Question

Suppose you have a sample of 50 and a sample proportion of 0.35. Construct a 99% confidence interval for the population proportion.

Answer 1

TRADITIONAL METHOD
given that,
possible chances (x)=17.5
sample size(n)=50
success rate ( p )= x/n = 0.35
I.
sample proportion = 0.35
standard error = Sqrt ( (0.35*0.65) /50) )
= 0.067
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
margin of error = 2.576 * 0.067
= 0.174
III.
CI = [ p ± margin of error ]
confidence interval = [0.35 ± 0.174]
= [ 0.176 , 0.524]
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DIRECT METHOD
given that,
possibile chances (x)=17.5
sample size(n)=50
success rate ( p )= x/n = 0.35
CI = confidence interval
confidence interval = [ 0.35 ± 2.576 * Sqrt ( (0.35*0.65) /50) ) ]
= [0.35 - 2.576 * Sqrt ( (0.35*0.65) /50) , 0.35 + 2.576 * Sqrt ( (0.35*0.65) /50) ]
= [0.176 , 0.524]
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interpretations:
1. We are 99% sure that the interval [ 0.176 , 0.524] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population proportion