Lesson 28b

Chi-Square, Goodness of Fit Test (continued)

Limitations of the Chi-Square Test
As a general rule of thumb, the expected value in any given cell should be at least five. In situations where the expected value is less than 5, cells are combined until this condition is satisfied. Consider the following problem.

Example 3: A study of the students who attend evening classes had the following breakdown in marital status. 60% were single, 30% were married, 8% were divorced and 2% were widowed. An evening statistics class is polled for their marital status. The results are 24 are single, 4 are married, 1 is divorced and 1 is widowed. At a 0.05 level of significance does this class fit the pattern previously mentioned?
Solution:
Step 1: Translate the claim into a hypothesis statement.
The claim translates into Single = 60%, Married = 30% Divorced = 8% and Widowed = 2%. These proportions are assigned to the Null Hypothesis. The corresponding diagram has only one "Reject H₀" region, so the 0.05 level of significance is assigned to this region.

       H₀: Stated proportions work.    claim
     H₁: At least one proportion differs from
               the stated model.

chisq1.jpg (4604 bytes)

Step 2: Use the level of significance, , to determine the size (area) of the "Reject H₀" region.

chisq2.jpg (5041 bytes)

Step 3: Unfortunately, the TI-83 does not perform the Chi-Square goodness-of-fit test directly. However, we can use the TI-83 to perform the Chi-Square goodness-of-fit test indirectly.
Since the claim is about multiple (more than two) proportions, the Chi-Square, , test is chosen as the test statistic.

The critical value taken from the Chi-Square table (page 764) is 7.815. The values in the Chi-Square table correspond to areas to the right. This means we use the 0.05 column. There are also k -1 degrees of freedom. With 4 marital categories in our data set, df = 4 -1 = 3. From the Chi-Square table, find the value in Row 3 and Column 0.05. This value is 7.815.

The test statistic is calculated in the following manner.
1. Find the total number of observations in your data set. 24 + 4 + 1 + 1 = 30

Marital Status	Number
Single	24
Married	4
Divorced	1
Widowed	1
Total	30

2. Calculate E, the expected frequency. Since the equal proportions 0.60, 0.30, 0.08, and
0.02, the expected frequencies are 0.60(30) = 18, 0.30(30) = 9, 0.08(30) = 2.4 and
0.02(30) = 0.6.

Marital Status	Number	Expected
Single	24	18
Married	4	9
Divorced	1	2.4
Widowed	1	0.6
Total	30	30

     WARNING: Two of the classes, divorced and widowed, do not have an
    expected frequency of at least 5. These classes must be combined with
    other classes to meet this requirement.

The data is reorganized in the following manner. The married, divorced and widowed
categories are combined into one class. There are now two classes of data:

Marital Status	Number
Never Married	24
Married at Some Time	6
Total	30

and the model becomes 60% never married and 40% married at some time.

4. The critical value changes because now there are two classes. The values in the Chi-Square table
      correspond to areas to the right. This means we use the 0.05 column. There are also k - 1 degrees
      of freedom. With 2 marital categories in our data set, df = 2 1 = 1. From the Chi-Square table,
      find the value in Row 1 and Column 0.05. This value is 3.841.

5. Calculate E, the expected frequency. Since the equal proportions 0.60, and 0.40, the
expected frequencies are 0.60(30) = 18, and 0.40(30) = 12.

Marital Status	Number	Expected
Never Married	24	18
Married at Some Time	6	12
Total	30	30

Use TI-83 to calculate the Chi-Square Test Statistic

Using the Stat function for editing data.
1. Enter the observed frequencies in List 1.
2. Enter the expected frequencies in List 2.
3. Subtract List 2 from List 1. Hit keys "2nd" "1" "-" "2nd" "2" "ENTER".
4. Store the differences in List 3. Hit keys "STO" "2nd" "3" "ENTER".
5. Square the differences. Hit keys "2nd" "3" "^" "2" "ENTER".
6. Store the results in List 4. Hit keys "STO" "2nd" "4" "ENTER".
7. Divide the squares by the expected values. Hit keys "2nd" "4" "

" "2nd" "2"
       "ENTER".
8. Store the quotients in List 5. Hit keys "STO" "2nd" "5" "ENTER".
9. Sum List 5. Hit keys "2nd" "STAT". Use the right arrow to select "MATH".
       Use the down arrow to select "SUM" and press "ENTER". Then press
       "2nd" "5" "ENTER".
       The Chi-Square Test Statistics will be displayed. In this case the Chi-Square test
       statistic is 5.

7. Label the critical value and test statistic on the diagram.

Step 4: Make two statements. Did you "Reject" or "Fail to reject" the Null Hypothesis, H₀. Does the data appear to support or refute the claim?
The test statistic falls in the "Reject H₀" region.

Conclusion

We reject the Null Hypothesis, H_0. The data does not support the model. Thus the students in the evening statistics class do not fit the marital status model.

Assignment:

Try problems Section 12.1: Use the Chi-Square table for the critical values and calculate the test statistic for each hypothesis test. Using critical values and test statistics will be the procedure required for testing claims about Goodness - of - fit on the exam and final.	MathXL: 5, 7, 11, 15 and 17.
	Textbook: 1 - 17 odd.

Lesson 27 Lesson 29

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