Concept explainers
Testing Claims About Proportions. In Exercises 7–22, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim.
15. Is Echinacea Effective for Colds? Rhinoviruses typically cause common colds. In a test of the effectiveness of echinacea, 40 of the 45 subjects treated with echinacea developed rhinovirus infections. In a placebo group, 88 of the 103 subjects developed rhinovirus infections (based on data from “An Evaluation of Echinacea Angustifolia in Experimental Rhinovirus Infections.” by Turner et al., New England Journal of Medicine, Vol. 353, No. 4). We want to use a 0.05 significance level to test the claim that echinacea has an effect on rhinovirus infections.
a. Test the claim using a hypothesis test.
b. Test the claim by constructing an appropriate confidence interval.
c. Based on the results, does echinacea appear to have any effect on the infection rate?
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ELEMENTARY STATS. 18 WEEK ACCESS CODE
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Identify the null hypothesis, alternative hypothesis, test statistic, P-value or critical value(s), conclusion about the null hypothesis, and final conclusion that addresses the original claim. Cardiac Arrest at Day and Night A study investigated survival rates for in-hospital patients who suffered cardiac arrest. Among 58,593 patients who had cardiac arrest during the day, 11,604 survived and were discharged. Among 28,155 patients who suffered cardiac arrest at night, 4139 survived and were discharged (based on data from “Survival from In-Hospital Cardiac Arrest During Nights and Weekends,” by Peberdy et al., Journal of the American Medical Association, Vol. 299, No. 7). We want to use a 0.01 significance level to test the claim that the survival rates are the same for day and night. a. Test the claim using a hypothesis test. b. Test the claim by constructing an appropriate confidence interval. c. Based on the results, does it appear that for in-hospital patients…arrow_forwardQuestion content area top Part 1 For the α and observed significance level (p-value) pair, indicate whether the null hypothesis would be rejected. α=0.01, p-value=0.10 Question content area bottom Part 1 Choose the correct conclusion below. A. Do not reject the null hypothesis since the p-value is not less than the value of α. B. Reject the null hypothesis since the p-value is less than the value of α. C. Reject the null hypothesis since the p-value is not less than the value of α. D. Do not reject the null hypothesis since the p-value is less than the value ofarrow_forward19% of deaths among male adults can be attributed to heart diseases. Is this percentage different among residents in Sonoma County? State the Null and Alternative hypothses. H0:H0: ? μ p Select an answer > = ≠ < Ha:Ha: ? μ p Select an answer < = > ≠arrow_forwardStudent performance across discussion sections: A professor who teaches a large introductory statistics class (197 students) with eight discussion sections would like to test if student performance differs by discussion section, where each discussion section has a different teaching assistant. The summary table below shows the average final exam score for each discussion section as well as the standard deviation the number of students in each section. Scores and Sec 1 Sec 2 Sec 3 Sec 4 Sec 5 Sec 6 Sec 7 Sec 8 ni 33 19 10 29 33 32 31 92.94 91.11 91.8 92.45 89.3 88.3 90.12 93.45 Si 4.21 5.58 3.43 5.92 9.32 7.27 6.93 4.57 The ANOVA output below can be used to test for differences between the average scores from the different discussion sections. df Sum Sq Mean Sq F value Pr(>F) section 525.01 75 1.87 0.0767 residuals 189 7584.11 40.13 Conduct a hypothesis test to determine if these data provide convincing evidence that the average score varies across some (or all) groups. Check conditions…arrow_forwardPCB contamination of a river by a manufacturer is being measured by amounts of the pollutant found in fish. A company scientist claims that the fish contain only 5 parts per million, but an investigator believes the true figure is higher. What is the conclusion if six fish are caught and show the following amounts of PCB ( in parts per million ) 6.8, 5.6, 5.2, 4.7, 6.3, and 5.4? Test at 5% significance level. In addition to your conclusions, you must provide the null and alternative hypothesis , alpha and the p-value.arrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
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