2. A study was conducted to examine the relationship between hypertension and ca

Question

2. A study was conducted to examine the relationship between hypertension and cardiovascular disease (CVD). A total of 4,434 individuals were recruited for this study and were asked about their hypertension status and whether they had ever been diagnosed with CVD. There were 1,182 individuals that were free of hypertension and 1,017 of them did not have a CVD diagnosis. There were 1,157 individuals that were diagnosed with CVD and 992 of them were living with hypertension. Let pi be the population proportion of individuals without hypertension who have a CVD diagnosis and p2 be the population proportion of individuals with hypertension who have a CVD diagnosis (20pts). a. Is the following study a cohort, case-control, or cross-sectional study (2pts). b. Provide estimates for p1 and p2 (4.5 pts). C. Construct a 95% confidence interval for the risk difference (4.5 pts). d. Construct a 95% confidence interval for the odds ratio (4.5 pts). e. Conduct the following hypothesis test: Hopi-P2 versus H1n # P2 Assume an alpha of .05. What is your conclusion? (do not simply state that the null hypothesis was accepted or rejected) (4.5 pts).

Explanation / Answer

a.
it is study of case control study
In contrast with cohort and cross sectional studies, case-control studies are usually retrospective. People with the outcome of interest are matched with a control group who do not.

b.
TRADITIONAL METHOD
given that,
sample one, x1 =165, n1 =1182, p1= x1/n1=0.1396
sample two, x2 =992, n2 =3252, p2= x2/n2=0.305
point estimates of proportion = p1 - p2 = 0.1396 - 0.305 = -0.1654

c.
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( (0.1396*0.8604/1182) +(0.305 * 0.695/3252))
=0.0129
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 * 0.0129
=0.0253
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.1396-0.305) ±0.0253]
= [ -0.1908 , -0.1401]
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DIRECT METHOD
given that,
sample one, x1 =165, n1 =1182, p1= x1/n1=0.1396
sample two, x2 =992, n2 =3252, p2= x2/n2=0.305
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 = [ (0.1396-0.305) ± 1.96 * 0.0129]
= [ -0.1908 , -0.1401 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ -0.1908 , -0.1401] 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, 95% of these intervals will contains the difference between
true population mean P1-P2

e.
Given that,
sample one, x1 =165, n1 =1182, p1= x1/n1=0.14
sample two, x2 =992, n2 =3252, p2= x2/n2=0.305
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 =(0.14-0.305)/sqrt((0.261*0.739(1/1182+1/3252))
zo =-11.093
| zo | =11.093
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =11.093 & | 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 != -11.0928 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: -11.093
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0

CVD NO CVD WITH HYPERTEN 992 2260 3252 WITHOUT HYPER 165 1017 1182 1157 3277 4434

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