Introduction to Probability and Statistics for Engineers and Scientists

*9.10Multiple Linear Regression 407

TABLE 9.6

Annealing Temperature

Hardness Copper Content (units of 1,000◦F)

79.2 .02 1.05

64.0 .03 1.20

55.7 .03 1.25

56.3 .04 1.30

58.6 .10 1.30

84.3 .15 1.00

70.4 .15 1.10

61.3 .09 1.20

51.3 .13 1.40

49.8 .09 1.40

whereb 0 ,...,bkare the values of the least squares estimatorsB 0 ,B 1 ,...,Bk, andssris the
value ofSSR.

EXAMPLE 9.10d A steel company is planning to produce cold reduced sheet steel consisting
of .15 percent copper at an annealing temperature of 1,150 (degrees F), and is interested
in estimating the average (Rockwell 30-T) hardness of a sheet. To determine this, they
have collected the data shown in Table 9.6 on 10 different specimens of sheet steel having
different copper contents and annealing temperatures. Estimate the average hardness and
determine an interval in which it will lie with 95 percent confidence.

SOLUTION To solve this, we first run Program 9.10, which gives the results shown in
Figures 9.18, 9.19, and 9.20.
Hence, a point estimate of the expected hardness of sheets containing .15 percent
copper at an annealing temperature of 1,150 is 69.862. In addition, sincet.025,7=2.365,
a 95 percent confidence interval for this value is

69.862±4.083 ■

When it is only a single experiment that is going to be performed at the input levels
x 1 ,...,xk, we are usually more concerned with predicting the actual response than its
mean value. That is, we are interested in utilizing our data setY 1 ,...,Ynto predict

Y(x)=

∑k

i= 0

βixi+e, wherex 0 = 1

A point prediction is given by

∑k
i= 0 BixiwhereBiis the least squares estimator ofβibased
on the set of prior responsesY 1 ,...,Yn,i=1,...,k.