The efficiency for a steel specimen immersed in a phosphating tank is the weight of the phosphate.
a) The equation of the estimated regression line is ŷ = - 16.592 + 0.1017x.
b) A point estimate is 1.9174 for true average efficiency ratio when tank temperature is 182.
c) The values of the residuals from the least squares line for the four observations for which temperature is 182 are -1.0774, -0.1874, 0.1226, 0.7826..
We have given the accompanying data on tank temperature (x) and efficiency ratio (y).
a) We have to find out the equation of the estimated regression line. Consider the calculations shown in the above table:
ΣΧ= 4308 ; ΣΥ= 39.99 ; ΣΧΥ = 7229.47 ;
ΣX² = 773790
Determine the slope and intercepts using the following formulas,
Slope b₁ = (n ΣΧΥ - ΣX ΣΥ)/(nΣΧ² - (ΣX)²)
=> b₁ = [24 (7229.47) - (4308) (39.99)]/(24(773790)-(4308)²)
=> b₁ = 1230.36/12096 = 0.10171626984
=> b₁ ≈ 0.1017
Intercept, b₀ = ΣY /n - b₁ΣΧ /n
=> b₀ = 39.99/24 - 0.1017×4308/24
= -16.592
Thus, the estimated regression equation of the form, ŷ = b₀ + b₁ x
ŷ = - 16.592 + 0.1017x
b) Calculate the point estimate for true average efficiency ratio when tank temperature is 182.
y = -16.592 + 0.1017x
=> y= -16.592 +0.1017 (182)
=> y= 1.9174
Therefore, the point estimate for true average efficiency ratio when tank temperature is 182 is 1.9174..
c)We have to find out the values of the residuals from the least squares line for the four observations for which temperature is 182.
From part (b), the value of ŷ is 1.9174.
The table below shows the residuals.
Vi ŷ e = y₁ - y
0.84 1.9174 -1.0774
1.73 1.9174 -0.1874
2.04 1.9174 0.1226
2.70 1.9174. 0.7826
Hence, we got all the required values.
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