Chi sequare

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Chi sequare

  1. 1. CHI-SQUARE To: Sir Shahid Mahmood By: Abuzar TabassumM.Sc Zoology 3rd semester 10040814-017 University Of Gujrat
  2. 2. Chi-Square Test• A fundamental problem in genetics is determining whether the experimentally determined data fits the results expected from theory (i.e. Mendel’s laws as expressed in the Punnett square).• A statistical method used to determine GOODNESS OF FIT – Goodness of fit refers to how close the observed data are to those predicted from a hypothesis
  3. 3. Goodness of Fit• Mendel has no way of solving this problem. Shortly after the rediscovery of his work in 1900, Karl Pearson and R.A. Fisher developed the “chi-square” test for this purpose.• The chi-square test is a “goodness of fit” test: it answers the question of how well do experimental data fit expectations.• We start with a theory for how the offspring will be distributed: the “null hypothesis”. We will discuss the offspring of a self-pollination of a heterozygote. The null hypothesis is that the offspring will appear in a ratio of 3/4 dominant to 1/4 recessive.
  4. 4. Formula• To calculate the chi-square statistic following formula is used. 2 (obs exp) 2 exp• The “Χ” is the Greek letter chi; the “∑” is a sigma; it means to sum the following terms for all phenotypes.• “obs” is the number of individuals of the given phenotype observed;• “exp” is the number of that phenotype expected from the null hypothesis.
  5. 5. • How can you tell if an observed set of offspring counts is legitimately the result of a given underlying simple ratio?• For example, you do a cross and see 290 purple flowers and 110 white flowers in the offspring. This is pretty close to a 3/4 : 1/4 ratio, but how do you formally define "pretty close"? What about 250:150?
  6. 6. Example• As an example, you count F2 offspring, and get 290 purple and 110 white flowers. This is a total of 400 (290 + 110) offspring.• We expect a 3/4 : 1/4 ratio. We need to calculate the expected numbers, this is done by multiplying the total offspring by the expected proportions. This we expect 400 * 3/4 = 300 purple, and 400 * 1/4 = 100 white.• Thus, for purple, obs = 290 and exp = 300. For white, obs = 110 and exp = 100.• Now its just a matter of plugging into the formula: 2 = (290 - 300)2 / 300 + (110 - 100)2 / 100 = (-10)2 / 300 + (10)2 / 100 = 100 / 300 + 100 / 100 = 0.333 + 1.000 = 1.333.• This is our chi-square value: now we need to see what it means and how to use it.
  7. 7. The Critical Question• Using the example here, how can you tell if your 290: 110 offspring ratio really fits a 3/4 : 1/4 ratio (as expected from selfing a heterozygote).You can’t be certain, but you can at least determine whether your result is reasonable.
  8. 8. Reasonable• What is a “reasonable” result is subjective and arbitrary.• For most work a result is said to not differ significantly from expectations if it could happen at least 1 time in 20. That is, if the difference between the observed results and the expected results is small enough that it would be seen at least 1 time in 20 over thousands of experiments, we “fail to reject” the null hypothesis.• For technical reasons, we use “fail to reject” instead of “accept”.• “1 time in 20” can be written as a probability value p = 0.05, because 1/20 = 0.05.• Another way of putting this. If your experimental results are worse than 95% of all similar results, they get rejected because you may have used an incorrect null hypothesis.
  9. 9. • The test statistic is compared to a theoretical probability distribution• In order to use this distribution properly you need to determine the degrees of freedom• If the level of significance read from the table is greater than .05 or 5% then your hypothesis is accepted and the data is useful• The hypothesis is termed the null hypothesis which states that there is no substantial statistical deviation between observed and expected data.
  10. 10. Degrees of Freedom• A critical factor in using the chi-square test is the “degrees of freedom”.• Degrees of freedom is the number of phenotypic possibilities in your cross minus one. Or• Degrees of freedom is simply the number of classes of offspring minus 1.• For our example, there are 2 classes of offspring: purple and white. Thus, degrees of freedom (d.f.) = 2 -1 = 1.
  11. 11. Critical Chi-Square• Critical values for chi-square are found on tables, sorted by degrees of freedom and probability levels. Be sure to use p = 0.05.• If your calculated chi-square value is greater than the critical value from the table, you “reject the null hypothesis”.• If your chi-square value is less than the critical value, you “fail to reject” the null hypothesis (that is, you accept that your genetic theory about the expected ratio is correct).
  12. 12. Chi-Square Table
  13. 13. Using the Table• In our example of 290 purple to 110 white, we calculated a chi-square value of 1.333, with 1 degree of freedom.• Looking at the table, 1 d.f. is the first row, and p = 0.05 is the sixth column. Here we find the critical chi-square value, 3.841.• Since our calculated chi-square, 1.333, is less than the critical value, 3.841, we “fail to reject” the null hypothesis. Thus, an observed ratio of 290 purple to 110 white is a good fit to a 3/4 to 1/4 ratio.
  14. 14. Chi-square with more than one degree of freedom
  15. 15. 9:3:3:1
  16. 16. phenotype observed expected expected proportion numberround 315 9/16 312.75yellowround 101 3/16 104.25greenwrinkled 108 3/16 104.25yellowwrinkled 32 1/16 34.75greentotal 556 556
  17. 17. Finding the Expected Numbers• You are given the observed numbers, and you determine the expected proportions from a Punnett square.• To get the expected numbers of offspring, first add up the observed offspring to get the total number of offspring. In this case, 315 + 101 + 108 + 32 = 556.• Then multiply total offspring by the expected proportion: --expected round yellow = 9/16 * 556 = 312.75 --expected round green = 3/16 * 556 = 104.25 --expected wrinkled yellow = 3/16 * 556 = 104.25 --expected wrinkled green = 1/16 * 556 = 34.75• Note that these add up to 556, the observed total offspring.
  18. 18. Calculating the Chi-Square Value• Use the formula.• X2 = (315 - 312.75)2 / 312.75 + (101 - 104.25)2 / 104.25 + (108 - 104.25)2 / 104.25 + (32 - 34.75)2 / 34.75 = 0.016 + 0.101 + 0.135 + 0.218 = 0.470. 2 (obs exp) 2 exp
  19. 19. D.F. and Critical Value• Degrees of freedom is 1 less than the number of classes of offspring. Here, 4 - 1 = 3 d.f.• For 3 d.f. and p = 0.05, the critical chi-square value is 7.815.• Since the observed chi-square (0.470) is less than the critical value, we fail to reject the null hypothesis. We accept Mendel’s conclusion that the observed results for a 9/16 : 3/16 : 3/16 : 1/16 ratio.• It should be mentioned that all of Mendel’s numbers are unreasonably accurate.
  20. 20. Chi-Square Table
  21. 21. Consider Another example: For Drosophila melanogaster• Gene affecting wing shape • Gene affecting body color – c+ = Normal wing – e+ = Normal (gray) – c = Curved wing – e = ebony• Note: – The wild-type allele is designated with a + sign – Recessive mutant alleles are designated with lowercase letters• The Cross: – A cross is made between two true-breeding flies (c+c+e+e+ and ccee). The flies of the F1 generation are then allowed to mate with each other to produce an F2 generation.
  22. 22. • The outcome – F1 generation • All offspring have straight wings and gray bodies – F2 generation • 193 straight wings, gray bodies • 69 straight wings, ebony bodies • 64 curved wings, gray bodies • 26 curved wings, ebony bodies • 352 total flies
  23. 23. Applying the chi square testStep 1: Propose a null hypothesis that allows us to calculate the expected values based on Mendel’s laws • The two traits are independently assorting
  24. 24.  Step 2: Calculate the expected values of the four phenotypes, based on the hypothesis • According to our hypothesis, there should be a 9:3:3:1 ratio on the F2 generation Phenotype Expected Expected Observed number probability number straight wings, 9/16 9/16 X 352 = 198 193 gray bodies straight wings, 3/16 3/16 X 352 = 66 64 ebony bodies curved wings, 3/16 3/16 X 352 = 66 62 gray bodies curved wings, 1/16 1/16 X 352 = 22 24 ebony bodies
  25. 25.  Step 3: Apply the chi square formula (O1 – E1)2 (O2 – E2)2 (O3 – E3)2 (O4 – E4)2 + + + E1 E2 E3 E4 (193 – 198)2 (69 – 66)2 (64 – 66)2 (26 – 22)2 + + + 198 66 66 22 0.13 + 0.14 + 0.06 + 0.73 Expected Observed number number 198 193 1.06 66 64 66 62 22 24
  26. 26.  Step 4: Interpret the chi square value – The calculated chi square value can be used to obtain probabilities, or P values, from a chi square table • These probabilities allow us to determine the likelihood that the observed deviations are due to random chance alone – Low chi square values indicate a high probability that the observed deviations could be due to random chance alone – High chi square values indicate a low probability that the observed deviations are due to random chance alone – If the chi square value results in a probability that is less than 0.05 (ie: less than 5%) it is considered statistically significant • The hypothesis is rejected
  27. 27.  Step 4: Interpret the chi square value – Before we can use the chi square table, we have to determine the degrees of freedom (df) • The df is a measure of the number of categories that are independent of each other • If you know the 3 of the 4 categories you can deduce the • df = n – 1 – where n = total number of categories • In our experiment, there are four phenotypes/categories – Therefore, df = 4 – 1 = 3 – Refer to Table
  28. 28. 1.06
  29. 29.  Step 4: Interpret the chi square value – With df = 3, the chi square value of 1.06 is slightly greater than 1.005 (which corresponds to P-value = 0.80) – P-value = 0.80 means that Chi-square values equal to or greater than 1.005 are expected to occur 80% of the time due to random chance alone; that is, when the null hypothesis is true. – Therefore, it is quite probable that the deviations between the observed and expected values in this experiment can be explained by random sampling error and the null hypothesis is not rejected. What was the null hypothesis?
  30. 30. If your hypothesis is supported by data •you are claiming that mating is random and so is segregation and independent assortment.If your hypothesis is not supported by data •you are seeing that the deviation between observed and expected is very far apart…something non-random must be occurring….
  31. 31. Let’s look at an other fruit fly cross x Black body, eyeless wild F1: all wild
  32. 32. F1 x F15610 1881 1896 622
  33. 33. Analysis of the results• Once the numbers are in, you have to determine the cross that you were using.• What is the expected outcome of this cross?• 9/16 wild type: 3/16 normal body eyeless: 3/16 black body wild eyes: 1/16 black body eyeless.
  34. 34. Now Conduct the Analysis: Phenotype Observed Hypothesis Wild 5610 Eyeless 1881 Black body 1896 Eyeless, black body 622 Total 10009To compute the hypothesis value take 10009/16 = 626
  35. 35. Now Conduct the Analysis: Phenotype Observed Hypothesis Wild 5610 5634 Eyeless 1881 1878 Black body 1896 1878 Eyeless, black body 622 626 Total 10009To compute the hypothesis value take 10009/16 = 626
  36. 36. 2 (obs exp) 2 exp• Using the chi square formula compute the chi square total for this cross:• (5610 - 5630)2/ 5630 = .07• (1881 - 1877)2/ 1877 = .01• (1896 - 1877 )2/ 1877 = .20• (622 - 626) 2/ 626 = .02• 2= .30• How many degrees of freedom?
  37. 37. 2 (obs exp) 2 exp• Using the chi square formula compute the chi square total for this cross:• (5610 - 5630)2/ 5630 = .07• (1881 - 1877)2/ 1877 = .01• (1896 - 1877 )2/ 1877 = .20• (622 - 626) 2/ 626 = .02• 2= .30• How many degrees of freedom? 3
  38. 38. CHI-SQUARE DISTRIBUTION TABLEAccept Hypothesis Reject Hypothesis Probability (p)Degrees of 0.95 0.90 0.80 0.70 0.50 0.30 0.20 0.10 0.05 0.01 0.001 Freedom 1 0.004 0.02 0.06 0.15 0.46 1.07 1.64 2.71 3.84 6.64 10.83 2 0.10 0.21 0.45 0.71 1.39 2.41 3.22 4.60 5.99 9.21 13.82 3 0.35 0.58 1.01 1.42 2.37 3.66 4.64 6.25 7.82 11.34 16.27 4 0.71 1.06 1.65 2.20 3.36 4.88 5.99 7.78 9.49 13.38 18.47 5 1.14 1.61 2.34 3.00 4.35 6.06 7.29 9.24 11.07 15.09 20.52 6 1.63 2.20 3.07 3.83 5.35 7.23 8.56 10.64 12.59 16.81 22.46 7 2.17 2.83 3.82 4.67 6.35 8.38 9.80 12.02 14.07 18.48 24.32 8 2.73 3.49 4.59 5.53 7.34 9.52 11.03 13.36 15.51 20.09 26.12 9 3.32 4.17 5.38 6.39 8.34 10.66 12.24 14.68 16.92 21.67 27.88 10 3.94 4.86 6.18 7.27 9.34 11.78 13.44 15.99 18.31 23.21 29.59
  39. 39. • When reporting chi square data use the following formula sentence…. With degrees of freedom, my chi square value is , which gives me a p value between % and %, I therefore my null hypothesis.• This sentence would go in the “reults” section of your formal lab.• Your explanation of the significance of this data would go in the “discussion” section of the formal lab.
  40. 40. • Looking this statistic up on the chi square distribution table tells us the following:• the P value read off the table places our chi square number of .30 close to .95 or 95%• This means that 95% of the time when our observed data is this close to our expected data, this deviation is due to random chance.• We therefore accept our null hypothesis.
  41. 41. • What is the critical value at which we would reject the null hypothesis?• For three degrees of freedom this value for our chi square is > 7.815• What if our chi square value was 8.0 with 4 degrees of freedom, do we accept or reject the null hypothesis?• Accept, since the critical value is >9.48 with 4 degrees of freedom.

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