This code creates the model
model.full <- lm(Criterion ~ CCTST + CSIT + Interview1, validdf)
summary(model.full)
Call:
lm(formula = Criterion ~ CCTST + CSIT + Interview1, data = validdf)
Residuals:
Min 1Q Median 3Q Max
-3.4659 -0.8431 0.0402 0.9387 2.8396
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -6.34295 2.22540 -2.850 0.00535 **
CCTST 0.06851 0.02330 2.940 0.00411 **
CSIT 0.07703 0.03899 1.976 0.05107 .
Interview1 0.71956 0.17047 4.221 5.52e-05 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 1.402 on 96 degrees of freedom
Multiple R-squared: 0.2374, Adjusted R-squared: 0.2135
F-statistic: 9.959 on 3 and 96 DF, p-value: 8.895e-06The CSIT was found not to be significant so it was removed from the model. The revised model is the following:
\[ \begin{aligned} \operatorname{\widehat{Criterion}} &= -4.03 + 0.07(\operatorname{CCTST}) + 0.72(\operatorname{Interview1}) \end{aligned} \]
Call:
lm(formula = Criterion ~ CCTST + Interview1, data = validdf)
Residuals:
Min 1Q Median 3Q Max
-3.1999 -0.9561 -0.0412 1.0142 2.7342
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -4.03104 1.92098 -2.098 0.03847 *
CCTST 0.06587 0.02361 2.790 0.00635 **
Interview1 0.72387 0.17299 4.184 6.28e-05 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 1.423 on 97 degrees of freedom
Multiple R-squared: 0.2064, Adjusted R-squared: 0.19
F-statistic: 12.61 on 2 and 97 DF, p-value: 1.355e-05mean(validdf$Criterion)The mean of the the criterion is 3.81 which is the minimum value we want predicted for our job applicants.
model.CCTST <- lm(Criterion ~ CCTST, validdf)
summary(model.CCTST)
Call:
lm(formula = Criterion ~ CCTST, data = validdf)
Residuals:
Min 1Q Median 3Q Max
-3.4826 -1.0237 0.0007 1.1893 3.0418
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -1.15543 1.93908 -0.596 0.5526
CCTST 0.06556 0.02552 2.569 0.0117 *
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 1.538 on 98 degrees of freedom
Multiple R-squared: 0.06309, Adjusted R-squared: 0.05353
F-statistic: 6.599 on 1 and 98 DF, p-value: 0.01171\[ \begin{aligned} \operatorname{\widehat{Criterion}} &= -1.155 + 0.066(\operatorname{CCTST}) \end{aligned} \]
The minimum score for the CCTST is 76 to be hired.
model.Interview <- lm(Criterion ~ Interview1, validdf)
summary(model.Interview)
Call:
lm(formula = Criterion ~ Interview1, data = validdf)
Residuals:
Min 1Q Median 3Q Max
-2.8533 -0.8533 0.1467 1.1467 3.1467
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.9639 0.7200 1.339 0.183714
Interview1 0.7224 0.1789 4.038 0.000107 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 1.471 on 98 degrees of freedom
Multiple R-squared: 0.1427, Adjusted R-squared: 0.1339
F-statistic: 16.31 on 1 and 98 DF, p-value: 0.000107\[ \begin{aligned} \operatorname{\widehat{Criterion}} &= 0.964 + 0.722(\operatorname{Interview1}) \end{aligned} \]
The minimum score for the Interview is 3.94 to be hired.
Even though it was not a significant predictor of the criterion, I wanted to calculate the minimum score anyway.
model.CSIT <- lm(Criterion ~ CSIT, validdf)
summary(model.CSIT)
Call:
lm(formula = Criterion ~ CSIT, data = validdf)
Residuals:
Min 1Q Median 3Q Max
-3.3443 -1.0721 0.2002 1.1820 3.0913
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 1.8034 1.2126 1.487 0.1402
CSIT 0.0726 0.0435 1.669 0.0984 .
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 1.567 on 98 degrees of freedom
Multiple R-squared: 0.02763, Adjusted R-squared: 0.01771
F-statistic: 2.785 on 1 and 98 DF, p-value: 0.09836\[ \begin{aligned} \operatorname{\widehat{Criterion}} &= 1.803 + 0.073(\operatorname{CSIT}) \end{aligned} \]
The minimum score for the Interview is 27.64 to be hired.