- Roulette Wheel (Different Proportions) - Roulette Wheel (Different Proportions)

Example: Roulette Wheel

Research Question: An American roulette wheel contains 38 slots: 18 red, 18 black, and 2 green.  A casino has purchased a new wheel and they want to know if there is any evidence that the wheel is unfair. They spin the wheel 100 times and it lands on red 44 times, black 49 times, and green 7 times.

Step 1: Check assumptions and write hypotheses

If the wheel is fair then \(p_{red}=\dfrac{18}{38}\), \(p_{black}=\dfrac{18}{38}\), and \(p_{green}=\dfrac{2}{38}\).

All of these proportions combined equal 1.

\(H_0: p_{red}=\dfrac{18}{38},\;p_{black}=\dfrac{18}{38}\;and\;p_{green}=\dfrac{2}{38}\)

\(H_a: At\;least\;one\;p_i\;is \;not\;as\;specified\;in\;the\;null\)

In order to conduct a chi-square goodness of fit test all expected values must be at least 5. 

For both red and black: \(Expected \;count=100(\dfrac{18}{38})=47.368\)

For green: \(Expected\;count=100(\dfrac{2}{38})=5.263\)

All expected counts are at least 5 so we can conduct a chi-square goodness of fit test. 

Step 2: Compute the test statistic

 \(\chi^2=\sum \dfrac{(Observed-Expected)^2}{Expected} \)

In the first step we computed the expected values for red and black to be 47.368 and for green to be 5.263.

 \(\chi^2= \dfrac{(44-47.368)^2}{47.368}+\dfrac{(49-47.368)^2}{47.368}+\dfrac{(7-5.263)^2}{5.263} \)


Step 3: Determine the p-value

Our sampling distribution will be a chi-square distribution.


We can find the p-value by constructing a chi-square distribution with 2 degrees of freedom to find the area to the right of \(\chi^2=0.868\)

Chi-squared distribution plot made using Minitab Express; degrees of freedom equal 2; area to the right of chi-squared value of 0.868 is shaded with a proportion of 0.647912

The p-value is 0.647912

Step 4: Make a decision

\(p>0.05\) therefore we should fail to reject the null hypothesis. 

Step 5: State a "real world" conclusion

There is not enough evidence that this roulette wheel is unfair.

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