Observed ratio: 315:85 Null hypothesis (a) The data fit a 3:1 ratio. Null hypothesis (b) The data fit a 1:1 ratio. Part A In assessing data that fell into two phenotypic classes, a geneticist observed values of 315:85. She decided to perform a ?² analysis by using two different null hypotheses (see table). Calculate the ?² value for hypothesis (a). Express your answer to one decimal point (example: 100.0).
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To do this, she would need to consult the table below: From the table, it can be seen that the population mean is 315.85, the standard deviation is 68.68, and there are a total of 315 observations. Show more…
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Problem Set 7 2. For the data set below, determine whether the genotype frequencies represent a population in Hardy-Weinberg Equilibrium. Use a x² test and show your work. Assume that this data allows you to imagine a perfectly representative sample of 1000 individuals. (The critical value at probability 0.05 with one degree of freedom is 3.84). a. Sickle-cell hemoglobin: AA, 51.4 percent; AS, 47.6 percent; SS, 1.0 percent AA = 0.514 * 1000 = 514 AS = 0.476 * 1000 = 476 SS = 0.01 * 1000 = 10 Frequency of A = p² + 0.5(2pq) = 0.514 + 0.5(0.476) = 0.752 Frequency of S = 1 - 0.752 = 0.248 Frequency of AA = p² * 1000 = (0.752)² * 1000 = 565 Frequency of AS = 2pq * 1000 = 2 * (0.752)(0.248) * 1000 = 373 Frequency of SS = q² * 1000 = (0.248)² * 1000 = 62 x² = Σ((observed - expected)² / expected) x² = (514 - 565)² / 565 + (476 - 373)² / 373 + (10 - 62)² / 62 = 76.66 76.66 • The chi-square value is widely deviated from the Hardy Weinberg equilibrium since its nowhere near the critical value of 3.84
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Sample Size and Critical Value In 1908 , William Gosset published the article "The Probable Error of a Mcan" under the pscudonym of "Student" (Biomatrika, Vol. 6, No. 1). He included the data listed below for yields from two different types of seed (regular and kiln dried) that were used on adjacent plots of land. The listed values are the yields of straw in cwt per acre, where cwt represents 100 lb. If the Wilcoxon signed-ranks test is used to tot the claim that there is no difference between the yields from the two types of seed, what is the sample size $n$ ? If the significance level is $0.05,$ what is the critical value? $$\begin{array}{l|lllllllllll}\text { Regular } & 19.25 & 22.75 & 23 & 23 & 22.5 & 19.75 & 24.5 & 15.5 & 18 & 14.25 & 17 \\\hline \text { Kiln dried } & 25 & 24 & 24 & 28 & 22.5 & 19.5 & 22.25 & 16 & 17.25 & 15.75 & 17.25\end{array}$$
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A cross in garden pea results in 302 yellow, round seeds: 112 yellow, wrinkled seeds: 95 green, round seeds: 25 green, wrinkled seeds. By means of chi-squared analysis, determine if a 9:3:3:1 dihybrid cross is an acceptable hypothesis to explain the observed results. Round off all your answers to the nearest hundredths (two decimal place). (10 points) a) Form your hypothesis Ho: _______________________________________________________________ Ha: _______________________________________________________________ b) Calculate your expected value and fill out the table below: Phenotype Observed value (O) Expected Ratio Expected value (E) (O-E)2 yellow, round seeds Yellow, wrinkled seeds green, round seeds green, wrinkled seeds total c) Calculate your X2 value. d) Use the chi square table to get the tabulated value of X2 (df, 0.05). e) Compare your calculated X2 value to the tabulated value of X2 (df, 0.05). f) conclusion:
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