释放63%所需时间(分)的对数值 实验次数 # 2#3# 0.910.650.820.98 2 0.960.490.820.98 3 130.610.820.89 280.810.660.78 5 1.230.310.720.77 x5.512.873.844,.4016.62 di 1.100.570.770.880.83 ∑x 6.1781.7872.9713.91414.85
释放63%所需时间（分）的对数值 实验次数 1 # 2 # 3 # 4 # 1 0.91 0.65 0.82 0.98 2 0.96 0.49 0.82 0.98 3 1.13 0.61 0.82 0.89 4 1.28 0.81 0.66 0.78 5 1.23 0.31 0.72 0.77 5.51 2.87 3.84 4.40 16.62 1.10 0.57 0.77 0.88 0.83 6.178 1.787 2.971 3.914 14.85 x. j xi 2 x. j
解:df=kn-1=4×5-1=19 df=k-1=4-1=3 dfe-dfrdf=19-3=16 Ss-=∑∑x-x)2=1.04 j=l i= S'=SSt/(k-1) 0.731/3 sS1=n∑(x7-x2=0.731 =0.2437 Se2=SSe/k(n-1) SS。=∑∑(x /14少-x)=0.309 =0.309/16 =0.0193
df 解: T=kn-1=4×5-1=19 dft=k-1=4-1=3 dfe=dfT -dft=19-3=16 St 2=SSt/(k-1) =0.731/3 =0.2437 Se 2=SSe/[k (n-1)] =0.309/16 =0.0193 SSe =0.309 = = = = = = = = − + − = − + − n i i i n i k j ij n i k j i i n i k j ij x x k x x x x x x 1 2 2 1 1 1 1 2 2 1 1 ( ) ( ) ( ) ( ) 1 2 1 1 (x x) n i k j ij − = = 2 1 1 ( x x ) n i k j ij − = = 2 1 1 (x x) n i k j ij − = = SST = =1.04 2 1 1 (x x) n i k j ij − = = 2 1 1 ( x x ) n i k j ij − = = 2 1 1 (x x) n i k j ij − = = 2 1 1 (x x) n i k j ij − = = 2 1 1 (x x) n i k j ij − = = SSt = =0.731 2 1 1 (x x) n i k j ij − = = 2 1 1 (x x) n i k j ij − = = 2 1 1 (x x) n i k j ij − = = = = = = = = = = − + − = − + − n i i i n i k j ij n i k j i i n i k j ij x x k x x x x x x 1 2 2 1 1 1 1 2 2 1 1 ( ) ( ) ( ) ( ) 1
方差分析表 变异来源df均方和S2FF05F01 组间307310243712.63*324529 组内1603090.0193 总变异191.04 结论:差异极显著。 即用崩解仪法测得4个不同厂生产的阿司匹林的释 放度有极显著差别
变异来源 df 均方和 S 2 F F0.05 F0.01 组 间 3 0.731 0.2437 12.63** 3.24 5.29 组 内 16 0.309 0.0193 总变异 19 1.04 结论:差异极显著。 即用崩解仪法测得4个不同厂生产的阿司匹林的释 放度有极显著差别。 方差分析表