1 00:00:07,550 --> 00:00:18,010 After flipping a coin 100 times, you tally up 42 heads and 58 tails. You expected a 50/50 2 00:00:18,010 --> 00:00:22,430 distribution. Does this mean the coin is unfair? 3 00:00:22,430 --> 00:00:29,810 This video is part of the Probability and Statistics video series. Many events and phenomena 4 00:00:29,810 --> 00:00:38,110 are probabilistic. Engineers, designers, and architects often use probability distributions 5 00:00:38,110 --> 00:00:41,780 to predict system behavior. 6 00:00:41,780 --> 00:00:49,120 Hi, my name is Lourdes Aleman, and I am a research scientist in the HHMI Education Group 7 00:00:49,120 --> 00:00:56,740 in the biology department at MIT. I also work in the MIT Office of Educational Innovation 8 00:00:56,750 --> 00:01:04,879 and Technology. I helped develop the Star Genetics software featured in this video. 9 00:01:04,879 --> 00:01:11,500 Before watching this video, you should know what a probability distribution is and be 10 00:01:11,500 --> 00:01:22,000 familiar with basic genetics vocabulary including genotype, phenotype, homozygous, and heterozygous. 11 00:01:22,000 --> 00:01:28,770 You should also know how to use Punnett squares to predict the expected results of Mendelian 12 00:01:28,770 --> 00:01:31,340 crosses. 13 00:01:31,340 --> 00:01:37,600 After watching this video, you will be able to apply Chi square hypothesis testing to 14 00:01:37,600 --> 00:01:44,600 experimental data obtained from genetic experiments. 15 00:01:47,240 --> 00:01:54,110 Going back to question of whether or not our coin is fair, it turns out there is a statistical 16 00:01:54,110 --> 00:02:02,590 tool that can help us. Statistical tests can be used to determine if sample data can be 17 00:02:02,590 --> 00:02:07,950 assumed to come from a population that has a certain distribution. 18 00:02:07,950 --> 00:02:17,350 Today, we will use a statistical test called the Chi Square Test for Goodness of Fit. 19 00:02:19,030 --> 00:02:28,370 The Chi Square test is used to analyze categorical data. In other words, the test compares a 20 00:02:28,370 --> 00:02:35,370 sample of data collected about an event to expected values. 21 00:02:35,380 --> 00:02:47,000 In our coin-flipping example, our categories are heads and tails. An event is one coin flip. 22 00:02:47,000 --> 00:02:50,960 Our sample is 100 events. 23 00:02:51,940 --> 00:02:56,819 We expect that the coin should have a 50/50 chance of 24 00:02:56,819 --> 00:03:05,479 being heads or tails in any given event. The question we are trying to answer is – does 25 00:03:05,480 --> 00:03:12,480 our sample come from the 50/50 population distribution we expect? 26 00:03:13,950 --> 00:03:22,160 What we have just done is state our null hypothesis about the coin flipping experiment. The Chi 27 00:03:22,160 --> 00:03:27,100 square statistic tests this null hypothesis. 28 00:03:27,100 --> 00:03:34,840 We also need to state an alternative hypothesis. The alternative hypothesis is simply that 29 00:03:34,840 --> 00:03:41,840 the data do not come from the specified 50/50 population distribution. 30 00:03:43,200 --> 00:03:50,620 Before we can test our hypothesis, we need to calculate a Chi square statistic. 31 00:03:51,600 --> 00:04:00,380 This statistic measures the discrepancy between observed counts and expected counts. 32 00:04:01,200 --> 00:04:09,780 Each term in our summation finds the deviation, or error, between observed and expected values 33 00:04:09,790 --> 00:04:20,269 for each category. This deviation is squared to obtain positive values. Otherwise, everything 34 00:04:20,269 --> 00:04:28,269 would sum to zero because both our observed and expected outcomes sum to the same value. 35 00:04:29,820 --> 00:04:36,720 Each term in the summation is divided by the expected value of the outcome as a way of 36 00:04:36,729 --> 00:04:40,729 normalizing each difference. 37 00:04:40,729 --> 00:04:49,229 The Chi square statistic lives in a distribution. This is the Chi square distribution for any 38 00:04:49,229 --> 00:04:58,840 sample that has two categories of data, or, in other words, one degree of freedom. 39 00:04:58,840 --> 00:05:07,220 If the outcome of an event can only fall into one of two categories, we only need to know 40 00:05:07,229 --> 00:05:14,499 how many events fell into one category in order to determine how many fell into the 41 00:05:14,499 --> 00:05:17,139 other category. 42 00:05:18,460 --> 00:05:25,480 The Chi square distribution is actually a set of distributions, each for a different 43 00:05:25,490 --> 00:05:29,909 number of degrees of freedom. 44 00:05:29,909 --> 00:05:39,249 If we look at these distributions, the x-axis is chi-squared values and the y-axis is relative 45 00:05:39,249 --> 00:05:49,169 frequency. You might also notice that these distributions are positively skewed. This 46 00:05:49,169 --> 00:05:52,430 makes sense for a couple of reasons: 47 00:05:52,430 --> 00:06:00,589 The Chi squared statistic is always positive. Larger and larger Chi squared values indicate 48 00:06:00,589 --> 00:06:08,319 larger discrepancies between observed and expected values. While possible, these large 49 00:06:08,319 --> 00:06:17,939 discrepancies are less probable if the null hypothesis is true. If we calculate the Chi 50 00:06:17,949 --> 00:06:26,400 square test statistic for the coin flipping experiment, we get 2.56. 51 00:06:26,400 --> 00:06:34,240 How do we interpret this number? First, we need to look at the appropriate Chi-squared distribution. 52 00:06:36,460 --> 00:06:43,460 In the coin flipping experiment, there are two possible outcomes: heads or tails. 53 00:06:44,860 --> 00:06:51,940 If we know how many heads were flipped, we can calculate the number of tails flipped. 54 00:06:52,060 --> 00:06:55,479 So, we only have one degree of freedom. 55 00:06:57,160 --> 00:07:02,480 Let’s refer to the Chi square distributions for one degree of freedom. 56 00:07:03,979 --> 00:07:12,379 We can compare our test statistic to the distribution to see if we have support for our null hypothesis 57 00:07:12,379 --> 00:07:16,789 or if we need to reject it. 58 00:07:16,789 --> 00:07:23,789 What will our criterion for support or rejection of our null hypothesis be? 59 00:07:24,260 --> 00:07:33,219 While this is a fairly arbitrary choice, in many fields, a result is said to not differ significantly 60 00:07:33,219 --> 00:07:40,050 from expectations if it has a 1 in 20 chance of happening. 61 00:07:40,050 --> 00:07:46,309 In other words, if the difference between the observed result and the expected result 62 00:07:46,309 --> 00:07:56,129 is small enough, we would expect to see it 1 in 20 times, which is a probability of 0.05. 63 00:07:57,920 --> 00:08:03,200 This probability is called the level of significance. 64 00:08:03,210 --> 00:08:08,919 If a Chi square value has a probability of occurring that is greater than our level of 65 00:08:08,919 --> 00:08:18,479 significance, it lends support to our null hypothesis. If a Chi square value has a probability 66 00:08:18,490 --> 00:08:25,490 of occurring that is less than our level of significance, we should reject the null hypothesis. 67 00:08:27,140 --> 00:08:35,120 Using the Chi square distribution for 1 degree of freedom, a chi square value of 2.56 has 68 00:08:35,130 --> 00:08:42,130 a greater than the 5% probability of occurring that we had set as our minimum. 69 00:08:42,969 --> 00:08:49,190 This means that we fail to reject our null hypothesis. In other words, what we observe 70 00:08:49,190 --> 00:08:56,000 is close enough to what we expect that we have confidence that our coin is fair. 71 00:09:00,620 --> 00:09:06,660 Now, suppose you are planning an experiment where you will cross mutant flies that are 72 00:09:06,660 --> 00:09:14,680 heterozygous for a wingless phenotype to each other. The wingless phenotype is dominant 73 00:09:14,690 --> 00:09:24,730 relative to the wild-type winged phenotype. Let’s use the notation Gg for the genotype 74 00:09:24,730 --> 00:09:27,839 of the parent mutant flies. 75 00:09:27,839 --> 00:09:34,839 In this case, the wild type, winged flies, are homozygous recessive. 76 00:09:34,980 --> 00:09:41,980 What percentage of flies resulting from a cross of two heterozygous wingless mutants 77 00:09:42,220 --> 00:09:50,779 would you expect to be wild type versus mutant? Pause the video here and use a Punnett square 78 00:09:50,779 --> 00:09:53,599 to justify your answer. 79 00:09:59,980 --> 00:10:08,620 Performing this cross using a Punnett square, you would expect a 3:1 ratio of mutant to 80 00:10:08,620 --> 00:10:10,850 wild-type flies. 81 00:10:10,850 --> 00:10:20,560 You perform the cross with the parental mutant flies. The cross produces 50 progeny. 32 are 82 00:10:20,560 --> 00:10:24,170 mutant and 18 are wild type. 83 00:10:24,170 --> 00:10:34,890 So, only 64% of our flies are mutant, even though you expected 75% of them to be mutant. 84 00:10:35,720 --> 00:10:42,449 Although it is not what you expected, it doesn’t seem that far off either. 85 00:10:42,449 --> 00:10:49,170 You decide to continue mating your parental mutant flies until you have 1000 progeny. 86 00:10:49,170 --> 00:10:55,399 Surely, with this large a sample size, the ratio of mutant to wild type flies will be 87 00:10:55,399 --> 00:10:57,879 closer to what you expect. 88 00:10:58,580 --> 00:11:08,260 Voila. Oh, wait a minute. Even with a 1000 progeny, our experiments still produce 68% 89 00:11:08,269 --> 00:11:12,600 mutant and 32% wildtype flies. 90 00:11:12,600 --> 00:11:18,310 Now you are really starting to wonder if you can call this result “close enough” to 91 00:11:18,310 --> 00:11:29,850 the expected outcome. We can use the Chi-square test to evaluate the hypothesis that the observed 92 00:11:29,850 --> 00:11:36,850 ratio of 68:32 is the same as the expected ratio of 3:1. 93 00:11:37,269 --> 00:11:43,410 Pause the video here, calculate the Chi square test statistic, and determine the degrees 94 00:11:43,410 --> 00:11:46,410 of freedom in your data. 95 00:11:51,769 --> 00:11:58,769 You should have obtained 29.2 for your Chi square value. 96 00:11:59,860 --> 00:12:06,160 Because there are only two phenotypic outcomes for your cross, there is only one degree of 97 00:12:06,170 --> 00:12:13,170 freedom. Any fly that isn’t a mutant is wild type and vice versa. 98 00:12:13,899 --> 00:12:22,319 The Chi square distribution can also be presented in tabular form. If you look at this table, 99 00:12:22,329 --> 00:12:31,529 our experiment with one degree of freedom and a Chi square value of 29.2 corresponds 100 00:12:31,529 --> 00:12:42,660 to a probability that is much smaller than 0.05. Do you have support for the null hypothesis 101 00:12:42,660 --> 00:12:49,660 or should you reject it? Pause the video here and take a moment to think about it. 102 00:12:56,460 --> 00:13:03,720 Because the ratio of progeny that you obtained in your experiment has a probability of happening 103 00:13:03,730 --> 00:13:10,730 that is less than our level of significance, you should reject our null hypothesis. 104 00:13:12,139 --> 00:13:17,800 This indicates that our discrepant results are due to something more than just random 105 00:13:17,800 --> 00:13:25,940 chance. A multitude of other factors could be contributing to our results. For example, 106 00:13:25,949 --> 00:13:32,610 there may be interactions with other genes that you are not aware of. Or, environmental 107 00:13:32,610 --> 00:13:40,860 factors could be playing a role. All we know is that our null hypothesis of a 3:1 ratio 108 00:13:40,860 --> 00:13:47,110 of mutant to wildtype flies is incorrect. 109 00:13:47,110 --> 00:13:53,680 As a next step, you decide to go back and review your notes to see if you can formulate 110 00:13:53,680 --> 00:13:57,519 another plausible hypothesis. 111 00:13:57,519 --> 00:14:05,759 Reviewing your Punnett square, you expected to obtain this genotypic ratio in the F1 progeny. 112 00:14:05,769 --> 00:14:15,669 If you were to select a GG mutant from the F1 progeny and cross it with a gg wild-type 113 00:14:15,670 --> 00:14:21,649 fly, you would expect that all of the progeny would be mutants. 114 00:14:21,649 --> 00:14:29,269 After attempting this cross with a variety of flies from the F1 progeny that have a mutant 115 00:14:29,269 --> 00:14:36,249 phenotype, it becomes clear that you cannot find a mutant fly that when crossed with a 116 00:14:36,249 --> 00:14:46,009 wild-type fly, produces all mutant progeny. Somehow, you keep selecting heterozygous mutants. 117 00:14:47,540 --> 00:14:55,840 You begin to suspect that the F1 population does not contain the expected number of GG 118 00:14:55,850 --> 00:14:57,459 flies! 119 00:14:57,459 --> 00:15:03,800 Checking your lab notebook, you see a note about an observation that you barely paid 120 00:15:03,800 --> 00:15:10,499 attention to at the time. When you mated the two parental mutant flies, there were a number 121 00:15:10,499 --> 00:15:18,479 of dead embryos in your vial. A new hypothesis is forming in your mind. 122 00:15:18,480 --> 00:15:24,389 Can you come up with a hypothesis that may explain the observations you have made through 123 00:15:24,389 --> 00:15:33,469 these experiments? Pause the video here. Discuss and justify your hypothesis with a classmate. 124 00:15:39,480 --> 00:15:47,120 While there are many possible hypotheses, one possible hypothesis is that the dead embryos 125 00:15:47,120 --> 00:15:56,820 in your vial were the missing GG mutants. It may be that the wingless trait when homozygous, 126 00:15:56,829 --> 00:15:58,759 results in lethality. 127 00:15:58,759 --> 00:16:10,039 This would result in a phenotypic ratio of 2:1 mutant to wild-type flies in the F1 progeny. 128 00:16:12,780 --> 00:16:21,800 Again, you can use the Chi square test for goodness of fit to test your new null hypothesis. 129 00:16:22,550 --> 00:16:31,309 Going back to your data of 1000 progeny, use the test to evaluate the null hypothesis. 130 00:16:32,420 --> 00:16:34,560 Pause the video. 131 00:16:40,980 --> 00:16:47,980 You should have obtained a Chi square value of 0.36. For one degree of freedom, you fail 132 00:16:48,249 --> 00:16:52,220 to reject your null hypothesis. 133 00:16:52,220 --> 00:16:58,129 While you have support for your null hypothesis, more experiments are needed to confirm the 134 00:16:58,129 --> 00:17:06,989 idea that the GG genotype is lethal as you suspect. The results from our Chi squared 135 00:17:06,990 --> 00:17:12,650 test have led us to a reasonable next experiment to perform. 136 00:17:16,580 --> 00:17:24,160 In this video, you saw how the Chi-square statistic can help us determine if our experimental 137 00:17:24,170 --> 00:17:33,390 genetic data came from the expected population distribution. Comparing observed values to 138 00:17:33,390 --> 00:17:42,320 expected values helped us determine if deviations in our data could be explained by random chance 139 00:17:42,320 --> 00:17:47,470 or if our hypothesis needed revision. 140 00:17:47,470 --> 00:17:54,090 The Chi-Square Test for Goodness of Fit is just one statistical test of many that allows 141 00:17:54,090 --> 00:18:00,860 us to determine how confident we are that a sample of data comes from a population with 142 00:18:00,860 --> 00:18:08,420 a certain distribution. It is important to understand what these tests mean and how to 143 00:18:08,420 --> 00:18:16,060 interpret their results, so that you know their limitations and can make informed decisions.