Complete the missing information for Table 1, using 0.50 as the cut value. Then complete the classification table (Table 2). Compute sensitivity, specificity, false positive rate, and false negative rate.

$\begin{array}{l}
\text { Table } 1 .\
\begin{array}{ccc}
\hline \begin{array}{c}
\text { Observed group } \
\text { membership }
\end{array} & \begin{array}{c}
\text { Predicted } \
\text { Probability }
\end{array} & \begin{array}{c}
\text { Predicted group } \
\text { membership }
\end{array} \
\hline 1 & 0.88 & \
1 & 0.72 & \
0 & 0.62 & \
1 & 0.49 & \
0 & 0.34 & \
1 & 0.40 & \
1 & 0.60 & \
0 & 0.21 & \
0 & 0.05 & \
1 & 0.57 & \
\hline
\end{array}
\end{array}$

$\begin{array}{l}
\text { Table } 2 .\
\begin{array}{cccc}
\hline & &&\underline { \text { Predicted }} \
& & .00 & &1.00 \
\hline{\text { Observed }} & .00 & & \
& 1.00 & & \
\hline
\end{array}
\end{array}$

Question

Complete the missing information for Table 1, using 0.50 as the cut value. Then complete the classification table (Table 2). Compute sensitivity, specificity, false positive rate, and false negative rate.

$\begin{array}{l}
\text { Table } 1 .\
\begin{array}{ccc}
\hline \begin{array}{c}
\text { Observed group } \
\text { membership }
\end{array} & \begin{array}{c}
\text { Predicted } \
\text { Probability }
\end{array} & \begin{array}{c}
\text { Predicted group } \
\text { membership }
\end{array} \
\hline 1 & 0.88 & \
1 & 0.72 & \
0 & 0.62 & \
1 & 0.49 & \
0 & 0.34 & \
1 & 0.40 & \
1 & 0.60 & \
0 & 0.21 & \
0 & 0.05 & \
1 & 0.57 & \
\hline
\end{array}
\end{array}$

$\begin{array}{l}
\text { Table } 2 .\
\begin{array}{cccc}
\hline & &&\underline { \text { Predicted }} \
& & .00 & &1.00 \
\hline{\text { Observed }} & .00 & & \
& 1.00 & & \
\hline
\end{array}
\end{array}$

Quizplus · Accepted Answer

The Answer is Assuming 0.5 is the cut value, cases with predicted probabilities at .5 or above are predicted as 1 and predicted probabilities below .5 are predicted as 0. There are four cases with observed value 1 and predicted value 1. There are three cases with observed value 0 and predicted value 0. There is one case with observed value 0 yet predicted value 1 (false positive).
There are two cases with observed value 1 yet predicted value 0 (false negative).

Sensitivity = 4/(2+4) = 0.67 = 67%
Specificity = 3/(3+1) = 0.75 = 75%
False positive rate = 1/(3+1) = 0.25 -25%
False negative rate = 2/(2+4) = 0.33 = 33%

\begin{array}{l} ext { Table } 1 .\\begin{array}{ccc}\hline \begin{array}{c} ext { Observed group } \ ext { membership }\end{array} & \begin{array}{c} ext { Predicted } \ ext { Probability }\end{array} & \begin{array}{c} ext { Predicted group } \ ext { membership }\end{array} \\hline 1 & 0.88 & 1 \1 & 0.72 & 1 \0 & 0.62 & 1 \1 & 0.49 & 0 \0 & 0.34 & 0 \1 & 0.40 & 0 \1 & 0.60 & 1 \0 & 0.21 & 0 \0 & 0.05 & 0 \1 & 0.57 & 1 \\hline\end{array}\end{array}

\begin{array}{l} ext { Table } 2 .\\begin{array}{cccc}\hline & &&\underline { ext { Predicted }} \& & .00 & &1.00 \\hline{ ext { Observed }} & .00 & 3&& 1\& 1.00 &2& & 4\\hline\end{array}\end{array}

Complete the Missing Information for Table 1, Using 0