Transient Absorption

The chi-square test statistic is an overall measure of how close the observed frequencies are to the expected frequencies. It has the form

χ² =

sum observed  frequency−expected  frequency ^! expected  frequency

The null hypothesis of independence is rejected if χ² is large, because this means that observed frequencies and expected frequencies are far apart. The chi-square curve is used to judge whether the calculated test statistic is large enough. We reject H0 if the test statistic is large enough so that the area beyond it (under the chi-square curve with (r-1)(c-1) degrees of freedom) is less than .05.

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:

I . You hypothesize that all the frequencies are equal in each category. For example, you might expect that half of the calves born in the National Animal Production Research Institute, Zaria are males and the other half will be females. You figure the expected frequency by dividing the number in the sample by the number of categories. In this example, where there are 100 new

calves born, and two categories male and female, you divide your sample of 100 by 2, the number of categories, to get 50 (expected frequencies) in each category.

2. You determine the expected frequencies on the basis of some prior knowledge. Someone told you that in Kaduna State 80% of the rural dwellers keeps poultry while 20% does not. You then administer a simple questionnaire randomly to a sample of 200 rural dwellers in Kaduna state.

You will calculate your expected number or frequency of poultry farmers and non-poultry farmer using your expected 80% and 20% respectively. When you analyse you data you will expect 80% of 200 or 160 respondents to be poultry farmers while 20% or 40 respondents to be non poultry farmers.

You read a newspaper that in Kwara state, 55% of the farmers keeps goat, 20% keeps sheep, 15 keeps pigs and 20% keeps cattle. As a student of Agricultural Science you were asked to confirm whether this is true.

Procedure

You will have to set up your null hypothesis which 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 E is sum of df is the "degree of freedom" (n-1) X2 is Chi Squ χ² = ^!!!_! ^!

where O is the Observed Frequency in each category

E is the Expected Frequency in the corresponding category χ² is the chi square

You are now ready to use the formula for χ² and find out if there is a significant difference between the observed and expected frequencies for the livestock farmers in Kwara State.

You will set up a worksheet; then you will follow the directions to form the columns and solve the formula.

Category O E O-E (O-E)² (O-E)²/E

Goat 95 80 15 225 2.813

Sheep 40 60 -20 400 6.667

Pig 30 20 10 100 5.000

Cattle 35 40 -5 25 0.625

χ² = 15.104

2. After calculating the Chi Square value, find the "Degrees of Freedom."

This calculated as number of categories less 1 df = N – 1

= 4 – 1 = 3

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 df and P=.05 level of significance is 7.815.

If the calculated chi-square value for the set of data you are analyzing (15.104) is equal to or greater than the table value (7.815), 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, then 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 what you read in the papers) and the observed frequencies (based upon the data you collected using your questionnaires) are not due to chance. That is, they are not due to chance variation in the sample you took; there is a real difference between them.

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

1. Write the observed frequencies in column O Test Summary 2. Calculate 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 using the df and α = 0.05)

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

For example, the reason observed frequencies in a fruit fly genetic breeding lab did not match expected frequencies could be due to such influences as:

• Mate selection (certain flies may prefer certain mates)

• Too small of a sample size was used

• Incorrect identification of male or female flies

• The wrong genetic cross was sent from the lab

• The flies were mixed in the bottle (carrying unexpected alleles)

Requirement for Chi Square analysis Chi-Square Test Requirements are as listed

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.

10.5 Further Reading

Devore, J. L., & Peck, R. (1986). Statistics: The exploration and analysis of data. St. Paul: West Pub. Co.

Donnelly R. A. (2004). The complete idiot’s guide to statistics (Vol. The complete idiot's guide).

Indianapolis, IN: Alpha.

Gomez, K. A., & Gomez, A. A. (1984). Statistical procedures for agricultural research (2nd ed.).

New York: Wiley.

Kaps, M., Lamberson, W. R., & Lamberson, W. (2004). Biostatistics for animal science.

Wallingford: CABI Publishing.

Jaisingh, L. R. (2006). Statistics for the utterly confused (2nd ed.). New York: McGraw-Hill.

McDonald, J.H. 2014. Handbook of Biological Statistics (3rd ed.). Sparky House Publishing, Baltimore, Maryland (http://www.biostathandbook.com/index.html )

Weiss N.A. 1999. Elementary Statistics, fourth edition. Reading, Massachusetts: Addison-Wesley Publishing Company

Unit 12. Goodness of fit