作直线相关分析时,当n=10,r=0.8531,α=0.05(双侧)查相关系数界值表r0.001,8=0.872,r0.002,8=0.847,r0.005,8=0.805,得P值范围为<0.001P>0.001r>rP<0.002连起来写即0.001