Chi Squared Statistical test used to see if

Chi Squared – in genetics. The Null Hypothesis is always : There is no

Phenotype Ratio observed Expected Normal wing 3 120 111 -9 81 0. 675 Vestigal

The Critical Value You compare the Chi Square value to the critical value which

The results In this example he Chi Square value is less than the Critical

Chi Squared test for independance For a contingency table that has r rows and

Incidence of Malaria in three tropical regions Null Hypothesis: There is no relationship between

Calculating expected values Sample A Sample B Sample C Column Totals Category III Row

Chi Squared Observed A 31 B 14 C 45 D 2 E 5 F

Chi Squared Observed Expected |O -E| (O — E)2/ E A 31 30. 96

Interpreting the result: Chi Square = 125. 516 Degrees of Freedom = (c -

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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.

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 theory is wrong.

Phenotype Ratio observed Expected Normal wing 3 120 111 -9 81 0. 675 Vestigal Wing 1 40 49 9 81 2. 025 |O -E| (O — E)2/ E 2. 7

The Critical Value You compare the Chi Square value to the critical value which is found in the following table. Df Degrees of freedom : the number of classes (phenotypes ) minus 1 Probability 0. 5 0. 10 0. 05 0. 02 0. 01 0. 001 1 0. 455 2. 706 3. 841 5. 412 6. 635 10. 827 2 1. 386 4. 605 5. 991 7. 824 9. 210 13. 815 3 2. 366 6. 251 7. 815 9. 837 11. 345 16. 268 4 3. 357 7. 779 9. 488 11. 668 13. 277 18. 465 5 4. 351 9. 236 11. 070 13. 388 15. 086 20. 517 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.

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 any difference was due to chance. This means that 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.

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.

Incidence of Malaria in three tropical regions Null Hypothesis: There is no relationship between the type of malaria and its geographical location Malaria A Malaria B Malaria C Totals Asia Africa South America Totals 31 14 45 90 2 5 53 60 53 45 2 100 86 64 100 250

Calculating expected values Sample A Sample B Sample C Column Totals Category III Row Totals a b c a+b+c d e f d+e+f g h i g+h+i c+f+i a+b+c+d+ e+f+g+h+i =N a+d+g b+e+h Now we need to calculate the expected values for each cell in the table and we can do that using 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.

Chi Squared Observed A 31 B 14 C 45 D 2 E 5 F 53 G 53 H 45 i 2 Expected |O -E| 30. 96 0. 04 (O — E)2/ E 0. 0016 0. 0000516 Chi squared=

Chi Squared Observed Expected |O -E| (O — E)2/ E A 31 30. 96 0. 04 0. 0016 0. 0000516 B 14 23. 04 9. 04 81. 72 3. 546 C 45 36. 00 9. 00 81. 00 2. 25 D 2 20. 64 18. 64 347. 45 16. 83 E 5 15. 36 107. 33 6. 99 F 53 24. 00 29. 00 841. 00 35. 04 G 53 34. 40 18. 60 345. 96 10. 06 H 45 25. 60 19. 40 376. 36 14. 70 i 2 40. 00 38. 00 1444. 00 36. 10 Chi squared = 125. 516

Interpreting the result: Chi Square = 125. 516 Degrees of Freedom = (c - 1)(r - 1) = 2(2) = 4 Df Probability 0. 5 0. 10 0. 05 0. 02 0. 01 0. 001 1 0. 455 2. 706 3. 841 5. 412 6. 635 10. 827 2 1. 386 4. 605 5. 991 7. 824 9. 210 13. 815 3 2. 366 6. 251 7. 815 9. 837 11. 345 16. 268 4 3. 357 7. 779 9. 488 11. 668 13. 277 18. 465 5 4. 351 9. 236 11. 070 13. 388 15. 086 20. 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.