6. Cardiologists use the short-range scaling exponent , which measures the rando

ID: 3313115 • Letter: 6

Question

6. Cardiologists use the short-range scaling exponent , which measures the randomness of heart rate patterns, as a tool to assess risk of heart attack. The article "Applying Fractal Analysis to Short Sets of Heart Rate Variability Data" (M. Pena et al., Med Biol Eng Comput, 2009:709-717) compared values of 1 computed from long series of measurements (approximately 40,000 heartbeats) with those estimated from the first 300 beats to determine how wel the long-term measurement (y) could be predicted the short-term one (x). Following are the data (obtained by digitizing a graph) Short Long Short Long Short Long Short LongShort LongShort Long 1.53 1.48 0.54 0.70 1.19.6 42 0.79 1.10 .34 1.31 1.02 0.79 0.81 1.40 1.19 1.66 1.422712 1.23 .33 1.48 1.47 1.20 1.461.42161.13 1.30 .33 .16 .48 0.81 0.81 0.88 0.90 .28 1.23 1.6 1.42 1.34 .1460 1.34 1.38 .52 1.68 1.05 .18 1.23 1.72 144 1.08 .14 0.92 1.34 1.36 1.52 1.05 0.71 1.24 1.49 .44141.15 1.42 .35 1.73 .55 1.16 0.82 1.05 1.10 1.27 1.65 145 0.91 0.93 1.26 1.29 1.33 1.30 0.98 1.16 .55 1.351.351.56 1.39 1.40 1.41 1.19 1.57 1.60 1.46 1.07 0.81 1.10 1.03 1.16 1.18 1.59 1.61 1.07 1.47 0.82 a. Compute the least-squares line for predicting the long-term measurement from the short-term measurement b. Compute the error standard deviation estimate s Compute a 95% confidence interval for the slope d. Find a 95% confidence interval for the mean long-term measurement for those with short-term measurements of 1.2 e. Can you conclude that the mean long-term measurement for those with short-term measurements of 1.2 is greater than 1.2? Perform a hypothesis test and report the P-value. For question 6 part e, use the confidence interval from part d. f Find a 95% prediction interval for the long-term measurement for a particular individual whose short-term measurement 1S g. The purpose of a short-term measurement is to substitute for a long-term measurement. For this purpose, which do you think is more relevant, the confidence interval or the prediction interval? Explain.

Explanation / Answer

Solution:

First of all we have to compute the regression model for the prediction of the long term measurement based on the short term measurements. Required regression model by using excel is given as below:

Regression Statistics

Multiple R

0.580306234

R Square

0.336755325

Adjusted R Square

0.32400062

Standard Error

0.169190521

Observations

ANOVA

Significance F

Regression

0.755781224

0.7557812

26.402439

4.23707E-06

Residual

1.488522479

0.0286254

Total

2.244303704

Coefficients

Standard Error

t Stat

P-value

Lower 95%

Upper 95%

Intercept

0.780948957

0.098693084

7.9129046

1.748E-10

0.5829068

0.978991114

Short

0.39857441

0.077568857

5.1383304

4.237E-06

0.242921115

0.554227704

Part a

Here, we have to find least squares regression line for prediction of the long term measurement from the short term measurement. Required regression line or equation is given as below:

Y = 0 + 1*X

Long term measurement = 0 + 1*Short term measurement

Where, 0 is the y-intercept of regression line and 1 is the slope for regression line.

Long term measurement = 0.780948957 + 0.39857441*Short term measurement

Y = 0.780948957 + 0.39857441*X

Part b

From above table for regression output, the value for error standard deviation or standard error is given as below:

Standard error = 0.169190521

Part c

Here, we have to find the 95% confidence interval for the slope 1.

Confidence interval formula is given as below:

Confidence interval = 1 -/+ t*SE(1)

We are given

1 = 0.39857441

SE(1) = 0.077568857

Sample size = n = 54

Degrees of freedom = n – 1 = 54

Confidence level = 95%,

So, by using t-table, critical t value is given as below:

Critical t value = 2.0057459

Confidence interval = 0.39857441 -/+ 2.0057459*0.077568857

Confidence interval = 0.39857441 -/+ 0.1555834

Lower limit = 0.39857441 - 0.1555834 = 0.24299101

Upper limit = 0.39857441 + 0.1555834 = 0.55415781

Confidence interval =(0.24299101, 0.55415781)

Part d

Here, we have to find 95% confidence interval for mean long term measurement for those with short term measurement of 1.2. This means we have to find prediction interval for X = 1.2.

Formula for prediction interval is given as below:

Yh -/+ t/2, n – 2 *SE*sqrt[(1 + (1/n) + ((x – xbar)^2)/(xi – xbar)^2))]

Where,

Yh = Predicted value of Y

t/2, n – 2 = Critical value

SE = standard error of estimate

n = sample size

xbar = sample mean of x

From the given regression model and excel calculations, we have

Yh = Predicted value of Y for x = 1.2 is given as below

Yh = 0.780948957 + 0.39857441*1.2 = 1.259238

n = 54

df = n – 2 = 54 – 2 = 52

Confidence level = 95%

t/2, n – 2 = Critical value = 2.006647 (by using t-table or excel)

SE = 0.169191

Xbar = 1.237222

(xi – xbar)^2 = Sum of Squared Differences from Xbar = 4.757483

Now, plug all values in the prediction interval formula given as below:

Prediction interval = 1.259238 -/+ 2.006647*0.169191*sqrt[1 + (1/54) + (1.2 - 1.237222)^2 / 4.757483]

Prediction interval = 1.259238 -/+ 0.3427

Lower limit = 1.259238 - 0.3427 = 0.916538

Upper limit = 1.259238 + 0.3427 = 1.601938