1. Guide to Using Minitab 14 For Basic Statistical Applications To Accompany Business Statistics: A Decision Making Approach, 8th Ed. Chapter 13: Goodness-of-Fit Tests and Contingency Analysis By Groebner, Shannon, Fry, & Smith Prentice-Hall Publishing Company Copyright, 2011

2. Chapter 13 Minitab Examples Goodness of Fit Test Goodness of Fit Test Goodness of Fit Test Goodness of Fit Test Woodtrim Products Contingency Analysis Contingency Analysis Contingency Analysis Contingency Analysis Benton Stone and Tile

3. Goodness-of-Fit - Woodtrim Products Issue: The company has recently become concerned the ripsaw may not be cutting to the manufacturer’s specifications Objective: Use Minitab to determine whether the data set’s distribution is consistent with the manufacturer’s specifications. Data file is Woodtrim.MTW

4. Open File Woodtrim.MTW Goodness-of-Fit – Woodtrim Products

5. Click on Stat, then Basic Statistics and then Normality Test. Minitab has a special option for testing Normality. Goodness-of-Fit – Woodtrim Products

6. Identify the Variable to test. Any of these Tests will work. Goodness-of-Fit – Woodtrim Products

7. The p-value indicates the hypothesis should be rejected. The hypotheses are: H 0 : Data are normally distributed. H A : Data are not normally distributed. Goodness-of-Fit – Woodtrim Products

8. Contingency Analysis - Benton Stone and Tile Issue: The company was interested in determining the relationship between absenteeism and marital status. Objective: Use Minitab to determine whether there is a statistically significant relationship between absenteeism and marital status. Data file is Benton.MTW

9. Open File Benton.MTW Contingency Analysis – Benton Stone and Tile

10. Click on Stat, then Tables and then Chi- Square Test. Contingency Analysis – Benton Stone and Tile

11. Specify the columns containing the table data. Contingency Analysis – Benton Stone and Tile

12. The Minitab output gives the expected frequencies and the row and column totals. Contingency Analysis – Benton Stone and Tile Since p-value > alpha = 0.05, do not reject the null hypothesis and therefore can’t conclude that a absenteeism and marital status are not independent.