The document discusses using a run test to determine if a sequence is random or not. A run is defined as a sequence of identical symbols or events. The number of runs in a random sequence will follow a normal distribution. For the given sequence of acceptable and damaged glass sculptures, the number of runs is calculated and compared to the expected number of runs for a random sequence to test the randomness of damage at the 0.05 significance level. The number of runs observed is found to not be significantly different than expected, so the damage is determined to occur randomly.

1. 04-09-2012
Understanding Equivalent calculations Understanding Equivalent calculations
n (n + 1)
Test Statistic U = n1n2 + 1 1 − R1
2
= # of ( X i , Y j ) pairs where X i < Y j
The p-value and hence the conclusion would be the same
n1n2 n1n2 (n1 + n2 + 1) irrespective of what form of the test statistic is used, i.e.
Under H 0 , µU = , σU =
2 12 R1 or R2 ɶ
or U or U
U-µ U
and has N(0,1) distribution But need to be careful on (a) left or right tailed, depending on the
σU
n ( n + 1)
form (b) Mean to be subtracted (S.E would be the same)
(n1 + n2 )( n1 + n2 + 1) ɶ
U = n1n2 + 2 2 − R2
R1 + R2 = 2
2
n ( n + n + 1)
= # of ( X i , Y j ) pairs where X i > Y j
Under H 0 : E ( R1 ) = 1 1 2
2 ɶ
U + U = n1n2
ɶ nn nn
⇒ U − 1 2 = 1 2 −U
2 2
Numerical illustration: small sample 14-34 in Aczel -Sounderpandyan
Q. Do Model B planes fly faster? (modified ex14.4)
Travel time in two models of copter-planes :
Test if the (average) current ratio for the 3 industries are the same.
Model A: 35 38 40 43 n1 = 4
Model B: 27 29 36 n2 = 3
R1 = 3 + 5 + 6 + 7 = 21 R2 = 1 + 2 + 4 = 7
4×5 mean sd
U = 12 + − 21 = 1 Easier to note this directly from pairs!
2
p-value=P(U ≤ 1) = 0.0571 (Table 9 in P798)
A 1.38 1.55 1.9 2 1.22 2.11 1.98 1.61 1.719 0.324
Note that the distribution of U is symmetric (about…?) under the null hypothesis
Let the rank of X obs. be r1 ,… rn1 . R1 = r1 + … + rn1
n1 ( n1 − 1)
B 2.33 2.5 2.79 3.01 1.99 2.45 2.512 0.356
Can you see why U = No. of (X i , Y j ) pairs with X i < Y j U = (n1 + n2 − r1 ) + … (n1 + n2 − rn1 ) −
n1 (n1 + 1) 2
= n1n2 + − R1 n1 ( n1 − 1)
2 = n1 ( n1 + n2 ) − R1 −
2 C 1.06 1.37 1.09 1.65 1.44 1.11 1.287 0.238
n1 (n1 + 1)
= n1n2 + − R1
2
Kruskal-Wallis Test Solving 14-34 using Kruskal-Wallis
Industry current ratio rank
• For comparing means of more than 2 A
A
1.38
1.55
6
8
populations – alternative to ANOVA A 1.9 11
A 2 14 ranksum sample size R^2/n
• Use if data is ordinal or the assumptions of A 1.22 4 A 79 8 780.125
ANOVA are violated A
A
2.11
1.98
15
12 B 103 6 1768.167
• Pull all observations and rank them A 1.61 9
B 2.33 16 C 28 6 130.6667
• Compute total of the ranks of observations B 2.5 18
B 2.79 19 total 210 20 2678.958
coming from 1st, 2nd ,3rd… populations B 3.01 20
B 1.99 13
• Null distribution is Chi-square with k-1 d.f B 2.45 17 12 × 2678.96
C 1.06 1 T.S. is − 3 × 21 = 13.54
12 R 2 C 1.37 5 20 × 21
T.S. is
n(n + 1 )
∑ n − 3(n + 1 ) i C
C
1.09
1.65
2
10 p − value = P(χ 2 df > 13.54) = 0.001
2
i
C 1.44 7
C 1.11 3
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2. 04-09-2012
Run test:
Problem 3
A test for randomness
• Which of the following sequences appear to be ‘random’?
– HTHTHTHTHTHTHTHTHTHTHTHTHTHTHTHT
– HHHHHHHHHHHHHHTTTTTTTTTTTTTTTTTT
A sequence of small glass sculptures was inspected for – HHHHHHHHHHHTTTTTTTTTTTHHHHHHHHH
shipping damage. The sequence of acceptable and damaged • NONE! How to determine objectively or statistically?
pieces was as follows: • Calculate the no. of runs
• A run is a sequence of identical symbols/events
D,A,A,A,D,D,D,D,D,A,A,D,D,A,A,A,A,D,A,A,D,D,D,D,D • Too many (or few) runs indicate lack of randomness
– HTHT HTHT HTHT HTHT HTHT HTHT HTHT HT
– HHHHHHHHHHHHHHTTTTTTTTTTTTTTTTTT
Test for the randomness of the damage to the shipment using – HHHHHHHHHHHTTTTTTTTTTTHHHHHHHHH
the 0.05 significance level.
Run test (cont.) Solution to Problem 3
• How to determine too many or too few?
• Acceptable no. of runs depend on n1, n2 nA = 11, nD = 14,
2 × 11× 14
If H 0 (the sequence is 'randomly' mixed) is true,
r-µr
has approximately µr = + 1 = 13.32,
σr 11 + 14
N(0,1) distribution, provided either n1or n 2 moderately large ( ≥ 10). 2 × 11×14(2 ×11× 14 − 25)
Small sample distributions are avaliable (Table 8 page 796-797).
σr = = 2.41
25 2 × 24
Atα = 0.05, the C.R. is | Z | > 1.96.
2n1n2 2n1n2 (2n1n2 − n1 − n2 )
9 - 13.32
µr = +1 σr = The observed r = 9, and value of the T.S. is = −1.79.
n1 + n2 (n1 + n2 )2 (n1 + n2 − 1) 2.41
So at 5% level we conclude that damages occur randomly
Data summarization Expected Value
And presentation in decision making
Decision trees Discrete: General, Binomial, Poisson
Probability Random variable
And its Distribution
Continuous: General, Normal,Exponential
T, Chi-square, F
Confidence interval/
Testing hypothesis Sampling
Sampling distribution of
π X, p, S 2
1 or 2
µ
sample
σ
Goodness of Fit
ANOVA NP
Test for indep/homogeneity
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