Question

a sample of 193 randomly selected students , found that the proportion of students planning to travel home for thanksgiving is 0.71

Answer 1

TRADITIONAL METHOD
given that,
possibile chances (x)=137.03
sample size(n)=193
success rate ( p )= x/n = 0.71
I.
sample proportion = 0.71
standard error = Sqrt ( (0.71*0.29) /193) )
= 0.033
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
margin of error = 1.96 * 0.033
= 0.064
III.
CI = [ p ± margin of error ]
confidence interval = [0.71 ± 0.064]
= [ 0.646 , 0.774]
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DIRECT METHOD
given that,
possibile chances (x)=137.03
sample size(n)=193
success rate ( p )= x/n = 0.71
CI = confidence interval
confidence interval = [ 0.71 ± 1.96 * Sqrt ( (0.71*0.29) /193) ) ]
= [0.71 - 1.96 * Sqrt ( (0.71*0.29) /193) , 0.71 + 1.96 * Sqrt ( (0.71*0.29) /193) ]
= [0.646 , 0.774]
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interpretations:
1. We are 95% sure that the interval [ 0.646 , 0.774] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion