Question

# Roulette USE SOFTWARE- In the casino game of roulette there is a wheel with 19 black...

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?

H0: p1 = 19/40, p2 = 19/40, and p3 = 2/40.

H0: The probabilities associated with this table do not fit those associated with a fair table.

H0: p1 = p2 = p3 = 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 (Oi) Probability (pi) Frequency Ei
 (Oi − Ei)2 Ei
1 black 0.475 71.25 3.264
2 red 83
3 green 11 0.050 7.50 1.633
Σ n = 150 χ2 =

(c) What is the value for the degrees of freedom?

(d) What is the critical value of χ2? Use the answer found in the χ2-table or round to 3 decimal places.
tα =

(e) What is the conclusion regarding the null hypothesis?

reject H0

fail to reject H0

(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 X2 = 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 X2 = 6.835

c) The value of degrees of freedom = n-1 = 3-1 = 2

d) The critical value at df = 2 and α =0.01     Xc2 = 9.21

e) Here X2 = 6.835 < Xc2 = 9.21. So we failed to reject H0.

f) We conclude that the “The results of this sample suggest the table is not fair.

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