A study was performed comparing two surveillance methods to detect patients with
ID: 3183705 • Letter: A
Question
A study was performed comparing two surveillance methods to detect patients with infections after caesarean section. Five hospitals were involved. Let’s call the two methods the standard Centers for Disease Control and Prevention (CDC) method and the enhanced method. A total of 197 patients were assessed in a hospital; 3 of the patients were identified as having an infection by the CDC method and 6 were identified as having an infection by the enhanced method. Two of the patients were identified as having an infection by both methods.
a. What test can be used to identify differences between the methods for identifying patients with infection in this hospital? Implement the method, and report a two-tailed p-value.
Explanation / Answer
Solution:-
x1 = 3, n1 = 197
p1 = 3/197 = 0.01523
x2 = 6, n2 = 197
p2 = 6/197 = 0.0305
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 0.05. 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 = 0.0228
SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = 0.01504
z = (p1 - p2) / SE
z = - 1.0153
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 - 1.02 or greater than 1.02.
Thus, the P-value = 0.3078
Interpret results. Since the P-value (0.3078) is greater than the significance level (0.05), we have to accept the null hypothesis.
From the above test we conclude that we have sufficient evidence that there is no differences between the methods for identifying patients with infection in this hospital.
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