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Belajar SPSS

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Oleh Syafdillah

29 April 2012

Nilai tengah sampel :

Varians sampel :

Hukum penambahan peluang :
$\dpi{120} P\left ( A\bigcup B \right )=P\left ( A \right )+\left P( B \right )-P\left ( A\bigcap B \right )$

Hukum penggandaan peluang :
$\dpi{120} P\left ( A\bigcap B \right )=P\left ( A \right )P\left ( B|A \right )$

Peluang Keadaan :
$\dpi{120} P\left ( B|A \right )=\frac{P\left ( A\bigcap B \right )}{P\left ( A \right )}$

Nilai tengah peubah acak diskret :
$\dpi{120} \mu =\sum xf(x)$

Sebaran binomial :
$\dpi{120} b(x; n,p)=\binom{n}{x}p^{x}q^{n-x}$

Uji $\dpi{120} \sigma _{2}^{2}=\sigma _{o}^{2}$
$\dpi{120} X^{2}=\frac{(n-1)s^{2}}{\sigma _{o}^{2}}$

Uji $\dpi{120} \sigma _{1}^{2}=\sigma _{2}^{2}$
$\dpi{120} f=\frac{s_{1}^{2}}{s_{2}^{2}}$

Uji $\dpi{120} \rho =\rho _{0}$
$\dpi{120} z=\frac{x-np_{0}}{\sqrt{np_{1}q_{o}}}$

Dugaan bagi $\dpi{120} \rho$
$\dpi{120} \hat{p}=\frac{x_{1}+x_{2}}{n_{1}+n_{2}}$

Uji $\dpi{120} \rho _{1}=\rho _{2}$
$\dpi{120} z=\frac{\hat{\rho} _{1}-\hat{\rho} _{2}}{\sqrt{\hat{\rho }}\hat{q}\left [(\frac{1}{N_{1}})(\frac{1}{N_{2}}) \right ]}$

Uji chi-kuadrat menggunakan tabel kontingensi
$\dpi{120} X^{2}=\sum \frac{(o-e)^{2}}{e}$

Garis regresi linear
$\dpi{120} \hat{y}=a+bx$

Slope garis regresi
$\dpi{120} b=\frac{n\sum xy-(\sum x)(\sum y)}{n\sum x^{2}-(\sum x)^{2}}$

Intersep y garis regresi
$\dpi{120} a=\bar{y}-b\bar{x}$

Dugaan $\dpi{120} \sigma ^{2}$ dalam garis regresi
$\dpi{120} s_{e}^{2}=\frac{n-1}{n-2}\left ( s_{y}^{2}-b^{2}s_{x}^{2} \right )$

Uji $\dpi{120} \beta =\beta _{o}$
$\dpi{120} t=\frac{s_{x}\sqrt{n-1}(b-\beta _{0})}{s_{e}}$

Koefisien korelasi
$\dpi{120} r=\frac{n\sum xy-(\sum x)(\sum y)}{\sqrt{\left [n\sum x^{2}-(\sum x)^{2} \right ]\left [n\sum y^{2}-(\sum y)^{2} \right ]}}$

Koefisien korelasi peringkat
$\dpi{120} r_{s}=1-\frac{6\sum d^{2}}{n(n^{2}-1)}$

Selang kepercayaan bagi $\dpi{120} \rho$
$\dpi{120} \hat{\rho }\pm z\tfrac{\alpha }{2}\sqrt{\frac{\hat{p}\hat{q}}{n}}$

Ukuran contoh bagi $\dpi{120} \rho$ dugaan
$\dpi{120} n=\frac{z^{2}_{\frac{\alpha }{2}}}{4e^{2}}$

Selang kepercayaan bagi $\dpi{120} \rho _{1}-\rho _{2}$
$\dpi{120} \left ( \hat{\rho }_{1} -\hat{\rho }_{2}\right )\pm z_{\frac{\alpha }{2}}\sqrt{{\frac{\hat{\rho }_{1}\hat{q}_{1}}{n_{1}}}+{\frac{\hat{\rho }_{1}\hat{q}_{2}}{n_{2}}}}$

Selang kepercayaan bagi $\dpi{120} \mu ^{2}$
$\dpi{120} \frac{\left ( n-1 \right )s^{2}}{X_{\frac{\alpha }{2}}^{2}}< \alpha ^{2}< \frac{\left ( n-1 \right )s^{2}}{X_{\frac{1-\alpha }{2}}^{2}}$

Selang kepercayaan bagi $\dpi{120} \frac{\sigma _{1}^{2}}{\sigma _{2}^{2}}$
$\dpi{120} \frac{s_{1}^{2}}{s_{2}^{2}}\frac{1}{f_{\frac{\alpha }{2}}(v_{1}v_{2})}< \frac{\sigma _{1}^{2}}{\sigma _{2}^{2}}< \frac{s_{1}^{2}}{s_{2}^{2}}{f_{\frac{\alpha }{2}}(v_{1}v_{2})}$

Uji $\dpi{120} \mu =\mu _{0}$ ( $\dpi{120} \sigma$ diketahui)
$\dpi{120} z=\frac{\bar{x}-\mu _{0}}{\frac{\alpha }{\sqrt{n}}}$

Uji $\dpi{120} \mu =\mu _{0}$ ( $\dpi{120} \sigma$ tidak diketahui)
$\dpi{120} t=\frac{\bar{x}-\mu _{0}}{\frac{s}{\sqrt{n}}}$

Uji $\dpi{120} \mu _{1}-\mu _{2}=d_{o}$
$\dpi{120} z=\frac{(\bar{x}_{1}-\bar{x}_{2})-d_{o}}{\sqrt{\frac{\sigma _{1}^{2}}{n_{1}}+\frac{\sigma _{2}^{2}}{n^{2}}}}$

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