Pembahasan Ujian PAI: A50 – No. 18 – November 2018

• \(Var\left[ {\hat S\left( t \right)} \right] = \frac{{S\left( t \right)F\left( t \right)}}{n} = \frac{{S\left( t \right)\left( {1 – S\left( t \right)} \right)}}{n}\)
  \(0,00056 \cdot 100 = S\left( t \right) – S{\left( t \right)^2}\)
  \(S{\left( t \right)^2} – S\left( t \right) + 0,056 = 0\)
  \(\left( {S\left( t \right) – 0,940454} \right)\left( {S\left( t \right) – 0,059545} \right) = 0\)

  \(S\left( t \right) = 0,940454\) atau \(S\left( t \right) = 0,059545\)

• \(Var\left[ {\hat S\left( r \right)} \right] = \frac{{S\left( r \right)F\left( r \right)}}{n} = \frac{{S\left( r \right)\left( {1 – S\left( r \right)} \right)}}{n}\)
  \(0,0004 \cdot 100 = S\left( r \right) – S{\left( r \right)^2}\)
  \(S{\left( r \right)^2} – S\left( r \right) + 0,04 = 0\)
  \(\left( {S\left( r \right) – 0,95825} \right)\left( {S\left( r \right) – 0,0417424} \right) = 0\)

  \(S\left( r \right) = 0,95825\) atau \(S\left( r \right) = 0,0417424\)

\(100Cov\left[ {\hat S\left( t \right),\hat S\left( r \right)} \right] = \frac{{100 \cdot {q_t} \cdot {q_r}}}{n}\)

\(= \frac{{100 \cdot 0,059545 \cdot 0,0417424}}{{100}}\)

\(= 0,00224\)