Chi-Square analysis This describes and gives a formula for a statistical analysis used for categorical data, called chi-square, denoted χ2. A chi-square table is given on the reverse. The chi-square is used to test hypotheses about the distribution of observations into categories (phenotypes). The null hypothesis (Ho) is that the observed frequencies are the same (except for chance variation) as the expected frequencies. If the frequencies you observe are different from expected frequencies, the value of χ2 goes up. If the observed and expected frequencies are exactly the same, χ2 = 0. You test whether a given χ2 is statistically significant by testing it against a table of chi-square distributions, according to the number of degrees of freedom for your sample, which is the number of categories minus 1. The chi-square assumes that you have at least 5 observations per category. The formula is: (observed frequency – expected frequency)2

2

χ = sum of (i.e., across categories) expected frequency

X = ∑ 2

(FO − FE)2 FE

df = Ncategories-1

Example You have a theory that people respond to sexy advertising. To test this you run ads for three products that are exactly the same, except you label the first one “bland,” the second one “ho-hum”, and the third “hot n’ steamy.” The null hypothesis is that people buy them with the same frequency. The experimental hypothesis is that a higher percentage will buy the third product, despite all three really being the same. You put them in a store, and watch the purchases of the first 24 shoppers who buy any one. Since there are 3 products and 24 shoppers, the expected frequencies (i.e., the Ho) are that 8 people would buy each of the three, and that any deviation from “8-each” would be by chance alone. You use χ2 to test this; the data are as follows: Expected frequency

Product name bland ho-hum hot n’ steamy

X2 =

(5 – 8)2 8

8 8 8

+

(5 – 8)2 8

Observed frequency 5 5 14

+

The statistical question is: do the frequencies you actually observe differ from the expected frequencies by more than chance alone?

(14 – 8)2 8

= 6.75, df = 3-1 = 2

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If accepted probability is p <.05, you would look up the value for 2 degrees of freedom in the 5% column on the table below to find out the critical value for this statistical test. Is the Ď‡2 you obtained greater than the critical value? Since it is (6.75 v. 5.99), you can assume that the choice of the hot nâ€™ steamy product occurred more often than you would expect by chance. That is, you reject the null hypothesis and conclude that, indeed, sex sells.

Example # 2

Data: There are FOUR different phenotypes with the data collected in the data table below: the phenotypes of parents that were heterozygous for both seed shape and seed color. Each of these traits is inherited INDEPENDENTLY (that is, they are located on separate chromosomes). (Check here for your understanding of the Law of Independent Assortment AND the Law of Segregation.) Hypothesis: Cross is between two dihybrids where the genes are on separated chtomosomes 9:3:3:1 ratio

Observed Values

Expected Values

315 Round, Yellow Seed

(9/16)(556) =

Round, Yellow Seed

108 Round, Green Seed

(3/16)(556) =

Round, Green Seed

101 Wrinkled, Yellow Seed (3/16)(556) = 32 Wrinkled, Green

(1/16)(556) =

Wrinkled, Yellow Wrinkled, Green

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556 Total Seeds

556.00 Total Seeds

CHI SQUARE = A mathematical means to determine if the results of an experiment agrees or did not agree with the hypothesis. Observed data – Expected Data = chi square X2 = THIS VALUE DETERMINES WHETHER OR NOT THE MATHEMATICAL RESULTS SUGGEST THAT THE HYPOTHESIS (1) IS CORRECT (2) IS NOT CORRECT (3) IS INCONCLUSIVE.

Number of classes or phenotypes(n) = 4 df = n-1 Degrees of freedom (df) = n-1 where n is the number of classes (categories) or phenotypes Determine the chi Square critical value with a probability of .05

X = ∑ 2

(FO − FE)2 FE

df = Nphenotypes-1

Enter the Chi-Square table at df = 3 and we see the probability of our chi-square value is = 0.47 . By statistical convention, we use the 0.05 probability level as our critical value to accept or reject hypothesis. If the calculated chi-square critical value is less than the 0.05 probability value, we accept the hypothesis. If the value is greater than the critical value, we reject the hypothesis. Therefore, because the calculated chi-square value is0.47 than we accept the hypothesis that the data fits a 9:3:3:1 ratio.

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chi-square-analysis-2011

X 2 = (5 – 8) 2 + (5 – 8) 2 + (14 – 8) 2 X 2 = Chi-Square analysis The statistical question is: do the frequencies you actually observe diff...

chi-square-analysis-2011

Published on Jul 12, 2012

X 2 = (5 – 8) 2 + (5 – 8) 2 + (14 – 8) 2 X 2 = Chi-Square analysis The statistical question is: do the frequencies you actually observe diff...