# 11.2.2 - Minitab: Goodness-of-Fit Test

11.2.2 - Minitab: Goodness-of-Fit Test## Example: Cards

**Research Question**: When randomly selecting a card from a deck with replacement, are we equally likely to select a heart, diamond, spade, and club?

I randomly selected a card from a standard deck 40 times with replacement. I pulled 13 Hearts (♥), 8 Diamonds (♦), 8 Spades (♠), and 11 Clubs (♣).

##
Minitab^{®}
– Conducting a Chi-Square Goodness-of-Fit Test

### Summarized Data, Equal Proportions

To perform a chi-square goodness-of-fit test in Minitab using summarized data we first need to enter the data into the worksheet. Below you can see that we have one column with the names of each group and one column with the observed counts for each group.

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

Suit | Count | |

1 | Hearts | 13 |

2 | Diamonds | 8 |

3 | Spades | 8 |

4 | Clubs | 11 |

- 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
*Suit*to enter it into the*Category names (optional)*box - Click
*OK*

This should result in the following output:

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

##### Observed and Expected Counts

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

Hearts | 13 | 0.25 | 10 | 0.9 |

Diamonds | 8 | 0.25 | 10 | 0.4 |

Spades | 8 | 0.25 | 10 | 0.4 |

Clubs | 11 | 0.25 | 10 | 0.1 |

##### Chi-Square Test

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

40 | 3 | 1.8 | 0.615 |

All expected values are at least 5 so we can use the chi-square distribution to approximate the sampling distribution. Our results are \(\chi^2 (3) = 1.8\). \(p = 0.615\). Because our p-value is greater than the standard alpha level of 0.05, we fail to reject the null hypothesis. There is not enough evidence to conclude that the proportions are different in the population.

## Note!

The example above tested equal population proportions. Minitab also has the ability to conduct a chi-square goodness-of-fit test when the hypothesized population proportions are not all equal. To do this, you can choose to test specified proportions or to use proportions based on historical counts.

# 11.2.2.1 - Example: Summarized Data, Equal Proportions

11.2.2.1 - Example: Summarized Data, Equal Proportions## Example: Tulips

A company selling tulip bulbs claims they have equal proportions of white, pink, and purple bulbs and that they fill customer orders by randomly selecting bulbs from the population of all of their bulbs.

You ordered 30 bulbs and received 16 white, 8 pink, and 6 purple.

Is there convincing evidence the bulbs you received were not randomly selected from a population with an equal proportion of each color?

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

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

\(H_0\colon p_{white}=p_{pink}=p_{purple}=\dfrac{1}{3}\)

\(H_a\colon\) at least one \(p_i\) is not \(\dfrac{1}{3}\)

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

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

All \(p_i\) are \(\frac{1}{3}\). \(30(\frac{1}{3})=10\), thus this assumption is met and we can approximate the sampling distribution using the chi-square distribution.

Let's use Minitab to calculate this.

First, enter the summarized data into a Minitab Worksheet.

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

Color | Count | |

1 | White | 16 |

2 | Pink | 8 |

3 | Purple | 6 |

- 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 - Click
*OK*

This should result in the following output:

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

##### Observed and Expected Counts

Category | Observed | Test | Expected | Contribution |
---|---|---|---|---|

White | 16 | 0.333333 | 10 | 3.6 |

Pink | 8 | 0.333333 | 10 | 0.4 |

Purple | 6 | 0.333333 | 10 | 1.6 |

##### Chi-Square Test

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

30 | 2 | 5.6 | 0.061 |

The test statistic is a Chi-Square of 5.6.

The p-value from the output is 0.061.

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

There is not enough evidence that the tulip bulbs were not randomly selected from a population with equal proportions of white, pink and purple.

# 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 convincing 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 | Expected | Contribution |
---|---|---|---|---|---|

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.

The p-value from the output is 0.648.

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

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