An experiment is conducted to compare the maximum load capacity in tons (the maximum weight that can be toleratedwithout breaking) for two alloys A and B. It is known that the two standard deviations in load capacity are equal at 6 tons each. Theexperiment is conducted on 40 specimens of each alloy (A and B) and the results are XA = 62.5, XB = 58.5, and XA -XB = 4. Themanufacturers of alloy A are convinced that this evidence shows conclusively that μA HB and strongly supports the claim that theiralloy is superior. Manufacturers of alloy B claim that the experiment could easily have given XA -XB = 4 even if the two populationmeans are equal. Complete parts (a) and (b) below.Click here to view page 1 of the standard normal distribution table.Click here to view page 2 of the standard normal distribution table.(a) Make an argument that manufacturers of alloy B are wrong. Do it by computing P(XA -XB >4 | μA = HBP(XA-XB > 4) =4

Sample sizes are: nA=40nB=40 Sample means are: x¯A=62.5x¯B=58.5 So, x¯A-x¯B=4 Population standard…

Answered: An experiment is conducted to compare…

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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EXERCISE 1. A random sample of six steel beams has mean compressive strength of 58.392 psi (pounds per square inch) with a standard deviation of s = 648 psi. Test the null hypothesis H, =H = 58,000 psi against the alternative hypothesis H,: H > 58,000 psi at 5% level of significance (value for t at 5 degree of freedom and 5% significance level is 2.0157). Here µ denotes the population mean (A.M.IE., Summer 2000)
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NOTE: PLEASE ANSWER THE LAST ITEMS D TO G.
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