General Statistics Chi-Square Analysis The Chi-square Test of Independence.

The Chi-square Test of Independence.

The chi-square test of independence is used to test whether two populations or variables are related or independent to each other with
respect to some characteristic.

1. Know when to use the chi-square test for independence.

The chi-square test of independence is used to test when two categories (each with many cells or groups) are related or not related
(independent).

Sometimes individual items are classified into categories in terms of two different criteria. For example, in studying the wear of tires, tires
may be classified into two categories: (1) front and rear as well as (2) left and right. A detail examination of both classification yields a
matrix or table of possible combinations of the classifications:

contingency table

 Category 1 Category 2 Row Total Front (F) Rear(B) Left (L) fLF fLB fL (Left Total) Right (R) fRF fRB fR (Right Total) Column Total fF (Front Total) fB (Rear Total) n (Total of All Categorized Cells)

The table above where categorical cells are combined to show their rows and column totals is called a contingency table.

Example 2 (from Textbook). For two categries: (1) Political candidates, A and B and (2) Voter Types, Male or Female a reporter
may summerize a survey of perference for either candidates as follows (contingency table):

 Voters Candidates A Candidates B Total Voter Types Female Female Votes for A Female Votes for B Female Votes Male Male Votes for A Male Votes for B Male Votes Total: Votes per Candidates Votes for A Votes for B All Votes

2. Know how to setup a Contingency Tables for the chi-square tests when studing independence (or homegeneity ) of
categorical data.

A contingency table organizes the results of a study involving two categoriies (with many classifications within each category) by
showing the frequency for each subtotals (groups, classes or cells) within each category.

A contingency table (Observed frequencies) is constructed as followed:

 Criterion A Criterion B Row Total 1 2 ..... k 1 f11 f12 ..... f1k Row1 2 f21 f22 ..... f2k. Row2 . . . ...... . . . . . ...... . . m fm1 fm2 ..... fmk Rowm Colum Total Column1 Column2 ..... Columnk n (Grand Total)

Where:
 m= The number of classes of grouping in category A (Row Classification) k= The number of classifications of category B (Column Classification) Rowi = Total of row frequencies (observed) Columnj = Total of column frequencies (observed) fij Number of items or results that belong to both the i-th and j-th categories For example, for f12 i=1 and j=2 so is the frequency observed for both criteria when Criterion A is 1 and Criterion B is 2. n= Grand total (total number of observations)

3. Know how to compute the expected freguencies for chi-square test of independent charateristics.

Each cell in the contingency table have some theoretical or expected frequency and unless otherwise known (explicitly) it is determined
from the information given in the row, column and grand totals.

The expected frequency, E for each cell is determined by taking the product of the corresponding row and column totals divided by the
grand total.

4. Know the relationship between the observed and expected frequencies and the chi-square test of independence.

The chi-sqaure test for independence evaluates the closeness of the observed frequencies (O) fij to the expected frequencies (E), Eij
good agreement between these two information favors the null hypothesis and large departure of the observed frequencies from the expected
frequencies favors the alternate hypothesis (larger values of ).

Remember the chi-square statistics is

5. Know the multiplication rule for independence characteristics and its role in chi-square analysis.

The cell frequencies found in the contingency table when divided by the grand total yield the proportions, p of observations or results out
of the total number of observations. This proportions is also a probability score of cells posibilities out of total possible outcome or score.

If p(m) is the proportion of rowm (frequency of rowm divided by the grand total and p(k) is the column proportion of columnk (probability, pk
then:

The multiplication rule for independent characteristics is p(m and k) = p(m) x p(k).

Where p(m and k) or p(mk) = is the probability or frequency of both m and k occuring (assumes that both m and k are indepedent events
or charateristics)

Example (Observed Proportion - from table below)

( Expected Proportion - from table below)

The chi-square test of independence attempts to compare the expected proportions using the multiplication rule of independent characteristics
with the actual or observed proportions in each cell of the contingency table.

If the multiplication rule does not hold for each cell, the two charateristics are not independent, they are related.

That is, the sum of square of the Observed frequencies minus the expected is small.

I is small.

6. Know how to compute and evaluate the chi-square test of independence statistics.

Problem: A researcher wants to study how 7 methods of preparations affects students getting over 80% on a aptitude test. The researcher
would also like to know how elapsed time after preparation affects student's performance on the test (> 80%) after 1, 2 and 3 months of final
preparation and if these two criteria are related or independent.

The following table is the observed results of the study: contingency table.

Use chi-square test for independent charateristics to evaluate this data.

 Number of Months (M) after Prep. Methods of Prep. (preperation) Row Totals 1 2 3 4 5 6 7 1. After 1 M 97 8 18 8 23 21 5 180 2. After 2 M 120 15 12 13 21 17 15 213 3. After 3 M 82 4 0 12 38 25 19 180 Column Totals 299 27 30 33 82 63 39 573

Procedure for calculating chi-square test of independence:

Step 1. Make a problems statement: (becomes the hypothesis statement, Ho ).

(1) Are the Methods of preperation and number of months after preperation related in terms of students achievement scores (> 80% on test)?

(2) Are the two criteria (methods of Preperation and number of months test is taken after preperation) independent or related to with respect
to students performance scores (>80% on test)?

Note: that the formulation of a problems statement or the question that the research would like to answer statistically may take on many forms
but similar expectations. For both problems statements above (1) and (2) are inversely related for when on is true the other is false and when on is
false the other is true. Either statement answers the larger question of indepedence: Yes of No?

Hypothesis:

So Ho (null hypothesis): Methods of Prep and Months after Prep are independent. (not related or interacts with respect to students scores).

Ha: Ho is not true. (alternate hypothesis): Methods of Prep and Months after Prep are related.

Step 2. Choose , the significance level of the test.

If you want the be 95 % certain that the test is true, then  = 0.05 =(100-95)/100

The df = (r-1)(c-1), or (m-1)(k-1) = (3-1)(7-1)=12

So df = 12

Step 3. Look up  from chi-square table:

 df 12 3.07 3.57 4.4 5.23 6.3 18.55 21.03 23.34 26.22 28.3

For d.f. = 12,

Step 4. (4) Determine or compute , The Expected Frequencies:

The following table is the Expected Frequencies, E of each cells in the study:

 Number of Months (M) after Prep. Methods of Prep. (preperation) Row Freq. Total 1 2 3 4 5 6 7 1 93.9267 8.4817 9.4241 10.3665 25.7592 19.7906 12.2513 180 2 111.1466 10.0366 11.1518 12.267 30.4817 23.4188 14.4974 213 3 93.9267 8.4817 9.4241 10.3665 25.7592 19.7906 12.2513 180 Column Freq. Totals 299 27 30 33 82 63 39 573

Step 5. Compute

Table of Expected Values (click for larger image)

(click for larger image)

= 45.5974 from computational table above.

Step 6 Perform test chi-square test:

Since , i.e. 45.60 > 21.03, Then we assume that the Null Hypothesis is not true (the types of preparations and number of weeks
after preperation are related each other).

Make Conclusion or inference:

There seems to be some relationship between types of students' preparation and number of weeks after preparation that tests are taken.

So Ha (alternate hypothesis) is favored by this test.

Formulate an hypothesis about the relationship between Political view and Opinion of Nuclear Power from the following survey of 100 students.
Use a chi-square test for independence showing contingency table to evaluate your hypothesis and make inference about the results of this study.

 Opinion Political Views Democrat Republican Independent Approve 10 15 20 Disapprove 9 2 16 Undecided 8 2 18

(a) What is your hypothesis?

(b) What is the value of the chi-square statistics?

(c) What conclusion did you make?

(d) What inference can you make?

Use use contingency table to help evaluate.