Question

Compute a 98% confidence interval, using method 1 in page 297 of your textbook, for the proportion of defective items in a process when it is found that a sample of size 100 yields 8 defectives.

Answer 1

TRADITIONAL METHOD
given that,
possible chances (x)=8
sample size(n)=100
success rate ( p )= x/n = 0.08
I.
sample proportion = 0.08
standard error = Sqrt ( (0.08*0.92) /100) )
= 0.027
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.02
from standard normal table, two tailed z α/2 =2.326
margin of error = 2.326 * 0.027
= 0.063
III.
CI = [ p ± margin of error ]
confidence interval = [0.08 ± 0.063]
= [ 0.017 , 0.143]
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DIRECT METHOD
given that,
possible chances (x)=8
sample size(n)=100
success rate ( p )= x/n = 0.08
CI = confidence interval
confidence interval = [ 0.08 ± 2.326 * Sqrt ( (0.08*0.92) /100) ) ]
= [0.08 - 2.326 * Sqrt ( (0.08*0.92) /100) , 0.08 + 2.326 * Sqrt ( (0.08*0.92) /100) ]
= [0.017 , 0.143]
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interpretations:
1. We are 98% sure that the interval [ 0.017 , 0.143] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 98% of these intervals will contains the true population proportion