Question

Consider again the conditions of Exercise 12, If a specimen of the alloy is to be produced at the temperature \({\bf{x}} = {\bf{3}}.{\bf{25}}\), what is the predicted value of the durability of the specimen, and what is the M.S.E. of this prediction?

Short answer

The value of the durability of the specimen is \(\hat y(3.25) = 42.673\) and\({\mathop{\rm M}\nolimits} .S.E(\hat y(3.25)) = 1.220{\sigma ^2}\)

Step-by-step solution

Obtain the regression line

The least-squares regression line for the problem mentioned in this exercise was found to be

\(y = 40.892 + 0.548x\)

The temperature at which the prediction is made is\(x = 3.25\).

Substituting \(x = 3.25\) in regression equation to obtain the predicted value:

\(\begin{array}{c}\hat y(3.25) = 40.892 + 0.548 \times 3.25\\ = 42.673.\end{array}\)

Determine the value of the durability of the specimen.

The MSE of this prediction is given as,

\({\bf{MS}}{{\bf{E}}_{{\bf{\hat y}}}}{\bf{ = }}{{\bf{\sigma }}^{\bf{2}}}\left( {{\bf{1 + }}\frac{{\bf{1}}}{{\bf{n}}}{\bf{ + }}\frac{{{{{\bf{(x - \bar x)}}}^{\bf{2}}}}}{{{\bf{s}}_{\bf{x}}^{\bf{2}}}}} \right)\)

From exercise 12, \(\bar x = 2.25\), \(s_x^2 = 10.5,n = 8\).

Substitute the values to obtain the MSE of this prediction as,

\(\begin{array}{c}MSE(\hat y(3.25)) = {\sigma ^2}\left( {1 + \frac{1}{n} + \frac{{{{(3.25 - \bar x)}^2}}}{{s_x^2}}} \right)\\ = {\sigma ^2}\left( {1 + \frac{1}{8} + \frac{{{{(3.25 - 2.25)}^2}}}{{10.5}}} \right)\\ = 1.220{\sigma ^2}\end{array}\)

Therefore, the value of the durability of the specimen is \(\hat y(3.25) = 42.673\)\({\rm{and }}{\mathop{\rm MSE}\nolimits} (\hat y(3.25)) = 1.220{\sigma ^2}\)