8.5.49-1T Question Help * Suppose a hospital would like to estimate the proporti
ID: 3334298 • Letter: 8
Question
8.5.49-1T Question Help * Suppose a hospital would like to estimate the proportion of patients who feel that physicians who care for them always communicated effectively when discussing their medical care A pilot sample of 50 patients found that 38 reported that their physician communicated effectively. Determine the additional number of patients that need to be sampled to construct a 95% confidence interval with a margin of error equal to 7% to estimate this proportion The additional number of patients that need to be sampled is (Round up to the nearest integer.)Explanation / Answer
8.5.49-T.
TRADITIONAL METHOD
given that,
possibile chances (x)=38
sample size(n)=50
success rate ( p )= x/n = 0.76
I.
sample proportion = 0.76
standard error = Sqrt ( (0.76*0.24) /50) )
= 0.06
II.B9
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 * 0.06
= 0.118
III.
CI = [ p ± margin of error ]
confidence interval = [0.76 ± 0.118]
= [ 0.642 , 0.878]
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DIRECT METHOD
given that,
possibile chances (x)=38
sample size(n)=50
success rate ( p )= x/n = 0.76
CI = confidence interval
confidence interval = [ 0.76 ± 1.96 * Sqrt ( (0.76*0.24) /50) ) ]
= [0.76 - 1.96 * Sqrt ( (0.76*0.24) /50) , 0.76 + 1.96 * Sqrt ( (0.76*0.24) /50) ]
= [0.642 , 0.878]
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interpretations:
1. We are 95% sure that the interval [ 0.642 , 0.878] 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
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.05 is = 1.96
Sample Proportion = 0.76
ME = 0.07
n = ( 1.96 / 0.07 )^2 * 0.76*0.24
= 143.002 ~ 144
additional no.of patients are needed its sample size is 144
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