# 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*

##### 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.