Differences

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--- two_sample_t-test [2026/04/05 23:54] – hkimscil
+++ two_sample_t-test [2026/04/07 22:39] (current) – hkimscil
@@ Line 5: / Line 5: @@
 가정
   * 두 모집단 p1, p2 가 있다
-  * 각 집단에서 샘플을 취해서 그 평균을 모아 본다 (sampling distribution)
+  * 각 집단에서 샘플을 취해서 그 평균을 구한 후
+  * 그 차이를 기록한다.
+  * 이것을 무한히 반복한다.
+  * 이렇게 해서 얻은 샘플평균차이를 모은 집합의 평균과 분산은 무엇이 될까?
 우리는 이미
+[[:central limit theorem]] 문서와
 [[:statistical review]] 문서의 [[:expected value and variance properties]] 와
 [[:mean and variance of the sample mean]] 문서를 통해서 아래를 알고 있다.
 \begin{eqnarray*}
-\overline{X} & \sim & \left( \mu, \frac{\sigma}{n} \right) \\
+\overline{X} & \sim & \left( \mu, \;\; \frac{\sigma^2}{n} \right) \\
 &  & \text{in other words, } \\
 E \left[ \overline{X} \right] & = & \mu \\
-Var \left[ \overline{X} \right] & = & \frac{\sigma}{n}
+Var \left[ \overline{X} \right] & = & \frac{\sigma^2}{n} \\
+& & \text {Assuming that X1 and X2 are independent } \\
+\overline{X_{1}} & \sim & \left( \mu_{1}, \frac{\sigma^2_{1}}{n_{1}} \right) \\
+\overline{X_{2}} & \sim & \left( \mu_{2}, \frac{\sigma^2_{2}}{n_{2}} \right) \\
+& & \text{note that } n_{1}, n_{2} \text{ are sample size.} \\
+& & \text{and } \\
+& & \frac{\sigma^2_{1}}{n_{1}} = Var \left[ \overline{X_{1}} \right] \\
 \end{eqnarray*}
 두 샘플 평균들의 차이를 모아 놓은 집합의 (distribution of sample mean difference) 성격은 아래와 같을 것이다.
 \begin{eqnarray*}
-& & \text {Assuming that X1 and X2 are independent } \\
+E \left[ \overline{X_{1}} - \overline{X_{2}} \right] & = &  \mu_{1} - \mu_{2} \;, \;\;\; \text{and} \\
-\overline{X_{1}} & \sim & \left( \mu_{1}, \frac{\sigma_{1}}{n_{1}} \right) \\
-\overline{X_{2}} & \sim & \left( \mu_{2}, \frac{\sigma_{2}}{n_{2}} \right) \\
-& & \text{note that } n_{1}, n_{2} \text{ are sample size.} \\
-& & \text{and } \\
-& & \frac{\sigma_{1}}{n_{1}} = Var \left[ \overline{X_{1}} \right]
-\\
-E \left[ \overline{X_{1}} - \overline{X_{2}} \right] & = &
-\mu_{1} - \mu_{2} \;, \;\;\; \text{and} \\
 Var \left[ \overline{X_{1}} - \overline{X_{2}} \right] & = &
 Var \left[ \overline{X_{1}} \right] + Var \left[ \overline{X_{2}} \right] \\
-& = & \frac{\sigma_{1}}{n_{1}} + \frac{\sigma_{2}}{n_{2}} \\
+& = & \frac{\sigma^2_{1}}{n_{1}} + \frac{\sigma^2_{2}}{n_{2}} \\
-\\
+\text{SE}_{\overline{X_{1}} - \overline{X_{2}}} & = & \text{SE}_{\text{diff}} \\
-\text{SE}_{\overline{X_{1}} - \overline{X_{2}}} & = & \sqrt { \frac{\sigma_{1}}{n_{1}} + \frac{\sigma_{2}}{n_{2}} } \\
+& = & \sqrt { \frac{\sigma^2_{1}}{n_{1}} + \frac{\sigma^2_{2}}{n_{2}} } \\
-\text{SE}_{\text{diff}} & = & \sqrt { \frac{\sigma_{1}}{n_{1}} + \frac{\sigma_{2}}{n_{2}} } \\
 \\
-& & \text{If variances of each population } \\
+& & \text{If variance of each population} \text{is unknown,} \\
-& & \text{are the same, }  \sigma_{1} = \sigma_{2} \\
+& & \text{we use sample variances, instead of using } \sigma \text{.} \\
-& & \text{We use poooled variance, } \text{S}^{2}_{\text{p}}\\
+& & \text{If degrees of freedom for each group is different} \\
-\text{S}^{2}_{\text{p}} & = & \dfrac {\text{SS}_{1} + \text{SS}_{2}} {\text{df}_{1} + \text{df}_{2} } \\
+& & \text{we use the following method to obtain pooled variance, } \; \text{s}^{2}_{\text{p}}\\
+\text{s}^{2}_{\text{p}} & = & \dfrac {\text{SS}_{1} + \text{SS}_{2}} {\text{df}_{1} + \text{df}_{2} } \\
 & & \text{Hence, } \\
-\text{SE}_{\text{diff}} & = & \sqrt {\frac{\text{S}^{2}_{\text{p}}}{n_1} +  \frac{\text{S}^{2}_{\text{p}}}{n_2} } \\
+\text{SE}_{\text{diff}} & = & \sqrt {\frac{\text{s}^{2}_{\text{p}}}{n_1} +  \frac{\text{s}^{2}_{\text{p}}}{n_2} } \\
 \end{eqnarray*}