7.31. Fluid Overload in Hemodialysis. The overload of fluid volume and hypertens

Question

7.31. Fluid Overload in Hemodialysis. The overload of fluid volume and
hypertension are known to contribute to high cardiovascular morbidity
and mortality seen in dialysis patients. The correct assessment of volume
status is especially important as only a small increase in extracellular
volume over prolonged periods of time can lead to a considerable cardiac
strain and, as a consequence, to left ventricular hypertrophy. In clinical
practice, volume overload is most often judged by a battery of clinical
signs such as edema, dyspnea, hypertension, and coughing. A study by
Ribitsch et al. (2012) compares volume overload in stable hemodialysis
(HD) patients assessed by standard clinical judgment with data obtained
from bioimpedance analysis.
Data set hemodialysis.dat|mat|xlsx provides measurements on 28 HD
patients (17 males and 11 females) from the dialysis unit of the University
Medical Center Graz. The variables are described in the following
table:
7.9 Exercises 327
Column Variable Unit Description
1 M0 (kg) Pre-dialytic body mass
2 BMI (kg/m2) Body mass index
3 P0 (mmHg) Pre-dialytic mean arterial pressure
4 P1 (mmHg) Post-dialytic mean arterial pressure
5 VE (L) Extracelular volume
6 VO (L) Volume overload
7 VU (L) Delivered ultrafiltration volume
8 B0 (pg/ml) Pre-dialytic NT-pro-BNP
9 B1 (pg/ml) Post-dialytic NT-pro-BNP
10 SW Wizemann’s clinical score
(a) Find the 95% CI for the population mean of the difference D = P1P0
in stable hemodialysis patients. Assume that this difference is normally
distributed.
(b) Find the 90% CI for the population variance of V0. Assume normality
of V0.
(c) Find the 99% CI for the population proportion of patients for which
B1 > B0.

DATASET: https://tinyurl.com/y89yuesl

Explanation / Answer

(a) Find the 95% CI for the population mean of the difference D = P1P0

in stable hemodialysis patients. Assume that this difference is normally
distributed.

USing minitab

Two-Sample T-Test and CI: Var4, Var3

Two-sample T for Var4 vs Var3

N Mean StDev SE Mean
Var4 28 93.1 18.2 3.4
Var3 28 102.6 15.8 3.0

Difference = (Var4) - (Var3)
Estimate for difference: -9.47
95% CI for difference: (-18.60, -0.34)
T-Test of difference = 0 (vs ): T-Value = -2.08 P-Value = 0.042 DF = 54
Both use Pooled StDev = 17.0366

Comment - the 95% confidence interval for mean difference D= P0-P1 is (-18.60, -0.34)

(b) Find the 90% CI for the population variance of V0. Assume normality
of V0.

Test and CI for One Variance: Var6 (VO)

Method

The chi-square method is only for the normal distribution.
The Bonett method is for any continuous distribution.

Statistics

Variable N StDev Variance
Var6 28 1.49 2.23

90% Confidence Intervals

CI for CI for
Variable Method StDev Variance
Var6 Chi-Square (1.22, 1.93) (1.50, 3.72)
Bonett (1.24, 1.91) (1.53, 3.66)

Comment - Here miinitab give two confidence interval but our population is normal so we use Chi-square method .

confidence interval for population variance is (1.50, 3.72)

(c) Find the 99% CI for the population proportion of patients for which B1 > B0.

Test and CI for Two Proportions

Sample X N Sample p
1 11 28 0.392857
2 17 28 0.607143

Difference = p (1) - p (2)
Estimate for difference: -0.214286
99% lower bound for difference: -0.517936
Test for difference = 0 (vs > 0): Z = -1.64 P-Value = 0.950

Fisher’s exact test: P-Value = 0.970

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