Question

Two different types of polishing solutions are being evaluated for possible use in a tumble-polish operation for manufacturing intraocular lenses used in the human eye following cataract surgery. 300 lenses were tumble-polished using the first polishing solution, and of this number, 255 had no polishing- induced defects. Another 300 lenses were tumble-polished using the second polishing solution, and 196 lenses were satisfactory upon completion. Is there any reason to believe that the two polishing solutions differ? Use ? 0.01, what is the P-value forthis test? There ? significant difference in the proportion of polishing-induced defects produced by the two polishing solutions at the 0.01 level of significa nce The P-value is . Round your answer to three decimal places (e.g. 98.765) Statistical Tables and Charts the tolerance is+/-2%

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P₁ = P₂
Alternative hypothesis: P₁ ? P₂

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the proportion from population 1 is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = (p₁ * n₁ + p₂ * n₂) / (n₁ + n₂)   
p = 0.75167
SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }
SE = 0.03528
z = (p₁ - p₂) / SE

z = 5.57

where p₁ is the sample proportion in sample 1, where p₂ is the sample proportion in sample 2, n₁ is the size of sample 1, and n₂ is the size of sample 2.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than -5.57 or greater than 5.57.

Thus, the P-value = less than 0.0001.

Interpret results. Since the P-value (almost 0) is less than the significance level (0.01), we cannot accept the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that there is significance difference in the proportion of polishing-induced defects produced by two polishing solutions at the 0.01 level of significance.

The p-value is almost 0.