A genetic experiment involving peas yielded one sample of offspring consisting of 435 green peas and 157 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. Ho p=0.25 H, p>0.25 O C. Ho p=0.25 H p<0.25 O D. Ho p#0.25 H, p=0 25 O E. Ho p=0.25 H, p#0.25 OF. Ho p#0.25 H, p>0 25 What is the test statistic? z = (Round to two decimal places as needed) What is the P.value? P-value = (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 greater than the significance level, a O B. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, a. O C. 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 not sufficient evidence to support the claim that less than 25% of offspring peas will be yellow O B. There is not sufficient evidence to warrant rejection of the claim that 25% of offspring peas will be yellow O C. There is sufficient evidence to support the claim that less than 25% of offspring peas will be yellow, O D. There is sufficient evidence to warrant rejection of the claim that 25% of offspring peas will be yellow.
A genetic experiment involving peas yielded one sample of offspring consisting of 435 green peas and 157 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. Ho p=0.25 H, p>0.25 O C. Ho p=0.25 H p<0.25 O D. Ho p#0.25 H, p=0 25 O E. Ho p=0.25 H, p#0.25 OF. Ho p#0.25 H, p>0 25 What is the test statistic? z = (Round to two decimal places as needed) What is the P.value? P-value = (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 greater than the significance level, a O B. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, a. O C. 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 not sufficient evidence to support the claim that less than 25% of offspring peas will be yellow O B. There is not sufficient evidence to warrant rejection of the claim that 25% of offspring peas will be yellow O C. There is sufficient evidence to support the claim that less than 25% of offspring peas will be yellow, O D. There is sufficient evidence to warrant rejection of the claim that 25% of offspring peas will be yellow.
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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![A genetic experiment involving peas yielded one sample of offspring consisting of 435 green peas and 157 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?
- A. \(H_0: p \neq 0.25\)
\(H_1: p = 0.25\)
- B. \(H_0: p = 0.25\)
\(H_1: p > 0.25\)
- C. \(H_0: p = 0.25\)
\(H_1: p < 0.25\)
- D. \(H_0: p \leq 0.25\)
\(H_1: p > 0.25\)
- E. \(H_0: p = 0.25\)
\(H_1: p \neq 0.25\)
- F. \(H_0: p = 0.25\)
\(H_1: p > 0.25\)
What is the test statistic?
\[ z = \]
(Round to two decimal places as needed.)
What is the P-value?
\[ \text{P-value} = \]
(Round to four decimal places as needed.)
What is the conclusion about the null hypothesis?
- A. Fail to reject the null hypothesis because the P-value is greater than the significance level \( \alpha \).
- B. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level \( \alpha \).
- C. Reject the null hypothesis because the P-value is greater than the significance level \( \alpha \).
- D. Reject the null hypothesis because the P-value is less than or equal to the significance level \( \alpha \).
What is the final conclusion?
- A. There is not sufficient evidence to support the claim that less than 25% of offspring peas will be yellow.
- B. There is not sufficient evidence to warrant rejection of the claim that 25% of offspring](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F81b0e6b6-ec7a-4eb0-91f9-72fff94e773d%2F5dff3b76-a079-47ef-a94a-51d214c0cc31%2Fa4x16e_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A genetic experiment involving peas yielded one sample of offspring consisting of 435 green peas and 157 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?
- A. \(H_0: p \neq 0.25\)
\(H_1: p = 0.25\)
- B. \(H_0: p = 0.25\)
\(H_1: p > 0.25\)
- C. \(H_0: p = 0.25\)
\(H_1: p < 0.25\)
- D. \(H_0: p \leq 0.25\)
\(H_1: p > 0.25\)
- E. \(H_0: p = 0.25\)
\(H_1: p \neq 0.25\)
- F. \(H_0: p = 0.25\)
\(H_1: p > 0.25\)
What is the test statistic?
\[ z = \]
(Round to two decimal places as needed.)
What is the P-value?
\[ \text{P-value} = \]
(Round to four decimal places as needed.)
What is the conclusion about the null hypothesis?
- A. Fail to reject the null hypothesis because the P-value is greater than the significance level \( \alpha \).
- B. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level \( \alpha \).
- C. Reject the null hypothesis because the P-value is greater than the significance level \( \alpha \).
- D. Reject the null hypothesis because the P-value is less than or equal to the significance level \( \alpha \).
What is the final conclusion?
- A. There is not sufficient evidence to support the claim that less than 25% of offspring peas will be yellow.
- B. There is not sufficient evidence to warrant rejection of the claim that 25% of offspring
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