Answer to Question #319033 in Statistics and Probability for zzzz

Question

Chinovac Labs tried a new vaccine on 189 randomly selected individuals. From the experiment, it was determined that 104 of them developed immunity. Calculate a 95% confidence interval for the proportion p of individuals in the population for whom the vaccine would help. What is the upper bound of this interval?

Answer 1

Confidence interval can be found the following way:

$p \in (a-Cr_{\alpha}*\sqrt{{\frac {a(1-a)} n}};a+Cr_{\alpha}*\sqrt{{\frac {a(1-a)} n}})$ , where a-sample proportion, $Cr_{\alpha}$ - critical value on ${\alpha}$ confidence level, n - sample size

Since the sample size is big(>30), then it is appropriate to use z-score as the critical value. So, 

$P(Z>Cr)={\frac {1+\alpha} 2}={\frac {1+0.95} 2}=0.975\implies Cr=1.96$ . So, the confidence interval is

$({\frac {104} {189}}-1.96*\sqrt{{\frac {{\frac {104} {189}}(1-{\frac {104} {189})}} {184}}},{\frac {104} {189}}+1.96*\sqrt{{\frac {{\frac {104} {189}}(1-{\frac {104} {189})}} {184}}})=(0.478,0.622)$

 its upper bound is 0.622