n 1 = 40 n 2 = 50 x ˉ 1 = 340 x ˉ 2 = 285 s 1 = 20 s 2 = 30 ν = ( s 1 2 n 1 + s 2 2 n 2 ) 2 1 n 1 − 1 ( s 1 2 n 1 ) 2 + 1 n 2 − 1 ( s 2 2 n 2 ) 2 = ( 2 0 2 40 + 3 0 2 50 ) 2 1 39 ( 2 0 2 40 ) 2 + 1 49 ( 3 0 2 50 ) 2 = 85.437 ≈ 85 n_1=40\\n_2=50\\\bar{x}_1=340\\\bar{x}_2=285\\s_1=20\\s_2=30\\\nu =\frac{\left( \frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2} \right) ^2}{\frac{1}{n_1-1}\left( \frac{{s_1}^2}{n_1} \right) ^2+\frac{1}{n_2-1}\left( \frac{{s_2}^2}{n_2} \right) ^2}=\frac{\left( \frac{20^2}{40}+\frac{30^2}{50} \right) ^2}{\frac{1}{39}\left( \frac{20^2}{40} \right) ^2+\frac{1}{49}\left( \frac{30^2}{50} \right) ^2}=85.437\approx 85\\ n 1 = 40 n 2 = 50 x ˉ 1 = 340 x ˉ 2 = 285 s 1 = 20 s 2 = 30 ν = n 1 − 1 1 ( n 1 s 1 2 ) 2 + n 2 − 1 1 ( n 2 s 2 2 ) 2 ( n 1 s 1 2 + n 2 s 2 2 ) 2 = 39 1 ( 40 2 0 2 ) 2 + 49 1 ( 50 3 0 2 ) 2 ( 40 2 0 2 + 50 3 0 2 ) 2 = 85.437 ≈ 85
Then the confidence interval for μ 1 − μ 2 \mu _1-\mu _2 μ 1 − μ 2 is
( x ˉ 1 − x ˉ 2 − s 1 2 n 1 + s 2 2 n 2 t ν , 1 + γ 2 , x ˉ 1 − x ˉ 2 − s 1 2 n 1 + s 2 2 n 2 t ν , 1 + γ 2 ) = = ( 340 − 285 − 2 0 2 40 + 3 0 2 50 ⋅ 1.663 , 340 − 285 + 2 0 2 40 + 3 0 2 50 ⋅ 1.663 ) = = ( 46.2002 , 63.7998 ) \left( \bar{x}_1-\bar{x}_2-\sqrt{\frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2}}t_{\nu ,\frac{1+\gamma}{2}},\bar{x}_1-\bar{x}_2-\sqrt{\frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2}}t_{\nu ,\frac{1+\gamma}{2}} \right) =\\=\left( 340-285-\sqrt{\frac{20^2}{40}+\frac{30^2}{50}}\cdot 1.663,340-285+\sqrt{\frac{20^2}{40}+\frac{30^2}{50}}\cdot 1.663 \right) =\\=\left( 46.2002,63.7998 \right) ( x ˉ 1 − x ˉ 2 − n 1 s 1 2 + n 2 s 2 2 t ν , 2 1 + γ , x ˉ 1 − x ˉ 2 − n 1 s 1 2 + n 2 s 2 2 t ν , 2 1 + γ ) = = ( 340 − 285 − 40 2 0 2 + 50 3 0 2 ⋅ 1.663 , 340 − 285 + 40 2 0 2 + 50 3 0 2 ⋅ 1.663 ) = = ( 46.2002 , 63.7998 )
Since the confidence interval doesn’t contain 0, the null hypothesis is rejected, the mean values differ.
We assume that two samples are independent, with normal distribution.
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