# 11.2.2.2 - Example: Summarized Data, Different Proportions

11.2.2.2 - Example: Summarized Data, Different Proportions## Example: Roulette

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.

Use Minitab to conduct a hypothesis test to address this question.

We'll go through each of the steps in the hypotheses test:

If the wheel is 'fair' then the probability of red and black are both 18/38 and the probability of green is 2/38.

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

\(H_a\colon\) at least one \(p_i\) is not as specified in the null

We can use the null hypothesis to check the assumption that all expected counts are at least 5.

\(Expected\;count=n (p_i)\)

With n = 100 we meet the assumptions needed to use Chi-square.

Let's use Minitab to calculate this.

First, enter the summarized data into a Minitab Worksheet.

C1 | C2 | |
---|---|---|

Color | Count | |

1 | Red | 44 |

2 | Black | 49 |

3 | Green | 7 |

- After entering the data, select
*Stat > Tables > Chi-Square Goodness of Fit Test (One Variable)* - Double-click
*Count*to enter it into the*Observed Counts*box - Double-click
*Color*to enter it into the*Category names (optional)*box - For
*Test*select*Input constants* - Select
*Proportions specified by historical counts*(this is what we would expect if the null was true) - Enter
*18/38*for*Black, 2/38*for*Green*and*18/38*for*Red* - Click
*OK*

This should result in the following output:

#### Chi-Square Goodness-of-Fit Test: Count

##### Observed and Expected Counts

Category | Observed | Historical Counts | Test Proportion |
Expected | Contribution to Chi-Sq |
---|---|---|---|---|---|

Red | 44 | 18 | 0.473684 | 47.3684 | 0.239532 |

Black | 49 | 18 | 0.473684 | 47.3684 | 0.056199 |

Green | 7 | 2 | 0.052632 | 5.2632 | 0.573158 |

##### Chi-Square Test

N | DF | Chi-Sq | P-Value |
---|---|---|---|

100 | 2 | 0.868889 | 0.648 |

The test statistic is a Chi-Square of 0.87.

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

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