Question

Required information Cardiologists use the short-range scaling exponent 1, 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 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 could be predicted the short-term one (X). Following are the data (obtained by digitizing a graph). Short Long 0.56 0.65 1.02 0.79 1.4 0.81 0.88 0.9 1.68 1.05 1.16 1.05 1.05 0.82 0.93 1.07 1.26 1.1 1.18 1.19 1.19 0.81 1.2 0.81 1.28 1.23 1.23 1.18 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.

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.

The 95% confidence interval is (  ,  ).

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.

The 95% prediction interval is (  ,  ).

Answer 1

S.No х Y (y-7) 1 2 3 4 0.56 1.02 1.4 0.88 1.68 5 6 7 8 9 10 11 12 13 14 15 Total 1.16 0.82 0.93 1.26 1.18 0.81 0.81 1.28 1.18 0.71 15.68 1.045 0.65 0.79 0.81 0.9 1.05 1.05 1.05 1.07 1.1 1.19 1.19 1.2 1.23 1.23 1.24 15.75 1.0500 (x-7) 0.2355 0.0006 0.1258 0.0273 0.4028 0.0131 0.0508 0.0133 0.0461 0.0181 0.0554 0.0554 0.0551 0.0181 0.1124 1.2300 SSX 0.16 0.07 0.06 0.02 0.00 0.00 0.00 0.00 0.00 0.02 0.02 0.02 0.03 0.03 0.04 0.4732 SSY (x-x)(y-7) 0.1941 0.0066 -0.0851 0.0248 0.0000 0.0000 0.0000 -0.0023 0.0107 0.0189 -0.0329 -0.0353 0.0422 0.0242 -0.0637 0.1022 SXY Mean sample size : 15 1.230 1.0453 y = 1.0500 SO (X-X) = S. 2(x-7)(y-7)= slope- B. 5./5.- intercept= . = V-B,X= Least square line equation: û =0.96314+0.08309*x 0.102 0.08309 co 0.96314

SSE =Syy-(Sxy)2/Sxx=

0.464708

	s2 =SSE/(n-2)=	0.0357
std error σ              =	=se =√s2=	0.1891

predicted value at X=1.2 is:0.083*1.2+0.963=

1.06

a)

std error of CI=s*√(1/n+(x0-x̅)2/Sxx)=		0.0555
for 95 % CI value of t=		2.160
margin of error E=t*std error=		0.120
lower confidence bound=xo-E=		0.943
Upper confidence bound=xo+E=		1.183

95% confidence interval =(0.943 , 1.183)

b)

std error of PI=s*√(1+1/n+(x0-x̅)2/Sxx)=		0.1970
for 95 % CI value of t=		2.160
margin of error E=t*std error=		0.426
lower confidence bound=xo-E=		0.637
Upper confidence bound=xo+E=		1.489

95% prediction interval =(0.637 , 1.489)