Question

Regression methods were used to analyze the data from a study investigating the relationship between roadway surface temperature (x) and pavement deflection (y). Summary quantities were n = 20, ∑?! = 12.75, ∑?!" = 8.86, ∑?! = 1478, ∑?!" = 143215.8, ∑?!?! = 1083.67.

Give a 95% confidence interval for the mean response of pavement deflection given that temperature is 90 F.

Answer 1

sample size ,   n =   20          
here, x̅ =Σx/n =   73.9000   ,   ȳ = Σy/n =   0.6375  
                  
SSxx =    Σx² - (Σx)²/n =   33991.600          
SSxy=   Σxy - (Σx*Σy)/n =   141.445          
SSyy =    Σy²-(Σy)²/n =   0.732          
estimated slope , ß1 = SSxy/SSxx =   141.445   /   33991.600   =   0.0042
                  
intercept,   ß0 = y̅-ß1* x̄ =   0.3300          
                  
so, regression line is   Ŷ =   0.33   +   0.0042   *x

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Predicted Y at X=   90   is                  
Ŷ =   0.330   +   0.004   *   90   =   0.704

standard error, S(ŷ)=Se*√(1/n+(X-X̅)²/Sxx) =    0.021              
margin of error,E=t*Std error=t* S(ŷ) =   2.1009   *   0.0214   =   0.0450
                  
Confidence Lower Limit=Ŷ +E =    0.704   -   0.0450   =   0.659
Confidence Upper Limit=Ŷ +E =   0.704   +   0.0450   =   0.749

Please revert in case of any doubt.

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