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Lesson 17

This document discusses using Minitab to perform a chi-square test on contingency tables. It provides an example using data from Problem 23 on page 575 of another source. The contingency table data is entered into Minitab, which calculates the expected counts, chi-square contributions, and outputs the chi-square test statistic, degrees of freedom, and p-value. It then shows how to write this up following the classical 5-step approach, including stating the null and alternative hypotheses, providing the sample contingency table with expected counts, stating the test statistic and degrees of freedom, comparing the test statistic to the critical value, and stating the conclusion about whether to reject the null hypothesis. The document concludes by assigning two additional problems

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83 views3 pages

Lesson 17

This document discusses using Minitab to perform a chi-square test on contingency tables. It provides an example using data from Problem 23 on page 575 of another source. The contingency table data is entered into Minitab, which calculates the expected counts, chi-square contributions, and outputs the chi-square test statistic, degrees of freedom, and p-value. It then shows how to write this up following the classical 5-step approach, including stating the null and alternative hypotheses, providing the sample contingency table with expected counts, stating the test statistic and degrees of freedom, comparing the test statistic to the critical value, and stating the conclusion about whether to reject the null hypothesis. The document concludes by assigning two additional problems

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zaenalkmi
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© Attribution Non-Commercial (BY-NC)
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Return to Cover Page

LESSON 17 - CONTINGENCY TABLES


In this lesson we will learn how to have Minitab do the computations necessary to do a
chi-square test on a contingency table. As an example we will do Problem 23 on page 575. We
will let Minitab do the computations for us, then write up the result using the classical approach.
Type the contingency table into C1, C2, and C3 in the data window. Since we have no
way of putting in row titles, we will dispense with the column titles also in this case. Now clear
the Session Window below the date/time stamp then type your name, Lesson 17, and Example.
Now click on Stat > Tables > Chi-square Test (Two-Way Table in Worksheet). Select C1, C2
and C3 into the box labeled "Columns containing the table:" and click on "OK". The results are
shown below.
————— 7/21/2007 1:02:02 PM ————————————————————
Jeonghun Kim
Lesson 17
Example

Chi-Square Test: C1, C2, C3

Expected counts are printed below observed counts


Chi-Square contributions are printed below expected counts

C1 C2 C3 Total
1 47 44 50 141
47.00 47.00 47.00
0.000 0.191 0.191

2 17 9 16 42
14.00 14.00 14.00
0.643 1.786 0.286

3 36 47 34 117
39.00 39.00 39.00
0.231 1.641 0.641

Total 100 100 100 300

Chi-Sq = 5.610, DF = 4, P-Value = 0.230

Notice that the expected value for each cell is typed below the observed value, but not in
parentheses as is customary. Below that appears the quantity (O - E)2/E for each cell. The row
and column totals for the observed values and the overall total appear where we would expect
them. The chi-square value, the degrees of freedom, and the P-Value are listed at the bottom.
When writing up the 5-step classical method, use what you need of this information for the "For
our Sample" step, but make it look the way we learned to do it in class. An example of the write-
up for this problem is on the next page.

77
SAMPLE WRITE-UP

Jeonghun Kim
July 20, 2007
Lesson 17
Example

Variables: Rows = Coauthorship.


Columns = Subject matter.

1. H0: Co-authorship and subject matter are independent.


Ha: The subject matters are dependent on Co-authorship.

2. α = 0.10
3. Assume H0 is true.

For our sample:

Engineering Psychology Biology


47 44 50
Coauthored 141
(47) (47) (47)
Internationally 17 9 16
42
coauthored (14) (14) (14)
36 47 34
Not coauthored 117
(39) (39) (39)

100 100 100 300

χ 2 (4) = 5.610

4. Reject H0 if χ 2 > 7.779 from Table 6.

5. Decision: Since χ 2 is not greater than 7.779, we fail to reject H0.


Conclusion: There is not a significant relationship between co-authorship and subject matters
at the 0.10 level of significance.

78
MINITAB ASSIGNMENT 17

See instructions on page 8.

1. Do Problem 14 on page 572 as the example was done above.

2. Do Problem 18 on page 573 as the example was done above.

Return to Cover Page

79

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