Example: Dog & Cat Ownership Section
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.