Question

The results of the blood pressure measurements taken of the same patient by fourteen different second-year medical students are listed in the table below.

                         Systolic        Diastolic

                            138              82

                            130              91

                            135              100

                            140              100

                            120              80

                            125              90

                            120              80

                            130              80

                            130              80

                            144              98

                            143              105

                            140              85

                            130              70

                            150              100

a. Use a 5% level of significance to test whether there is a significant linear correlation between the two variables {use hypothesis test for the slope}

b. For a given systolic blood pressure of 125, identify the 95% prediction interval estimate for the slope (diastolic blood pressure). Write a statement interpreting your interval in the context of this example.

Answer 1

x	y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
138	82	16.58	44.13	-27.05
130	91	15.43	5.56	-9.26
135	100	1.15	128.98	12.17
140	100	36.86	128.98	68.95
120	80	194.01	74.70	120.38
125	90	79.72	1.84	-12.12
120	80	194.01	74.69898	120.3827
130	80	15.43	74.69898	33.954
130	80	15.43	74.70	33.95
144	98	101.43	87.56	94.24
143	105	82.29	267.56	148.38
140	85	36.86	13.27	-22.12
130	70	15.43	347.56	73.24
150	100	258.29	128.98	182.53

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	1875.00	1241.00	1062.93	1453.21	817.6
mean	133.93	88.64	SSxx	SSyy	SSxy

a)

sample size ,   n =   14          
here, x̅ = Σx / n=   133.929   ,     ȳ = Σy/n =   88.643  
                  
SSxx =    Σ(x-x̅)² =    1062.9286          
SSxy=   Σ(x-x̅)(y-ȳ) =   817.6          
                  
estimated slope , ß1 = SSxy/SSxx =   817.6   /   1062.929   =   0.76924
                  
intercept,   ß0 = y̅-ß1* x̄ =   -14.37981          
                  
so, regression line is   Ŷ =   -14.380   +   0.769   *x
                  

SSE=   (SSxx * SSyy - SS²xy)/SSxx =    824.2540
      
std error ,Se =    √(SSE/(n-2)) =    8.2878

b)

X Value=   125              
Confidence Level=   95%              
                  
                  
Sample Size , n=   14              
Degrees of Freedom,df=n-2 =   12              
critical t Value=tα/2 =   2.179   [excel function: =t.inv.2t(α/2,df) ]          
                  
X̅ =    133.93              
Σ(x-x̅)² =Sxx   1062.93              
Standard Error of the Estimate,Se=   8.2878              
                  
Predicted Y at X=   125   is          
Ŷ =   -14.3798   +   0.7692   *125=   81.7747

For Individual Response Y                  
standard error, S(ŷ)=Se*√(1+1/n+(X-X̅)²/Sxx) =   8.874              
margin of error,E=t*std error=t*S(ŷ)=    2.179   *   8.874   =   19.3345
                  
Prediction Interval Lower Limit=Ŷ -E =   81.775   -   19.335   =   62.4402
Prediction Interval Upper Limit=Ŷ +E =   81.775   +   19.335   =   101.1092