Question

An article in the Journal of Sound and Vibration (Vol. 151, 1991, pp. 383–394) described a study investigating the relationship between noise exposure and hypertension. The following data are representative of those reported in the article.

y	x		y	x
1	60		5	85
0	63		4	89
1	65		6	90
2	70		8	90
5	70		4	90
1	70		5	90
4	80		7	94
6	90		9	100
2	80		7	100
3	80		6	100

Draw a scatter diagram of y (blood pressure rise in millimeters of mercury) versus x (sound pressure level in decibels). Fit the simple linear regression model. Round your answers to two decimal places (e.g. 98.76).

(a) Find the predicted mean rise in blood pressure level associated with a sound pressure level of 69 decibels.

(b) Compute a 99% CI on this mean response.
,

(c) Compute a 99% PI on a future observation when the sound pressure level is 69 decibels.
,

Answer 1

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	1656	86	3059.2	124.2	533.20
mean	82.80	4.30	SSxx	SSyy	SSxy

sample size ,   n =   20          
here, x̅ = Σx / n=   82.80   ,     ȳ = Σy/n =   4.30  
                  
SSxx =    Σ(x-x̅)² =    3059.2000          
SSxy=   Σ(x-x̅)(y-ȳ) =   533.2          
                  
estimated slope , ß1 = SSxy/SSxx =   533.2   /   3059.200   =   0.1743
                  
intercept,   ß0 = y̅-ß1* x̄ =   -10.1315          
                  
so, regression line is   Ŷ =   -10.13   +   0.17   *x

................

a)

Predicted Y at X=   69   is                  
Ŷ =   -10.13154   +   0.174294   *   69   =   1.895

b)

standard error, S(ŷ)=Se*√(1/n+(X-X̅)²/Sxx) =    0.442                      
margin of error,E=t*Std error=t* S(ŷ) =   2.8784   *   0.4416   =   1.2710      
                          
Confidence Lower Limit=Ŷ +E =    1.895   -   1.271   =   0.62   
Confidence Upper Limit=Ŷ +E =   1.895   +   1.271   =   3.17   

c)

For Individual Response Y                  
standard error, S(ŷ)=Se*√(1+1/n+(X-X̅)²/Sxx) =   1.3900              
margin of error,E=t*std error=t*S(ŷ)=    2.8784   *   1.39   =   4.0009
                  
Prediction Interval Lower Limit=Ŷ -E =   1.895   -   4.00   =   -2.11
Prediction Interval Upper Limit=Ŷ +E =   1.895   +   4.00   =   5.90

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THANKS

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