SOLUTION: Two extrusion machines that manufacture steel rods are being compared. In a sample
of 1000 rods taken from machine 1, 960 met specifications regarding length and
diameter. In a s
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of 1000 rods taken from machine 1, 960 met specifications regarding length and
diameter. In a s
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Question 1204252: Two extrusion machines that manufacture steel rods are being compared. In a sample
of 1000 rods taken from machine 1, 960 met specifications regarding length and
diameter. In a sample of 600 rods taken from machine 2, 582 met the specifications.
Are the two machines equally effective at producing rods that meet the specifications?
Conduct a hypothesis test at 0.01 significance level to reach a conclusion. Use the Pvalue method Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! p1 = 960/1000 = .96
p2 = 582/600 = .97
average of p1 + p2 = (960 + 582) / (1000 + 600) = .96375
call that p0.
standard error = sqrt((p0 * (1-p0) * (1/1000 + 1/600)) = .009652
z-score = (.96 - .97) / .009652 = -1.036055
area to the left of that = .150088
that's the test p-value on the left of the confidence interval.
same area ia on the right for a total test p-value of .300176.
critical p-value is .01.
since the test p-value is greater than the critical p-value, the results are not significant.
the conclusion is that there is not enough evidence to show that the two samples are not equally effectivfe at meeting the specifications.
this means that they can be considered to be equally effective at producing rods that meet the specifications.
online calculator gives the same results plus or minus a few because of rounding, as shown below.
