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The paper “Linkage Studies of the Tomato” (Transactions of the Canadian Institute, 1931) reported the following data on phenotypes resulting from crossing tall cut-leaf tomatoes with dwarf potato-leaf tomatoes. We wish to investigate whether the following frequencies are consistent with genetic laws, which state that the phenotypes should occur in the ratio 9:3:3:1. Dwarf potato 104 Phenotype Frequency Tall cut 926 Tall potato 288 Dwarf cut 293

The paper “Linkage Studies of the Tomato” (Transactions of the Canadian Institute, 1931) reported the following data on phenotypes resulting from crossing tall cut-leaf tomatoes with dwarf potato-leaf tomatoes. We wish to investigate whether the following frequencies are consistent with genetic laws, which state that the phenotypes should occur in the ratio 9:3:3:1. Dwarf potato 104 Phenotype Frequency Tall cut 926 Tall potato 288 Dwarf cut 293

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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The paper "Linkage Studies of the Tomato" (Transactions of the Canadian Institute, 1931) reported the following data on phenotypes resulting from crossing tall cut-leaf tomatoes with dwarf potato-leaf tomatoes. We wish to investigate whether the following frequencies are consistent with genetic laws, which state that the phenotypes should occur in the ratio 9:3:3:1. Dwarf potato 104 Phenotype Frequency Tall cut 926 Tall potato 288 Dwarf cut 293 We want to test Ho: The proposed 9:3:3:1 genetic model is correct versus Ha: The proposed 9:3:3:1 genetic model is not correct. Use a chi-square goodness-of-fit test to compute a test statistic and P-value. At the α = 0.05 significance level, what can be concluded about the proposed genetic model? Because the P-value of 0.6895 > 0.05, we cannot reject the null hypothesis. We do not have convincing evidence that the distribution of phenotypes 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 phenotypes 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 phenotypes is different from what the biologists predict. Because the P-value of 0.6895 > 0.05, we cannot reject the null hypothesis. We do have convincing evidence that the distribution of phenotypes is the same as what the biologists predict. Because the P-value of 0.6895 > 0.05, we reject the null hypothesis in favor of the alternate. We do have convincing evidence that the distribution of phenotypes is different from what the biologists predict.
Transcribed Image Text:The paper "Linkage Studies of the Tomato" (Transactions of the Canadian Institute, 1931) reported the following data on phenotypes resulting from crossing tall cut-leaf tomatoes with dwarf potato-leaf tomatoes. We wish to investigate whether the following frequencies are consistent with genetic laws, which state that the phenotypes should occur in the ratio 9:3:3:1. Dwarf potato 104 Phenotype Frequency Tall cut 926 Tall potato 288 Dwarf cut 293 We want to test Ho: The proposed 9:3:3:1 genetic model is correct versus Ha: The proposed 9:3:3:1 genetic model is not correct. Use a chi-square goodness-of-fit test to compute a test statistic and P-value. At the α = 0.05 significance level, what can be concluded about the proposed genetic model? Because the P-value of 0.6895 > 0.05, we cannot reject the null hypothesis. We do not have convincing evidence that the distribution of phenotypes 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 phenotypes 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 phenotypes is different from what the biologists predict. Because the P-value of 0.6895 > 0.05, we cannot reject the null hypothesis. We do have convincing evidence that the distribution of phenotypes is the same as what the biologists predict. Because the P-value of 0.6895 > 0.05, we reject the null hypothesis in favor of the alternate. We do have convincing evidence that the distribution of phenotypes is different from what the biologists predict.
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