Solution:
Required margin of error = 0.05
Estimated population proportion p = 0.8
Significance level = 0.10
The [tex]\text{provided estimate population proportion}[/tex] p is 0.8
The significance level, α = 0.1 is [tex]z_c=1.645[/tex], which is obtained by looking into a standard normal probability table.
The number of patients surveyed to estimate the population proportion p within the required margin of error :
[tex]$n \geq p(1-p)\left(\frac{z_c}{E}\right)^2$[/tex]
[tex]$=0.8\times (1-0.8)\left(\frac{1.64}{0.05}\right)^2$[/tex]
= 173.15
Therefore, the number of patients surveyed to satisfy the condition is n ≥ 173.15 and it must be an integer number.
Thus we conclude that the number of patients surveyed so the margin of error of 90% confidence interval is within 0.05 are n= 174.
