COURSE SYLLABUS: STAT 1201(2200)
APPLIED NONPARAMETRIC STATISTICS
TH 11:00-12:15
PM
Instructor: Henry W. Block
Department of Statistics
2703 CL
Phone: 624-8369
email: hwb@stat.pitt.edu
Office Hours: TH 2:15-3:30 PM or by
appointment
Grader: Ms. Hyewook Jeong (hyj2@pitt.edu)
Grader’s Office (Hours): 2504 CL (1-3 PM, Wednesdays)
Required Text: Hollander and Wolfe, Nonparametic
Statistical
Methods
(2nd Ed.),Wiley Interscience
Prerequisite: STAT 1000 OR STAT 0200 (grade B- or better) including
familiarity with MINITAB
Description: This is an undergraduate
course on
nonparametric
statistics and is also appropriate
for
graduate students in departments other than
statistics.
It focuses on various
nonparametric
methods encountered in business and
the
social sciences. Topics include: one- and two-
sample location problems,
two- sample dispersion
problems, one- and two-way
layouts and dependence
problems. The statistical
package MINITAB will be
used throughout the course
Website:www.stat.pitt.edu/hwb/s1201/index.htm
Course
Requirements and Grading:
1) 2 Tests 40%
2) Homeworks 20%
3) Final 40%
Final Exam: Friday, December 12, 4:00-5:50 PM
Test #2: Tuesday, November 18, Open Book and Notes, covering Chapters 6-9
Homework Assignments Show all work including saying which tables were used.
If MINITAB is used include output with handwritten annotation and answers
circled.
Number Date Due Problems
1 9/2 2.4 2.5 2.7 2.9 2.11 2.13 2.15
2 9/9 3.2 3.3(Grads only) 3.4 3.5 3.7 3.8 3.9(Use of Table A.4
only)
Extra Credit: 3.9 (Use of exact null distribution as in
Comment 11)
3 9/16 3.18 3.20 3.21 3.23 3.25 3,28 3.46 3.59 3.71
4 9/23 4.2(Grads only) 4.5 4.6 4.7 4.8 4.10 4.12
5 9/30 4.18 4.19 4.21 4.28 4.33(Do two ways: 1)large sample and 2)
MINITAB) 5.3
6 10/16 6.4 6.7 6.8(without MINITAB) 6.10
7 10/21 7.2(Grads only) 7.3 7.5 7.6 7.9
8 10/28 7.30 7.31 7.32 7.34 8.3
9 11/4 8.7 8.22 8.23 8.28 8.30 8.41 8.49(Grads only)
10 11/11 9.3 9.4 9.8 9.9 9.14 9.21
11 12/2 10.2 10.3 10.7 10.13 10.16 11.1
PRACTICE PROBLEMS FOR TEST #1: 2.6 2.18 3.10 3.73(MINITAB output will be provided on the test)
4.17 (MINITAB output such as Table 4.5 will be given) 5.5
PRACTICE PROBLEMS FOR TEST #2: 6.12 7.10 8.8(Grad students only) 8.16 8.26 8.32 9.6 (MINITAB output will be given where appropriate)
PRACTICE PROBLEMS FOR FINAL: (Appropriate MINITAB output will be given on the exam)
3.15; Apply Wilcoxin Test to data of Table 4.6 (p.137); 4.14; 5.9;
7.10 Do this problem but include the 'Peak Psychiatric Period' column; 10.6; 11.1
Homework Solutions:
STAT 1201/2200 Homework #11 Solution
10.2 To test H0 ∶ p1 =p2 vs H1 ∶ p1 ≠p2 p1 =2338=0.605, p2 =1663=0.254, p =39101=0.386 SD p1 − p2 = p (1−p )n1.+p (1−p )n2.= 39101×6210138+39101×6210163=0.100001 A=p1 − p2 SD (p1 − p2 )=0.605−0.2540.100001=3.5129 for α=0.05 ⇒ zα2=z0.025=1.96 Since |A|=3.5129>1.96=z0.025, we reject H0 at the α=0.05 level. p−value=2P0 A ≥3.5129 ≤2×0.0002=0.0004
10.3 p1 =67110=0.609, p2 =3498=0.347, p =101208=0.486 SD p1 − p2 = p1 (1−p1 )n1.+p2 (1−p2 )n2.= 67110×43110110+3498×649898=0.0669 for α=0.01 ⇒ zα2=z0.005=2.575 Pd,L=p1 − p2 −zα2×SD p1 − p2 =0.609−0.347−2.575×0.0669=0.0898 Pd,U=p1 – p2 +zα2×SD p1 – p2 =0.609−0.347+2.575×0.0669=0.4345 ∴99% confidence limits for the difference in true rates∶(0.0898,0.4345)
10.7 for α=0.05 ⇒ zα2=z0.025=1.96, we have D=0.2 By Comment 1, m= zα2 22D2=1.9622×0.22=48.018 Therefore,sample size should be 49.
10.13 Using the approximate test based on χ2,we have E11=1.25, E12=3.75, E21=0.75, E22=2.25, and χ2=4.444. The P−value=P χ2≥4.444 =0.035. We reject the null hypothesis at α=0.05,which is the opposite of the conclusion based on the exact test in Problem 10.12.
10.16 To test H0 ∶ p1 =p2 vs H1 ∶ p1 >p2 ,we have nL =0,nU =5,and the observed X=2. Using the Fisher′s exact test,we find the P−value is: P X=2 +P X=3 +P X=4 +P X=5 =1−P X=0 −P X=1 =1− 50 155 205 − 51 154 205 =1−0.1936−0.4402=0.3662.
11.1 With n=12,to test H0 ∶ F is exponential versus H1 ∶ F is IFR. We use procedure 11.10 ,and obtain the Si ′s as:
| 𝑋(𝑖) | 3 | 5 | 7 | 18 | 43 | 85 | 91 | 98 | 100 | 130 | 230 | 487 |
| 𝐷𝑖 | 36 | 22 | 20 | 99 | 200 | 294 | 36 | 35 | 8 | 90 | 200 | 257 |
| 𝑆𝑖 | 36 | 58 | 78 | 177 | 377 | 671 | 707 | 742 | 750 | 840 | 1040 | 1297 |
Therefore,ε=4.222,and the P−value is larger than 0.10. We do not reject H
0 .STAT 1201/2200 Practice Problem for Final Solution
3.15
| i | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | |||||||||||
| 𝑍𝑖 | 2.5 | 3.7 | 0 | -0.6 | 4.7 | 0 | 1.4 | 0 | 1.9 | 5.2 | |||||||||||
| (1) | 𝑅𝑖 | 4 | 5 | . | 1 | 6 | . | 2 | . | 3 | 7 | ||||||||||
| 𝜓𝑖 | 1 | 1 | . | 0 | 1 | . | 1 | . | 1 | 1 | |||||||||||
| (2) | 𝑅𝑖 | 7 | 8 | 2 | 4 | 9 | 2 | 5 | 2 | 6 | 10 | ||||||||||
| 𝜓𝑖 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | |||||||||||
1 T+=4+5+6+2+3+7=27 Since there are 3 zero Z values redifining n=7. From Table A.4,for n=7,P−value=P T+≥27 =0.016 2 By Comment 9, since alternative hypothesis H1∶ θ>0,𝑡𝑟𝑒𝑎𝑡 𝑧𝑒𝑟𝑜 𝑍 𝑣𝑎𝑙𝑢𝑒𝑠 𝑎𝑠 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒.
T+=7+8+9+5+6+10=45 From Table A.4,for n=10,P−value=P T+≥45 =0.042 P−value of procedure based on discarding the zero Z values is smaller than treating the zero Z values in conservative manner,as presented in Comment 9.
*Apply wilcoxon test to data of Table 4.6 W=7+6+14+15+13+9=68 To test H0 ∶ Δ=0 vs H1 ∶ Δ>0 at the α level of significance,Reject H0 if W≥Wα P−value=P W≥68 =0.095 is obtained by entering Table A.6 at n=7,m=8. For α=0.047, Wα=71 , For α=0.116, Wα=67 Since P−value 0.095 is between 0.05 and 0.10, thus we would reject H0 for α=0.10,but not for α=0.05.
4.14 With m=n=7 and α=0.049,we have from Table A.6 that the exact level α=0.049 test is to reject H0 if W≥66. Also,the approximate level α=0.049 test based on the large−sample approximation is to reject H0 if W∗≥ z0.049=1.655 or if W≥n m+n+1 2+1.655 mn m+n+1 12=52.5+1.655 61.25=65.4524.
Hence,from Table A.6 the exact significance level of the test based on the large−sample approximation is α∗=P W≥65.4524 ∈ 0.049,0.064 ,
which is bigger than 0.049 above.
5.19 In Example 4.2 H0 ∶ Δ=0 is rejected. So the assumption that median of X is equal to median of Y is not satisfied. By Comment 13,we need to modify the test procedures. X =median X =1165.5, Y =median Y =798 Xi′=Xi−X , i=1,⋯,m, Yj′=Yj−Y , j=1,⋯,n C=9+3+7+12+3+7+4+10+5+10+1=71 𝐸0 𝐶 =𝑛 𝑁+1 24𝑁=11×2424×23=68.87 ,𝑉𝑎𝑟0 𝐶 =𝑚𝑛 𝑁+1 3+𝑁2 48𝑁2=66.37 C∗=71−68.87 66.37=0.26,P−value=2P C∗≥0.26 =2×0.3974=0.7948
7.10 To test H0 ∶ τ1=τ2=τ3=τ4 vs H1 ∶ τ1 ,⋯,τ4 not all equal R1= ri114i=1 =21 , R2= ri214i=1 = 30 , R3= ri314i=1 =36 , R4= ri414i=1 =53 S= 12𝑛𝑘(𝑘+1) 𝑅𝑗2𝑘𝑗=1 − 3𝑛(𝑘+1)= 1214×4×5 212+ 302+362+532 − 3×41×5 = 314×5 ×5446−210=233.4−210=23.4 Using large−Sample approximation,for α=0.05,χ k−1,α2= χ 3,0.052=7.81 Since S=23.4>7.81=χ 3,0.052 thus we can reject H0 at α=0.05 level.
10.6 To test H0 ∶ p1.=p.1 vs H1 ∶ p1.>p.1 p.1 =proportion of standard ELISA with positive=82+6101 p1. =proportion of ABC− ELISA with positive=82+13101 d=p1. −p.1 =82+13101−82+6101=13−6101, SD d = 13+6101 approximate large−sample test, D=dSD d =13−6101 13+6101=1.606, P−value=P D≥1.606 =0.0540
Therefore, we aceept H0 ∶ p1.=p.1 at α=0.05 level. There is evidence of equal proportions positive for both ELISA.
11.1 With n=12,to test H0 ∶ F is exponential versus H1 ∶ F is IFR. We use procedure 11.10 ,and obtain the Si ′s as:
| 𝑋(𝑖) | 3 | 5 | 7 | 18 | 43 | 85 | 91 | 98 | 100 | 130 | 230 | 487 |
| 𝐷𝑖 | 36 | 22 | 20 | 99 | 200 | 294 | 36 | 35 | 8 | 90 | 200 | 257 |
| 𝑆𝑖 | 36 | 58 | 78 | 177 | 377 | 671 | 707 | 742 | 750 | 840 | 1040 | 1297 |
Therefore,ε=4.222,and the P−value is larger than 0.10. We do not reject H0 .