Full 2L QBank
Social & Preventive Medicine General e4ccb23b

In a population of 100 prevalence of candida glabrata was found to be 80%. If the investigator has to repeat the prevalence with 95% confidence what will the prevalence be

A
78-82%
B
72-88%
C
76-84%
D
74-86%
High-Yield Explanation
Confidence Intervals for Population proportions (For 95% Confidence) CI = P + 2 SEP = P + 2 √pq/n Here, P=0.80 (80%); p=0.80; q= 1-p = 1-0.80 = 0.20; n=100 CI = 0.80 + 2 √0.8*0.2/100 = 0.80 + 0.08 = 0.72, 0.88 (72%, 88%)

Related Social & Preventive Medicine MCQs

Practice 2,00,000+ NEET PG Questions Free

Timed mock tests, mistake queue analytics, audio lectures & zero attempt limits on i❤️Exams.

Start Free Mock Test Now