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Chi Squared Statistical test used to see if the results of an experiment support a theory or to check that categorical data is independent of each other.

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Presentation on theme: "Chi Squared Statistical test used to see if the results of an experiment support a theory or to check that categorical data is independent of each other."— Presentation transcript:

1 Chi Squared Statistical test used to see if the results of an experiment support a theory or to check that categorical data is independent of each other.

2 Chi Squared – in genetics. The Null Hypothesis is always : There is no significant difference between the observed an expected results. There will always be a slight difference from what you expect but the test allows us to determine whether this difference is due to chance or whether the theory is wrong.

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7 PhenotypeRatioobservedExpected |O -E| (O — E) 2 (O — E) 2 / E Normal wing 3120111-9810.675 Vestigal Wing 140499812.025 2.7

8 The Critical Value You compare the Chi Square value to the critical value which is found in the following table. Accept Null Hypothesis if the Chi Square value is lower than the critical Value Reject Null Hypothesis if the chi square value is greater than the Critical Value. DfProbability 0.50.100.050.020.010.001 10.4552.7063.8415.4126.63510.827 21.3864.6055.9917.8249.21013.815 32.3666.2517.8159.83711.34516.268 43.3577.7799.48811.66813.27718.465 54.3519.23611.07013.38815.08620.517 Degrees of freedom : the number of classes (phenotypes ) minus 1

9 The results In this example he Chi Square value is less than the Critical Value, this means we can accept the null hypothesis that there is no significant differences between the observed and expected and any difference was due to chance. This means that the theory that wing length in fruit flies is controlled by monohybrid inheritance. If the Chi square was greater than the critical value then there would have been a significant difference between the Observed and expected meaning something other than monohybrid inheritance was occurig......possibly sex linkage/ epistasis or codominance.

10 Chi Squared test for independance For a contingency table that has r rows and c columns, the chi square test can be thought of as a test of independence. In a test of independence the null hypothesis is Null hypothesis: The two categorical variables are independent.

11 AsiaAfricaSouth America Totals Malaria A 31144590 Malaria B 255360 Malaria C 53452100 Totals 8664100250 Incidence of Malaria in three tropical regions Null Hypothesis: There is no relationship between the type of malaria and its geographical location

12 Calculating expected values Category ICategory IICategory III Row Totals Sample A abca+b+c Sample B defd+e+f Sample C ghig+h+i Column Totals a+d+gb+e+hc+f+i a+b+c+d +e+f+g+h +i=N Now we need to calculate the expected values for each cell in the table and we can do that using the the row total times the column total divided by the grand total (N). For example, for cell a the expected value would be (a+b+c)(a+d+g)/N.

13 Chi Squared Observed Expected|O -E| (O — E)2 (O — E)2/ E A 31 30.96 0.04 0.0016 0.0000516 B 14 C 45 D 2 E 5 F 53 G H 45 i 2 Chi squared=

14 Chi Squared Observed Expected|O -E| (O — E)2 (O — E)2/ E A 31 30.96 0.04 0.0016 0.0000516 B 14 23.04 9.0481.723.546 C 45 36.00 9.0081.002.25 D 2 20.64 18.64347.4516.83 E 5 15.36 10.36107.336.99 F 53 24.00 29.00841.0035.04 G 53 34.40 18.60345.9610.06 H 45 25.60 19.40376.3614.70 i 2 40.00 38.00 1444.0036.10 Chi squared = 125.516

15 Chi Square = 125.516 Degrees of Freedom = (c - 1)(r - 1) = 2(2) = 4 Interpreting the result: DfProbability 0.50.100.050.020.010.001 10.4552.7063.8415.4126.63510.827 21.3864.6055.9917.8249.21013.815 32.3666.2517.8159.83711.34516.268 43.3577.7799.48811.66813.27718.465 54.3519.23611.07013.38815.08620.517 Reject Null H because 125.516 is greater than 9.488 (for alpha = 0.05) The Null hypothesis states that there is no relationship between location and type of malaria. Our data tell us there is a relationship between type of malaria and location, but that's all it says.

16 Chi Square = 125.516 Degrees of Freedom = (c - 1)(r - 1) = 2(2) = 4 Interpreting the result: DfProbability 0.50.100.050.020.010.001 10.4552.7063.8415.4126.63510.827 21.3864.6055.9917.8249.21013.815 32.3666.2517.8159.83711.34516.268 43.3577.7799.48811.66813.27718.465 54.3519.23611.07013.38815.08620.517


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