CHAPTER
EIGHT
Linear
Regression
8.1 Scatter Diagram
Example 8.1 A chemical engineer is
investigating the effect of process operating temperature (
) on product yield (
). The study results in the following data:
|
|
100 |
110 |
120 |
130 |
140 |
150 |
160 |
170 |
180 |
190 |
|
|
45 |
51 |
54 |
61 |
66 |
70 |
74 |
78 |
85 |
89 |
(Hines and Montgomery, 1990, p
457) Check if there is any linear relationship between temperature and product
yield.
Solution Make a data file with 
as Var1 and 
as Var2. Then follow
the steps in Statistica to get a scatter diagram.
- Graphs
- Scatterplot
- Click Variables (Select Var1 and Var2) / OK
- Click Advanced (In Graph Type click Regular &
in Fit click Off)
- OK
Since
the scatter diagram (See Figure 8.1) between temperature (Var1) and product
yield (Var2) shows a linear trend, one recommends estimating the line of best
fit.
Figure 8.1 Scatterplot
8.2
The Correlation Coefficient
The strength of linear relationship between 
is measured by the
correlation coefficient, defined by
where
In Example 8.1, we have
, 
, 
8.3 Estimating the Line of
Best Fit
The
simple linear regression model is of the form
where
is the conditional
mean of y at x,
y and x
are respectively the dependent (response) and independent (explanatory) variables,
is the random error
component,
and 
are the 
intercept and the slope of the regression line respectively.
The least squares
estimators of the regression parameters are given by
and 
Once the parameters are estimated, the
equation 
will be called the estimated
regression line, the prediction line, the line of best fit or the least squares line. It should be noted that 
can be used as a point estimate of 
the conditional mean of y at 
,
or a predictor of the response at .
For
the data in Example 8.1,
and 
so
that the line of best fit is given by
.
At
a temperature of 1400 C, we predict the yield to be
Estimating
and 
Using Statistica
To estimate and 
by using Basic Statistics and Tables
Module, we can find the estimates of and by simply plotting a
scatter graph of the dependent variable against the independent variable with a
linear fit. While you are in Basic
Statistics and Tables Module follow the steps:
- Enter the values of
in one column, say
Var1 and the corresponding 
values in another
column, say Var2 - Graphs / Stats 2D Graphs / Scatterplots
- Variables / = Var1 and = Var2 / OK
- In advance, Select Regular for Graph
Type and Linear for Fit
- OK
For the data in Example 8.1
you will get 
(Figure 8.2) which
means that the estimates of and are given by 
and 
respectively. Thus, the predicted simple linear regression
model is given by
Figure 8.2 A Graph of Y versus X
8.4 Sources of
Variation
The
variation in the dependent variable say 
is attributed partly
to that in the independent variable. The rest is attributed to what is called the
Sums of Squares Due to Errors defined by 
and can be calculated
by the following table:
|
|
|
|
|
|
|
100 |
45 |
45.5636 |
|
0.3176 |
|
110 |
51 |
50.3939 |
|
0.3674 |
|
120 |
54 |
55.2242 |
|
1.4987 |
|
130 |
61 |
60.0545 |
0.9455 |
0.8940 |
|
140 |
66 |
64.8848 |
|
1.2437 |
|
150 |
70 |
69.7151 |
|
0.0812 |
|
160 |
74 |
74.5454 |
|
0.2975 |
|
170 |
78 |
79.3757 |
|
1.8926 |
|
180 |
85 |
84.2060 |
|
0.6304 |
|
190 |
89 |
89.0363 |
|
0.0013 |
|
|
|
|
|
7.2244 |
The
SSE = 7.2244, here compare these errors (or residuals) with that
obtained by Statistica.
Predicted
and Residual Values Using Statistica
Follow
the steps:
- Statistics / Multiple regression
- Variables (select the dependent and independent variables) / OK / OK
- Click Residual / assumption / prediction
- Click perform residual analysis
- In advanced click Summary: Residual & predicted (get
Figure 8.3)
Figure 8.3 Predicted
and Residual Values
Decomposition
of the Sum of Squares
It can be proved that 
where 
is the Total Sum of
Squares, 
is the Sum of Squares
due to Regression and 
is the Sum of Squares
of Errors, also known as the residual sum of squares.
The coefficient of determination is defined by
In Example 8.1,
, 
so that 
= 7.2243. Note that the expression 
may not be
computationally efficient.
The coefficient of determination 
Calculation
of Sums of Squares Using Statistica
To calculate the sum of squares using Statistica follow
the steps:
- Enter the values of in one column, say
Var1 and the corresponding values in another
column, say Var2.
- Statistics / Multiple Regression
- Variables ( select the dependent and
independent variables) / OK /OK
- In Advanced click ANOVA ( Overall
Goodness of Fit)
The resulting spreadsheet of result shows the Total Sum
of Squares (TSS), the sum of squares due to regression (SSR) and the sum of
squares due to errors (SSE), the mean squares and the 
value.
For the data in Example 8.1 above, we have , 
and 
(Figure 8.4).
Figure 8.4 Analysis of Variance
8.5
Confidence Interval Estimation of Regression Parameters
Confidence Interval (CI) for the Slope Parameter
A 
CI for 
is given by 
where 
is the 100(
)th percentile of the t-distribution with 
, and
, the estimate for 
For the data in Example 8.1, 
, and thus a 95% CI for 
is given by
In other words,
.
Confidence Interval for the Conditional Mean
A 100
% CI for the conditional mean at is given by
In example 8.1, a 95% CI for the conditional mean at
140 
is given by
i.e.
The above problem can be solved using Statistica
following the steps:
- Statistics
- Multiple Regression
- Click variables (Choose Dependent and Independent Variable say
Var2 and Var1)
- OK/OK
- Click Residuals/Assumptions/Prediction
- Click Compute Confidence Limits (Checked by default )
- Click Predict Dependent Variable (Under Common value put 140)
Click Apply, then OK
We find that 95%
confidence interval for 
is given by
[64.18146,
65.58823] (See Figure 8.5).
Figure
8.5 Confidence Interval for Mean
8.6 Prediction Interval (PI) for a Future
Observation 
A 
PI for a future
observation at is given by
For the data in Example 8.1 a 95% prediction
interval for the yield at 140 is given by
i.e, 
The
problem can be solved using Statistica following the steps:
- Statistics
- Multiple Regression
- Click variables (Choose Dependent and Independent Variable say
Var2 and Var1)
- OK / OK
- Click Residuals/Assumptions/Prediction
- Click Compute Prediction Limits
- Click Predict Dependent Variable to enter fixed say 140
- Click Apply, then OK
It is predicted
with 95% confidence that product yield will be in the interval
[62.58338,
67.18632] (See Figure 8.6).
Figure: 8.6 Prediction Interval for
Product Yield
8.7 Testing the Slope of the Regression
Line
The
following table has a list of possible null hypotheses involving the slope , the critical
region and the p-value in each case.
Hypotheses
about and their respective
rejection regions and p-values
|
|
Rejection
Region |
p-value |
|
|
|
|
|
|
|
|
|
|
|
|
The test Statistic for these hypotheses is
The hypothesis 
is known as the
hypothesis of the significance of the regression.
In example 8.1, test the hypothesis at significance level 
.
The value of the test statistics is 
Since t = 46.1689 >
= 2.306, we reject 
in favor of the
alternative hypothesis 
, and conclude that the regression is significant.
8.8 Testing the
Significance of the Regression by Analysis
of Variance
In order to test the
hypothesis 
at the 5% significant level,
using an F test, we reproduce here the ANOVA table of example 8.1 shown
in figure 8.4
|
SV |
SS |
DF |
MS |
|
|
Regression |
1924.875758 |
1 |
1924.8757 |
2131.5738 |
|
Error |
7.22424243 |
8 |
0.90303 |
|
|
Total |
1932.10 |
9 |
|
|
The test statistic for the above hypothesis is
.
The observed value of the test statistic is 
. Since > 
, the critical value from the F distribution with 1 and n
– 2 degrees of freedom, we reject 
in favor of the
alternative hypothesis at 5% level of
significance.
Testing
the Significance of the Regression Using Statistica
Using a t-test we
follow the steps:
- Enter the values of in one column,
say Var1 and the corresponding values in another
column, say Var2.
- Statistics / Multiple Regression, to get
Figure 8.7,click Advanced
Figure 8.7 Multiple Regression Setup
- Variables (select the dependent and
independent variables) / OK
- OK ( to get Figure 8.8)
Figure 8.8 Regression Results
- In Advanced, click Summary: Regression results.You will get
Figure 8.9
Figure 8.9 Regression Summary
The regression parameters
are given by and (See Column B of Figure 8.9).
Since the 
for testing the null
hypothesis 
against the alternative hypothesis that 
is given by 0 (See the
column labeled by p-level), we reject the null hypothesis at any 
in favor of the
alternative hypothesis.
Using an F test, follow the steps:
- Enter the values of in one column, say
Var1 and the corresponding values in another
column, say Var2.
- Statistics / Multiple Regression, to get
Figure 8.7, click Advanced
- Variables ( select the dependent and
independent variables) / OK /OK
- In Advanced click ANOVA ( Overall
Goodness of Fit)
For the data in Example 8.1, we have 
(See Figure 8.4) so
that we reject null hypothesis at any and accept the
alternative hypothesis , indicating that the regression of on is significant whether
we are testing at 1% or 5% level of significance.
8.9
Checking Model Assumptions
We now discuss how to verify the assumptions that
the random errors are normally distributed and that they have a constant
variance.
Checking
the Assumption of Normality
To check the assumption that the errors follow a normal
distribution, a normal probability plot of residuals is drawn. If the plot is approximately linear, then the
assumption is justified, otherwise, the assumption is not justified.
In Statistica, to get the normal probability plot of
residuals, we follow the steps below assuming that we have the data of Example 8.1
in the Multiple Regression Module.
- Statistics / Multiple Regression
- Variable / Select the dependent and independent variables / OK
/ OK
- In Residual / assumption / prediction, click Perform residual analysis
- In Quick, click
Since the normal probability plot of residuals (See
Figure 8.10) for the data in Example 8.1 exhibits a linear trend. Thus, the normality assumption is valid.
Figure 8.10
Checking
the Assumption of Constancy of Variance
To check the assumption that the errors have a
constant variance, a graph of the residuals versus the independent variable is plotted. If the graph shows no pattern, then the
assumption is justified. Otherwise, the
assumption is not justified.
In Statistica, to plot the graph of the residuals
versus the independent variable, we proceed as follows assuming that we have
the data in multiple regression module:
- Statistics / Multiple Regression
- Variable / Select the dependent and independent variables / OK
- In Residuals / assumption / prediction, click Perform residual
analysis
- under Type of residual click Raw residuals
- In Residuals, click Residuals vs. independent Var, “select the
independent variable”
The graph in Figure 8.11 is the residual plot for
example 8.1. Since it does not exhibit any pattern, we conclude that the
constant variance assumption is justified. The value 
states that raw
residuals 
and temperatures 
are almost
uncorrelated.
Figure 8.11 Graph of Residuals vs
Independent Variable
8.10
Multiple Linear Regression
The
multiple regression model is a mathematical model which explains the
relationship between the dependent variable and two or more independent
variables. For example, a manufacturer
wants to model the quality of a product 
as a function of temperature 
and pressure 
at which it is
produced.
The
multiple linear regression model with 
independent variables 
is given by
where
and are the intercept and
the random error term respectively. We
shall refer to the 
in the model as the regression parameters.
Example
8.2 Consider the problem of
predicting gasoline mileage (in miles per gallon), where the independent
variables are fuel octane rating 
and average speed (mile
per hour) 
. The sample data
obtained from 20 test runs with cars at various speeds are as follows:
|
|
|
|
24.8 |
88 |
52 |
|
30.6 |
93 |
60 |
|
31.1 |
91 |
28 |
|
28.2 |
90 |
52 |
|
31.6 |
90 |
55 |
|
29.9 |
89 |
46 |
|
31.5 |
92 |
58 |
|
27.2 |
87 |
46 |
|
33.3 |
94 |
55 |
|
32.6 |
95 |
62 |
|
30.6 |
88 |
47 |
|
28.1 |
89 |
58 |
|
25.2 |
90 |
63 |
|
35.0 |
93 |
54 |
|
29.2 |
91 |
53 |
|
31.9 |
92 |
52 |
|
27.7 |
89 |
52 |
|
31.7 |
94 |
53 |
|
34.2 |
93 |
54 |
|
30.1 |
91 |
58 |
Estimate
the linear regression model and interpret your results.
Solution
To solve this problem by using
Statistica, you must be in Multiple Regression module and follow
the steps:
- Enter the values of
each variable in a separate column ( or variable)
- Statistics / Multiple
Regression
- Variables ( select the
dependent variable from the list on the left)
- Hold down the Ctrl key
and select the independent variables from the list on the right.
- OK
- In Advanced Click
Summary: Regression Results
For the data in Example 8.2, one can read the estimates of the regression parameters as : 
(See column labeled B in Figure 8.12).
Figure 8.12 Regression Summary
Thus,
the predicted multiple linear regression model for the given data is
If 'average
speed' ( or Var3) is held fixed, it is estimated that a 1-unit
increase in octane ( or Var2) would result in 1.0193 unit increase in the expected 'gasoline mileage'. Similarly if 'octane' ( or Var2) is held
fixed, it is estimated that a 1-unit increase in average speed ( or Var3) would result in,
unit decrease in the expected 'gasoline mileage'.
Exercises
8.1 (cf.
Devore, J. L., 2000, 510). The following data represent the burner area
liberation rate 
and emission rate
(Nox) 
:
|
|
100 125
125 150 150
200 200 250
250 300 300
350 400 |
|
|
150 140
180 210 190
320 280 400
430 440 390
600 610 |
(a) Assuming that the simple linear
regression model is valid, obtain the least square estimate of the true
regression line.
(b) What is the estimate of the expected Nox
emission rate when burner area liberation rate equals 225?
(c) Estimate the amount by which you expect
Nox emission rate to change when burner area liberation rate is decreased by
50.
8.2 (cf.
Devore, J. L., 2000, 510). The following data represent the wet deposition
(NO3) and lichen N (% dry
weight) :
|
|
0.05 |
0.10 |
0.11 |
0.12 |
0.31 |
0.42 |
0.58 |
0.68 |
0.68 |
0.73 |
0.85 |
|
|
0.48 |
0.55 |
0.48 |
0.50 |
0.58 |
0.52 |
0.86 |
1.0 |
0.86 |
0.88 |
1.04 |
(a) What are the least square estimates of 
?
(b) Predict lichen N for an NO3 deposition
value of 0 .5.
(c) Test the significance of regression at
5% level of significance.
8.3 (Devore,
J. L., 2000, 510). The following data represent 
available travel space
in feet, and 
separation distance:
|
|
12.8 12.9
12.9 13.6 14.5
14.6 15.1 17.5
19.5 20.8 |
|
|
5.5
6.2 6.3 7.0
7.8 8.3 7.1
10.0 10.8 11.0 |
(a) Derive the equation of the estimated
line.
(b) What separation distance would you
predict if available travel space value is 15.0?
8.4 (Devore,
J. L., 2000, 511). Consider the following data set in which the variable of
interest are commuting distance and
commuting time:
|
|
15 16
17 18 19
20 5 10
15 20 25
50 5 10
15 20 25
50 |
|
|
42 45
35 42 49
46 16 32
44 45 63
115 8 16
22 23 31
60 |
Obtain the least square estimate of the regression
model.
8.5
(cf.
Devore, J. L., 2000, 584). Soil and sediment adsorption, the extent to which
chemicals collect in a condensed form on
the surface, is an important characteristic influencing the effectiveness of
pesticides and various agricultural chemicals, The article “Adsorption of
Phosphate, Arsenate, Methancearsonate, and Cacodylate by 
= amount of
extractable iron, and 
= amount of extractable aluminum.
|
x1 |
61 |
175 |
111 |
124 |
130 |
173 |
169 |
169 |
160 |
244 |
257 |
333 |
199 |
|
x2 |
13 |
21 |
24 |
23 |
64 |
38 |
33 |
61 |
39 |
71 |
112 |
88 |
54 |
|
|
4 |
18 |
14 |
18 |
26 |
26 |
21 |
30 |
28 |
36 |
65 |
62 |
40 |
(a) Find the least square estimates of the
parameters and write the equation of the estimated model.
(b) Make a prediction of Adsorption index
resulting from an extractable iron = 250 and extractable aluminum = 55.
(c) Test the null hypothesis that 
against the
alternative hypothesis that 
at 5% level of
significance.
8.6 (Johnson,
R. A., 2000, 345). The following table shows how many weeks a sample of 6
persons have worked at an automobile inspection station and the number of cars
each one inspected between noon and 2 P.M. on a given day:
|
Number of weeks employed (x) |
2 |
7 |
9 |
1 |
5 |
12 |
|
Number of cars inspected (y) |
13 |
21 |
23 |
14 |
15 |
21 |
(a) Find the equation of the least squares
line, which will enable us to predict 
in terms of
.
(b) Use the result of part (a) to estimate
how many cars someone who has been working at the inspection station for 8
weeks can be expected to inspect during the given 2-hour period.
8.7 (cf.
Devore, J. L., 2000, 590). An investigation of die casting process resulted in
the accompanying data on = on furnace temperature, = die close time and y = temperature difference on the die
surface (A Multiple Objective Decision Making Approach for Assessing
Simultaneous Improvement in Die Life and Casting Quality in a Die Casting
Process,” Quality Engineering, 1994: 371-383).
|
x1 |
1250 |
1300 |
1350 |
1250 |
1300 |
1250 |
1300 |
1350 |
1350 |
|
x2 |
6 |
7 |
6 |
7 |
6 |
8 |
8 |
7 |
8 |
|
|
80 |
95 |
101 |
85 |
92 |
87 |
96 |
106 |
108 |
(a) Write the equation of the estimated
model.
(b) Test the null hypothesis that 
against the
alternative hypothesis that 
at 5% level of
significance.
8.8 (cf.
Johnson, R. A., 2000, 334). The following are measurements of the air velocity
and evaporation coefficient of burning fuel droplets in an impulse engine:
|
Air Velocity (cm/sec) |
20 |
60 |
100 |
140 |
180 |
220 |
160 |
300 |
340 |
380 |
|
Evaporation coefficient mm2/sec)
|
0.18 |
0.37 |
0.35 |
0.78 |
0.56 |
0.75 |
1.18 |
1.36 |
1.17 |
1.65 |
(a) Fit a straight line to the data by the
method of least square and use it to estimate the evaporation coefficient of a
droplet when the air velocity is 190 cm/s.
(b) Test the null hypothesis that against the
alternative hypothesis at the 0.05 level of significance.
8.9 (Johnson,
R. A., 2000, 344). A chemical company, wishing to study the effect of
extraction time on the efficiency on an extraction operation, obtained the data
shown in the following table;
|
Extraction time (minutes) (x) |
27 |
45 |
41 |
19 |
35 |
39 |
19 |
49 |
15 |
31 |
|
Extraction efficiency (%) (y) |
57 |
64 |
80 |
46 |
62 |
72 |
52 |
77 |
57 |
68 |
(a) Draw a scattergram to verify that a
straight line will provide a good fit to the data.
(b) Draw a straight line to predict the
extraction efficiency one can expect when the extraction time is 35 minutes.
8.11 (cf.
Johnson, R. A., 2000, 347). The cost of manufacturing a lot of certain product
depends on the lot size, as shown by the following sample data:
|
Cost (Dollars) |
30 |
70 |
140 |
270 |
530 |
1010 |
2500 |
5020 |
|
|
1 |
5 |
10 |
25 |
50 |
100 |
250 |
500 |
(a) Draw a scattergram to verify the
assumption that the relationship is linear, letting lot size be and cost .
(b) Fit a straight line to these data by the
method of least squares, using lot size as the independent variable, and draw
its graph on the diagram obtained in part (a).
8.12 (Johnson, R. A., 2000, 345). The
following table, is the tensile force
applied to a steel specimen in thousands of pounds, and is the resulting
elongation thousands of an inch:
|
|
1 |
2 |
3 |
4 |
5 |
6 |
|
|
14 |
33 |
40 |
63 |
76 |
85 |
(a) Graph the data to verify that it is
reasonable to assume that the regression of
on is linear.
(b) Find the equation of the least square
line, and use it to predict the elongation when the tensile force is 3.5
thousand pounds.
8.13 (Johnson,
R. A., 2000, 385). The following are the data on the number of twists required
to break a certain kind of forged alloy bar and the percentage of two alloying
elements present in the metal;
|
No. of twists ( |
41 |
49 |
69 |
65 |
40 |
50 |
58 |
57 |
31 |
36 |
44 |
57 |
19 |
31 |
33 |
43 |
|
%age Element A ( |
1 |
2 |
3 |
4 |
1 |
2 |
3 |
4 |
1 |
2 |
3 |
4 |
1 |
2 |
3 |
4 |
|
%age Element B ( |
5 |
5 |
5 |
5 |
10 |
10 |
10 |
10 |
15 |
15 |
15 |
15 |
20 |
20 |
20 |
20 |
Fit a least square regression plane and use its
equation to estimate the number of twists required to break one of the bars
when = 2.5 and = 12.
8.14 (Johnson, R. A., 2000, 263). Twelve
specimens of cold-reduced sheet steel, having different copper contents and
annealing temperatures, are measured for hardness with the following results:
|
Hardness |
78.9 |
65.1 |
55.2 |
56.4 |
80.9 |
69.7 |
57.4 |
55.4 |
85.3 |
71.8 |
60.7 |
58.9 |
|
Copper
content |
0.02 |
0.02 |
0.02 |
0.02 |
0.10 |
0.10 |
0.10 |
0.10 |
0.18 |
0.18 |
0.18 |
0.18 |
|
Annealing
Temp. |
1000 |
1100 |
1200 |
1300 |
1000 |
1100 |
1200 |
1300 |
1000 |
1100 |
1200 |
1300 |
Fit an equation of the form 
, where represents the copper content,
represents the
annealing temperature, and y represents the hardness.
8.15 Suppose the following data gives the mass
of adults in kilograms sampled from
three villages 
|
|
|
|
|
71.5
68.0 |
62.0
65.0 |
60.5
62.8 |
|
63.3
73.5 |
71.3
73.1 |
58.4
58.7 |
|
73.6
74.1 |
64.7
72.5 |
65.5
58.1 |
|
78.1
73.6 |
66.0
73.0 |
50.6
58.5 |
|
66.5
76.5 |
72.6
66.8 |
62.1
52.6 |
|
74.3
73.3 |
71.1
71.9 |
64.5
58.8 |
|
76.3
76.0 |
65.4
65.0 |
60.6
63.7 |
|
62.0
64.0 |
59.1
69.9 |
62.9 61.6 |
|
62.8
69.6 |
69.7
69.8 |
60.2
67.2 |
|
72.9
69.2 |
77.1
78.5 |
63.4
58.2 |
(a) Assuming that these samples are
independent, run t-tests to determine which of the villages have
identical mean mass of adults stating clearly the hypotheses you are
testing. State your conclusions based on
the 
as well as the t-value.
(b) State the assumption under which your
tests are valid.
8.16
(cf.
Dougherty, 1990, 595) When smoothing a surface with an abrasive, the roughness
of the finished surface decreases as the abrasive grain becomes finer. The
following data give measurements of surface roughness (in micrometers) in terms of the grit numbers
of the grains, finer grains possessing larger grit numbers.
|
|
24 |
30 |
36 |
46 |
54 |
60 |
|
|
0.34 |
0.30 |
0.28 |
0.22 |
0.19 |
0.18 |
(a) Draw a scatter diagram. Do you recommend fitting a
linear regression model?
(b) How strong is the linear correlation between the
two variables?
(c) Do you think that there is strong nonlinear
correlation between the two variables?
8.17
(cf.
Johnson, R. A., 2000, 578). The article “ How to optimize and Control the Wire
Bonding Process: Part II” (Solid State Technology, Jan. 1991: 67-72) described
on experiment carried out to assess the impact of the variable = force (gm), = power (mw), 
= temperature (
), and 
= time (ms) on y =
ball bond share strength (gm). The following data generated to be consistent
with the information given in the article:
|
Observations |
Force |
Power |
Temperature |
Time |
Strength |
|
1 |
30 |
60 |
175 |
15 |
26.2 |
|
2 |
40 |
60 |
175 |
15 |
26.3 |
|
3 |
30 |
90 |
175 |
15 |
39.8 |
|
4 |
40 |
90 |
175 |
15 |
39.7 |
|
5 |
30 |
60 |
225 |
15 |
38.6 |
|
6 |
40 |
60 |
225 |
15 |
35.5 |
|
7 |
30 |
90 |
225 |
15 |
48.8 |
|
8 |
40 |
90 |
225 |
15 |
37.8 |
|
9 |
30 |
60 |
175 |
25 |
26.6 |
|
10 |
40 |
60 |
175 |
25 |
23.4 |
|
11 |
30 |
90 |
175 |
25 |
38.6 |
|
12 |
40 |
90 |
175 |
25 |
52.1 |
|
13 |
30 |
60 |
225 |
25 |
39.5 |
|
14 |
40 |
60 |
225 |
25 |
32.3 |
|
15 |
30 |
90 |
225 |
25 |
43.0 |
|
16 |
40 |
90 |
225 |
25 |
56.0 |
|
17 |
25 |
75 |
200 |
20 |
35.2 |
|
18 |
45 |
75 |
200 |
20 |
46.9 |
|
19 |
35 |
45 |
200 |
20 |
22.7 |
|
20 |
35 |
105 |
200 |
20 |
58.7 |
|
21 |
35 |
75 |
150 |
20 |
34.5 |
|
22 |
35 |
75 |
250 |
20 |
44.0 |
|
23 |
35 |
75 |
200 |
10 |
35.7 |
|
24 |
35 |
75 |
200 |
30 |
41.8 |
|
25 |
35 |
75 |
200 |
20 |
36.5 |
|
26 |
35 |
75 |
200 |
20 |
37.6 |
|
27 |
35 |
75 |
200 |
20 |
40.3 |
|
28 |
35 |
75 |
200 |
20 |
46.0 |
|
29 |
35 |
75 |
200 |
20 |
27.8 |
|
30 |
35 |
75 |
200 |
20 |
40.3 |
(a) Find the least square estimates of the
parameters and write the equation of the estimated model.
(b) Make a prediction of strength resulting
from a force of 35 gm, power of 75 mw, temperature of 200 degrees and time of
20 ms.
(c) Test the null hypothesis that 
against the
alternative hypothesis that 
at 5% level of
significance.