One thousand children were given immunization procedure A and one child develope

Question

One thousand children were given immunization procedure A and one child developed a particular neurological condition called NDC within one month of the immunization. One thousand different but comparable children were given immunization procedure B and eight children developed NDC within one month of the immunization.

A. For procedure A estimate the probability of NDC and give its 95% confidence limits.

B. For procedure B estimate the probability of NDC and give its 95% confidence limits.

C. Does the proportion of children with NDC differ significantly between these two groups? Show your work and give the p-value.

Explanation / Answer

PART A.
TRADITIONAL METHOD
given that,
possibile chances (x)=1
sample size(n)=1000
success rate ( p )= x/n = 0.001
I.
sample proportion = 0.001
standard error = Sqrt ( (0.001*0.999) /1000) )
= 0.001
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 * 0.001
= 0.002
III.
CI = [ p ± margin of error ]
confidence interval = [0.001 ± 0.002]
= [ -0.001 , 0.003]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possibile chances (x)=1
sample size(n)=1000
success rate ( p )= x/n = 0.001
CI = confidence interval
confidence interval = [ 0.001 ± 1.96 * Sqrt ( (0.001*0.999) /1000) ) ]
= [0.001 - 1.96 * Sqrt ( (0.001*0.999) /1000) , 0.001 + 1.96 * Sqrt ( (0.001*0.999) /1000) ]
= [-0.001 , 0.003]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 95% sure that the interval [ -0.001 , 0.003] 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

PART B.
TRADITIONAL METHOD
given that,
possibile chances (x)=8
sample size(n)=1000
success rate ( p )= x/n = 0.008
I.
sample proportion = 0.008
standard error = Sqrt ( (0.008*0.992) /1000) )
= 0.003
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 * 0.003
= 0.006
III.
CI = [ p ± margin of error ]
confidence interval = [0.008 ± 0.006]
= [ 0.002 , 0.014]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possibile chances (x)=8
sample size(n)=1000
success rate ( p )= x/n = 0.008
CI = confidence interval
confidence interval = [ 0.008 ± 1.96 * Sqrt ( (0.008*0.992) /1000) ) ]
= [0.008 - 1.96 * Sqrt ( (0.008*0.992) /1000) , 0.008 + 1.96 * Sqrt ( (0.008*0.992) /1000) ]
= [0.002 , 0.014]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 95% sure that the interval [ 0.002 , 0.014] 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

PART C.
Given that,
sample one, x1 =1, n1 =1000, p1= x1/n1=0.001
sample two, x2 =8, n2 =1000, p2= x2/n2=0.008
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.001-0.008)/sqrt((0.005*0.996(1/1000+1/1000))
zo =-2.339
| zo | =2.339
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =2.339 & | z | =1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -2.3386 ) = 0.0194
hence value of p0.05 > 0.0194,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: -2.339
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.0194
proportion of children with NDC differ significantly between these two groups

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