Question

q5. A random sample of 500 persons was taken in a certain area if test is positive for the covid-19. let 60 persons were found with positive test. obtain 98% confidence limit for the proportion of the covid 19 positive cases

Answer 1

Let n = sample size = 500
x = total number of Covid 19 positive = 60
= the point estimate of the proportion of Covid 19 positive.

We want to find 98% confidence interval for population proportion(p).
The formula of confidence interval for population proportion(p) is as follow:
(Lower limit, Upper limit) =
Where E is the margin of error
the formula of margin of error is as follows:

Let's find the value of Zc.
Zc = critical z-value for confidence level = c
Here c = 0.98
therefore, level of significance = α = 1 - c = 1 - 0.98 = 0.02
therefore 1 - (α/2 ) = 1 - (0.02/2) = 1 - 0.01 = 0.99
Let's use excel to find z-score for probability = p = 0.99
The general command to find z-value for given probability is "=NORMSINV(p)"
Here p = 0.99
therefore, Zc = "=NORMSINV(0.99)" = 2.326348
So we have, n = 500, = 0.12, and Zc = 2.326348
Plugging these values in the formula of E, we get


Lower limit = - E = 0.12 - 0.034 = 0.086
Upper limit = + E = 0.12 + 0.034 = 0.154
Answer: The 98% confidence interval to estimate the population proportion(p) is (0.086, 0.154).