A prospective cohort study was conducted to investigate the association between

Question

A prospective cohort study was conducted to investigate the association between smoking at study entry and cardiovascular death during the follow-up period. Participants, categorized at the start of the study as smokers (exposed - yes) and non- smokers (exposed - no), were followed over time to see if they ultimately died from a cardiovascular cause. The data are summarized in the following table: Death from cardiovascular cause:? Y es 0 Exposed?Yes 1586 2100 35 0 a) Compare the proportion of those who died from a cardiovascular cause between the two groups (where group is based on smoking exposure status) using a 99% confidence interval. Provide a written conclusion of the interval Compare the proportion of those who died from a cardiovascular cause between the two groups using a hypothesis test with significance level.05. Provide a written conclusion of the test Generate a relative risk point estimate. Also, calculate a relative risk interval estimate using a 98% confidence interval. Provide a written conclusion of the interval Generate an odds ratio point estimate. Comment on the magnitude of this estimated odds ratio as compared to the estimated relative risk. Also, calculate an odds ratio interval estimate using a 98% confidence interval. Provide a written conclusion of the interval Conduct a chi-square test of independence for this data at the.02 significance level. Provide a written conclusion of the test. Comment on your findings in total from (a) through (e) on specifically whether it appears there is an association between smoking status and future death from cardiovascular causes. The total length of this entire answer to (f) should be no more than 4 sentences b) c) d) e) f)

Explanation / Answer

PART A.
TRADITIONAL METHOD
given that,
sample one, x1 =35, n1 =1586, p1= x1/n1=0.0221
sample two, x2 =18, n2 =2100, p2= x2/n2=0.0086
I.
standard error = sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where
p1, p2 = proportion of both sample observation
n1, n2 = sample size
standard error = sqrt( (0.0221*0.9779/1586) +(0.0086 * 0.9914/2100))
=0.0042
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.58
margin of error = 2.58 * 0.0042
=0.0108
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.0221-0.0086) ±0.0108]
= [ 0.0027 , 0.0243]
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DIRECT METHOD
given that,
sample one, x1 =35, n1 =1586, p1= x1/n1=0.0221
sample two, x2 =18, n2 =2100, p2= x2/n2=0.0086
CI = (p1-p2) ± sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where,
p1, p2 = proportion of both sample observation
n1,n2 = size of both group
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ (0.0221-0.0086) ± 2.58 * 0.0042]
= [ 0.0027 , 0.0243 ]
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interpretations:
1) we are 99% sure that the interval [ 0.0027 , 0.0243] contains the difference between
true population proportion P1-P2
2) if a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the difference between
true population mean P1-P2

PART B.
Given that,
sample one, x1 =35, n1 =1586, p1= x1/n1=0.022
sample two, x2 =18, n2 =2100, p2= x2/n2=0.009
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 =(0.022-0.009)/sqrt((0.014*0.986(1/1586+1/2100))
zo =3.408
| zo | =3.408
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =3.408 & | 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 != 3.408 ) = 0.0007
hence value of p0.05 > 0.0007,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 3.408
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.0007

there is significance difference between both

PART E.

Given table data is as below

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calculation formula for E table matrix

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expected frequecies calculated by applying E - table matrix formulae

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calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above

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set up null vs alternative as

null, Ho: data is independent

alternative, H1: exists a dependent data

level of significance, = 0.02

from standard normal table, chi square value at right tailed, ^2 /2 =5.4119

since our test is right tailed,reject Ho when ^2 o > 5.4119

we use test statistic ^2 o = (Oi-Ei)^2/Ei

from the table , ^2 o = 11.2645

critical value

the value of |^2 | at los 0.02 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 2 - 1 ) = 1 * 1 = 1 is 5.4119

we got | ^2| =11.2645 & | ^2 | =5.4119

make decision

hence value of | ^2 o | > | ^2 | and here we reject Ho

^2 p_value =0.0008

ANSWERS

---------------

null, Ho: data is independent

alternative, H1: exists a dependent data

test statistic: 11.2645

critical value: 5.4119

p-value:0.0008

decision: reject Ho

PART F.

association exists between smoking status and future death

MATRIX col1 col2 TOTALS row 1 35 1586 1621 row 2 18 2100 2118 TOTALS 53 3686 N = 3739

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