$解：(1) \bar{x}_{甲}=\frac {10+9+8+8+10+9}6=9(环)$
$\bar{x}_{乙}=\frac {10+10+8+10+7+9}6=9(环)$
$(2)\ \mathrm {s}²_{甲}=\frac 16×［(10-9)²×2+(9-9)²×2+(8-9)²×2］=\frac 23(环²)$
$s²_{乙}=\frac 16×［(10-9)²×3+(8-9)²+(7-9)²+(9-9)²］=\frac 43(环²)$
$(3) \because \bar{x}_{甲 }=\bar{x}_{乙 }， s_{甲 }^2＜s_{乙}^2$
$∴推荐甲参加省比赛更合适$
