11.3.2 - Minitab Express: Test of Independence

11.3.2 - Minitab Express: Test of Independence

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.

MinitabExpress  – 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 Express using raw data:

  1. Open Minitab data set
  2. On a PC: Select STATISTICS > Cross Tabulation and Chi-Square
    On a Mac: Select Statistics > Tables > Cross Tabulation and Chi-Square
  3. Double-click on the variable Seating to insert it into the Rows box
  4. Double-click on the variable Ever Cheat to insert it into the Columns box
  5. Click the Display tab and check the boxes Chi-square test for association and Expected cell counts
  6. Click OK

This should result in the following output:

Cell contents grouped by Back, Front, Middle, All; First row: count, Next row: expected count

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 1.54 2 0.4633
Likelihood Ratio 1.63 2 0.4435
Video Walkthrough

Select your operating system below to see a step-by-step guide for this example.

All expected values are at least 5 so we can use the Pearson chi-square test statistic. Our results are \(\chi^2 (2) = 1.54\). \(p = 0.4633\). Because our \(p\) value is greater than the standard alpha level of 0.05, we fail to reject the null hypothesis. There is not evidence of a relationship in the population between seat location and whether a student has cheated.


11.3.2.1 - Video Example: Dog & Cat Ownership (Raw Data)

11.3.2.1 - Video Example: Dog & Cat Ownership (Raw Data)

This example uses the dataset:

1. Check any necessary assumptions and write null and alternative hypotheses.

 \(H_0:\) 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

Cell contents grouped by No, Yes, Missing; First row: count, Next row: expected count

Rows: Dog | Columns: Cat
  No Yes All
BaNock 183 69 252
  176.02 75.98  
Yes 183 89 272
  189.98 82.02  
Missing 1 0  
All 366 158 524

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.

2. Calculate an appropriate test statistic.
Chi-Square Test
  Chi-Square DF P-Value
Pearson 1.77 1 0.1833
Likelihood Ratio 1.77 1 0.1828

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

\(Pearson\;\chi^2 = 1.77\)

3. Determine a p value associated with the test statistic.

\(p = 0.1833\)

4. Decide between the null and alternative hypotheses.

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

5. State a "real world" conclusion.

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


11.3.2.2 - Video Example: Coffee and Tea (Summarized Data)

11.3.2.2 - Video Example: Coffee and Tea (Summarized Data)

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