Question 1184274
sample size is 150
60 are positive for malaria.
the mean proportion is 60/150 = 6/15 = 2/5 = .4
the mean proportion standard deviation is sqrt(.4 * .6) / 150) = .04
critical z-score for 99% two-tailed confidence level is plus or minus 2.5758.


the z-score formula is z = (x - m) / s
z is the z-score.
x is the sample raw score proportion.
m is the population mean proportion.
s is the standard error = sqrt(p*q/n), where:
p is the population proportion.
q = 1 - p
n is the sample size.


when z = -2.5758, the formula becomes:
-2.5758 = (x - .4) / .04
solve for x to get:
x = -2.5758 * .04 + .4 = .297


when z = 2.5758, the formula becomes:
2.5758 = (x - .4) / .04
solve for x to get:
x = 2.5758 * .04 + .4 = .503.


i believe that's your two-tailed confidence interval.


with a sample size of 150, between 30% and 505 will show positive for malaria.


that a very high percentage, showing that the population is at high risk for malaria.


here's a world malaria report that shows where the highest risks are.


<a href = "https://www.gs.international/learning-centre-articles/malaria-world-map-risk" target = "_blank">https://www.gs.international/learning-centre-articles/malaria-world-map-risk</a>


here's a reference on mean proportion and standard deviation.


<a href = "https://stats.libretexts.org/Bookshelves/Introductory_Statistics/Book%3A_Introductory_Statistics_(Shafer_and_Zhang)/06%3A_Sampling_Distributions/6.03%3A_The_Sample_Proportion#:~:text=at%20least%2030.-,The%20Sampling%20Distribution%20of%20the%20Sample%20Proportion,interval%20%5B0%2C1%5D." target = "_blank">https://stats.libretexts.org/Bookshelves/Introductory_Statistics/Book%3A_Introductory_Statistics_(Shafer_and_Zhang)/06%3A_Sampling_Distributions/6.03%3A_The_Sample_Proportion#:~:text=at%20least%2030.-,The%20Sampling%20Distribution%20of%20the%20Sample%20Proportion,interval%20%5B0%2C1%5D.</a>


i'm never 100% certain that i'm right with these types of problems, but i'm reasonably certain that what i gave you is accurate.