1. Case: The Roulette Wheel Goodness of Fit Tests
Class 06 07
Case: The Roulette Wheel
Goodness of Fit Tests
EMBS 11.2

2. What we learned last class
Probability Distributions have characteristics
Descriptive Statistics are used to estimate those characteristics.
Location (mean, median, mode)
Variability (variance and standard deviation)
Shape (skewness)
The MEAN is important.
Sample mean “value” times n is total value.
Measures of variability are under-appreciated

3. Descriptive Statistics Matter

4. Baseball Statistics A batter came to the plate five times Got a hit
Struck Out
Walked
Flied Out
Grounded Out
WALK is the fault of the pitcher: 1 success in 4 trials
Batting Average = ¼ = 0.250
WALK is the fault of the batter: 2 successes in 5 trials
On Base Percentage = 2/5 = 0.400

5. The Roulette wheel.
Surveillance video of 18 hours of play of a roulette wheel in a Reno, Nevada casino
904 spins of the wheel
22,527 bets places

6. Outcome
Frequency
Number of Bets 00 22
354 18 23
518 25
442 19 30
595 1
362 20 24
983 2
450 21 26
447 3 28
357 32
576 4 15
375
746 5
636
461 6
363
521 7
682
703 8
633 27
490 9
503
827 10
484 29
878 11
783 33
695 12
360 31
664 13
525
925 14
649 17
613
340 34
597 16
643 35
627
1,079 36
641
Total
904
22,527

7. First We Examine the Wheel
H0: The wheel works properly
H0: All 38 outcomes have equal probability of occurring
H0: P0=P28=P9=…=P2=1/38
HA: they are not all equal
Like before, the Hypothesis is about a parameter of a probability distribution

8. Goodness-of-fit Test Calculate the expected counts under H0
Outcome
Observed
Expected
Distance 00 22
23.79
0.13 25
0.06 1 23
0.03 2 30
1.62 3 28
0.75 4 15
3.25 5 6 20
0.60 7 8 26
0.21 9 10 24
0.00 11 12 21
0.33 13 14 27
0.43 16 17 18 19 32
2.83
1.41
0.96 29 33
3.57 31
1.14
1.94 34 35 36
Total
904
31.20
Calculate the expected counts under H0
Calculate the Distances as (O-E)2/E
The sum of the distances is the test statistic. We call it the calculated chi-squared.
We reject H0 (and say the results are statistically significant) if the calculated chi-squared statistic is too big.
The calculated chi-squared statistic
The sum of the distances.

9. We need a p-value! For the lady tasting tea For a GOODNESS OF FIT TEST
Number correct depends on n and P
For the lady tasting tea
P(X≥8 │ H0) = 1-binomdist(7,10,.5,true)
Pvalue = 0.055
For a GOODNESS OF FIT TEST
P(χ2≥ calculated χ2 │H0) = chidist(calculated χ2, dof)
dof stands for “degrees of freedom”
dof is the parameter of the chi-squared distribution
dof here is 37, the number of cells - 1.
P(χ2≥ 31.2│H0) = chidist(31.2,37)
Pvalue = 0.74
χ2 depends on number of cells - 1.
Can also use
=chisq.dist.rt(31.2,37)
NOT statistically significant.

10. WARNING
The chi-squared test does not work well when some cells have low expected counts.
If some cells have expected counts < 5, combine then with neighboring cells.

11. Roulette Wheel Demonstration
H0: All 38 are equally likely to get bet on.
Ha: The p’s are not equal. (Some segments are more popular than others)

12. H0: The Fill Amounts are Normally Distributed with μ=10.2 and σ=0.16
Lorex Pharma
H0: The Fill Amounts are Normally Distributed with μ=10.2 and σ=0.16
Ha: They are not…

13. Assignment 08 Due Monday, Feb 13
Youth Soccer (football) teams from several countries compete annually in an important international tournament.
The birth months (Jan=1, Feb=2, .. Dec=12) of the 288 boys competing in the 2005 under 16 division showed higher counts for the early months and lower counts for the later months.
Formulate and test a relevant hypothesis
If you find statistical significance, offer an option about how it came to be that early months are more prevalent.
opinion
Helsen, W.F., Van Winckel, J., and WIlliams, M., The relative age effect in youth soccer across Europe, Journal of Sports Sciences, June 2005; 23(6):

14. The Data look like….. ID Birth Month 1 5 2 3 285 286 287 288 .