HW#4 (CHP. 8 #'s: 12, 13, 14, 19 & CHP. 9 #'s: 17, 20, 29, 33) Rheanne Berg RSM 612 CHAPTER 8: 12. The white \"Spirit\" black bear, Ursys anerucabys kermodei, differs from the ordinary black bear by a single amino acid change in the melanocortin 1 receptor gene. In this population, the gene has two forms (or alleles): the \"white: allele b and the \"black\" allele B. The trait is recessive: white bears have two copies of the while allele of this gene (bb), whereas a bear is black if it has one ore two copies of the black allele (Bb or BB). Both colors morphs and all three genotypes are found together in the bear population of the northwest coast of British Columbia. If processing the white allele has no effect on growth, survival, reproductive success, or mating patterns of individual bears, then the frequency of individuals with 0, 1, or 2 copies of the white allele (b) in the population will follow a binomial distribution. To investigate, Hendrick and Ritland sampled and genotyped 87 bears from the northwest coast: 42 were BB, 24 were Bb, and 21 were bb. Assume that this is a random sampled of bears. a. Calculate the fraction of b alleles in the population (remember, each bear has two copies of the gene). proportion from the data b. With your estimate of the fraction of b alleles, and assuming a binomial distribution, calculate the expected frequency of bears with 0, 1, and 2 copies. c. Compare the observed and expected frequencies in a graph. Describe how they differ. 13. Refer to Assingment Problem 12. A formal hypothesis test was carried out to compare the observed and expected frequencies of genotypes. The procedure obtained P= 0.0001. Answer the following questions. . What are the null and alternative hypothesis? Null hypothesis H0 = Binomial distribution is a good fit to the data Alternative hypothesis H = Binomial distribution is not a good fit to the data. . What are the degrees of freedom for the test statistic? Here chi-squared test applied Therefore, the degrees of free is (n - k -1) = (3 -1 - 1) = 1 N is the number of observation K is the number of parameters estimated Degrees of freedom = 1 . The observed frequencies are compatible with a binomial distribution. The MS-Excel output to the information is shown below: Observed 42 24 21 87 Expected 33.518 40.965 12.517 87.000 O-E 8.482 -16.965 8.483 0.000 (O-E)2/E 2.146 7.026 5.750 14.922 % of CHISQ 14.38 47.08 38.53 100.00 14.92 chi square 1 df 0.0001 p-value True. Yes, the observed frequencies are compatible with binomial distribution. d. The difference between the observed and expected frequencies is statistically significant. True. Since the value of p is less than = 0.05, the difference between observed and expected frequencies is statistically significant. e. The test statistic exceeds the critical value corresponding to = 0.05. True. The critical value for the test statistic at = 0.05 is 3.8415 hence, this exceeds the test statistic value. f. The test statistic exceeds the critical value corresponding to = 0.01. True. The critical value for the test statistic at = 0.01 is 6.63349 hence, this exceeds the test statistic value at 0.01. g. The difference is large between the true genotype frequencies in the bear population and that expected under the binomial distribution. True. It can be observed from the output. 14. In North America, between 100 million and 1 billion birds die each year by crashing into windows on buildings, more than any other human-related cause. This figure represents up to 5% of all birds in the area. One possible solution is to construct windows angled downward slightly, so that they reflect the ground rather than an image of the sky to a flying bird. An experiment by Klem et al. compared the number of birds that died as result of vertical windows, windows angled 20 degrees off vertical, and windows angled 40 degrees off vertical. The angles were randomly assigned with equal probability to six windows and changed daily; assume for this exercise that windows and window locations were identical in every respect expect angle. Over the course of the experiment, 30 birds were killed by windows in the vertical orientation, 15 were killed by windows set at 20 degrees off vertical, and 8 were killed by windows set at 40 degrees off vertical. FIGURE a. Clearly state an appropriate null hypothesis and an alternative hypothesis H0: Windows kill the same number of birds per time period at any angle. HA: Windows angled towards the ground kill a different number of birds than windows at the vertical. b. What proportion of deaths occurred while the windows were set at a vertical orientation? (30)/(30+15+18) = 30/53 c. What statistical test would you use to test the null hypothesis? According to the null hypothesis, this is a goodness of fit test. So our test statistic should be computed as follows Window Angle Observed Deaths Expected Deaths (vertical) 0 30 17.67 20 15 17.67 40 8 17.67 O iE i 2 = 14.30 3 X 2= s i =1 d. Carry out the statistical test from part (c). Is there evidence that window angle affects the mortality rates of birds? The critical value for = 0.05 is 5.99. Since 14.30 > 5.99, we reject H0 at 0.05. There is enough evidence to say that window angle influences bird mortality. Test statistic is greater than the critical value ; reject the null hypothesis. 19. Hurricanes hit the United Sates often and hard, causing some loss of life and enormous economic costs. They are ranked in severity by the Saffir-Simpson scale, which ranges from Category 1 to Category 5, with 5 being the worst. In some years, as many as three hurricanes that rate a Category 3 or higher hit the U.S. coastline. In other years, no hurricane of this severity hits the United States. The following table lists the numbers of years that had 0, 1, 2, 3, or more hurricanes of at least Category 3 in severity, over the 100 years of the 20th century: Number of Hurricanes Category 3 or Number of years higher 0 50 1 39 2 7 3 4 >3 0 a. What is mean number of severe hurricanes to hit the United States per year? x = (0*50+1*39+2*7+3*4) / (100) =0.65 x = 0.65 b. What model would describe that distribution of hurricanes per year, if they were to hit independently of each other and if the probability of a hurricane were the same in every year? Poisson Distribution . Test the fit of the model from part (b) to the data. Pr [ X ]= e X! x Poisson Formula For X= 0 Pr[X=0]= e x X! = e x X! = e0.65 0.650 0! = 0.522 For X=1 Pr[X=1] = For X=2 Pr[X=2]= e X! x = e0.65 0.651 1! 0.65 e 0.65 2! 2 =0.339 = 0.110 For X=3 Pr[X=3] = e x X! = e0.65 0.653 3! = 0.024 For X > 3 All the probabilities solved for Pr[X= 0, 1, 2, & 3] all sum to one. Pr[X>3]=1-Pr[X3]=1-Pr[X=3]-Pr[X=2]-Pr[X=1]-Pr[X=0] = 0.522+0.339+0.110+0.024 = 0.995 Pr [X>3] = 1-0.995 = 0.005 Number of Hurricanes Category 3 or higher Number of years (observed) Pr[X] Expected Pr[X]*n 0 50 0.522 52.2 1 39 0.339 33.9 >1 11 0.139 13.9 Calculate chi-squared statistic Chi-squared statistic formula =(50-52.2)^2/(52.2) = (39-33.9)^2/(33.9) + (11-13.9)^2/(13.9) = 0.09 + 0.77 + 0.61 = 1.47 Calculate P-value Degrees of Freedom = n-1-0= 3-1 =2 = 0.05 and 2 d.f. = 5.991 (i.e. chart in book - critical value) Our test statistic, 1.47, is less than the critical value, so we fail to reject the null hypothesis. CHAPTER 9: 17. Some people feel that they have good intuition about when others are lying, while others do not feel they have this ability. Are the more \"intuitive\" people better able to detect lies? Each of 100 people who thought they had good intuitive abilities was shown a video clip of a person stating the name of his or her favorite movie. The person was truthful in some of the clips shows, whereas in others the person was lying. Another 100 people who claimed not have intuitive abilities were shown video clips. Fifty-nine of the 100 \"intuitive\" subjects correctly identified whether the person in the video was lying, whereas 69 of the 100 \"non-intuitive\" subjects correctly identified whether the person in the video was lying. a. Draw a graph that best presents these data. What association is suggested? b. Test whether the success rates of the two groups are different. c. Are \"intuitive\" people better at detecting lies than \"non-intutive\" people? Calculate an odds ratio and confidence interval for your answer. Interpret your result. 20. Day care centers expose children to a wider variety of germs than the children would be exposed to if they stayed at home more often. This has the obvious downside of more frequent colds and other illnesses, but it also serves to challenge the immune system of children at a critical stage in their development. A study tested whether social activity outside the house in young children affected their probability of later developing the white blood cell disease acute lymphoblastic leukemia, the most common cancer of all children. They compared 1272 children with ALL to 6238 children without ALL. Of the ALL kids, 1020 has significant social activity outside the home (including day care) when younger. Of the kids without ALL, 5343 has significant social activity out side the home. The rest of the children in both groups lacked regular contact with children who were not in their immediate families. a. Is this an experiment or observational study? Observational Study b. What are the proportions of children with significant social activity in children with and without ALL? Proportion of children with acute leukemia = 0.8018 Proportion of children without acute leukemia = 0.8565 c. What is the odds ratio for ALL, comparing the groups with and without significant social activity? Odds ratio = ad/bc = 0.6780 d. What is the 95% confidence interval for this odds ratio? Confidence interval is (0.5807, 0.7917) e. Does this confidence interval indicate that amount of social activity is associated with ALL? If so, did the children with more social activity have a higher or lower occurrence of ALL? Yes, it indicates that there is association. The children with more social activity have a higher occurrence of acute lymphoblastic leukemia. f. The researchers interpreted their study results in terms of the differing immune system exposure of the children, but gave several alternative explanations for the pattern. Can you think of any possible confounding variables? Confounding factors are viral infections, genders, smoking, exposure to certain chemicals and electromagnetics fields, etc. 29. 33