Lanjutan Contoh Soal 5 Distribusi Normal

$\begin{array}{l}\\ 21.& \text{Pada suatu kelas seorang guru matematika}\\ & \text{menyatakan bahwa nilai ulangan mapel }\\ & \text{yang diampunya tidak kurang dari}68\\ & \text{Untuk menentukan uji tersebut, maka guru}\\ & \text{yang bersangkutan memilih 10 anak secara}\\ & \text{acak untuk ditanyai nilai hasil ulangannya}\\ & \text{Hipotesis}{\text{H}}_0\text{dan}{\text{H}}_1\text{yang tepat dari kondisi}\\ & \text{kondisi di atas adalah}....\\ & \text{a}.\begin{array}{c}{\text{H}}_0:\mu =68\\ {\text{H}}_1:\mu \ne 68\end{array}\\ & \text{b}.\begin{array}{c}{\text{H}}_0:\mu =68\\ {\text{H}}_1:\mu >68\end{array}\\ & \text{c}.\begin{array}{c}{\text{H}}_0:\mu \ge 68\\ {\text{H}}_1:\mu <68\end{array}\\ & \text{d}.\begin{array}{c}{\text{H}}_0:\mu \le 68\\ {\text{H}}_1:\mu >q68\end{array}\\ & \text{e}.\begin{array}{c}{\text{H}}_0:\mu >68\\ {\text{H}}_1:\mu \ne 68\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \text{Cukup jelas}\\ & \text{Dan ini contoh uji satu pihak, yaitu kiri}\end{array}$.

$\begin{array}{l}\\ 22.& \text{Untuk menguji hipotesis}\\ & \{\begin{array}{ll}{H}_0& :\mu =800\\ H_1& :\mu \ne 800\end{array}\\ & \text{Jika dalam penelitian didapatkan}\bar{\text{x}}=795\\ & \text{dan}n=100\text{serta simpangan baku}\\ & \text{populasi}55,\text{maka perhitungan nilai}\\ & \text{statistik ujinya adalah}....\\ & \text{a}.z=-1,10\\ & \text{b}.z=-0,91\\ & \text{c}.t=-0,91\\ & \text{d}.t=0,91\\ & \text{e}.z=0,91\\ \\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \underset{―}{\text{Hipotesis}}\\ & {\text{H}}_0:\mu =800\\ & {\text{H}}_0:\mu \ne 800\\ & \text{dengan},n=100,\bar{\text{x}}=795,\sigma =55,\text{maka}\\ & \text{z}=\left({\displaystyle \frac{\bar{\text{x}}-\mu }{\sigma }}\right)\sqrt{n}={\displaystyle \frac{795-800}{55}\sqrt{100}}\\ & =\left({\displaystyle \frac{-5}{55}}\right)\times 10=-0,91\end{array}\end{array}$.

$\begin{array}{l}\\ 23.& \text{Pada soal no.22 di atas, jika taraf nyata}\\ & 5\mathrm{\%},\text{maka kesimpulannya adalah}...\\ & \text{a}.H_0\text{ditolak}\\ & \text{b}.H_0\text{diterima}\\ & \text{c}.H_1\text{diterima}\\ & \text{d}.H_1\text{ditolak}\\ & \text{e}.\text{tidak dapat ditarik kesimpulan}\\ \\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \text{Kasus uji dua pihak}\\ & \begin{array}{rl}& \underset{―}{\text{Hipotesis}}\\ & {\text{H}}_0:\mu =800\\ & {\text{H}}_0:\mu \ne 800\\ & \text{dengan}\alpha =5\mathrm{\%}=0,05\Rightarrow z_{\frac{1}{2}(1-0,05)}\\ & =z_{0,475}=1,96\end{array}\end{array}$.

$\begin{array}{l}\\ 24.& \text{Untuk menguji hipotesis}\\ & \{\begin{array}{ll}{H}_0& :\mu =20\\ H_1& :\mu <20\end{array}\\ & \text{Jika dalam penelitian terhadap sampel}n=21\\ & \text{diperoleh}\bar{\text{x}}=19,8\text{dengan simpangan baku}\\ & 0,4,\text{maka perhitungan nilai statistik ujinya}\\ & \text{adalah}....\\ & \text{a}.z=-2,29\\ & \text{b}.t=-2,29\\ & \text{c}.z=2,29\\ & \text{d}.t=2,29\\ & \text{e}.z=2,99\\ \\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \underset{―}{\text{Hipotesis (penelitian sampel)}}\\ & {\text{H}}_0:\mu =20\\ & {\text{H}}_0:\mu <20\\ & \text{dengan},n=21,\bar{\text{x}}=19,8,s=0,4,\text{maka}\\ & \text{t}=\left({\displaystyle \frac{\bar{\text{x}}-\mu }{s}}\right)\sqrt{n}={\displaystyle \frac{19,8-20}{0,4}\sqrt{21}}\\ & =\left({\displaystyle \frac{-0,2}{0,4}}\right)\times 4,5828=-2,29\end{array}\end{array}$.

$\begin{array}{l}\\ 25.& \text{Pada soal no.24 di atas, jika taraf nyata}\\ & 0,01,\text{maka kesimpulannya adalah}...\\ & \text{a}.H_0\text{ditolak}\\ & \text{b}.H_0\text{diterima}\\ & \text{c}.H_1\text{diterima}\\ & \text{d}.H_1\text{ditolak}\\ & \text{e}.\text{tidak dapat ditarik kesimpulan}\\ \\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \text{Kasus uji satu pihak}\\ & \begin{array}{rl}& \underset{―}{\text{Hipotesis}}\\ & {\text{H}}_0:\mu =20\\ & {\text{H}}_0:\mu <20\\ & \text{dengan}\alpha =0,01=1\mathrm{\%}\Rightarrow t_{0,99}=2,53\\ & \text{Kriteria pengambilan keputusan}:\\ & \text{Tolak}{H}_0\text{jika}t\le -2,53\end{array}\end{array}$.