a. Test the claim using a hypothesis test. Consider the first sample to be the sample of subjects treated with echinacea and the second sample to be the sample of subjects treated with a placebo. What are the null and alternative hypotheses for the hypothesis test? OA. Ho: P1 P2 H₁: P₁ P₂ OD. Ho: P₁ P₂ H₁: P₁ P₂ Identify the test statistic. (Round to two decimal places as needed.) Identify the P-value. P-value= (Round to three decimal places as needed.) What is the conclusion based on the hypothesis test? The P-value is the significance level of a = 0.05, so b. Test the claim by constructing an appropriate confidence interval. The 95% confidence interval is < (P₁-P₂)< (Round to three decimal places as needed.) What is the conclusion based on the confidence interval? Because the confidence interval limits the null hypothesis. There 0, there c. 2. Based on the results, does echinacea appear to have any effect on the infection rate? OB. Ho: P1 P2 H₁: P₁ P₂ OE. Ho: P₁ P2 H₁: P₁ P₂ sufficient evidence to support the claim that echinacea treatment has an effect. appear to be a significant difference between the two proportions. There OA. Echinacea does appear to have a significant effect on the infection rate. There is evidence that it increases the infection rate. OB. Echinacea does not appear to have a significant effect on the infection rate. OC. Echinacea does appear to have a significant effect on the infection rate. There is evidence that it lowers the infection rate. R. The results are inconclusiv OC. Hg: P₁ P₂ H₁: P₁ P₂ OF. Hg: Py SP₂ H₁: P₁ P₂ evidence to support the claim that echinacea treatment has an effect.
a. Test the claim using a hypothesis test. Consider the first sample to be the sample of subjects treated with echinacea and the second sample to be the sample of subjects treated with a placebo. What are the null and alternative hypotheses for the hypothesis test? OA. Ho: P1 P2 H₁: P₁ P₂ OD. Ho: P₁ P₂ H₁: P₁ P₂ Identify the test statistic. (Round to two decimal places as needed.) Identify the P-value. P-value= (Round to three decimal places as needed.) What is the conclusion based on the hypothesis test? The P-value is the significance level of a = 0.05, so b. Test the claim by constructing an appropriate confidence interval. The 95% confidence interval is < (P₁-P₂)< (Round to three decimal places as needed.) What is the conclusion based on the confidence interval? Because the confidence interval limits the null hypothesis. There 0, there c. 2. Based on the results, does echinacea appear to have any effect on the infection rate? OB. Ho: P1 P2 H₁: P₁ P₂ OE. Ho: P₁ P2 H₁: P₁ P₂ sufficient evidence to support the claim that echinacea treatment has an effect. appear to be a significant difference between the two proportions. There OA. Echinacea does appear to have a significant effect on the infection rate. There is evidence that it increases the infection rate. OB. Echinacea does not appear to have a significant effect on the infection rate. OC. Echinacea does appear to have a significant effect on the infection rate. There is evidence that it lowers the infection rate. R. The results are inconclusiv OC. Hg: P₁ P₂ H₁: P₁ P₂ OF. Hg: Py SP₂ H₁: P₁ P₂ evidence to support the claim that echinacea treatment has an effect.
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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![### Evaluating the Effectiveness of Echinacea on Rhinovirus Infections
Rhino viruses typically cause common colds. A study was conducted to test the effectiveness of echinacea in preventing rhinovirus infections. In this study, 38 out of 45 subjects treated with echinacea developed rhinovirus infections, while 96 out of 109 subjects in a placebo group developed rhinovirus infections. A significance level of 0.05 was used to test the claim that echinacea affects rhinovirus infections. Complete parts (a) through (c) below.
#### a. Hypothesis Test
To test the claim, we employed a hypothesis test.
Consider two samples:
- Sample 1: Subjects treated with echinacea
- Sample 2: Subjects treated with placebo
Formulate the null and alternative hypotheses:
- \( H_0: p_1 = p_2 \)
- \( H_1: p_1 \neq p_2 \)
Where:
- \( p_1 \) is the proportion of infections in the echinacea group.
- \( p_2 \) is the proportion of infections in the placebo group.
Identify the test statistic formula:
\[ z = \frac{(\hat{p_1} - \hat{p_2}) - 0}{\sqrt{\hat{p}(1-\hat{p}) (\frac{1}{n1} + \frac{1}{n2})}} \]
(Round to two decimal places as needed.)
#### Identify the P-value:
(P-value to be rounded to three decimal places as needed.)
#### Conclusion Based on Hypothesis Test:
- If the P-value is less than the significance level (α = 0.05), we reject the null hypothesis, thereby providing sufficient evidence to support the claim that echinacea treatment has an effect.
#### b. Confidence Interval
Construct a 95% confidence interval for the difference between the two proportions.
\[ \text{The 95% confidence interval} = (\hat{p_1} - \hat{p_2}) \pm Z_{\alpha/2} \sqrt{\hat{p}(1-\hat{p})(\frac{1}{n1} + \frac{1}{n2})} \]
(Round to three decimal places as needed.)
#### Conclusion Based on Confidence Interval:
If the confidence interval does not contain zero, there is evidence suggesting](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5e4e5146-969d-4643-8e1a-977cd9815aba%2Ff0fad29f-2558-42f0-8082-a2a58c9b0600%2F367or9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Evaluating the Effectiveness of Echinacea on Rhinovirus Infections
Rhino viruses typically cause common colds. A study was conducted to test the effectiveness of echinacea in preventing rhinovirus infections. In this study, 38 out of 45 subjects treated with echinacea developed rhinovirus infections, while 96 out of 109 subjects in a placebo group developed rhinovirus infections. A significance level of 0.05 was used to test the claim that echinacea affects rhinovirus infections. Complete parts (a) through (c) below.
#### a. Hypothesis Test
To test the claim, we employed a hypothesis test.
Consider two samples:
- Sample 1: Subjects treated with echinacea
- Sample 2: Subjects treated with placebo
Formulate the null and alternative hypotheses:
- \( H_0: p_1 = p_2 \)
- \( H_1: p_1 \neq p_2 \)
Where:
- \( p_1 \) is the proportion of infections in the echinacea group.
- \( p_2 \) is the proportion of infections in the placebo group.
Identify the test statistic formula:
\[ z = \frac{(\hat{p_1} - \hat{p_2}) - 0}{\sqrt{\hat{p}(1-\hat{p}) (\frac{1}{n1} + \frac{1}{n2})}} \]
(Round to two decimal places as needed.)
#### Identify the P-value:
(P-value to be rounded to three decimal places as needed.)
#### Conclusion Based on Hypothesis Test:
- If the P-value is less than the significance level (α = 0.05), we reject the null hypothesis, thereby providing sufficient evidence to support the claim that echinacea treatment has an effect.
#### b. Confidence Interval
Construct a 95% confidence interval for the difference between the two proportions.
\[ \text{The 95% confidence interval} = (\hat{p_1} - \hat{p_2}) \pm Z_{\alpha/2} \sqrt{\hat{p}(1-\hat{p})(\frac{1}{n1} + \frac{1}{n2})} \]
(Round to three decimal places as needed.)
#### Conclusion Based on Confidence Interval:
If the confidence interval does not contain zero, there is evidence suggesting
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