An article in Optical Engineering1 Read the article by Operating Curve Extraction of a Correlator’s Filter(Letendre, Bergeron, and Arsenault 2004Letendre, Mathias, Alain Bergeron, and Henri H Arsenault. 2004. “Operating Curve Extraction of a Correlator’s Filter.”Optical Engineering 43 (11): 2775–79.) reported on the use of an optical correlator to perform an experiment by varying brightness and contrast. The resulting modulation is characterized by the useful range of gray levels. The data follow:
Table 1.1 Useful Range Data
Useful Range (ng) \(y_i\)
Brightness % (\(x_i1\))
Contrast %(\(x_{i2}\))
96
54
56
50
61
80
50
65
70
112
100
50
96
100
65
80
100
80
155
50
25
144
57
35
255
54
26
Matrix of Scatter plots for the Useful Range Data in Table 1.1
## Loading required package: scatterplot3d
Questions
Fit a multiple linear regression model to these data.
Estimate \(σ^2\).
Compute the standard errors of the regression coefficients.
Predict the useful range when brightness = 80 and contrast = 75.
Test for significance of regression using α=0.05. What is the P-value for this test?
Construct a t-test on each regression coefficient. What conclusions can you draw about the variables in this model? Use α=0.05.
Answer to Question 1:
Question: Fit a multiple linear regression model to these data.
Extending the table using the values of Brightness percentage, contrast percentage and useful range of gray levels:
Table 1.2 Extended Version of Table 1.1 Useful Range Data
Observation Number
Useful Range (ng) \(y_i\)
Brightness % (\(x_i1\))
Contrast %(\(x_{i2}\))
\(x_{i1}x_{i2}\)
\({x_{i1}}^2\)
\({x_{i2}}^2\)
\(x_{i1}y_i\)
\(x_{i2}y_i\)
1
96
54
56
3024
2916
3136
5184
5376
2
50
61
80
4880
3721
6400
3050
4000
3
50
65
70
4550
4225
4900
3250
3500
4
112
100
50
5000
10000
2500
11200
5600
5
96
100
65
6500
10000
4225
9600
6240
6
80
100
80
8000
10000
6400
8000
6400
7
155
50
25
1250
2500
625
7750
3875
8
144
57
35
1995
3249
1225
8208
5040
9
255
54
26
1404
2916
676
13770
6630
From the data in the Table 1.2, we could calculate that:
\(n=9\)
\(\sum_{i=1}^{9}y_{i} = 1038\)
\(\sum_{i=1}^{9}x_{i1} = 641\)
\(\sum_{i=1}^{9}x_{i2} = 487\)
\(\sum_{i=1}^{9}x_{i1}x_{i2} = 36603\)
\(\sum_{i=1}^{9}x_{i1}^2 = 49527\)
\(\sum_{i=1}^{9}x_{i2}^2 = 30087\)
\(\sum_{i=1}^{9}x_{i1}y_i = 70012\)
\(\sum_{i=1}^{9}x_{i2}y_i = 46661\)
For the model \(Y=\beta_0\ + \beta_1x_1 +\beta_2x_2 + \epsilon\), the least square normal equations on the book of2 Equations on Chap 12.1(Montgomery and Runger 2010Montgomery, Douglas C, and George C Runger. 2010. Applied Statistics and Probability for Engineers. John Wiley & Sons.) are:
Practical Interpretation: The fitted equation above can be used in predicting the userful range of gray level, given that the value for the regressor variables, which are the brightness percentage \((x_1)\) and contrast percentage, are known.
# Answer to Question 2:
Estimate \(σ^2\).
The mean squared error of the useful range data points from the fitted linear model is represented by \(σ^2\). This unbiased estimator, \(σ^2\), is given by the equation:
\(\sigma^2=\frac{\sum_{i=1}^{n}{e_i}^2}{n-p}\)
where the \(\sum_{i=1}^{n}{e_i}^2\) or the sum of squares of the residual error can also be represented as \(SS_E\)
\(\sigma^2=\frac{SS_E}{n-p}\)
Note that n is the number of observations and p is the number of regressors plus one. Thus, in this problem, n = 9 and p=2+1=3.
The residual error \((e_i)\) is the difference of the observed value \((y)\) of the useful range and the fitted value \((\hat{y_i})\) of it from the fitted linear equation:
\(\hat{y}=238.5569\ +\ 0.3339x_1-2.7167x_2\)
Manual Computation
Table 1.3 displays all 9 fitted values of useful range \((\hat{y_i})\) and the corresponding residuals \(e\) and its squares \(e^2\).
Table: Table 1.3 Observations, Fitted Values and Residuals for Useful Range Data
Practical Interpretation: This implies that the average of the squared difference between each predicted values and the actual values for the useful range is \(\sigma^2\approx1321.3\).
Answer to Question 3:
Question: Compute the standard errors of the regression coefficients.
This means that the larger the standard error of the regression coefficients, the less fit the regression model is to our observed values.
MATRIX-APPROACH COMPUTATION
In matrix-approach computation, the variances of the regression coefficients is given by the diagonal elements of the matrix \(\sigma^2\left(X'X\right)^{-1}\)
Our matrix \(X\) is given by:
library(knitr)
x.data <- c(1,54,56,1,61,80,1,65,70,1,100,50,1,100,
65,1,100,80,1,50,25,1,57,35,1,54,26
)
x <- matrix(x.data,nrow=9,ncol=3,byrow=TRUE) #Matrix of X-Values
x
Finally, multiply the derived mean square error from question 2, \(σ^2=1321.3\), to the inverse of the product of the two matrices \(\left(X'X\right)^{-1}\):
From the matrix \(\sigma^2\left(X'X\right)^{-1}\), knowing that the variances of the regression coefficients are given by the diagonal elements of it means that:
The Standard error for the intercept of the regression model is \(S_E({\hat{B}}_0)=\sigma^2C_{00}=\sqrt{2045.696527}=45.22937861\)
The Standard error for the Brightness Percentage, one of the two regressors of the model \(\beta_1\) is \(S_E({\hat{B}}_1)=\sigma^2C_{11}=\sqrt{0.4573925}=0.676307992\)
The Standard error for the Contrast Percentage, one of the two regressors of the modelis \(S_E({\hat{B}}_1)=\sigma^2C_{11}=\sqrt{0.4573925}=0.6887483575\)
