A random sample of 481 nonsmoking women of normal weight (body mass index betwee

Question

A random sample of 481 nonsmoking women of normal weight (body mass index between 19.8 and 26.0) who had given birth at a large metropolitan medical center was selected. It was determined that 6.9% of these births resulted in children of low birth weight (less than 2500 g).

(a)   Calculate a 99% confidence interval for the proportion of all such births that result in children of low birth weight. Use the long formula.

(b)   Calculate a 99% confidence interval for the proportion of all such births that result in children of low birth weight. Use the traditional formula.

Explanation / Answer

TRADITIONAL METHOD
given that,
sample size(n)=481
success rate ( p )= x/n = 0.069
I.
sample proportion = 0.069
standard error = Sqrt ( (0.069*0.931) /481) )
= 0.0116
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
margin of error = 1.96 * 0.0116
= 0.0227
III.
CI = [ p ± margin of error ]
confidence interval = [0.069 ± 0.0227]
= [ 0.0463 , 0.0917]
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DIRECT METHOD
given that,
sample size(n)=481
success rate ( p )= x/n = 0.069
CI = confidence interval
confidence interval = [ 0.069 ± 1.96 * Sqrt ( (0.069*0.931) /481) ) ]
= [0.069 - 1.96 * Sqrt ( (0.069*0.931) /481) , 0.069 + 1.96 * Sqrt ( (0.069*0.931) /481) ]
= [0.0463 , 0.0917]
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interpretations:
1. We are 95% sure that the interval [ 0.0463 , 0.0917] 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

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