A light bulb manufacturer wants to compare the mean lifetimes of two of its light bulbs, model A and model B. Independent random samples of the two modelswere taken. Analysis of 15 bulbs of model A showed a mean lifetime of 1348 hours and a standard deviation of 81 hours. Analysis of 11 bulbs of model Bshowed a mean lifetime of 1351 hours and a standard deviation of 109 hours. Assume that the populations of lifetimes for each model are normally distributedand that the variances of these populations are equal. Construct a 99% confidence interval for the difference H-, between the mean lifetime u, of model Abulbs and the mean lifetime u, of model B bulbs. Then find the lower limit and upper limit of the 99% confidence interval,Carry your intermediate computations to at least three decimal places. Round your responses to at least two decimal places. (If necessary, consult a list offormulas.)Lower limit:Upper limit:ExplanationCheck2021 McGraw-Hill Education. Al Rights Reserved Terms of UsePrivacy I AccessiblMacBook Air888 4FSF2F3@232$%&234589QWRYSDFH.KCVBaltcommandcommandoptionV

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A light bulb manufacturer wants to compare the mean lifetimes of two of its light bulbs, model À and model B. Independent random samples of the two model were taken. Analysis of 15 bulbs of model A showed a mean lifetime of 1348 hours and a standard deviation of 81 hours. Arnalysis of 11 bulbs of model B showed a mean lifetime of 1351 hours and a standard deviation of 109 hours. Assume that the populations of lifetimes for each model are normally distributed and that the variances of these populations are equal. Construct a 99% confidence interval for the difference u, -h between the mean lifetime p, of model A bulbs and the mean lifetime u, of model B bulbs. Then find the lower limit and upper limit of the 99% confidence interval. Carry your intermediate computations to at least three decimal places. Round your responses to at least two decimal places. (If necessary, consult a list of formulas.) Lower limit: x 5 ? Upper limit:

A light bulb manufacturer wants to compare the mean lifetimes of two of its light bulbs, model À and model B. Independent random samples of the two model were taken. Analysis of 15 bulbs of model A showed a mean lifetime of 1348 hours and a standard deviation of 81 hours. Arnalysis of 11 bulbs of model B showed a mean lifetime of 1351 hours and a standard deviation of 109 hours. Assume that the populations of lifetimes for each model are normally distributed and that the variances of these populations are equal. Construct a 99% confidence interval for the difference u, -h between the mean lifetime p, of model A bulbs and the mean lifetime u, of model B bulbs. Then find the lower limit and upper limit of the 99% confidence interval. Carry your intermediate computations to at least three decimal places. Round your responses to at least two decimal places. (If necessary, consult a list of formulas.) Lower limit: x 5 ? Upper limit:

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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Suppose there are two different vaccines for Covid, Vaccine X and Vaccine Y. An interesting question is which vaccine has a higher 6-month antibody effectiveness quotient (6AEQ). To examine this we randomly select 78 recipients of vaccine X and 93 recipients on vaccine Y. The vaccine X recipients had a mean 6AEQ of x = 151. The vaccine Y recipients had a mean 6AEQ of y = 148. It is recognized that the true standard deviation of 6AEQ for vaccine X recipients is o, = 9.7 while it is recognized that the true standard deviation of 6AEQ for vaccine Y recipients is o, = 10.5. The true (unknown) mean SAEQ for vaccine X recipients is ly, while the true (unknown) mean 6AEQ for vaccine Y recipients is 4,. 6AEQ measurements are known to be a normally distributed. summary: Sample Size Sample Mean Standard Deviation 9.7 Гуре Vaccine 78 151 Vaccine Y 93 148 10.5 What is the length of your 96% confidence interval for uy Hy ? ) If we used this data to test Hg: Hy - Hy =0 against the alternative H: uy…