Question

Cardiologists use the short-range scaling exponent α₁, 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” compared values of α₁ computed from long series of measurements (approximately 40,000 heartbeats) with those estimated from the first 300 beats to determine how well the long-term measurement (y) could be predicted the short-term one (x). Following are the data (obtained by digitizing a graph).

Short	Long
0.5	0.6
1.02	0.79
1.4	0.81
0.88	0.9
1.68	1.05
1.16	1.05
0.82	1.05
0.93	1.07
1.26	1.1
1.18	1.19
0.81	1.19
0.81	1.2
1.28	1.23
1.18	1.23
0.71	1.24

Note: This problem has a reduced data set for ease of performing the calculations required. This differs from the data set given for this problem in the text.

A) Find a 95% confidence interval for the mean long-term measurement for those with short-term measurements of 1.2. Round the answers to three decimal places.

B) Find a 95% prediction interval for the long-term measurement for a particular individual whose short term measurement is 1.2. Round the answers to three decimal places.

For A:

(0.938, 1.188) = Incorrect

(0.936, 1.185) = Incorrect

Answer 1

	x	y	(x-xbar)^2	(y-ybar)^2	(x-xbar)*(y-ybar)
	0.5	0.6	0.2930418	0.1995111	0.241795556
	1.02	0.79	0.0004551	0.0658778	0.005475556
	1.4	0.81	0.1286418	0.0560111	-0.084884444
	0.88	0.9	0.0260284	0.0215111	0.023662222
	1.68	1.05	0.4078951	1.111E-05	0.002128889
	1.16	1.05	0.0140818	1.111E-05	0.000395556
	0.82	1.05	0.0489884	1.111E-05	-0.000737778
	0.93	1.07	0.0123951	0.0005444	-0.002597778
	1.26	1.1	0.0478151	0.0028444	0.011662222
	1.18	1.19	0.0192284	0.0205444	0.019875556
	0.81	1.19	0.0535151	0.0205444	-0.033157778
	0.81	1.2	0.0535151	0.0235111	-0.035471111
	1.28	1.23	0.0569618	0.0336111	0.043755556
	1.18	1.23	0.0192284	0.0336111	0.025422222
	0.71	1.24	0.1097818	0.0373778	-0.064057778
sum	15.62	15.7	1.2915733	0.5155333	0.153266667
mean	1.041333333	1.046666667	sxx	syy	sxy
	slope=b1=sxy/sxx	0.118666639		0.4746666	0.665849106
	intercept=b0=ybar-(slope*xbar)	0.92309514		1.3977617	0.18782777
	SST	SYY	0.5155333
	SSR	sxy^2/sxx	0.0181876		0.18782777
	SSE	syy-sxy^2/sxx	0.4973457		0.035279271
	r^2	SSR/SST	0.0352793
	error variance s^2	SSE/(n-2)	0.0382574
	S^2b1	s^2/sxx	0.0296207
	standard error b1=se(b1)	sqrt(s^2b1)	0.1721068

• estimated slope , ß1 = SSxy/SSxx =   0.1533 /   1.2916 =   0.1187
                    
  intercept,   ß0 = y̅-ß1* x̄ =   0.9231
                    
  so, regression line is   Ŷ =  0.9230   +   0.1187   *x
                    
  SSE=   (SSxx * SSyy - SS²xy)/SSxx =   0.4973
                    
  std error ,Se =    √(SSE/(n-2)) =    0.1956
  ========

  A)

  X Value=   1.2                      
  Confidence Level=   95%                      
                            
                            
  Sample Size , n=   15                      
  Degrees of Freedom,df=n-2 =   13                      
  critical t Value=tα/2 =   2.160   [excel function: =t.inv.2t(α/2,df) ]                  
                            
  X̅ =    1.04   
  Σ(x-x̅)² =Sxx   1.29                      
  Standard Error of the Estimate,Se=   0.1956
                            
  Predicted Y at X=   1.2   is                  
  Ŷ =   0.9230   +   0.1187   *   1.2   =   1.065
                            
  standard error, S(ŷ)=Se*√(1/n+(X-X̅)²/Sxx) =    0.0574   
  margin of error,E=t*Std error=t* S(ŷ) =   2.1604   *   0.0574   =   0.1240  
                            
  Confidence Lower Limit=Ŷ +E =    1.065   -   0.1240   =   0.941   
  Confidence Upper Limit=Ŷ +E =   1.065   +   0.1240 =   1.189
  ==============

  B)

  For Individual Response Y                  
  standard error, S(ŷ)=Se*√(1+1/n+(X-X̅)²/Sxx) =   0.20   
  margin of error,E=t*std error=t*S(ŷ)=    2.1604   *   0.20 =   0.4404
                    
  Prediction Interval Lower Limit=Ŷ -E =   1.065    -   0.44   =   0.625
  Prediction Interval Upper Limit=Ŷ +E =   1.065    +   0.44   =   1.505