Question

1. Until recently, an average of 60 out of every 100 patients has survived a particularly severe infection. When a new drug was administered to patients from a random sample, 12/15 survived.

(10 pts)

1. Please identify the statistical test to be used.
2. Please compute the test statistic and p-value.
3. Does the new drug appear to be effective? Why or why not?

Answer 1

Sol:

p=60/100=0.6

Ho:p=0.6

Ha:p>0.6

statistical test to be used is z test for proprtion

alpha=0.05

Test statistic

Z=p^-p/sqrt(p*(1-p)/n

p^=sample proprtion of patients survived after new drug

p^=12/15=0.8

z=(0.8-0.6)/sqrt(0.6*(1-0.6)/15)

z= 1.5811

z=1.58

P value in excel is

left tail prob in excel

=NORMSDIST(1.5811)

=0.943072

1-left tail prob=right tail prrob which is required

1-0.943072=0.0569

p-value=0.0569

p>0.05

Do not reject Ho

Accept Ho

Conclusion:
There is no suffcient statistical evidence at 5% level of significance to conclude that the new drug is effective