X 1 , X 2 , ⋯ , X 11 X_{1} \, , \, X_{2} \, , \, \cdots , \, X_{11} X 1 , X 2 , ⋯ , X 11 N ( μ 1 , σ 2 ) N ( \mu_{1} , \quad \sigma^{2} ) N ( μ 1 , σ 2 ) Y 1 , Y 2 , ⋯ , Y 9 Y_{1} \, , Y_{2} \, , \, \cdots , \, Y_{9} Y 1 , Y 2 , ⋯ , Y 9 N ( μ 2 , σ 2 ) N ( \mu_{2} , \quad \sigma^{2} ) N ( μ 2 , σ 2 ) X 1 , X 2 , ⋯ , X 11 X_{1} \, , \, X_{2} \, , \, \cdots , \, X_{11} X 1 , X 2 , ⋯ , X 11 Y 1 , Y 2 , ⋯ , Y 9 Y_{1} \, , Y_{2} \, , \, \cdots , \, Y_{9} Y 1 , Y 2 , ⋯ , Y 9 S 1 2 S_{1}^{2} S 1 2 S 2 2 S_{2}^{2} S 2 2 μ 1 − μ 2 \mu_{1} - \mu_{2} μ 1 − μ 2 t α ( k ) t_{\alpha} ( k ) t α ( k ) k k k t t t 100 ( 1 − α ) 100 ( 1 - \alpha ) 100 ( 1 − α ) 1 2 × t 0.025 ( 18 ) ( 10 S 1 2 + 8 S 2 2 18 ) ( 1 11 + 1 9 ) 2 \times t_{0.025} ( 18 ) \sqrt{\left( \frac{10 S_{1}^{2} + 8 S_{2}^{2}}{18} \right) \left( \frac{1}{11} + \frac{1}{9} \right)} 2 × t 0.025 ( 18 ) ( 18 10 S 1 2 + 8 S 2 2 ) ( 11 1 + 9 1 ) 2 t 0.025 ( 18 ) ( 10 S 1 2 + 8 S 2 2 18 ) ( 1 11 + 1 9 ) t_{0.025} ( 18 ) \sqrt{\left( \frac{10 S_{1}^{2} + 8 S_{2}^{2}}{18} \right) \left( \frac{1}{11} + \frac{1}{9} \right)} t 0.025 ( 18 ) ( 18 10 S 1 2 + 8 S 2 2 ) ( 11 1 + 9 1 ) 3 2 × t 0.025 ( 18 ) ( 10 S 1 2 + 8 S 2 2 18 ) ( 1 10 + 1 8 ) 2 \times t_{0.025} ( 18 ) \sqrt{\left( \frac{10 S_{1}^{2} + 8 S_{2}^{2}}{18} \right) \left( \frac{1}{10} + \frac{1}{8} \right)} 2 × t 0.025 ( 18 ) ( 18 10 S 1 2 + 8 S 2 2 ) ( 10 1 + 8 1 ) 4 t 0.025 ( 18 ) ( 10 S 1 2 + 8 S 2 2 18 ) ( 1 10 + 1 8 ) t_{0.025} ( 18 ) \sqrt{\left( \frac{10 S_{1}^{2} + 8 S_{2}^{2}}{18} \right) \left( \frac{1}{10} + \frac{1}{8} \right)} t 0.025 ( 18 ) ( 18 10 S 1 2 + 8 S 2 2 ) ( 10 1 + 8 1 ) 해설
등분산 가정의 t t t 2 × t 0.025 ( 18 ) ( 10 S 1 2 + 8 S 2 2 18 ) ( 1 11 + 1 9 ) 2 \times t_{0.025}(18) \sqrt{\left(\dfrac{10 S_{1}^{2} + 8 S_{2}^{2}}{18}\right)\left(\dfrac{1}{11} + \dfrac{1}{9}\right)} 2 × t 0.025 ( 18 ) ( 18 10 S 1 2 + 8 S 2 2 ) ( 11 1 + 9 1 )
두 모분산이 같으므로 자유도로 가중한 합동분산 을 쓴다.S p 2 = 10 S 1 2 + 8 S 2 2 18 S_{p}^{2} = \dfrac{10 S_{1}^{2} + 8 S_{2}^{2}}{18} S p 2 = 18 10 S 1 2 + 8 S 2 2
신뢰구간은 다음과 같다.( X ˉ − Y ˉ ) ± t 0.025 ( 18 ) S p 1 11 + 1 9 (\bar{X} - \bar{Y}) \pm t_{0.025}(18) \, S_{p} \sqrt{\dfrac{1}{11} + \dfrac{1}{9}} ( X ˉ − Y ˉ ) ± t 0.025 ( 18 ) S p 11 1 + 9 1
구간의 길이는 이 오차한계의 2배이다.
2배가 없고 1 11 + 1 9 \dfrac{1}{11} + \dfrac{1}{9} 11 1 + 9 1 2배가 있고 1 10 + 1 8 \dfrac{1}{10} + \dfrac{1}{8} 10 1 + 8 1 2배가 없고 1 10 + 1 8 \dfrac{1}{10} + \dfrac{1}{8} 10 1 + 8 1 오류나 개정된 내용이 보이면 알려주세요