I need help with this TnT
1 answer:
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Answer:
sample size if he uses a previous estimate of 32% is 523
sample size if he does not use any prior estimates is 601
Step-by-step explanation:
estimate E = 4 percentage = 0.04
confidence Cl = 95% = 0.95
previous estimate p = 32% = 0.32
q = 1 - 0.32 = 0.68
to find out
size of sample
solution
first we calculate z value for 95% confidence E = 0.04 is 1.96
from probability P(-1.96 < z < 1.96) = 0.95)
here z = 1.96
so in 1st part size of sample we know
E = z ×
put all value E, z p and q and we get n
n = (z/E)² ×p×q
n = (1.96/0.04)² ×0.32×0.68
n = 523
sample size if he uses a previous estimate of 32% is 523
now in 2nd part we take p = 0.5
so n will be
n = (z/E)² ×p×q
n = (1.96/0.04)² ×0.5×0.5
n = 601
sample size if he does not use any prior estimates is 601