王静龙《非参数统计分析》课后习题计算题参考答案习题一
1.One Sample t-test for a Mean
Sample Statistics for x
N Mean Std. Dev. Std. Error
-------------------------------------------------
26 1.38 8.20 1.61
Hypothesis Test
Null hypothesis: Mean of x = 0
Alternative: Mean of x ^= 0
t Statistic Df Prob > t
---------------------------------
0.861 25 0.3976
95 % Confidence Interval for the Mean
Lower Limit: -1.93
Upper Limit: 4.70
则接受原假设认为一样
习题二
1.描述性统计
习题三
1.1
{}+01=1339
:6500:650013=BINOMDIST(13,39,0.5,1)
=0.026625957
S n H me H me P S +
==<≤
另外:在excel2010中有公式 BINOM.INV(n,p,a) 返回一个数值,它使得累计二项式分布的函数值大于或等于临界值a 的最小整数
*
**0*0+1inf :2BINOM.INV(39,0.5,0.05)=14
1sup :113
2S 1313
n m i n d i n m m i n d d m i d αα==??????
??=≥??
? ?????????
??????
??≤=-=?? ? ?????????
=≤=∑∑= 以上两种都拒绝原假设,即中位数低于6500
1.2
****01426201inf :221inf :122BINOM.INV(40,0.5,1-0.025)=26d=n-c=40-26=14580064006200
n
n i c n m i n c c i n m m i x x me x αα==????????=≤??
? ???????????????
??=≥-??
? ?????????
====∑∑
2.
{}+01=4070
:6500:6500
2402*(1-BINOMDIST(39,70,0.5,1))=0.281978922
S n H me H me P S +==≠≥=
则接受原假设,即房价中位数是6500
3.1
{
}+01=15521552527207911
::22
n 1552=5.33E-112
S n H p H p P S φ+=+==
>?≥≈ ?比较大,则用正态分布近似
**
+**0:=155215525272079
1inf :221inf :122m=BINOM.INV(2079,0.5,0.975)=1084
n
n i c n m i S n n c c i n m m i αα===+=????????=≤??
? ?????????
??????
??=≥-??
? ?????????
∑∑另外
则拒绝原假设,即相信孩子会过得更好的人多
3.2
P 为认为生活更好的成年人的比例,则
1522
=0.7465132079
p 的比估计是:
4.
{}00.90610.90618154157860:65:6510.9060.094~(,)
181541BINOMDIST(18153,157860,0.094,1)=0
S n H P H P p S b n p P S +++===>=-=≥=-
因为0〈0.05则拒绝原假设
习题四
1.
()()++0.025+W =6+8+10+1+4+12+9+11+2+7=70p 2P W 70n=12c =65p 2P W 65=0.05≥≥符号秩和检验统计量:
值为,当得所以值小于即拒绝原假设
2.
()()++0.025+W =2.5+2.5+7+7+7+7+10.5+14+14+14+14+14+17.5+17.5+19+20+23+24=234.5
p 2P W 234.5n=25 c =236p 2P W 236=0.05≥≥符号秩和检验统计量:
值为,当得所以值小于即接受原假设
{}011826
:0:0
2182*(1-BINOMDIST(17,25,0.5,1))=0.043285251S n H me H me P S +===≠≥=+符号检验:
则拒绝原假设
t t =0.861df=25 p=0.3976检验:
统计量接受原假设
3.