An experiment is conducted to compare the maximum load capacity in tons (the max

Question

An experiment is conducted to compare the maximum load capacity in tons (the maximum weight that can be tolerated without breaking) for two alloys A and B. It is known that the two standard deviations in load capacity are equal at 5 tons each. The experiment is conducted on 50 specimens of each alloy (A and B) and the results are x,-46.3, xB-43.3, and A-XB-3. The manufacturers of alloy A are convinced that this evidence shows conclusively that A B and strongly supports the claim that their alloy is superior. Manufacturers of alloy B claim that the experiment could easily have given XA-XB-3 even if the two population means 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. weightthat can be thatt eir46.3,h" 43.3, and, % ons Theamanpacitya e eq alat (a) Make an argument that manufacturers of alloy B are wrong. Do it by computing P |XA-XB-3 I ??-?? (b) Do you think these data strongly support alloy A? Since the probability in (a) negligible, the experiment alloy A

Explanation / Answer

let XA denotes the maximum load capacity of alloy A

XB denotes the same for alloy B

assumption is XA~N(uA,52) and XB~N(uB,52) independently

there is a random sample of n=50 from each alloy

XAbar be the mean of those 50 samples for alloy A

XBbar be the mean of those 50 samples for alloy B

so XAbar~N(uA,52/50) XBbar~N(uB,52/50) independently

so XAbar-XBbar~N(uA-uB,52/50+52/50)

lets take the null hypothesis is H0: uA=uB and alternative hypothesis is H1: uA>uB

a) P[XAbar-XBbar>3|uA=uB] now under uA=uB XAbar-XBbar~N(0,1)

=0.001349898 [answer]

b) the above probaility is very low. so under uA=uB , the chance that XAbar-XBbar>3 is very low.

so the data strongly suggests that uA=uB is less feasible than uA>uB

so

SInce the probability in a is negligible, the experiment supports the claim of manufacturer of allow A

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