Question

I have a sample size of 58 for a process improvement on reducing spending. My improvement lasted for 9 weeks, 5 weeks I collected 40 receipts of expenses before and after my process improvement I collected 18 receipts. Can someone help me calculate a confidence interval?

Answer 1

TRADITIONAL METHOD
given that,
possible chances (x)=18
sample size(n)=58
success rate ( p )= x/n = 0.31
I.
sample proportion = 0.31
standard error = Sqrt ( (0.31*0.69) /58) )
= 0.061
II.
margin of error = Z a/2 * (standard 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 * 0.061
= 0.119
III.
CI = [ p ± margin of error ]
confidence interval = [0.31 ± 0.119]
= [ 0.191 , 0.429]
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DIRECT METHOD
given that,
possible chances (x)=18
sample size(n)=58
success rate ( p )= x/n = 0.31
CI = confidence interval
confidence interval = [ 0.31 ± 1.96 * Sqrt ( (0.31*0.69) /58) ) ]
= [0.31 - 1.96 * Sqrt ( (0.31*0.69) /58) , 0.31 + 1.96 * Sqrt ( (0.31*0.69) /58) ]
= [0.191 , 0.429]
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interpretations:
1. We are 95% sure that the interval [ 0.191 , 0.429] 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