eenThe P-value isthe significance level of a = 0.05, sothe null hypothesis. Theresufficient 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< (P1 - P2) <]:(Round to three decimal places as needed.)eenWhat is the conclusion based on the confidence interval?Because the confidence interval limits0, thereappear to be a significant difference between the two proportions. Thereevidence to support the claim that echinacea treatment hasan effect.c. Based on the results, does echinacea appear to have any effect on the infection rate?eenO A. Echinacea does appear to have a significant effect on the infection rate. There is evidence that it lowers the infection rate.B. Echinacea does not appear to have a significant effect on the infection rate.O C. Echinacea does appear to have a significant effect on the infection rate. There is evidence that it increases the infection rate.O D. The results are inconclusive.Click to soloot VOurpoworlo) Rhino viruses typically cause common colds. In a test of the effectiveness of echinacea, 37 of the 42 subjects treated with echinacea developed rhinovirus infections. In a placebo group, 84 of the 101 subjectsdeveloped rhinovirus infections. Use 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 thehypothesis test?O A. Ho: P1 P2H4: P1 =P2O B. Ho: P1 2 P2H,:P, #P2C. Ho: P1 =P2%3DH4: P, # P2O D. Ho: P1 = P2H1: P1 P2Identify the test statistic.(Round to two decimal places as needed.)Identify the P-value.P-value =(Round to three decimal places as needed.)

Name: eenThe P-value isthe significance level of a = 0.05, sothe null hypot
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Description: eenThe P-value isthe significance level of a = 0.05, sothe null hypothesis. Theresufficient evidence to support the claim that echinacea treatment has an effect.b. Test the claim by constructing an appropriate confidence interval.The 95% confidence

State the hypotheses. Let p1 denotes the population proportion of the subjects treated with Echinacea developed rhinovirus infections. Let p1 denotes the population proportion of the subjects treated with placebo group developed rhinovirus infections. Correct option: Option CObtain the value of the test statistic. The value of test statistic is obtained below: The value of sample proportion1, Thus, the value of test statistic is 0.74. Obtain the P-value. Use Excel to obtain the probability value for Z=0.74 Follow the instruction to obtain the P-value: Open EXCEL Go to Formula bar. In formula bar enter the function as“=NORMSDIST” Enter the test statistic Z as 0.74. Click enter EXCEL output: From the Excel output, the P-value is 0.7704 The P-value in right tail is 0.2296(=1–0.7704) The P-value in two tail is 0.4592(=2*0.2296) Thus, the P-value is 0.459. The conclusion is obtained as shown below: Use the level of significance is 0.05. The p-value is greater than level of significance. By rejection rule, do not reject the null hypothesis. That is, there is no evidence to conclude that Echinacea has an effect on rhinovirus infections. The P-value is greater than the significance level of α=0.05, so fail to reject the null hypothesis. There is no sufficient evidence to support the claim that the Echinacea treatment has an effect.b. Obtain the 95% confidence interval for the difference between the two population proportions. The 95% confidence interval for the difference between the two proportions is obtained below: The formula for the confidence interval is, Obtain the critical value of z at the 95% confidence level. Here, confidence level is 0.95. Use EXCEL Procedure for finding the critical value of z. Follow the instruction to obtain the critical value of z: Open EXCEL Go to Formula bar. In formula bar enter the function as“=NORMINV” Enter the probability as 0.975 Enter the mean as 0 and the Standard deviation as 1. Click enter EXCEL output: From the EXCEL output, the critical value of z at the 95% confidence level is 1.96. The required confidence interval is, Thus, the 95% confidence interval for the difference between the two proportions lies between ––0.073 and 0.171. Because the confidence interval limits contains 0, there it do not appear to be a significant difference between the two proportions. There is no evidence to support the claim that Echinacea treatment has an effect. (c ) Determine whether the Echinacea appears to have any effect on the infection rate or not. Since, the confidence interval contains 0, it can be concluded that Echinacea does not appear to have a significant effect on the infection rate. Correct option: Option B

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Statistics

Rhino viruses typically cause common colds. In a test of the effectiveness of echinacea, 37 of the 42 subjects treated with echinacea developed rhinovirus infections. In a placebo group, 84 of the 101 subjects developed rhinovirus infections. Use 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? OA. Ho: P1 #P2 H: P, =P2 O B. Ho: P1 2 P2 H: P, #P2 OC. Ho: P1 =P2 H4: P1 + P2 OD. Ho: P1 = P2 H: P, 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.)

Rhino viruses typically cause common colds. In a test of the effectiveness of echinacea, 37 of the 42 subjects treated with echinacea developed rhinovirus infections. In a placebo group, 84 of the 101 subjects developed rhinovirus infections. Use 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? OA. Ho: P1 #P2 H: P, =P2 O B. Ho: P1 2 P2 H: P, #P2 OC. Ho: P1 =P2 H4: P1 + P2 OD. Ho: P1 = P2 H: P, 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.)

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

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Concept explainers

2-Proportion Test

In a 2-proportion test, two separate proportions are measured at the same time, and their similarity is evaluated. It is also known as a 2-proportion z-test, and a comparison of two different proportions is done in this method.

Topic Video

Question

een
The P-value is
the significance level of a = 0.05, so
the null hypothesis. There
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< (P1 - P2) <]:
(Round to three decimal places as needed.)
een
What is the conclusion based on the confidence interval?
Because the confidence interval limits
0, there
appear to be a significant difference between the two proportions. There
evidence to support the claim that echinacea treatment has
an effect.
c. Based on the results, does echinacea appear to have any effect on the infection rate?
een
O A. Echinacea does appear to have a significant effect on the infection rate. There is evidence that it lowers the infection rate.
B. Echinacea does not appear to have a significant effect on the infection rate.
O C. Echinacea does appear to have a significant effect on the infection rate. There is evidence that it increases the infection rate.
O D. The results are inconclusive.
Click to soloot VOur
poworlo)

Rhino viruses typically cause common colds. In a test of the effectiveness of echinacea, 37 of the 42 subjects treated with echinacea developed rhinovirus infections. In a placebo group, 84 of the 101 subjects
developed rhinovirus infections. Use 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?
O A. Ho: P1 P2
H4: P1 =P2
O B. Ho: P1 2 P2
H,:P, #P2
C. Ho: P1 =P2
%3D
H4: P, # P2
O D. Ho: P1 = P2
H1: P1 <P2
O E. Ho: P1 SP2
H,:P, #P2
O F. 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.)

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