Question

In: Statistics and Probability

The results of the blood pressure measurements taken of the same patient by fourteen different second-year...

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.

Solutions

Expert Solution

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


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