A type of rice, which is resistant to an herbicide was crossed with another type

Question

A type of rice, which is resistant to an herbicide was crossed with another type of rice, which is susceptible to the herbicide. It is supposed that their offspring will have a 1: 2: 1 ratio measuring the number of plants that are resistant moderately resistant and susceptible to the herbicide. A crop of 3120 was comprised of 772 resistant, 1611 moderately resistant and 737 susceptible plants. The null hypotheses for this chi-square goodness-of-fit test is A. H_0: Pr = 1/4 and Pm = 2/4 = 1/2 and Ps = 1/4 B. H_0: Pr = 1 and Pm = 2 and Ps = 1 C. H_0: The distribution of the proportions is not uniform D. H_0: The distribution of the proportions is uniform: Pr = Pm = Ps Refer to question number 5: The alternative hypothesis is A. H_a: The distribution of the proportions is not uniform B. H_a: the proportions are not al the same C. H_a: Pr notequalto Pm notequalto Ps D. H_a: The distribution specified by H_0 is not true Refer to question number 5: The expected count of rice that is to the herbicide is A. 1560 B. 184.25. C. 737. D. 780. Refer to question number 5: The contribution to the chi-square statistic for the rice that is susceptible is to the herbicide is about A. 2.51 B. 0.055. C. 0.893. D. 2.371. Refer to question number 5: The degrees of freedom are ______ Refer to question number 5: The Chi-Square test statistic is ________ (give your answer to two decimal places.) Refer to question number 5: The p-value for the test is 0.127 (check that you can get this). Your conclusion based on this is test is therefore that A. the test results prove that ratio 1: 2: 1 is correct. B. the best results are not consistent with the resistant model determined by the 1: 2: 1 radio. C. the test results are consistent with the resistant model determined by the 1: 2: 1 ratio. D. the test results show the resistant and moderately resistant part of the model is correct, but the part of the model is incorrect.

Explanation / Answer

5) option A is correct

6)option D is correct

7) expected count =np=3120/4=780

option D

8)

chi square contribution =(O-E)2/E =(737-780)2/780 =2.37

option D

9)

degree of freedom =number of categories -1 =2

10)chi square test stat =4.12

11)

option C is correct

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