In a randomized controlled trial, insecticide-treated bednets were tested as a way to reduce malaria. Among 310 infants using bednets, 10 developed malaria. Among 286 infants not using bednets, 33 developed malaria. Use a 0.01 significance level to test the claim that the incidence of malaria is lower for infants using bednets. a. Test the claim using a hypothesis test. b. Test the claim by constructing an appropriate confidence interval. c. Based on the results, do the bednets appear to be effective? a. What are the null and alternative hypotheses? Let the infants using bednets be sample 1 and let the infants not using bednets be sample 2. Choose the correct hypotheses below. OA. Ho: P₁ P₂ H₁: P₁ P2 O D. Ho: P₁ P₂ H₁: P₁ P2 Identify the test statistic. (Round to two decimal places as needed.) Identify the P-value. OB. Ho: P₁ P2 H₁: P₁ P₂ OE. Ho: P₁ P₂ H₁: P₁
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YES, because 0.0332% is not included in the confidence interval b. no, because 0.0332% is not included in the confidence interval c. yes, because 0.0332% is included in the confidence interval d. no, because 0.0332% is included in the confidence intervalTraffic engineers compared rates of traffic collisions at intersections with raised medians and rates at intersections with two-way left-turn lanes. They found that out of 4,512 collisions at intersections with raised medians, 2,202 were rear-end collisions, and out of 4,277 collisions at two-way left-turn lanes, 1,903 were rear-end collisions.Assuming these to be random samples of collisions from the two types of intersections, construct a 95% confidence interval for the difference between the proportions of collisions that are of the rear-end type at the two types of intersection. Interpret the results. Does the confidence interval contradict the claim that the proportion of rear-end collisions is the same at both types of intersection?A study of 420,092 cell phone users found that 134 of them developed cancer of the brain or nervous system. Prior to this study of cell phone use, the rate of such cancer was found to be 0.0322% for those not using cell phones. Complete parts (a) and (b). a. Use the sample data to construct a 95% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system. %An online site presented this question, "Would the recent norovirus outbreak deter you from taking a cruise?" Among the 34,006 people who responded, 65% answered "yes." Use the sample data to construct a 90% confidence interval estimate for the proportion of the population of all people who would respond "yes" to that question. Does the confidence interval provide a good estimate of the population proportion?A study of 420,082 cell phone users found that 136 of them developed cancer of the brain or nervous system. Prior to this study of cell phone use, the rate of such cancer was found to be 0.0415% for those not using cell phones. Complete parts (a) and (b). a. Use the sample data to construct a 90% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system. %A retail store is considering adding a new product line to its inventory. The store wants to know whether the new product line will be popular with customers. To determine this, they take a random sample of customers and ask if they would be interested in buying the new product line. Of the 125 people they asked, 70 were interested in this new product line. The manager is interested in creating a 90 % confidence interval regarding the population parameter.The Salesforce company of a unicorn has various clients associated .The company wants to estimate the proportion of its clients who are associated with more than one company. A random sample of 600 clients is selected without replacement and 125 say that they are associated with more than one company .a. Construct a 99% confidence interval estimate of the population proportion of the clients who have been associated with more than one companyUse the technology of your choice to obtain the required Tukey multiple-comparison confidence intervals with a family confidence level of 0.95. Hospital Stays. The U.S. National Center for Health Statistics collects data on length of stay in noninstitutional, short-stay hospitals by sex and age. Results are published in Vital and Health Statistics. Independent random samples of Americans were taken to compare the lengths of stay (in days) for males and females in four age groups (15–24 years, 25–34 years, 35–44 years, and 45–64 years). Obtained for a two-way ANOVA on these data and also supplies a table of cell and marginal means.A researcher is interested in estimating the average amount of sleep obtained by first-year students at MacEwan University. The researcher obtains a random sample of 100 first-year students from MacEwan from which she obtains an average of 7.1 hours of sleep. a. Suppose the researcher obtains a 95% confidence interval of (6.9, 3). What is the margin of error? b. It is recommended that young adults sleep at least 7.5 hours per night. Does the interval from (a) provide evidence that, on average, first-year students at MacEwan are under sleeping? Explain c. Is it necessary for the population of interest to be normally distributed for the interval in (a) to be valid? Explain. d. Briefly explain why the interval estimate from (c) is superior to the point estimate.A study of 420,031 cell phone users found that 136 of them developed cancer of the brain or nervous system. Prior to this study of cell phone use, the rate of such cancer was found to be 0.0316% for those not using cell phones. Complete parts (a) and (b). a. Use the sample data to construct a 90% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system. %An October 2011 CBS News Poll on illegal immigrants interviewed 1012 randomly selected American adults. Of those in the sample, 688 said that they “oppose allowing the children of illegal immigrants to attend state college at the lower tuition rate of state residents.” Although the samples in national polls are not SRSs, they are similar enough that our method gives approximately correct confidence intervals. Find the 99% confidence interval for the proportion of adults who “oppose allowing the children of illegal immigrants to attend state college at the lower tuition rate of state residents.” Interpret the confidence interval in a clear sentence. Find the 95% confidence interval. Is the 95% confidence interval longer or shorter than the 99% confidence interval? Find the margin of error for each confidence interval. (Hint: you do have enough information from the calculator output! Margin of error = (width of interval)/2.) Margin of error for 99% confidence…Recommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons IncProbability and Statistics for Engineering and th…StatisticsISBN:9781305251809Author:Jay L. DevorePublisher:Cengage LearningStatistics for The Behavioral Sciences (MindTap C…StatisticsISBN:9781305504912Author:Frederick J Gravetter, Larry B. 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