Rhino viruses typically cause common colds. In a test of the effectiveness of echinacea, 40 of the 46 subjects treated with echinacea developed rhinovirus infections. In a placebogroup, 93 of the 110 subjects developed rhinovirus infections. Use a 0.01 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 andalternative hypotheses for the hypothesis test?OA H₂: P₁ = P₂H₁: P₁ P₂O D. Ho: P₁ P₂H₁: P₁ P2Identify the test statistic.Z=(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 istreatment has an effect.OB. Ho: P₁ P₂H₁: P₁ = P₂O E. Ho: P₁ SP₂H₁: P₁ #P₂the significance level of x = 0.01, sob. Test the claim by constructing an appropriate confidence interval.The 99% confidence interval is < (P₁-P₂) <.(Round to three decimal places as needed.)What is the conclusion based on the confidence interval?0, therethe null hypothesis. ThereBecause the confidence interval limitsthe claim that echinacea treatment has an effect.c. Based on the results, does echinacea appear to have any effect on the infection rate?OC. Ho: P₁ = P₂H₁: P₁

Given that 40 of the 46 subjects treated with echinacea develop infection and 93 of the 110 placebo…

Answered: a. Test the claim using a hypothesis…

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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You wish to test the following claim (Ha) at a significance level of a = 0.02. H.:µ1 – µ2 =0 Ha:µ1 – P2 > 0 || You obtain the following two samples of data. Sample #2 Sample #1 61.3 63.5 47.7 51.2 52 39.2 59.7 53.3 54.7 51 54.2 45.5 55.6 46.6 65.3 33 54 48.4 46.6 49.2 65.8 58.6 53.4 49.4 54 54.7 67.7 41.4 47.1 46.1 34 55.4 57.7 55.5 40.7 54.2 Do not test the sample variances using Fcdf. Assume the sample variances are not equal and DO NOT POOL. What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = What is the p-value for this sample? For this calculation, use the degrees of freedom reported from the technology you are using. (Report answer accurate to four decimal places.) p-value = The p-value is... Oless than (or equal to) a Ogreater than a
P VALUE FOR THE TEST HYPOTHESIS
Passed Failed White Results from a civil servant exam are shown in the table to the right. Is there sufficient evidence to support the claim that the results from the test are discriminatory? Use a 0.05 significance level. 16 14 candidates Minority candidates 7 21 Determine the null and alternative hypotheses. O A. Ho: White and minority candidates do not have the same chance of passing the test. H4: White and minority candidates have the same chance of passing the test. B. Ho: A white candidate is more likely to pass the test than a minority candidate. H,:A white candidate is not more likely to pass the test than a minority candidate. Ho: A white candidate is not more likely to pass the test than a minority candidate. H,: A white candidate is more likely to pass the test than a minority candidate. D. Ho: White and minority candidates have the same chance of passing the test. H,: White and minority candidates do not have the same chance of passing the test. Determine the test…
Lillie conducts a hypothesis test and calculates that her sample results have a p-value of 0043, what decision can be made about the null hypothesis? O fail to reject the null O Reject the null
Test the claim that the mean GPA of night students is larger than 2 at the 0.10 significance level. The null and alternative hypothesis would be: Ho p= 0.5 Ho:p ≥ 0.5 Ho:p ≤ 0.5 H₁:p 0.5 H₁:p 0.5 The test is: right-tailed left-tailed two-tailed Based on a sample of 25 people, the sample mean GPA was 2.02 with a standard deviation of 0.07 The p-value is: Based on this we: Ho:μ = 2 Ho:μ ≥ 2 Ho: ≤2 H₁:μ ‡ 2 H₁:μ 2 Reject the null hypothesis O Fail to reject the null hypothesis (to 2 decimals)
Looking at youth depression scores from a general sample of random children in the US, we want to determine if our sample is significantly different than the national average of scores of children with depression in the US. The national average of scores on a measure of depression is 5 points. We have no reason to assume it will be higher or lower than the US national average. The sample: 11 9 12 3 3 4 13 1. What is the null and alternative hypotheses? Do not specify any directionality. 2. What did you discover (using 95% Cl)? Provide an interpretation.
Use technology to find the P-value for the hypothesis test described below. The claim is that for 12 AM body temperatures, the mean is mu greater than 98.6 degrees F. The sample size is n equals 5 and the test statistic is t equals 1.486. P-value equals (Round to three decimal places as needed.)
Please help I need in 1 hour thankyou
The Test statistic test: The P value: The Critical Value: The Decision Rule: The Decision: The Conclusion:
Consider a drug testing company that provides a test for marijuana usage. Among 335 tested subjects, results from 27 subjects were wrong (either a false positive or a false negative). Use a 0.05 significance level to test the claim that less than 10 percent of the test results are wrong. Identify the null and alternative hypotheses for this test. Choose the correct answer below. O A. Ho: p=0.1 H1:p0.1 OC. Ho: p<0.1 H1:p=0.1 O D. Ho: p=0.1 H1:p+0.1
Consider the hypothesis test below. Ho : P1 – P2 0 The following results are for independent samples taken from the two populations. Sample 1 Sample 2 100 n2 = 300 P1 = 0.27 P2 = 0.15 Use pooled estimator of p. a. What is the p-value (to 4 decimals)? Use Table 1 from Appendix B. b. With a = 0.05, what is your hypothesis testing conclusion? Select your answer -

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