Question 1182750
In other words, how likely is it that of 1118 people, 68 will get flu symptoms given that the probability is 0.057?
68/1118=0.0608
1-sample proportion test
Ho: true proportion is < =0.057
Ha: true proportion is >0.057 (in other words, something has changed the proportion, like the drug)
alpha=0.05 p{reject Ho|Ho true}
critical value is z>1.645
z=(0.0608-0.057)/sqrt(p*(1-p)/n)
=0.0038/sqrt(0.057*0.943/1118); sqrt term=0.0069
=0.0038/0.0069
=0.548
probability z>0.548 is 0.2918
There is insufficient evidence to conclude that the drug is producing these side effects. We failed to reject the null hypothesis, and the p-value is 0.2918, meaning that if the null hypothesis were true, we would have a 29.2% chance of finding a result this much or more extreme.
==
normal approximation
np=63.726=mean
np(1-p)=60.093=variance
sd= sqrt(V)=7.75
probability the value is above 68
z=(68-63.7)/7.75=0.5548; that probability is 0.2895