Question

Listed below are systolic blood pressure measurements​ (in mm​ Hg) obtained from the same woman. Find the regression​ equation, letting the right arm blood pressure be the predictor​ (x) variable. Find the best predicted systolic blood pressure in the left arm given that the systolic blood pressure in the right arm is

8585

mm Hg. Use a significance level of

0.050.05.

Right Arm	100	99	92	79	80
Left Arm	176	170	145	144	146

n	alphaαequals=0.05	alphaαequals=0.01	​NOTE: To test Upper H 0H0​: rhoρequals=0 against Upper H 1H1​: rhoρnot equals≠​0, reject Upper H 0H0 if the absolute value of r is greater than the critical value in the table.
4	0.950	0.990
5	0.878	0.959
6	0.811	0.917
7	0.754	0.875
8	0.707	0.834
9	0.666	0.798
10	0.632	0.765
11	0.602	0.735
12	0.576	0.708
13	0.553	0.684
14	0.532	0.661
15	0.514	0.641
16	0.497	0.623
17	0.482	0.606
18	0.468	0.590
19	0.456	0.575
20	0.444	0.561
25	0.396	0.505
30	0.361	0.463
35	0.335	0.430
40	0.312	0.402
45	0.294	0.378
50	0.279	0.361
60	0.254	0.330
70	0.236	0.305
80	0.220	0.286
90	0.207	0.269
100	0.196	0.256

PrintDone

What is the regression equation?

Answer 1

Solution:

In excel, we use data analysis in data menu and then regression. Right arm as x and left arm as y.

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.8582
R Square	0.7365
Adjusted R Square	0.6487
Standard Error	9.1865
Observations	5
ANOVA
	df	SS	MS	F	Significance F
Regression	1	707.6256	707.6256	8.3850	0.0627
Residual	3	253.1744	84.3915
Total	4	960.8000
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
Intercept	37.3823	41.2377	0.9065	0.4315	-93.8546	168.6191
X Variable 1	1.3202	0.4559	2.8957	0.0627	-0.1307	2.7711

The regression equation is Y = 37.3823 + 1.3202X

When X = 85, the predicted systolic blood pressure in the left arm would be

Y = 37.3823 + 1.3202 (85)

Y = 149.60

---------------------------------------------------------------------------------------------------------------------------------------

Null Hypothesis (Ho): = 0

Alternative Hypothesis (Ha): 0

Degrees of freedom, = n - 2 = 5 - 2 = 3

Test Statistics  

2.90

Using t-tables, the critical value at df = 3 and a = 0.05/2 = 0.025 is 3.182

Since test statistics lie within the critical values, we fail to reject Ho.

Hence, we can conclude that the correlation coefficient is zero.