Rhino viruses typically cause common colds. In a test of the effectiveness of echinacea, 43 of the 48 subjects treated with echinacea developed rhinovirus infections. In a placebo group, 80 of the 96 subjects developed rhinovirus infectic a 0.05 significance level to test the claim that echinacea has an effect on rhinovirus infections. Complete parts (a) through (c) below. IT 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? OC. Ho: P₁ P2 OA. Ho: P₁ P2 P2 P2 H₁: P₁ P₂ OB. Ho: P₁ H₁: P₁ OE Ho: P₁ H₁: P₁ P2 D. Ho: P₁ P2 P₂ H₁: P₁ P₂ OF. Ho: P₁ SP₂ H₁: P₁ P2 H₁: P₁ P₂ Identify the test statistic. z = 1.01 (Round to two decimal places as needed.) Identify the P-value. P-value=312 (Round to three decimal places as needed.) What is the conclusion based on the hypothesis test? The P-value is greater than the significance level of a=0.05, so fail to reject the null hypothesis. There is not sufficient evidence to support the claim that echinacea treatment has an effect. b. Test the claim by constructing an appropriate confidence interval. The 95% confidence interval is -499 (P₁-P2) <.624. (Round to three decimal places as needed.) Help me solve this Media. View an example Get more help. [0 Clear all Fin
Rhino viruses typically cause common colds. In a test of the effectiveness of echinacea, 43 of the 48 subjects treated with echinacea developed rhinovirus infections. In a placebo group, 80 of the 96 subjects developed rhinovirus infectic a 0.05 significance level to test the claim that echinacea has an effect on rhinovirus infections. Complete parts (a) through (c) below. IT 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? OC. Ho: P₁ P2 OA. Ho: P₁ P2 P2 P2 H₁: P₁ P₂ OB. Ho: P₁ H₁: P₁ OE Ho: P₁ H₁: P₁ P2 D. Ho: P₁ P2 P₂ H₁: P₁ P₂ OF. Ho: P₁ SP₂ H₁: P₁ P2 H₁: P₁ P₂ Identify the test statistic. z = 1.01 (Round to two decimal places as needed.) Identify the P-value. P-value=312 (Round to three decimal places as needed.) What is the conclusion based on the hypothesis test? The P-value is greater than the significance level of a=0.05, so fail to reject the null hypothesis. There is not sufficient evidence to support the claim that echinacea treatment has an effect. b. Test the claim by constructing an appropriate confidence interval. The 95% confidence interval is -499 (P₁-P2) <.624. (Round to three decimal places as needed.) Help me solve this Media. View an example Get more help. [0 Clear all Fin
MATLAB: An Introduction with Applications
6th Edition
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Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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Find part B

Transcribed Image Text:What is the conclusion based on the confidence interval?
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Transcribed Image Text:esc
Rhino viruses typically cause common colds. In a test of the effectiveness of echinacea, 43 of the 48 subjects treated with echinacea developed rhinovirus infections. In a placebo group, 80 of the 96 subjects developed rhinovirus infectia
a 0.05 significance level to test the claim that echinacea has an effect on rhinovirus infections. Complete parts (a) through (c) below.
...
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?
OC. Ho: P₁ P2
OA. Ho: P₁ = P2
H₁: P₁ P2
OB. Ho: P1
H₁: P₁
P2
P2
H₁: P₁ P₂
Ho: P₁ = P2
OE. Ho: P₁
P2
H₁: P₁ = P2
OF. Ho: P₁ ≤P2
H₁: P₁ P2
H₁: P₁ P2
Identify the test statistic.
z = 1.01
(Round to two decimal places as needed.)
Identify the P-value.
P-value = .312
(Round to three decimal places as needed.)
What is the conclusion based on the hypothesis test?
The P-value is greater than the significance level of a = 0.05, so fail to reject the null hypothesis. There is not sufficient evidence to support the claim that echinacea treatment has an effect.
b. Test the claim by constructing an appropriate confidence interval.
The 95% confidence interval is.499 < (P₁-P2) <.624
(Round to three decimal places as needed.)
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Transcribed Image Text:### Hypothesis Test Conclusion
#### What is the conclusion based on the hypothesis test?
The P-value is **greater than** the significance level of \(\alpha = 0.05\), so we **fail to reject** the null hypothesis. There **is not** sufficient evidence to support the claim that echinacea treatment has an effect.
#### Confidence Interval Construction
##### b. Test the claim by constructing an appropriate confidence interval.
The 95% confidence interval is \(-0.052 < (p_1 - p_2) < 0.177\). (Round to three decimal places as needed).
#### Conclusion Based on Confidence Interval
What is the conclusion based on the confidence interval?
Because the confidence interval limits **include** 0, there **does not** appear to be a significant difference between the two proportions. There **is not** evidence to support the claim that echinacea treatment has an effect.
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This conclusion implies that when evaluating the impact of echinacea treatment, statistical evidence is insufficient under the given conditions to confirm its effectiveness. The hypothesis test's P-value and confidence interval suggest that the observable differences might be due to random variation rather than a real effect.
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