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Question 2

An article in Optical Engineering [“Operating Curve Extraction of a Correlator’s Filter” (2004, Vol. 43, pp. 2775-2779)] 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:

A. Fit a multiple linear regression model to these data.

## 
## Call:
## lm(formula = Useful_Range ~ Contrast + Brightness, data = QUES2)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -32.334 -20.090  -8.451   8.413  69.047 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)   
## (Intercept) 238.5569    45.2285   5.274  0.00188 **
## Contrast     -2.7167     0.6887  -3.945  0.00759 **
## Brightness    0.3339     0.6763   0.494  0.63904   
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 36.35 on 6 degrees of freedom
## Multiple R-squared:  0.7557, Adjusted R-squared:  0.6742 
## F-statistic: 9.278 on 2 and 6 DF,  p-value: 0.01459

Using the values of estimated coefficients found in the lm function, we can get the regression model: \[ Useful ~Range~=~\beta_0~+~\beta_1X_1~+~\beta_2X_2 \] where \(\beta_0\) = 238.5569, \(\beta_1\) = -2.7167, \(\beta_2\) = 0.3339

Substituting the values, our fitted regression model is: \[ Useful ~Range~=~238.5569~-~2.7167X_1~+~0.3339X_2 \]

B. Estimate \(\sigma^2\)

\(\sigma^2\) is found at \[ \sigma^2 ~=~\frac{SS_E}{n-p} \] \(SS_E\) can be obtained using the function:

SSE <- sum((fitted(multipleregression)-Useful_Range)^2)
SSE 
## [1] 7927.636

Substituting the value obtained for \(SS_E\) and the value n = 9 and p = k + 1 = 2 + 1 = 3 We get

\[ \sigma^2~=~\frac{7927.636}{6}~=~1321.273 \]

Hence, the estimated \(\sigma^2\) is 1321.273.

C. Compute the standard errors of the regression coefficients.

Looking at the function above, we take the values of the the standard errors of the regression coefficients:

\(\hat{\beta_1}\) = Contrast = 0.6887

\(\hat{\beta_2}\) = Brightness = 0.6763

D. Predict the useful range when brightness = 80 and contrast = 75.

We let:

\(X_1\) = Contrast = 75 and \(X_2\) = Brightness = 80

and substituting them to the regression model \[ \hat{y}~=~238.5569 ~-~2.7167X_1 ~+~ 0.3339X_2 \] We get: \[ \hat{y}~=~238.5569 ~-~2.7167(75) ~+~ 0.3339(80)~=~61.5164 \] The useful range is predicted to be 61.5164

E. Test for significance of regression using \(\alpha\) = 0.05. What is the P-value for this test?

Null Hypothesis: \(H_0\):\(\beta_1\) = \(\beta_2\) = 0

Alternative Hypothesis: \(H_1\):\(\beta_1\) \(\ne\) \(\beta_2\)

Test Statistic: \[ f_0~=~\frac{SS_R/k}{SS_E/(n-p)} \] Reject \(H_0\) if P-value > 0.05

Computations: We can obtain \(SS_R\) using the function

## [1] 24518.36

So our values are:

\(SS_E\) = 7927.636, k = 2, \(SS_R\) = 24518.36, n - p = 6

\[ f_0~=~\frac{24518.36/2}{7927.636/6} ~=~9.2783 \] To get the p-value

  x = 9.2783
  v1 = 2
  v2 = 6
  pf(x, df = v1, df2 = v2, lower.tail = FALSE)
## [1] 0.01458643

\[ P(9.2783) > ~0.05~=~ 0.01458643~>~0.05 \]

Hence, the p-value of the test is 0.01458643 and we can reject \(H_0\).

F. Construct a t-test on each regression coefficient. What conclusions can you draw about the variables in this model? Use \(\alpha\) = 0.05.

Null Hypothesis: \(H_0\):\(\beta_j\) = 0 Alternative Hypothesis: \(H_0\):\(\beta_j\) \(\ne\) 0 Test Statistic: \[ t_0~=~\frac{\hat{\beta_j}}{se(\beta_j)} \] Reject \(H_0\) if: \(|t_0|\) > \(t_{a/2,n-p}\)

Computations:

t-test for Contrast \(\beta_1\):

\[ t_0~=~\frac{\hat{\beta_1}}{se(\hat{\beta_1})} \]

when \(\hat{\beta_1}\) = -2.7167 and se(\(\hat{\beta_1}\)) = 0.6887

we get: \[ t_0~=~\frac{-2.7167}{0.6887} = -3.9447 \] With \(\alpha\) = 0.05, \(\alpha/2\) = 0.025 and n - p = 6 we get:

\[ |t_0|~>~t_{\alpha/2,n-p}~=~3.9447~>~t_{0.025,6}~=~2.447 \] so we reject \(H_0\) for this coefficient

t-test for Brightness \(\beta_2\): \[ t_0~=~\frac{\hat{\beta_2}}{se(\hat{\beta_2})} \] when \(\beta_2\) = 0.3339 and se(\(\hat{\beta_2}\)) = 0.6763

we get: \[ t_0~=~\frac{0.3339}{0.6763}~=~0.4937 \]

With \(\alpha\) = 0.05, \(\alpha/2\) = 0.025 and n - p = 6 we get: \[ |t_0|~>~t_{\alpha/2,n-p}~=~0.4937~>~t_{0.025,6}~=~2.447 \] with this, we fail to reject \(H_0\)

What do these test mean?

As we reject \(H_0\) for the Contrast \(\hat{\beta_1}\), we can conclude that it is a good predictor of the regression model while as we fail to reject \(H_0\) for the Brightness \(\hat{\beta_2}\), we can conclude that it is not a good predictor of the regression model.