Two different types of polishing solutions are being evaluated for possible use
ID: 3134565 • Letter: T
Question
Two different types of polishing solutions are being evaluated for possible use in a tumble-polish operation for manufacturing interocular lenses used in the human eye following cataract surgery. Three hundred lenses were tumble polished using the first polishing solution, and of this number 246 had no polishing-induced defects. Another 300 lenses were tumble-polished using the second polishing solution, and 191 lenses were satisfactory upon completion. Is there any reason to believe that the two polishing solutions differ? Use alpha = 0.01. what is the P -value for this test? significant difference in the fraction of polishing-induced defects produced by the two polishing solutions at the 0.01 level of significance. The P -value is Round your answer to four decimal places (e.g. 98.7654).Explanation / Answer
Formulating the hypotheses
Ho: p1 - p2 = 0
Ha: p1 - p2 =/= 0
Here, we see that pdo = 0 , the hypothesized population proportion difference.
Getting p1^ and p2^,
p1^ = x1/n1 = 0.82
p2 = x2/n2 = 0.636666667
Also, the standard error of the difference is
sd = sqrt[ p1 (1 - p1) / n1 + p2 (1 - p2) / n2] = 0.035539753
Thus,
z = [p1 - p2 - pdo]/sd = 5.158542635
As significance level = 0.01 , then the critical z is
zcrit = 2.575829304
Also, the p value is
P = 2.48879*10^-7 [ANSWER, P VALUE]
As P < 0.05, then we REJECT THE NULL HYPOTHESIS.
Hence,
There [[IS]] significant evidence in the fraction of polishing induced defects produced by the two polishing solutions at the 0.01 level of significance. The P value is [[2.48879*10^-7 or 0.0000002488]]. [ANSWER]
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