Chi-Square(Unit -4)

Pearson's chi-square is used to assess two types of comparison: tests of goodness of fit and tests of independence. A test of goodness of fit establishes whether or not an observed frequency distribution differs from a theoretical distribution. A test of independence assesses whether paired observations on two variables, expressed in a contingency table, are independent of each other – for example, whether people from different regions differ in the frequency with which they report that they support a political candidate.

The first step in the chi-square test is to calculate the chi-square statistic. In order to avoid ambiguity, the value of the test-statistic is denoted by Χ² rather than χ² (i.e. uppercase chi instead of lowercase); this also serves as a reminder that the distribution of the test statistic is not exactly that of a chi-square random variable. However some authors do use the χ² notation for the test statistic. An exact test which does not rely on using the approximate χ² distribution is Fisher's exact test: this is significantly more accurate in evaluating the significance level of the test, especially with small numbers of observation.

The chi-square statistic is calculated by finding the difference between each observed and theoretical frequency for each possible outcome, squaring them, dividing each by the theoretical frequency, and taking the sum of the results. A second important part of determining the test statistic is to define the degrees of freedom of the test: this is essentially the number of observed frequencies adjusted for the effect of using some of those observations to define the "theoretical frequencies".

The chi-square (I) test is used to determine whether there is a significant difference between the expected frequencies and the observed frequencies in one or more categories. Do the number of individuals or objects that fall in each category differ significantly from the number you would expect? Is this difference between the expected and observed due to sampling error, or is it a real difference?

Ø     Chi-Square Test Requirements

1. Quantitative data.

2. One or more categories.

3. Independent observations.

4. Adequate sample size (at least 10).

5. Simple random sample.

6. Data in frequency form.

7. All observations must be used.

           

Ø     Expected Frequencies

When you find the value for chi square, you determine whether the observed frequencies differ significantly from the expected frequencies. You find the expected frequencies for chi square in three ways:

1 . You hypothesize that all the frequencies are equal in each category. For example, you might expect that half of the entering freshmen class of 200 at Tech College will be identified as women and half as men. You figure the expected frequency by dividing the number in the sample by the number of categories. In this example, where there are 200 entering freshmen and two categories, male and female, you divide your sample of 200 by 2, the number of categories, to get 100 (expected frequencies) in each category.

2. You determine the expected frequencies on the basis of some prior knowledge. Let's use the Tech College example again, but this time pretend we have prior knowledge of the frequencies of men and women in each category from last year's entering class, when 60% of the freshmen were men and 40% were women. This year you might expect that 60% of the total would be men and 40% would be women. You find the expected frequencies by multiplying the sample size by each of the hypothesized population proportions. If the freshmen total were 200, you would expect 120 to be men (60% x 200) and 80 to be women (40% x 200).Now let's take a situation, find the expected frequencies, and use the chi-square test to solve the problem.

Ø     Situation

Thai, the manager of a car dealership, did not want to stock cars that were bought less frequently because of their unpopular color. The five colors that he ordered were red, yellow, green, blue, and white. According to That, the expected frequencies or number of customers choosing each color should follow the percentages of last year. She felt 20% would choose yellow, 30% would choose red, 10% would choose green, 10% would choose blue, and 30% would choose white. She now took a random sample of 150 customers and asked them their color preferences. The results of this poll are shown in Table 1 under the column labeled _observed frequencies."

Table 1 - Color Preference for 150 Customers for Thai's Superior Car Dealership

Category Color	Observed Frequencies	Expected Frequencies
Yellow	35	30
Red	50	45
Green	30	15
Blue	10	15
White	25	45

last year, we would expect 20% to choose yellow. Figure the expected frequencies for yellow by taking 20% of the 150 customers, getting an expected frequency of 30 people for this category. For the color red we would expect 30% out of 150 or 45 people to fall in this category. Using this method, Thai figured out the expected frequencies 30, 45, 15, 15, and 45. Obviously, there are discrepancies between the colors preferred by customers in the poll taken by Thai and the colors preferred by the customers who bought their cars last year. Most striking is the difference in the green and white colors. If Thai were to follow the results of her poll, she would stock twice as many green cars than if she were to follow the customer color preference for green based on last year's sales.In the case of white cars, she would stock half as many this year. What to do??? Thai needs to know whether or not the discrepancies between last year's choices (expected frequencies) and this year's preferences on the basis of his poll (observed frequencies) demonstrate a real change in customer color preferences. It could be that the differences are simply a result of the random sample she chanced to select. If so, then the population of customers really has not changed from last year as far as color preferences go. The null hypothesis states that there is no significant difference between the expected and observed frequencies. The alternative hypothesis states they are different. The level of significance (the point at which you can say with 95% confidence that the difference is NOT due to chance alone) is set at .05 (the standard for most science experiments.) The chi-square

formula used on these data is

X2 = (O - E)2 where O is the Observed Frequency in each category

E E is the Expected Frequency in the corresponding category

df is the "degree of freedom" (n-1)

X2 is Chi Square

Ø     PROCEDURE

We are now ready to use our formula for X2 and find out if there is a significant difference between the observed and expected frequencies for the customers in choosing cars. We will set up a worksheet; then you will follow the directions to form the columns and solve the formula.

2. After calculating the Chi Square value, find the "Degrees of Freedom." (DO NOT SQUARE THE NUMBER YOU GET, NOR FIND THE SQUARE ROOT – THE NUMBER YOU GET FROM COMPLETING THE CALCULATIONS AS ABOVE IS _CHI SQUARE.)

Degrees of freedom (df) refers to the number of values that are free to vary after restriction has been placed on the data. For instance, if you have four numbers with the restriction that their sum has to be 50, and then three of these numbers can be anything, they are free to vary, but the fourth number definitely is restricted. For example, the first three numbers could be 15, 20, and 5, adding up to 40; then the fourth number has to be 10 in order that they sum to 50. The degrees of freedom for these values are then three. The degrees of freedom here is defined as N - 1, the number in the group minus one restriction (4 - I ).

3. Find the table value for Chi Square. Begin by finding the df found in step 2 along the left hand side of the table. Run your fingers across the proper row until you reach the predetermined level of significance (.05) at the column heading on the top of the table. The table value for Chi Square in the correct box of 4 df and P=.05 level of significance is 9.49.

4. If the calculated chi-square value for the set of data you are analyzing (26.95) is equal to or greater than the table value (9.49 ), reject the null hypothesis. There IS a significant difference between the data sets that cannot be due to chance alone. If the number you calculate is LESS than the number you find on the table, than you can probably say that any differences are due to chance alone.

In this situation, the rejection of the null hypothesis means that the differences between the expected frequencies (based upon last year's car sales) and the observed frequencies (based upon this year's poll taken by Thai) are not due to chance. That is, they are not due to chance variation in the sample Thai took; there is a real difference between them. Therefore, in deciding what color autos to stock, it would be to Thai's advantage to pay careful attention to the results of her poll!

Ø     The steps in using the chi-square test may be summarized as follows:

ü     Chi-square

I. Write the observed frequencies in column O

2. Figure the expected frequencies and write them in column E.

3. Use the formula to find the chi-square value:

4. Find the df. (N-1)

5. Find the table value (consult the Chi Square Table.)

6. If your chi-square value is equal to or greater than the table value,

    reject the null

                    Hypothesis: differences in your data are not due to chance alone

Ø     Problem

            The approximation to the chi-square distribution breaks down if expected frequencies are too low. It will normally be acceptable so long as no more than 20% of the events have expected frequencies below 5. Where there is only 1 degree of freedom, the approximation is not reliable if expected frequencies are below 10. In this case, a better approximation can be obtained by reducing the absolute value of each difference between observed and expected frequencies by 0.5 before squaring; this is called Yates' correction for continuity.

In cases where the expected value, E, is found to be small (indicating either a small underlying population probability, or a small number of observations), the normal approximation of the multinomial distribution can fail, and in such cases it is found to be more appropriate to use the G-test, a likelihood ratio-based test statistic. Where the total sample size is small, it is necessary to use an appropriate exact test, typically either the binomial test or (for contingency tables) Fisher's exact test; but note that this test assumes fixed and known marginal totals.