The number of surveys must be 600
Explanation:
Proposition a is between 0.4 and 0.6
Confidence 95%
Error level 4%
Incident rate 0.8
Response rate 0.6
Surveys should they send out are:
E = 0.04 (this is the margin error of 4%)
P = (0.4+0.6)/2=0.5
Alpha = 1 - 0.95 = 0.05
So, the z value that we need to search in the table of z distribution is alpha/2 = 0.025
Z-value = 1.96
P (z>1.96) = 0.025 nearly
The number of surveys is
n = p*q (2 * half-alpha) ^2
=0.5 * 0.5 * (1.96/0.04)^2
=(0.5^2) * (49^2)
=0.25 * 2.401
= 600.25
The number of surveys must be 600
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Explanation:
https://us05web.z.oom.us/j/86531212630?pwd=SStKR2s0aE50SUNMZG4wNkpsbFpaZz09