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
Uploaded: 2024-03-20T07:07:00.000Z
Channel: Bartleby
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…

Math

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

Problem 1P

See similar textbooks

Similar questions

For testing the second figure in the report, let p be the proportion of email users who feel that spam has increased in their work email. The second set of hypotheses that the students wants to test, is then: Ho: p = 0.29 Ha: p ≠ 0.29 Based upon the data collected by this student, a 95% confidence interval for p was found to be: (0.273, 0.304). At the 0.05 significance level, which conclusion can be made? a. The student should not reject Ho, since 0.29 falls within the confidence interval. b. The student should reject Ho, since 0.29 falls within the confidence interval. c. The student should not reject Ho, since 0.29 does not fall within the confidence interval d. The student should reject Ho, since 0.29 does not fall within the confidence interval.
A drug company claims that less than 8% of patients who take drug A experience dizzy spells. To check this claim, scientists give the drug to a SRS of 400 patients. In all, 35 patients in the sample had dizzy spells. Test this claim at the 0.05 significance level. When considering a confidence interval, remember PHANTOMS! Parameter: - p (the true proportion of) = _____ Hypotheses: H0 = _____ Ha = _____ Assess Conditions: Random - text says SRS Normal: Is np≥10andn(1−p)≥10 Yes or No? _____ Independent: It is fair to assume there are more than 4000 patients taking drug A Name: One-sample z-test for p Test Statistic: z = _____ Obtain your p-value (Use Desmos): p−value = Make a decision: Because our p-value is = _____ State your conclusion: _____
Is there a relationship between confidence intervals and two-tailed hypothesis tests? Let c be the level of confidence used to construct a confidence interval from sample data. Let ? be the level of significance for a two-tailed hypothesis test. The following statement applies to hypothesis tests of the mean. For a two-tailed hypothesis test with level of significance ? and null hypothesis H0: μ = k, we reject H0 whenever k falls outside the c = 1 – α confidence interval for μ based on the sample data. When k falls within the c = 1 – α confidence interval, we do not reject H0. (A corresponding relationship between confidence intervals and two-tailed hypothesis tests also is valid for other parameters, such as p, μ1 − μ2, or p1 − p2, which we will study in later sections.) Whenever the value of k given in the null hypothesis falls outside the c = 1 – α confidence interval for the parameter, we reject H0. For example, consider a two-tailed hypothesis test with ? = 0.01 and H0: μ = 21H1:…
According to a study completed in 2006 by Pew Internet, 42% of all Americans had a broadband Internet connection at home. This same statistics student wanted to see if this percentage is different for students at his university. Ho: p = 0.42 Ha: p ≠ 0.42 Based upon the data the student collected, a 95% confidence interval for p was found to be: (0.439, 0.457). At the 0.05 level of significance, which conclusion can be made? a. The student should reject the null hypothesis, because 0.42 is a possible value for p given the data collected. b. The student should not reject the null hypothesis, because 0.42 is not a possible value for p given the data collected. c. The student should reject the null hypothesis, because 0.42 is not a possible value for p given the data collected. d. The student should not reject the null hypothesis, because 0.42 is a possible value for p given the data collected.
Here are the summary statistics for randomly selected weights of newborn girls: n =247, x= 32.2 hg, s= 6.8hg . Construct a confidence interval estimate of the mean. Use 90% confidence level. Are these results very different from the confidence interval 31.0 hg < u < 33.2 hg with only 18 sample values, x= 32.1 hg and s=. 2.7 hg ? What is the confidence interval for the population mean u ? _ hg < u< hg _ (round to one decimal places as needed ? Are the results between the two confidence intervals very different ? a. Yes, because the confidence interval limits are not similar. b. No. because the confidence interval limits are similarc. Yes, because one confidence interval does not contain the mean of the other confidence intervald.
Researchers polled residents in Kent and Sussex counties and asked if they were in favor of increasing funding for the arts. In a random sample of 100 residents from Sussex County, 22% supported the increase and in a random sample of 100 residents from Kent County, 26% supported the increase. A 95% confidence interval for the difference (ps - pk) was calculated as -0.04 ± 0.12. Which of the following is the best interpretation of the interval? ± −0.04±0.12 a We are 95% confident that the difference in the sample proportions of voters in districts Sussex County and Kent County who favor an increase in state spending for the arts is between −0.16 and 0.08. b We are 95% confident that the difference in the proportions of all voters in Sussex County and Kent County who favor an increase in state spending for the arts is between −0.16 and 0.08. c We are 95% confident that the majority of all voters in the state favor an increase in state spending for the arts.…
Researchers polled residents in Kent and Sussex counties and asked if they were in favor of increasing funding for the arts. In a random sample of 100 residents from Sussex County, 22% supported the increase and in a random sample of 100 residents from Kent County, 26% supported the increase. A 95% confidence interval for the difference (ps - pk) was calculated as -0.04 ± 0.12. Which of the following is the best interpretation of the interval? ± −0.04±0.12 a We are 95% confident that the difference in the sample proportions of voters in districts Sussex County and Kent County who favor an increase in state spending for the arts is between −0.16 and 0.08. b We are 95% confident that the difference in the proportions of all voters in Sussex County and Kent County who favor an increase in state spending for the arts is between −0.16 and 0.08. c We are 95% confident that the majority of all voters in the state favor an increase in state spending for the arts.…
Given poll results from 1000 people of whom 14% indicated their favorite pie was chocolate, compare the width of a 95% confidence interval with the width of an 80% confidence interval. 1. What is the sample size n? 2. What is the sample proportion p-hat? 3. For the 95% confidence interval, a = 0.05; a/2=0.025. What is the critical value Z a/2 ? 4. Calculate the margin of error E. What is it? 5. p-hat plus or minus E gives us the confidence interval. What is it? How wide is it?
A November 2018 CBS News Poll found that 35% of its sample planned to use online retailers for most holiday shopping. Give a 95% confidence interval for the proportion of all adults who plan to shop in this way, assuming that the result ˆp=0.35 comes from a sample size n= 650 n=1800 n=500 Explain briefly what your results show about the effect of increasing the size of a sample on the width of the confidence interval.
A small college is concerned that their online classes are less successful than their face-to-face classes. To test this claim, the college randomly selects students from their online and face-to-face classes, gathers their average grades, and finds the summary statistics in the table below. Construct and interpret a 99% confidence interval on the difference between the means of the two courses using the information below. Online Grades Face-to-Face Grades MEAN (?) 75.2 77.8 SD (s) 3.99 4.32 COUNT (n) 27 29
A random sample of 100 people from Country S had 15 people with blue eyes. A separate random sample of 100 people from Country B had 25 people with blue eyes. Assuming all conditions are met, which of the following is a 95 percent confidence interval to estimate the difference in population proportions of people with blue eyes (Country S minus Country B) ? (−0.01,0.21) A (−0.15,−0.05) B (−0.19,−0.01) C (−0.21,0.01) D (−0.24,0.04)
3. In a random sample of 300 persons eating lunch at a Lexington pizzeria, 270 ordered pizza. a) Find and interpret a 95% confidence interval for the true proportion of people who eat pizza at this Lexington pizzeria? b) If we take 270/300 = 0.90 as an estimate of the corresponding true proportion, with what confidence can we assert that our error is less than 0.04?