Question

12) Particle removing techniques are of two types gravity settling and filtration. 90 of 100 samples have more than 90 percent removal rate via gravity settling technique, 75 of 100 samples have more than 95 percent removal rate via filtration technique. a) Can you conclude that gravity settling technique is better than filtration at 1% significance? b) What is the 99% confidence interval of the difference between the two techniques?

Answer 1

Solution:-

12

a)

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 one-tailed test.

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 = 0.825

SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }
SE = 0.053735
z = (p₁ - p₂) / SE

z = 2.79

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 one-tailed test, the P-value is the probability that the z-score is greater than 2.79

Thus, the P-value = 0.017.

Interpret results. Since the P-value (0.0003) is less than the significance level (0.01), we have to to reject the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that gravity settling technique is better than filtration at 1% significance.

b) 99% confidence interval of the difference between the two techniques is C.I = ( 0.0116, 0.2884).

Pl (1-PI) . Po(1-P2) 7l2

C.I = (0.90 - 0.75) + 2.576 × 0.053735

C.I = 0.15 + 0.13842

C.I = ( 0.0116, 0.2884)