Example: Coffee and Tea Preference Section
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 convincing evidence of a relationship between between liking coffee and liking tea in the population.