Question

Problem 4: Woburn MA., A Civi Action' Example A major court case on the health effects of drinking contaminated water took place in the town of Wobu Massachusetts. A town wel in Wobu was contaminated by industrial chemicals. During the period that residents drank water from this we there were 16 birth defects aong 414 births. In years when the contaminated well was shut off and water was supplied from other wells, there were 3 birth defects among 228 births. The plaintiffs suing the firm responsible for the contamination claimed that these data show that the rate of birth defects was higher when the contaminated well was in use. Problem is real and is motivation for the film: 'A Civil Action' (John Travolta, Robert DuVall) phaton<- 16/414 phatoff<-3/228 phatpooled<-(16+3)/(414+228) sediff<-sqrt (phatpooled (1-phatpooled) ((1/414)+(1/228))) zstatistic <- (phaton-phatoff)/sediff #prop, test (c(X1,X2),c(n1,n2), correct = FALSE) (a) Analyze these data with a Z-based hypothesis test. Give all steps and reasoning. ANSWER Hypotheses: . Test Statistic Value: . P-Value . Conclusion: (part b) What assumptions does your analysis require? Do these assumptions seem reasonable in this case? ANSWER-Assumptions: ANSWER-Are assumptions reasonable?:

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 one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. 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 = 0.02960

SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }
SE = 0.01398
z = (p₁ - p₂) / SE

z = 1.82

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 1.82.

Thus, the P-value = 0.0344.

Interpret results. Since the P-value (0.0344) is less than the significance level (0.05), we can reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that rate of birth defects was higher when contaminated well was in use.