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- 71 y = 23.397 + 0.65027x 70 69 68 67 66 65 64 66 68 70 72 MidParent height (x) Mean height of adult Offspring (y)The frequency of the A1 allele is p = 0.4, and that of the A2 allele is q = 0.6. The observed genotype frequencies are as follows: f(A1A1) = 0.25 f(A1A2) = 0.36 f(A2A2) = 0.39 Is there a Hardy-Weinberg equilibrium in the population?Which is the correct equation for the chisquare statistic? Group of answer choices a)x^2 = [(observed-expected)^2]/expected b)x^2 = sum[((observed-expected)^2)/observed] c)x^2 = sum[((observed-expected)^2)/expected] d)x^2 = sum[(observed-expected)/expected] What is a "p-value"? Group of answer choices a)the threshold at which you will reject the null hypothesis b)the probability of the null hypothesis being false c)the percentage of days you are sick during the year d)the probability of the null hypothesis being true
- Which is the correct equation for the chisquare statistic? Group of answer choices a)x^2 = [(observed-expected)^2]/expected b)x^2 = sum[((observed-expected)^2)/observed] c)x^2 = sum[((observed-expected)^2)/expected] d)x^2 = sum[(observed-expected)/expected] What is a "p-value"? Group of answer choices a)the threshold at which you will reject the null hypothesis b)the probability of the null hypothesis being false c) the percentage of days you are sick during the year d) the probability of the null hypothesis being true(1 point) Humans with the genotypes DD and Dd show the Rh+ blood phenotype, whereas those with the genotype dd show the Rh- blood phenotype. In a sample of 400 Basques from Spain, 230 people were Rh+ and 170 people were Rh-. Assuming that this population is in Hardy-Weinberg proportions, what is the allele frequency of the allele D? (a) (a) 0.348 (answer) (b) (b) 0.652 (c) (c) 0.425 (d) (d) 0.575 (e) (e) 0.288 2. (2 points) In the Basque population mentioned above, what proportion of the Rh+ individuals would be expected to be heterozygote? (a) (a) 0.454 (b) (b) 0.789 (answer) (c) (c) 0.516 (d) (d) 0.250 (e) (e) 0.500 How is the answer for #2, b? please explainCompare the 95% confidence interval for the difference between two true means with the 95% confidence interval for an individual true kean. Describe one way in which these two equations are similar and state four ways in which these equations differ.
- The typical tail size for elephants is 150 cm, with a standard deviation of 10 cm. The typical tail size for mice is 7 cm, with a standard deviation of 3 cm. Do elephants or mice exhibit greater variability in tail size? Elephants, because CV = 15 > 2.3 Elephants, because SD = 10 cm > SD = 3 cm %3D Elephants, because = 150 cm > 7 cm O Mice, because CV = 0.43 > CV = 0.07 %3D %3DIn a human population of 1000, 840 are tongue rollers (360 TT and 480 Tt), and 160 are not tongue rollers (tt). What is the frequency of the dominantallele (T) in the population?You perform a chi-square test to compare observed and expected values and obtain a chi-square value of 9.4 with 3 degrees of freedom. What do you conclude? it is not likely that the difference between observed and expected values is due to random chance, since p>0.05 it is impossible to conclude anything from this information it is likely that the difference between observed and expected values is due to random chance, since p>0.05 it is likely that the difference between observed and expected values is due to random chance, since p<0.05 the experiment was done incorrectly and must be repeated it is not likely that the difference between observed and expected values is due to random chance, since p<0.05
- Cerebellar abiotrophy (CA) is a recessive, genetic neurological disease found in the Arabian horse breed. If the incidence of this disease is about 1 in 400 Arabian horses, estimate the frequency of carriers in a population in Hardy-Weinberg equilibrium. Value of q²: 0.0025 Value of q: 0.05 Value of p: 0.95 Carrier Frequency: 0.095 How many horses would you expect to be carriers for cerebellar abiotrophy if the population consists of 582 Arabian horses? (Round to the nearest whole number and enter 2 digits.)Suppose that you are interested in estimating a population mean. You select a random sample of items, and compute the sample mean and the sample standard deviation. You then compute a 95% confidence interval to be LCL=28.4 - UCL=37.9. So what does that mean? It means that you are 95% confident that the unknown population mean that you are estimating is between the LCL and UCL. So what does that mean? It means that if you were to iterate this sampling process many times, say 100, and calculate 100 confidence intervals, then 95 of those intervals will contain the unknown population mean, and 5 will not. Give me an example of how CI can be used in your work. FYI I work in Endocrinology dept. Specific diabetesPhenylketonuria is a recessive autosomal genetic disorder that can cause mental retardation. Phenylketonuria is particularly common in Turkey, where 1 in 2600 children are born with the disease. As of 2011, the population of Turkey was 73.6 million. Based on this data, and assuming that the Turkish population is in Hardy-Weinberg equilibrium. (a) Estimate the number of carriers of phenylketonuria in Turkey (b) Estimate the proportion of unaffected individuals who are carriers.