# 11.3.2 - Minitab: Test of Independence

11.3.2 - Minitab: Test of Independence## Raw vs Summarized Data

If you have a data file with the responses for individual cases then you have "raw data" and can follow the directions below. If you have a table filled with data, then you have "summarized data." There is an example of conducting a chi-square test of independence using summarized data on a later page. After data entry the procedure is the same for both data entry methods.

##
Minitab^{®}
– Chi-square Test Using Raw Data

**Research question**: Is there a relationship between where a student sits in class and whether they have ever cheated?

**Null hypothesis**: Seat location and cheating are not related in the population.**Alternative hypothesis**: Seat location and cheating are related in the population.

To perform a chi-square test of independence in Minitab using raw data:

- Open Minitab file: class_survey.mpx
- Select
*Stat > Tables > Chi-Square Test for Association* - Select
*Raw data (categorical variables)*from the dropdown. - Choose the variable
*Seating*to insert it into the*Rows*box - Choose the variable
*Ever_Cheat*to insert it into the*Columns*box - Click the
*Statistics*button and check the boxes*Chi-square test for association*and*Expected cell counts* - Click
*OK*and*OK*

This should result in the following output:

##### Rows: Seating Columns: Ever_Cheat

No | Yes | All | |
---|---|---|---|

Back | 24 | 8 | 32 |

24.21 | 7.79 | ||

Front | 38 | 8 | 46 |

34.81 | 11.19 | ||

Middle | 109 | 39 | 148 |

111.98 | 36.02 | ||

All | 1714 | 55 | 226 |

##### Chi-Square Test

Chi-Square | DF | P-Value | |
---|---|---|---|

Pearson | 1.539 | 2 | 0.463 |

Likelihood Ratio | 1.626 | 2 | 0.443 |

#### Interpret

All expected values are at least 5 so we can use the Pearson chi-square test statistic. Our results are \(\chi^2 (2) = 1.539\). \(p = 0.463\). 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 of a relationship in the population between seat location and whether a student has cheated.

# 11.3.2.1 - Example: Raw Data

11.3.2.1 - Example: Raw Data## Example: Dog & Cat Ownership

Is there a relationship between dog and cat ownership in the population of all World Campus STAT 200 students? Let's conduct an hypothesis test using the dataset: fall2016stdata.mpx

\(H_0:\) There is not a relationship between dog ownership and cat ownership in the population of all World Campus STAT 200 students

\(H_a:\) There is a relationship between dog ownership and cat ownership in the population of all World Campus STAT 200 students

Assumption: All expected counts are at least 5. The expected counts here are 176.02, 75.98, 189.98, and 82.02, so this assumption has been met.

Let's use Minitab to calculate the test statistic and p-value.

- After entering the data, select
*Stat > Tables > Cross Tabulation and Chi-Square* - Enter
*Dog*in the*Rows*box - Enter
*Cat*in the*Columns*box - Select the
*Chi-Square*button and in the new window check the box for the*Chi-square test*and*Expected cell counts* - Click
*OK*and*OK*

##### Rows: Dog Columns: Cat

No | Yes | All | |
---|---|---|---|

No | 183 | 69 | 252 |

176.02 | 75.98 | ||

Yes | 183 | 89 | 272 |

189.98 | 82.02 | ||

Missing | 1 | 0 | |

All | 366 | 158 | 524 |

##### Chi-Square Test

Chi-Square | DF | P-Value | |
---|---|---|---|

Pearson | 1.771 | 1 | 0.183 |

Likelihood Ratio | 1.775 | 1 | 0.183 |

Since the assumption was met in step 1, we can use the Pearson chi-square test statistic.

\(Pearson\;\chi^2 = 1.771\)

\(p = 0.183\)

Our p value is greater than the standard 0.05 alpha level, so we fail to reject the null hypothesis.

There is not enough evidence of a relationship between dog ownership and cat ownership in the population of all World Campus STAT 200 students.

# 11.3.2.2 - Example: Summarized Data

11.3.2.2 - Example: Summarized Data## Example: Coffee and Tea Preference

Is there a relationship between liking tea and liking coffee?

The following table shows data collected from a random sample of 100 adults. Each were asked if they liked coffee (yes or no) and if they liked tea (yes or no).

Likes Coffee | |||
---|---|---|---|

Yes | No | ||

Likes Tea | Yes | 30 | 25 |

No | 10 | 35 |

Let's use the 5 step hypothesis testing procedure to address this research question.

\(H_0:\) Liking coffee an liking tea are not related (i.e., independent) in the population

\(H_a:\) Liking coffee and liking tea are related (i.e., dependent) in the population

Assumption: All expected counts are at least 5.

Let's use Minitab to calculate the test statistic and p-value.

- Enter the table into a Minitab worksheet as shown below:
C1 C2 C3 Likes Tea Likes Coffee-Yes Likes Coffee-No 1 Yes 30 25 2 No 10 35 - Select
*Stat > Tables > Cross Tabulation and Chi-Square* - Select
*Summarized data in a two-way table*from the dropdown - Enter the columns
*Likes Coffee-Yes*and*Likes Coffee-No*in the Columns containing the table box - For the row labels enter
*Likes Tea*(leave the column labels blank) - Select the
*Chi-Square*button and check the boxes for*Chi-square test*and*Expected cell counts*. - Click
*OK*and*OK*

**Output**

##### Rows: Likes Tea Columns: Worksheet columns

No | Yes | All | |
---|---|---|---|

Yes | 30 | 25 | 55 |

22 | 33 | ||

No | 10 | 35 | 45 |

18 | 27 | ||

All | 40 | 60 | 100 |

##### Chi-Square Test

Chi-Square | DF | P-Value | |
---|---|---|---|

Pearson | 10.774 | 1 | 0.001 |

Likelihood Ratio | 11.138 | 1 | 0.001 |

Since the assumption was met in step 1, we can use the Pearson chi-square test statistic.

\(Pearson\;\chi^2 = 10.774\)

\(p = 0.001\)

Our p value is less than the standard 0.05 alpha level, so we reject the null hypothesis.

There is evidence of a relationship between between liking coffee and liking tea in the population.