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Sagot :
Solution :
Given :
[tex]$n_1 = n_2 = 40$[/tex]
[tex]$\overline X_1 = 18$[/tex]
[tex]$S_1 = 4$[/tex]
[tex]$\overline X_2 = 12$[/tex]
[tex]$S_2 =10$[/tex]
So we want to test :
[tex]$H_0 : \mu_1=\mu_2 $[/tex] vs [tex]$H_1 : \mu_1 \neq \mu_2 $[/tex]
a). For 98% confidence interval :
[tex]$=\left((\overline X_1 - \overline X_2) \pm Tn_1+n_2-2, \alpha / 2 \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\right)$[/tex]
[tex]$=\left((18-12) \pm 2.375 \sqrt{\frac{4^2}{40}+\frac{10^2}{40}}\right)$[/tex]
[tex]$=(6 \pm 4.0445)$[/tex]
= (1.956, 10.045)
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