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A genetic experiment involving peas yielded one sample of offspring consisting of 444 green peas and 126 yellow peas. Use a 0.01 significance level to test the claim that under the same circumstances, 23% 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=0.23 OB. Ho: p=0.23 H₁: p=0.23 H₁: p>0.23 OC. Ho: p=0.23 O D. Ho: p=0.23 H₁: p<0.23 H₁: p<0.23 O E. Ho: p0.23 OF. Ho: p0.23 H₁: p=0.23 H₁: p>0.23 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. Reject the null hypothesis because the P-value is greater than the significance level, a. OB. Fail to 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 less than or equal to the significance level, c. 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 support the claim that less than 23% of offspring peas will be yellow. OB. There is sufficient evidence to warrant rejection of the claim that 23% of offspring peas will be yellow. OC. There is not sufficient evidence to warrant rejection of the claim that 23% of offspring peas will be yellow. O D. There is not sufficient evidence to support the claim that less than 23% of offspring peas will be yellow.

A genetic experiment involving peas yielded one sample of offspring consisting of 444 green peas and 126 yellow peas. Use a 0.01 significance level to test the claim that under the same circumstances, 23% 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=0.23 OB. Ho: p=0.23 H₁: p=0.23 H₁: p>0.23 OC. Ho: p=0.23 O D. Ho: p=0.23 H₁: p<0.23 H₁: p<0.23 O E. Ho: p0.23 OF. Ho: p0.23 H₁: p=0.23 H₁: p>0.23 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. Reject the null hypothesis because the P-value is greater than the significance level, a. OB. Fail to 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 less than or equal to the significance level, c. 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 support the claim that less than 23% of offspring peas will be yellow. OB. There is sufficient evidence to warrant rejection of the claim that 23% of offspring peas will be yellow. OC. There is not sufficient evidence to warrant rejection of the claim that 23% of offspring peas will be yellow. O D. There is not sufficient evidence to support the claim that less than 23% of offspring peas will be yellow.

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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**Statistical Hypothesis Testing Using P-Values: A Case Study with Pea Plants** ### Experimental Background A genetic experiment involving peas yielded a sample of offspring consisting of 444 green peas and 126 yellow peas. The experiment aims to test the claim that, under the same circumstances, 23% of offspring peas will be yellow. A significance level (α) of 0.01 is used to evaluate this claim. By employing the P-value method and normal distribution for binomial approximation, we will: 1. Identify the null and alternative hypotheses. 2. Calculate the test statistic. 3. Determine the P-value. 4. Draw conclusions about the null hypothesis. 5. Formulate the final conclusion concerning the original claim. ### Hypothesis Formulation Identify the null (H₀) and alternative (H₁) hypotheses: - **Option A**: H₀: p = 0.23, H₁: p > 0.23 - **Option B**: H₀: p = 0.23, H₁: p ≠ 0.23 - **Option C**: H₀: p ≠ 0.23, H₁: p < 0.23 - **Option D**: H₀: p = 0.23, H₁: p < 0.23 - **Option E**: H₀: p ≠ 0.23, H₁: p > 0.23 - **Option F**: H₀: p ≠ 0.23, H₁: p = 0.23 ### Test Statistic Calculate the test statistic (z) and round to two decimal places: - \( z = \_\_\_\_ \) ### P-Value Calculation Determine the P-value, rounding to four decimal places: - P-value = \_\_\_\_ ### Null Hypothesis Conclusion Draw a conclusion about the null hypothesis: - **Option A**: Reject the null hypothesis because the P-value is greater than the significance level, α. - **Option B**: Fail to reject the null hypothesis because the P-value is greater than the significance level, α. - **Option C**: Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, α. - **Option D**: Reject the null hypothesis because the
Transcribed Image Text:**Statistical Hypothesis Testing Using P-Values: A Case Study with Pea Plants** ### Experimental Background A genetic experiment involving peas yielded a sample of offspring consisting of 444 green peas and 126 yellow peas. The experiment aims to test the claim that, under the same circumstances, 23% of offspring peas will be yellow. A significance level (α) of 0.01 is used to evaluate this claim. By employing the P-value method and normal distribution for binomial approximation, we will: 1. Identify the null and alternative hypotheses. 2. Calculate the test statistic. 3. Determine the P-value. 4. Draw conclusions about the null hypothesis. 5. Formulate the final conclusion concerning the original claim. ### Hypothesis Formulation Identify the null (H₀) and alternative (H₁) hypotheses: - **Option A**: H₀: p = 0.23, H₁: p > 0.23 - **Option B**: H₀: p = 0.23, H₁: p ≠ 0.23 - **Option C**: H₀: p ≠ 0.23, H₁: p < 0.23 - **Option D**: H₀: p = 0.23, H₁: p < 0.23 - **Option E**: H₀: p ≠ 0.23, H₁: p > 0.23 - **Option F**: H₀: p ≠ 0.23, H₁: p = 0.23 ### Test Statistic Calculate the test statistic (z) and round to two decimal places: - \( z = \_\_\_\_ \) ### P-Value Calculation Determine the P-value, rounding to four decimal places: - P-value = \_\_\_\_ ### Null Hypothesis Conclusion Draw a conclusion about the null hypothesis: - **Option A**: Reject the null hypothesis because the P-value is greater than the significance level, α. - **Option B**: Fail to reject the null hypothesis because the P-value is greater than the significance level, α. - **Option C**: Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, α. - **Option D**: Reject the null hypothesis because the
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