t-test
Differences
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| Both sides previous revisionPrevious revisionNext revision | Previous revision | ||
| t-test [2026/03/31 23:04] – hkimscil | t-test [2026/03/31 23:18] (current) – hkimscil | ||
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| \begin{eqnarray} | \begin{eqnarray} | ||
| - | z\; | + | z\; |
| - | t & = & \frac{ \overline{X}-\mu}{s_{\overline{X}} }, \;\;\; \text{where } \;\; s_{\overline{X}} = \frac{s}{\sqrt{n}} \\ | + | t & = & \frac{ \overline{X}-\mu}{s_{\overline{X}} }, \; \text{where } \;\; s_{\overline{X}} = \frac{s}{\sqrt{n}} \\ |
| - | t & = & \frac{(\overline{X_a}-\overline{X_b})-(\mu_a-\mu_b)}{\sigma_{\text{diff} }} , \; | + | t & = & \frac{(\overline{X_a}-\overline{X_b})-(\mu_a-\mu_b)}{\sigma_{\text{diff} }}, \\ |
| - | & & \text{where } \;\; \sigma_{\text{diff} } = \sqrt{ \frac{s^2_{\text{pooled}}}{n_a} | + | & & \qquad \qquad |
| - | & & s^2_{\text{pooled}} = \frac {\text{SS}_a + \text{SS}_b} {df_a + df_b} \nonumber \\ | + | & & \qquad \qquad |
| - | t & = & \frac{\overline{D}-0}{s_{ \overline{D} }}, \;\;\; \text{where } \;\; s_{ \overline{D} } = \frac {s_D}{\sqrt{n}} \\ | + | t & = & \frac{\overline{D}-0}{s_{ \overline{D} }}, \; \text{where } \;\; s_{ \overline{D} } = \frac {s_D}{\sqrt{n}} \\ |
| \end{eqnarray} | \end{eqnarray} | ||
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| 위의 4 경우의 공통점은 : | 위의 4 경우의 공통점은 : | ||
| - | + | \begin{eqnarray*} | |
| - | $t = \displaystyle \frac{\text{obtained difference}}{\text{difference by random error}}$ | + | t & = & \displaystyle \frac{\text{obtained difference}}{\text{difference by random error}} |
| + | \end{eqnarray*} | ||
| </ | </ | ||
t-test.1774998287.txt.gz · Last modified: by hkimscil
