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A genetic experiment involving peas yielded one sample of offspring consisting of 450 green peas and 175 yellow peas. Use a 0.01 significance level to test the claim that under the same circumstances, 25% of offspring peas will be yellow. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method and the normal distribution as an approximation to the binomial distribution. What are the null and alternative hypotheses? O A. Ho:p= 0.25 H,:p>0.25 O B. H,: p= 0.25 H,: p#0.25 O D. Ho: p#0.25 Ос. Но: р#0.25 H1: p<0.25 H1: p>0.25 O E. Ho: p+0.25 H1: p= 0.25 O F. Ho: p= 0.25 H1: p<0.25 What is the test statistic? z = 1.73 (Round to two decimal places as needed.) What is the P-value? P-value = 0.042 (Round to four decimal places as needed.) What is the conclusion about the null hypothesis? O A. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, a. B. Reject the null hypothesis because the P-value is greater than the significance level, a. O C. Fail to reject the null hypothesis because the P-value is greater than the significance level, a. O D. Reject the null hypothesis because the P-value is less than or equal to the significance level, a. What is the final conclusion? O A. There is sufficient evidence to warrant rejection of the claim that 25% of offspring peas will be yellow. O B. There is sufficient evidence to support the claim that less than 25% of offspring peas will be yellow. OC. There is not sufficient evidence to warrant rejection of the claim that 25% of offspring peas will be yellow. O D. There is not sufficient evidence to support the claim that less than 25% of offspring peas will be yellow.

A genetic experiment involving peas yielded one sample of offspring consisting of 450 green peas and 175 yellow peas. Use a 0.01 significance level to test the claim that under the same circumstances, 25% of offspring peas will be yellow. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method and the normal distribution as an approximation to the binomial distribution. What are the null and alternative hypotheses? O A. Ho:p= 0.25 H,:p>0.25 O B. H,: p= 0.25 H,: p#0.25 O D. Ho: p#0.25 Ос. Но: р#0.25 H1: p<0.25 H1: p>0.25 O E. Ho: p+0.25 H1: p= 0.25 O F. Ho: p= 0.25 H1: p<0.25 What is the test statistic? z = 1.73 (Round to two decimal places as needed.) What is the P-value? P-value = 0.042 (Round to four decimal places as needed.) What is the conclusion about the null hypothesis? O A. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, a. B. Reject the null hypothesis because the P-value is greater than the significance level, a. O C. Fail to reject the null hypothesis because the P-value is greater than the significance level, a. O D. Reject the null hypothesis because the P-value is less than or equal to the significance level, a. What is the final conclusion? O A. There is sufficient evidence to warrant rejection of the claim that 25% of offspring peas will be yellow. O B. There is sufficient evidence to support the claim that less than 25% of offspring peas will be yellow. OC. There is not sufficient evidence to warrant rejection of the claim that 25% of offspring peas will be yellow. O D. There is not sufficient evidence to support the claim that less than 25% of offspring peas will be yellow.

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter13: Probability And Calculus
Section13.CR: Chapter 13 Review
Problem 6CR
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A genetic experiment involving peas yielded one sample of offspring consisting of 450 green peas and 175 yellow peas. Use a 0.01 significance level to test the claim that under the same circumstances, 25% of offspring peas will be yellow. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method and the normal distribution as an approximation to the binomial distribution. What are the null and alternative hypotheses? O A. Ho:p= 0.25 H,:p>0.25 O B. H,: p= 0.25 H,: p#0.25 O D. Ho: p#0.25 Ос. Но: р#0.25 H1: p<0.25 H1: p>0.25 O E. Ho: p+0.25 H1: p= 0.25 O F. Ho: p= 0.25 H1: p<0.25 What is the test statistic? z = 1.73 (Round to two decimal places as needed.) What is the P-value? P-value = 0.042 (Round to four decimal places as needed.) What is the conclusion about the null hypothesis? O A. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, a. B. Reject the null hypothesis because the P-value is greater than the significance level, a. O C. Fail to reject the null hypothesis because the P-value is greater than the significance level, a. O D. Reject the null hypothesis because the P-value is less than or equal to the significance level, a. What is the final conclusion? O A. There is sufficient evidence to warrant rejection of the claim that 25% of offspring peas will be yellow. O B. There is sufficient evidence to support the claim that less than 25% of offspring peas will be yellow. OC. There is not sufficient evidence to warrant rejection of the claim that 25% of offspring peas will be yellow. O D. There is not sufficient evidence to support the claim that less than 25% of offspring peas will be yellow.
Transcribed Image Text:A genetic experiment involving peas yielded one sample of offspring consisting of 450 green peas and 175 yellow peas. Use a 0.01 significance level to test the claim that under the same circumstances, 25% of offspring peas will be yellow. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method and the normal distribution as an approximation to the binomial distribution. What are the null and alternative hypotheses? O A. Ho:p= 0.25 H,:p>0.25 O B. H,: p= 0.25 H,: p#0.25 O D. Ho: p#0.25 Ос. Но: р#0.25 H1: p<0.25 H1: p>0.25 O E. Ho: p+0.25 H1: p= 0.25 O F. Ho: p= 0.25 H1: p<0.25 What is the test statistic? z = 1.73 (Round to two decimal places as needed.) What is the P-value? P-value = 0.042 (Round to four decimal places as needed.) What is the conclusion about the null hypothesis? O A. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, a. B. Reject the null hypothesis because the P-value is greater than the significance level, a. O C. Fail to reject the null hypothesis because the P-value is greater than the significance level, a. O D. Reject the null hypothesis because the P-value is less than or equal to the significance level, a. What is the final conclusion? O A. There is sufficient evidence to warrant rejection of the claim that 25% of offspring peas will be yellow. O B. There is sufficient evidence to support the claim that less than 25% of offspring peas will be yellow. OC. There is not sufficient evidence to warrant rejection of the claim that 25% of offspring peas will be yellow. O D. There is not sufficient evidence to support the claim that less than 25% of offspring peas will be yellow.
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