Question

An article published in USA today stated that “in a study of colorectal surgery patients; 104 were kept warm with blankets & 96 were kept cool. The results show that only 6 of those “warmed patients” developed wound infection vs. 18 who were kept cool. Let x = number of colorectal surgery patients who developed wound infection At the 0.02 level of significance, test the claim of the article’s headline: Warmer Surgical Patients Recover Better (higher recovery rate for Warmer Surgical Patients)

Sample 1

Sample 2

Claim: H0: H1:

2- Test Statistic:

3- Critical Region/Critical Value:

4- Decision about H0:

Answer 1

Solution:-

P₁ = Proportion of Warmer Surgical Patients developed infection

P₁ = Proportion of Cold Surgical Patients developed infection

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 0.02. 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 = 0.12
SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }
SE = 0.04599
z = (p₁ - p₂) / SE

z = - 2.82

z_Critical = - 2.054

Rejection region is z < - 2.054

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.

Interpret results. Since the z-value (-2.82) lies in the rejection region, hence we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that the claim of the article’s headline: Warmer Surgical Patients Recover Better.