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Biologists wish to mate pairs of fruit flies having genetic makeup RrCc, indicating that each has one dominant gene (R) and one recessive gene (r) for eye color, along with one dominant (C) and one recessive (c) gene for wing type. Each offspring will receive one gene for each of the two traits from each parent, so the biologists predict that the following phenotypes should occur in a ratio of 9:3:3:1. Phenotype Frequency Red eyes and Red eyes and straight wings curly wings 99 42 White eyes and straight wings 49 White eyes and curly wings 10 Use a chi-square goodness-of-fit test to compute a test statistic and P-value. At the a = 0.05 significance level, what can be concluded about the proposed genetic model? Because the P-value of 0.1029 > 0.05, we cannot reject the null hypothesis. We do not have convincing evidence that the distribution of eye color and wing shape is different from what the biologists predict. Because the P-value of 0.5177 > 0.05, we cannot reject the null hypothesis. We do not have convincing evidence that the distribution of eye color and wing shape is different from what the biologists predict. Because the P-value of 0.1854 > 0.05, we cannot reject the null hypothesis. We do not have convincing evidence that the distribution of eye color and wing shape is different from what the biologists predict. Because the P-value of 0.1029 > 0.05, we cannot reject the null hypothesis. We do have convincing evidence that the distribution of eye color and wing shape is the same as what the biologists predict. Because the P-value of 0.1029 > 0.05, we reject the null hypothesis in favor of the alternate. We do have convincing evidence that the distribution of eye color and wing shape is different from what the biologists predict.

Biologists wish to mate pairs of fruit flies having genetic makeup RrCc, indicating that each has one dominant gene (R) and one recessive gene (r) for eye color, along with one dominant (C) and one recessive (c) gene for wing type. Each offspring will receive one gene for each of the two traits from each parent, so the biologists predict that the following phenotypes should occur in a ratio of 9:3:3:1. Phenotype Frequency Red eyes and Red eyes and straight wings curly wings 99 42 White eyes and straight wings 49 White eyes and curly wings 10 Use a chi-square goodness-of-fit test to compute a test statistic and P-value. At the a = 0.05 significance level, what can be concluded about the proposed genetic model? Because the P-value of 0.1029 > 0.05, we cannot reject the null hypothesis. We do not have convincing evidence that the distribution of eye color and wing shape is different from what the biologists predict. Because the P-value of 0.5177 > 0.05, we cannot reject the null hypothesis. We do not have convincing evidence that the distribution of eye color and wing shape is different from what the biologists predict. Because the P-value of 0.1854 > 0.05, we cannot reject the null hypothesis. We do not have convincing evidence that the distribution of eye color and wing shape is different from what the biologists predict. Because the P-value of 0.1029 > 0.05, we cannot reject the null hypothesis. We do have convincing evidence that the distribution of eye color and wing shape is the same as what the biologists predict. Because the P-value of 0.1029 > 0.05, we reject the null hypothesis in favor of the alternate. We do have convincing evidence that the distribution of eye color and wing shape is different from what the biologists predict.

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Biologists wish to mate pairs of fruit flies having genetic makeup RrCc, indicating that each has one dominant gene (R) and one recessive gene (r) for eye color, along with one dominant (C) and one recessive (c) gene for wing type. Each offspring will receive one gene for each of the two traits from each parent, so the biologists predict that the following phenotypes should occur in a ratio of 9:3:3:1. Phenotype Frequency Red eyes and Red eyes and straight wings curly wings 99 42 White eyes and straight wings 49 White eyes and curly wings 10 Use a chi-square goodness-of-fit test to compute a test statistic and P-value. At the a = 0.05 significance level, what can be concluded about the proposed genetic model? Because the P-value of 0.1029 > 0.05, we cannot reject the null hypothesis. We do not have convincing evidence that the distribution of eye color and wing shape is different from what the biologists predict. Because the P-value of 0.5177 > 0.05, we cannot reject the null hypothesis. We do not have convincing evidence that the distribution of eye color and wing shape is different from what the biologists predict. Because the P-value of 0.1854 > 0.05, we cannot reject the null hypothesis. We do not have convincing evidence that the distribution of eye color and wing shape is different from what the biologists predict. Because the P-value of 0.1029 > 0.05, we cannot reject the null hypothesis. We do have convincing evidence that the distribution of eye color and wing shape is the same as what the biologists predict. Because the P-value of 0.1029 > 0.05, we reject the null hypothesis in favor of the alternate. We do have convincing evidence that the distribution of eye color and wing shape is different from what the biologists predict.
Transcribed Image Text:Biologists wish to mate pairs of fruit flies having genetic makeup RrCc, indicating that each has one dominant gene (R) and one recessive gene (r) for eye color, along with one dominant (C) and one recessive (c) gene for wing type. Each offspring will receive one gene for each of the two traits from each parent, so the biologists predict that the following phenotypes should occur in a ratio of 9:3:3:1. Phenotype Frequency Red eyes and Red eyes and straight wings curly wings 99 42 White eyes and straight wings 49 White eyes and curly wings 10 Use a chi-square goodness-of-fit test to compute a test statistic and P-value. At the a = 0.05 significance level, what can be concluded about the proposed genetic model? Because the P-value of 0.1029 > 0.05, we cannot reject the null hypothesis. We do not have convincing evidence that the distribution of eye color and wing shape is different from what the biologists predict. Because the P-value of 0.5177 > 0.05, we cannot reject the null hypothesis. We do not have convincing evidence that the distribution of eye color and wing shape is different from what the biologists predict. Because the P-value of 0.1854 > 0.05, we cannot reject the null hypothesis. We do not have convincing evidence that the distribution of eye color and wing shape is different from what the biologists predict. Because the P-value of 0.1029 > 0.05, we cannot reject the null hypothesis. We do have convincing evidence that the distribution of eye color and wing shape is the same as what the biologists predict. Because the P-value of 0.1029 > 0.05, we reject the null hypothesis in favor of the alternate. We do have convincing evidence that the distribution of eye color and wing shape is different from what the biologists predict.
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