ouppose that x1 N(1, O1) and x2 * N(2, 02), and that x1 and X2 are independent. Develop a procedure For constructing a 100(1 – a)% confidence interval on µ1 – H2, assuming that o, and o2 are unknown and cannot be assumed equal.
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- A manufacturer of MP3 players conducts a set of comprehensive tests on the electrical functions of its product. All MP3 players must pass all tests prior to being sold. Of a random sample of 900 MP3 players, 18 failed one or more tests. Find a 95% confidence interval for the proportion of MP3 players from the population that pass all tests.Q2: Let x1,X2, . , Xn and y1, y2, ..., Ym represent two independent random samples from the respective normal distributions N(H1,07) and N (H2, 03). It is given that of = 30, but ožis unknown. Then 1. A random variable that can be used to find a 95% confidence interval for - is (x- y) - (H1 - H2) А. ns3 + ms n+m. 3(n+m-2)"tm, nm O A (x – y) – (41 – H2) В. ns? + ms; n + m. (n + m – - 2) пт (x – y) – (41 - H2) С. ns3 + mS;_n+3 3(n+ m-2) ("+3m nm ОсIn random, independent samples of 225 adults and 200 teenagers who watched a certain television show, 112 adults and 138 teens indicated that they liked the show. Let p1 be the proportion of all adults watching the show who liked it, and let p2 be the proportion of all teens watching the show who liked it. Find a 95% confidence interval for −p1p2. Then find the lower limit and upper limit of the 95% confidence interval. Carry your intermediate computations to at least three decimal places. Round your responses to at least three decimal places. a.) The Lower Limit is: b.) The Upper Limit is:11. Consider a random sample Y₁, Y2, ..., Yn from a normal population Y~N(μ, o²) where the population variance and mean are unknown. We want to construct a Σ(X-X)² Show 100(1 a)% confidence interval for the population variance if g² whether or not is a pivotal quantity and construct a 100(1-a) confidence interval.A manufacturer of MP3 players conducts a set of comprehensive tests on the electrical functions of its product. All MP3 players must pass all tests prior to being sold. Of a random sample of 700 MP3 players, 14 failed one or more tests. Find a 99% confidence interval for the proportion of MP3 players from the population that pass all tests.Assume that you have a sample of n₁ = 8, with the sample mean X₁ = 42, and a sample standard deviation of S₁ = 7, and you have an independent sample of n₂ = 13 from another population with a sample mean of X₂ = 33, and the sample standard deviation S₂ = 8. Construct a 99% confidence interval estimate of the population mean difference between μ₁ and μ₂. Assume that the two population variances are equal. ≤11-1₂5 (Round to two decimal places as needed.) (...Let X equal the length of life of a 60-watt light bulb marketed by a certain manufacturer. We do not know the distribution of X except that Var(X) = 1158. Let u be the mean of X. Suppose a random sample of n = 25 bulbs is tested until they burn out, yielding a sample mean of = 1368 hours. (i) Compute an approximate 95% confidence interval for u. (ii) Compute an approximate 95% one-sided confidence interval for that provides a lower bound for u, i.e. compute & such that ɛ P(X - ≤ μµ)≈ 0.95.In random, independent samples of 200 adults and 300 teenagers who watched a certain television show, 124 adults and 153 teens indicated that they liked the show. Let p, be the proportion of all adults watching the show who liked it, and let p, be the proportion of all teens watching the show who liked it. Find a 90% confidence interval for p,-p,. Then find the lower limit and upper limit of the 90% confidence interval. Carry your intermediate computations to at least three decimal places. Round your responses to at least three decimal places. (If necessary, consult a list of formulas.) Lower limit: Upper limit: O OSuppose a marketing company randomly surveyed 404 households and found that in 214 of them, the woman made the majority of the purchasing decisions. Construct a 90% confidence interval for the population proportion of households where the women make the majority of the purchasing decisions.p'=α2=zα2=Margin of Error: E=We are 90% confident that the proportion of households in the population where women make the majority of purchasing decisions is between___ and ___.SEE MORE QUESTIONSRecommended 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. WallnauPublisher:Cengage LearningElementary Statistics: Picturing the World (7th E…StatisticsISBN:9780134683416Author:Ron Larson, Betsy FarberPublisher:PEARSONThe Basic Practice of StatisticsStatisticsISBN:9781319042578Author:David S. Moore, William I. Notz, Michael A. 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