Question 2

figure 1.

A.

## 
## Call:
## lm(formula = UsefulRange ~ Brightness + Contrast, data = data)
## 
## 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 **
## Brightness    0.3339     0.6763   0.494  0.63904   
## Contrast     -2.7167     0.6887  -3.945  0.00759 **
## ---
## 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

\[\hat{y} = b_0 + b_1 x_1 + b_2 x_2\] \[\hat{y} = 238.5569 + 0.3339 x_1 + (-2.7167) x_2\]

B.

We can get the unbiased estimate σ2 by using the following formula:

\[σ^2=\frac{\Sigma e^2}{n-2}=\frac{SS_e}{n-p}\]

\[SS_e=\Sigma (y-\hat{y})^2\]

Pulling these values from the function above, we get:

## [1] 1321.273

The unbiased estimator, σ2, is 1,321.273.

C.

From the summary above, we can get the following standard errors of the regression coefficients:

\[Brightness = 0.6763\] \[Contrast = 0.6887\]

D.

From the equation above, we can get the predicted useful range when brightness = 80 and contrast = 75: \[\hat{y} = 238.5569 + 0.3339 x_1 + (-2.7167) x_2\] Plugging in these values: \[\hat{y} = 238.5569 + 0.3339 x 80 + (-2.7167) x 75 = 61.5164\] The predicted useful range will be 61.5164.

E.

From the p-value in the summary above, 0.01459 and comparing it to α = 0.05: \[0.01459 < 0.05\] Therefore, the regression is significant.

F.

We can get the t-value from the following equation:

\[t_\hat{β}=\frac{\hat{β} - β_0}{SE(\hat{β})}\]

However, since the t-value was already provided from the summary above, we can get the following:

\[β_{1} - Brightness\]

\[H_0: β_1 = 0\]

\[H_1: β_1 > 0\]

\[t_1=0.494\]

Using the pt function, we get a p-value of:

## [1] 0.6827085

A p-value of 0.6827085 which is greater than 0.05, thus we Fail to Reject the Null Hypothesis and there is no relationship with the model.

\[β_{2} - Contrast\]

\[H_0: β_2 = 0\] \[H_1: β_2 > 0\]

\[t_2=-3.945\] Using the pt function, we get a p-value of:

## [1] 0.002132922

A p-value of 0.002132922 is less than 0.05, thus we Accept the Alternative Hypothesis and there is a strong relationship with the model.