... u. NLWKW‘ «« designA = c(l.8, 1.9, 1.1, 1.4) designB = c(l 9, 2.1, 1.5, 1.5) y = t.test(designA , designB) Y n1=0 hl=NULL y1= c(l.8, 1.9, 1.1, 1. c1=c(rep("A", 4), rep( diff=1.75—1.55 d1 = rep(O, 10000) for(i in 1:10000){ c2=sample(c1); x1=y1[c2==“A"]; x2=y1[c2::"B"] m1 : mean(x1); m2 = mean(x2); d1[i] = m2—m1; hl=c(h1,d1[i]); if(abs(d1[i]) >= diff)n1=n1+1 } n1 hist(h1, main=“Randomization Distribution") pvalue= nl/lOOOO , 1.9, 2.1, 1.5, 1.5) pvalue OUTPUT > designA : c(l.8, 1.9, 1.1, 1.4) > designB = c(1.9, 2.1, 1.5, 1.5) > > y = t.test(designA , designB) > Y Welch Two Sample t—test data: designA and designB t = —0.8402, df = 5.756, p—value = 0.4343 alternative hypothesis: true difference in means is not equal to O 95 percent confidence interval: —0.7885191 0.3885191 sample estimates: mean of x mean of y 1.55 1.75 n1=0 hl=NULL y1= c(l.8, 1.9, 1.1, 1. c1=c(rep("A“, 4), rep( diff=1.75—1.55 d1 = rep(O, 10000) for(i in 1:10000){ , 1.9, 2.1, 1.5, 1.5) VVVVVVV + c2=sample(cl); + Xl=yl[C2=="A"]; x2=yl[c2=="B“] + ml 2 mean(xl); m2 = mean(x2); + dl[i] = mZ—ml; + hl=c(hl,dl[i]); + if(abs(dl[i]) >= diff)nl=nl+l + } > n1 [1] 4454 > hist(hl, main="Randomization Distribution“) > pvalue= nl/lOOOO > pvalue [1] 0.4454 Randomization Distribution 1500 Frequency 1000 500 ~06 43.4 43.2 {3.0 8.2 8.4 {3.6 m Chapter 3.13 (:0 +3) Assumptions: Results of production from each diet are both approximately normal distributed. Two samples, designs A and B, are independent. Results in each diet are i.i.d. Summary: Will try to test the hypothesis to see if there exists a significant difference between the mean values for the power attainable for the two designs. The null hypothesis assumes there is no difference in the mean values. I used a t-test to check if the difference in means is not equal to zero. The p-Value obtained 0.07842. Therefore there is not sufficient evidence to reject the null hypothesis. The assumptions were needed to conduct test. Using a randomization distribution, 1 also tested the above hypothesis. Here, I did not make assumptions about the distributions of the ratings. The mean of diet A: (166.5) and brand B: (156.6667), to obtain a absolute value of the difference of 9.83 The p—value obtained after a large number of observations should be approximately equal to the p-Value obtained from the t—test above. I obtained the p—Value: 0.0913. This also leads to the conclusion that one cannot reject the null hypothesis. A 95% confidence interval for the mean difference: [4.8649, 14.79508] Here, the 95% confidence interval for the difference in mean hen production between diet A and diet B numbers above. Thus, not only do we estimate the difference to be 9.83 mg/dl, but we are 95% confident it is no less than lower bound or greater than upper bound. m»wmw»wwwwmwmmwwwwmx : -- ‘memesawflvwwwm 992E WWWNWWWWWWWWWWK»WWMWMWWWWWWWWWV dietA dietB c(l66,l74,150,166,165,l78) C(158,159,l42,163,161,157) ll 1! y = t.test(dietA , dietB) Y nl=0 hl =NULL yl= C(l66,174,150,166,l65,l78, 158,159,142,l63,l6l,157) cl=c(rep("A", 6), rep("B", 6)) diff=l66.5 — 156.6667 d1 = rep(0,10000) for(i in l:lOOOO){ C2=sample(cl); xl=yl[c2==“A“]; X2=yl[c2=="B"]; ml 2 mean(xl); m2 = mean(x2); dl[i] = m2—ml; hl=c<hl,dl[i]); if<abs(dl[i])>=9.83)n1=n1+l } nl hist(hl, main="Randomization Distribution“) pvalue= nl/10000 pvalue > dietA = c(166,174,150,166,165,l78) > dietB = c(l58,159,142,163,161,157) > > y = t.test(dietA , dietB) > y Welch Two Sample t—test data: dietA and dietB 1:: 1.9735, df= 9.436, p-Value = 0.07842 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -1.359600 21.026267 sample estimates: mean of X mean of y 166.5000 156.6667 > nl=0 > h1=NULL > y1= c(166,l74,150,166,165,178,158,159,142,163,161,157) > cl=c(rep("A", 6), rep("B", 6)) > diff=166.5 ~ 156.6667 > d1 = rep(0,10000) > for(i in 1:10000){ + 02=sample(cl); + x1=y1[02=="A"]; + X2=y1[02=="B"]; + m1 = mean(x1); + m2 = mean(x2); + d1[i] = m2—m1; + h1=c(h1,d1[i]); + if(abs(d1[i])>=9.83)n1=n1+1 +} >n1 [1] 913 > hist(h1, main="Randomization Distribution") > pvalue= n1/ 10000 > pvalue [1] 0.0913 1500 1000 Frequency 500 Randomizatien Distribution ...
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This note was uploaded on 12/14/2011 for the course STAT 5166 taught by Professor Staff during the Fall '11 term at FSU.

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StatAppHW3 - Homework 3 for STA 5166 (Assigned, Oct. 8)...

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