Roulette USE SOFTWARE In the casino game of roulette there is a wheel with 19 black slots, 19 red slots, and 2 green slots. In the game, a ball is rolled around a spinning wheel and it lands in one of the slots. It is assumed that each slot has the same probability of getting the ball. This results in the table of probabilities below.
Fair Table Probabilities
black  red  green  
Probability  19/40  19/40  2/40 
You watch the game at a particular table for 150 rounds and count
the number of black, red, and green results. Your observations are
summarized in the table below.
Outcomes (n = 150)
black  red  green  
Counts  56  83  11 
The Test: Test the claim that this roulette table
is not fair. That is, test the claim that the distribution of
colors for all spins of this wheel does not fit the expected
distribution from a fair table. Test this claim at the 0.01
significance level.
(a) What is the null hypothesis for this test?
H_{0}: p_{1} = 19/40, p_{2} = 19/40, and p_{3} = 2/40.
H_{0}: The probabilities associated with this table do not fit those associated with a fair table.
H_{0}: p_{1} = p_{2} = p_{3} = 1/3
_{H0}: The probabilities are not all equal to 1/3.
(b) The table below is used to calculate the test statistic.
Complete the missing cells.
Round your answers to the same number of decimal places as
other entries for that column.
Observed  Assumed  Expected  
i  Color  Frequency (O_{i})  Probability (p_{i})  Frequency E_{i} 


1  black  0.475  71.25  3.264  
2  red  83  
3  green  11  0.050  7.50  1.633  
Σ  n = 150  χ^{2} =  
(c) What is the value for the degrees of freedom?
(d) What is the critical value of χ^{2}?
Use the answer found in the
χ^{2}table or round to 3 decimal
places.
t_{α} =
(e) What is the conclusion regarding the null hypothesis?
reject H_{0}
fail to reject H_{0}
(f) Choose the appropriate concluding statement.
We have proven that this table is fair.
The results of this sample suggest the table is not fair.
There is not enough data to conclude that this table is not fair.
b)
Observered 
Assumed 
Expected 

i 
Color 
Frequency (Oi) 
Probability (pi) 
Frequency Ei 
(Oi Ei)^2/Ei 
1 
black 
56 
0.475 
71.25 
3.264 
2 
red 
83 
0.475 
71.25 
1.938 
3 
green 
11 
0.05 
7.5 
1.633 
∑ 
n = 150 
X^{2} = 6.835 
From the given information, we fill the value for frequency and probabilities.
The Expected frequency is calculated as probability * Total outcomes
i.e. 0.475*150 = 71.25
Chi square value using the formula is X^{2} = 6.835
c) The value of degrees of freedom = n1 = 31 = 2
d) The critical value at df = 2 and α =0.01 X_{c}^{2} = 9.21
e) Here X^{2} = 6.835 < X_{c}^{2} = 9.21. So we failed to reject H_{0}.
f) We conclude that the “The results of this sample suggest the table is not fair.
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