Question

detail for solution

Dr. Oz then pointed out the part that "really caught my attention." "They did CT scans of the chest in all the patients. All the patients had pneumonia when they started. Over the course of the five-day treatment with the hydroxychloroquine and 55 percent of the control population where they just got the normal therapy there was resolve and resolution of the pneumonia in 81 percent of the patients on the hydroxychloroquine, there was improvement in the lung's images," he pointed out Dr. Oz noted that these results are "statistically significant." He went on to say that even though the study only monitored a small group of people, "they still got the measures that we like to see." Dr. Oz acknowledged that a bigger clinical trial is still needed, adding that the Chinese study "is an early effort to try to show a lot of people whether this is the right way or the wrong way to go." "I should point out in the 31 patients that were the control group, four patients had bad outcomes, they got significantly worse. None of the patients in the hydroxychloroquine group got significantly worse," he said. Here is the question: Dr Oz told us that control group had 31 patients. Likely the test group also had 31 patients. Thus 17 out of 31 (55%) in control group improved, and 25 out of 31 (81%) in test group improved (1) Given the data, what is the p-value for the hypothesis test where the null hypothesis is "hydroxychloroquine makes no difference"? (2) If you want to argue that Dr. Oz made the correct conclusion that these results were "statistically significant", what is your argument? (3) If you want to argue that Dr. Oz made the wrong conclusion, what is your argument?

Answer 1

Given that,
sample one, x1 =17, n1 =31, p1= x1/n1=0.55
sample two, x2 =25, n2 =31, p2= x2/n2=0.81
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (p1-p2)/√(p^q^(1/n1+1/n2))
zo =(0.548-0.81)/sqrt((0.677*0.32(1/31+1/31))
zo =-2.17
| zo | =2.17
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =2.173 & | z α | =1.96
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -2.1734 ) = 0.0297
hence value of p0.05 > 0.0297,here we reject Ho
ANSWERS
---------------
1.
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: -2.17
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.0297
2.
yes,
statistically significant
3.
we have enough evidence to support the claim that difference in proportion to show a lot of people whether this is the right way or the wrong way to go.