# 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.