I need the complete working solutions and accurate answers to this question. Tha
ID: 2934186 • Letter: I
Question
I need the complete working solutions and accurate answers to this question. Thank you.
(10 pts. + 1 pt. BONUS) 1. An abdominal aortic aneurysm (AAA) is often signaled by inflammation. A cardiovascular magnetic resonance study was conducted to identity wall edema as a marker for inflammation. Independent random samples of AAA and normal patients were obtained, and the MR-STIR intensity values were recorded for each. Please assume that the distribution is normal. The summary data are given in the following table. Please give all answers to three decimal places. Measure MR-STIR intensity for AAA patients MR-STIR intensity for normal patients Differences in intensity (AAA - normal) 10 10 10 1.4860.3404 1.1670.2579 0.3190.3314 (1 pt.) a) Should this situation be analyzed via a 2-sample independent or paired method? Please explain your answer. If this is a pairs situation, please state the common characteristic that makes these data paired. (6 pts.) b) Is there any evidence to suggest the true mean intensity values for the AAA patients 0.05. Remember to include the complete are greater than for normal patients? Use 4-step method in the hypothesis test. (2 pts.) c) Calculate and interpret the appropriate confidence interval or bound. (1 pt.) d) In practical terms, does the data imply that the true MR-STIR intensity is greater for AAA patients than normal patients? This part uses the information from parts b) and c). If no addition reasoning is provided, you will receive 0 points (1 pt. BONUS) e) Explain why the results in parts b) and c) are consistent with each other.Explanation / Answer
a.
it is t test for 2 sample independent test. since the invidual data is not
given and the mesaurements are independent to each other it fall under
independent test
b.
Given that,
MR_STIR intensity for AAA patients data
mean(x)=1.486
standard deviation , s.d1=0.3404
number(n1)=10
MR_STIR intensity for normal patients data
y(mean)=1.167
standard deviation, s.d2 =0.2579
number(n2)=10
null, Ho: u1 < u2
alternate, H1: u1 > u2
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.833
since our test is right-tailed
reject Ho, if to > 1.833
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =1.486-1.167/sqrt((0.11587/10)+(0.06651/10))
to =2.362
| to | =2.362
critical value
the value of |t | with min (n1-1, n2-1) i.e 9 d.f is 1.833
we got |to| = 2.36209 & | t | = 1.833
make decision
hence value of | to | > | t | and here we reject Ho
p-value:right tail - Ha : ( p > 2.3621 ) = 0.02123
hence value of p0.05 > 0.02123,here we reject Ho
ANSWERS
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null, Ho: u1 < u2
alternate, H1: u1 > u2
test statistic: 2.362
critical value: 1.833
decision: reject Ho
p-value: 0.02123
we have evidence that true mean intensy values for the AAA patients are greter than for normal patioents.
c.
given that,
mean(x)=1.486
standard deviation , s.d1=0.3404
sample size, n1=10
y(mean)=1.167
standard deviation, s.d2 =0.2579
sample size,n2 =10
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 1.486-1.167) ± t a/2 * sqrt((0.116/10)+(0.067/10)]
= [ (0.319) ± t a/2 * 0.135]
= [0.071 , 0.567]
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interpretations:
1. we are 95% sure that the interval [0.071 , 0.567] 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
we have evidence that true mean intensy values for the AAA patients are greter than for normal patients.
d.
yes, MR_STIR is relavantly has high intensity rate as compared to the normal patients.