你这里的用法就是：
ks.test(data,pt,df=df) #data是样本的数据，df是要检验的t分布的自由度
我们可以用很多方法分析一个单变量数据集的分布。最简单的办法就是直接看数
字。利用函数summary 和fivenum 会得到两个稍稍有点差异的汇总信息。此外，stem
(\茎叶"图)也会反映整个数据集的数字信息。
>attach(faithful)
>summary(eruptions)
Min. 1st Qu. Median Mean 3rd Qu. Max.
1.600 2.163 4.000 3.488 4.454 5.100
>fivenum(eruptions)
[1] 1.6000 2.1585 4.0000 4.4585 5.1000
>stem(eruptions)
The decimal point is 1 digit(s) to the left of the |
16 | 070355555588
18 | 000022233333335577777777888822335777888
20 | 00002223378800035778
22 | 0002335578023578
24 | 00228
26 | 23
28 | 080
30 | 7
32 | 2337
34 | 250077
36 | 0000823577
38 | 2333335582225577
40 | 0000003357788888002233555577778
42 | 03335555778800233333555577778
44 | 02222335557780000000023333357778888
46 | 0000233357700000023578
48 | 00000022335800333
50 | 0370
茎叶图和柱状图相似，R 用函数hist 绘制柱状图。
>hist(eruptions)
>## 让箱距缩小，绘制密度图
>hist(eruptions, seq(1.6, 5.2, 0.2), prob=TRUE)
>lines(density(eruptions, bw=0.1))
>rug(eruptions) # 显示实际的数据点
更为精致的密度图是用函数density 绘制的。在这个例子中，我们加了一条
由density 产生的曲线。你可以用试错法（trial-and-error）选择带宽bw（bandwidth）
因为默认的带宽值让密度曲线过于平滑（这样做常常会让你得到非常有\意思"的密度
分布）。(现在已经有一些自动的带宽挑选方法2，在这个例子中bw = "SJ"给出的结
果不错。)
我们可以用函数ecdf 绘制一个数据集的经验累积分布（empirical cumulative
distribution）函数。
>plot(ecdf(eruptions), do.points=FALSE, verticals=TRUE)
显然，这个分布和其他标准分布差异很大。那么右边的情况怎么样呢，就是火山
爆发3分钟后的状况？我们可以拟合一个正态分布，并且重叠前面得到的经验累积密
度分布。
>long <- eruptions[eruptions >3]
>plot(ecdf(long), do.points=FALSE, verticals=TRUE)
>x <- seq(3, 5.4, 0.01)
>lines(x, pnorm(x, mean=mean(long), sd=sqrt(var(long))), lty=3)
分位比较图（Quantile-quantile (Q-Q) plot）便于我们更细致地研究二者的吻合
程度。
par(pty="s") # 设置一个方形的图形区域
qqnorm(long)qqline(long)
上述命令得到的QQ图表明二者还是比较吻合的，但右侧尾部偏离期望的正态分布。
我们可以用t 分布获得一些模拟数据以重复上面的过程
x <- rt(250, df = 5)
qqnorm(x)qqline(x)
这里得到的QQ图常常会出现偏离正态期望的长尾区域(如果是随机样本)。我们可以用
下面的命令针对特定的分布绘制Q-Q图
qqplot(qt(ppoints(250), df = 5), x, xlab = "Q-Q plot for t dsn")
qqline(x)
最后，我们可能需要一个比较正规的正态性检验方法。R提供了Shapiro-Wilk 检
验
>shapiro.test(long)
Shapiro-Wilk normality test
data: long
W = 0.9793, p-value = 0.01052
和Kolmogorov-Smirnov 检验
>ks.test(long, "pnorm", mean = mean(long), sd = sqrt(var(long)))
One-sample Kolmogorov-Smirnov test
data: long
D = 0.0661, p-value = 0.4284
alternative hypothesis: two.sided
(注意一般的统计分布理论（distribution theory）在这里可能无效，因为我们用同样
的样本对正态分布的参数进行估计的。)
转载于：
http://www.biostatistic.net/thread-2413-1-1.html
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