Question

#2)    Fourteen second year medical students took blood pressure measurements of the same
         patient and the results are listed below.

Systolic   |    138    130    135    140   120    125     120    130     130     144    143     140     130     150

Diastolic |    82      91      100    100    80      90       80      80       80       98     105      85       70      100

1. Calculate the correlation coefficient.

2. Is there a significant correlation between systolic and diastolic blood pressure?
   Conduct an appropriate test to justify your answer (use a = .05).

3. Find the regression line that predicts diastolic pressure from systolic pressure.

4. Find the standard error

Answer 1

2.

a.

( X)	( Y)	X^2	Y^2	X*Y
138	82	19044	6724	11316
130	91	16900	8281	11830
135	100	18225	10000	13500
140	100	19600	10000	14000
120	80	14400	6400	9600
125	90	15625	8100	11250
120	80	14400	6400	9600
130	80	16900	6400	10400
130	80	16900	6400	10400
144	98	20736	9604	14112
143	105	20449	11025	15015
140	85	19600	7225	11900
130	70	16900	4900	9100
150	100	22500	10000	15000

calculation procedure for correlation
sum of (x) = ∑x = 2155
sum of (y) = ∑y = 1411
sum of (x^2)= ∑x^2 = 291579
sum of (y^2)= ∑y^2 = 126359
sum of (x*y)= ∑x*y = 191123
to caluclate value of r( x,y) = covariance ( x,y ) / sd (x) * sd (y)
covariance ( x,y ) = [ ∑x*y - N *(∑x/N) * (∑y/N) ]/n-1
= 191123 - [ 16 * (2155/16) * (1411/16) ]/16- 1
= 67.434
and now to calculate r( x,y) = 67.434/ (SQRT(1/16*191123-(1/16*2155)^2) ) * ( SQRT(1/16*191123-(1/16*1411)^2)
=67.434 / (9.109*10.973)
=0.675
value of correlation is =0.675
coefficient of determination = r^2 = 0.455
properties of correlation
1. If r = 1 Corrlation is called Perfect Positive Correlation
2. If r = -1 Correlation is called Perfect Negative Correlation
3. If r = 0 Correlation is called Zero Correlation
& with above we conclude that correlation ( r ) is = 0.6747> 0 ,perfect positive correlation              
b.

Given that,
value of r =0.675
number (n)=14
null, Ho: ρ =0
alternate, H1: ρ!=0
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.179
since our test is two-tailed
reject Ho, if to < -2.179 OR if to > 2.179
we use test statistic (t) = r / sqrt(1-r^2/(n-2))
to=0.675/(sqrt( ( 1-0.675^2 )/(14-2) )
to =3.169
|to | =3.169
critical value
the value of |t α| at los 0.05% is 2.179
we got |to| =3.169 & | t α | =2.179
make decision
hence value of | to | > | t α| and here we reject Ho
ANSWERS
---------------
null, Ho: ρ =0
alternate, H1: ρ!=0
test statistic: 3.169
critical value: -2.179 , 2.179
decision: reject Ho
we have enough evidence to support the claim that significant correlation between systolic and diastolic blood pressure.
c.

Line of Regression Y on X i.e Y = bo + b1 X
X	Y	(Xi - Mean)^2	(Yi - Mean)^2	(Xi-Mean)*(Yi-Mean)
138	82	16.576	44.128	-27.046
130	91	15.434	5.556	-9.26
135	100	1.148	128.984	12.168
140	100	36.862	128.984	68.953
120	80	194.006	74.7	120.383
125	90	79.72	1.842	-12.117
120	80	194.006	74.7	120.383
130	80	15.434	74.7	33.954
130	80	15.434	74.7	33.954
144	98	101.433	87.555	94.239
143	105	82.29	267.555	148.382
140	85	36.862	13.271	-22.118
130	70	15.434	347.558	73.24
150	100	258.29	128.984	182.524

calculation procedure for regression
mean of X = ∑ X / n = 133.9286
mean of Y = ∑ Y / n = 88.6429
∑ (Xi - Mean)^2 = 1062.929
∑ (Yi - Mean)^2 = 1453.22
∑ (Xi-Mean)*(Yi-Mean) = 817.639
b1 = ∑ (Xi-Mean)*(Yi-Mean) / ∑ (Xi - Mean)^2
= 817.639 / 1062.929
= 0.769
bo = ∑ Y / n - b1 * ∑ X / n
bo = 88.6429 - 0.769*133.9286 = -14.379
value of regression equation is, Y = bo + b1 X
Y'=-14.379+0.769* X          
d.

Standard Error of Y on X i.e Y = bo + b1 X
Xi	Yi	Y'=-14.38+0.77*X	Y-Y'	(Y-Yi)^2
138	82	91.743	-9.743	94.926
130	91	85.591	5.409	29.257
135	100	89.436	10.564	111.598
140	100	93.281	6.719	45.145
120	80	77.901	2.099	4.406
125	90	81.746	8.254	68.129
120	80	77.901	2.099	4.406
130	80	85.591	-5.591	31.259
130	80	85.591	-5.591	31.259
144	98	96.357	1.643	2.699
143	105	95.588	9.412	88.586
140	85	93.281	-8.281	68.575
130	70	85.591	-15.591	243.079
150	100	100.971	-0.971	0.943

Standard error = Sqrt( ( ∑ Y -Yi )^2/ n-2 )
∑ Y -Yi )^2 = 824.267
Standard Error = 8.288
Standard Error^2 = 66.304