In 2006, a study was conducted to compare two methods for assisting people exper

Question

In 2006, a study was conducted to compare two methods for assisting people experiencing trouble breathing as a result of an airway blockage. In Pittsburg, 350 patients were assigned randomly to one of two treatments: 150 to treatment 1 and 200 to treatment 2. A total of 250 patients survived the obstruction: 95 in the group receiving treatment 1 and 155 in the group receiving treatment 2. A test of significance was conducted on these hypotheses:

H0: The survival rates for the two treatments are equal.
Ha: Treatment 2 produces a higher survival rate.

The statistical test resulted in a p-value of 0.0018.

Part A: What does the p-value measure in the context of this study? (3 points)

Part B: Based on this p-value and the study design, what conclusion can be drawn in the context of this study? Use a significance level of = 0.05. (4 points)

Part C: Based on your conclusion in part B, which type of error—Type I or Type II—could have been made? What is one potential consequence of this error?

Explanation / Answer

Ans:

a)The P value, or calculated probability, is the probability of finding the observing p1<p2 when the null hypothesis (H 0) of a study is true i.e. p1=p2.

probability of sample proprtion for treatement 1 i.e. survival rate being less than survival rate for treatment 2,when null hypothesis is true.

b)As,p-value=0.0018<0.05,we reject null hypothesis.

There is enough evidence to conclude that treatment 2 produces a higher survival rate.

c)As,we are rejecting null hypothesis in part b),so there are chances of making type I error.

(we make type I error,when we reject null hypothesis and null hypothesis is true.)

As,we have concluded that treatment 2 have higher survival rate,we might be preferring treatment 2 in future i.e. we might be biased towards treatment 2 unnecessarily ,but actually both treatments have equal survival rates.

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