Question

1. Ten steel rods of nominally the same composition and diameter were subjected to various tensile forces (x, in thousands of pounds), and elongation (y, in thousands of an inch) of the steel rods was observed. The following data was observed: Σ!! Xi-458, Σ i X- 260.46, 201y630, 1 48735.1, 21 ry 3558.42. (a) Fit the least squares line that will enable us to predict elongation in terms of tensile force. For (b) and (c), assume independent normally distributed errors with mean zero and common variance. (b) Find a 99% confidence interval for the expected change in elongation corresponding to a 1000 pounds increase in force (notice, that this means a unit change in x) Hint: A CI for the slope parameter is actually requested (c) Find a 95% confidence interval for the expected elongation corresponding to a force of x* = 5 thousands of pounds. Also, find a prediction interval for elongation corresponding to the same value of the force. Which interval is wider? Does it make sense?

Answer 1

a)

	X	Y	XY	X²	Y²
total sum	45.800	630.000	3558.42	260.460	48735.10
mean	4.5800	63.0000

sample size ,   n =   10          
here, x̅ =   4.5800       ȳ =   63  
                  
SSxx =    Σx² - (Σx)²/n =   50.696          
SSxy=   Σxy - (Σx*Σy)/n =   673.020          
SSyy =    Σy²-(Σy)²/n =   9045.100          
slope ,    ß1 = SSxy/SSxx =   13.2756          
                  
intercept,   ß0 = y̅-ß1* x̄ =   2.1977          
                  
so, regression line is   Ŷ =   2.1977   +   13.2756   *x

b)

SSE=   (Sx*Sy - S²xy)/Sx =    110.353
      
std error ,Se =    √(SSE/(n-2)) =    3.714

X Value   1
Confidence Level   99%
  
Intermediate Calculations  
Sample Size , n=   10
Degrees of Freedom,df=n-2 =   8
critical t Value   3.355
X̅ =    4.580
Σ(x-x̅)² =Sxx   50.696
Standard Error of the Estimate,Se=   3.7140

Predicted Y (YHat) at x=1 is 15.4733

standard error, S(ŷ)=Se*√(1/n+(X-X̅)²/Sxx) = 2.206

margin of error,E=t*std error=t*S(ŷ)= 7.4022

Confidence Lower Limit=Ŷ -E =   8.0712
Confidence Upper Limit=Ŷ +E =   22.8755

c)

Predicted Y (YHat) aat x=5 is 68.5758
margin of error,E=t*std error=t*S(ŷ)= 7.4022

Confidence Lower Limit=Ŷ -E =   65.8207
Confidence Upper Limit=Ŷ +E =   71.3308

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prediction interval

standard error, S(ŷ)== 2.206

margin of error,E=t*Se*√(1+1/n+(X-X̅)²/Sxx) = 8.9968

Prediction Interval Lower Limit=Ŷ -E =   59.5789
Prediction Interval Upper Limit=Ŷ +E =   77.5726

prediction interval is wider than confidence interval