It is a common requirement to efficiently estimate population parameters based on simple random sample data.
基于简单随机样本来estimate全局参数, 由于是预估, 一般是预估一个区间, 所以称为区间预估.
For any particular random sample, we can always compute its sample mean.
> library(MASS) # load the MASS package > height.survey = survey$Height
> mean(height.survey, na.rm=TRUE) # skip missing values [1] 172.38
直接用样本mean来作为全局mean, 显然这个方法比较简陋
Here, we discuss the case where the population variance σ2 is assumed known.
Let us denote the 100(1 −α∕2) percentile of the standard normal distribution as zα∕2. For random sample of sufficiently large size, the end points of the interval estimate at (1 − α) confidence level is given as follows:
这个就比直接用样本平均值高级点, 虽然我不知道为什么是这样来用全局方差
Assume the population standard deviation σ of the student height in survey is 9.48. Find the margin of error and interval estimate at 95% confidence level(1 − α).
> library(MASS) # load the MASS package > height.response = na.omit(survey$Height)
> n = length(height.response) > sigma = 9.48 # population standard deviation > sem = sigma/sqrt(n); sem # standard error of the mean [1] 0.65575
> E = qnorm(.975)∗sem; E # margin of error [1] 1.2852
> xbar = mean(height.response) # sample mean > xbar + c(−E, E) [1] 171.10 173.67
Here, we discuss the case where the population variance is not assumed.
Let us denote the 100(1 −α∕2) percentile of the Student t distribution with n− 1 degrees of freedom as tα∕2. For random samples of sufficiently large size, and with standard deviation s, the end points of the interval estimate at (1 −α) confidence level is given as follows:
不知道全局方差, 就通过样品标准偏差(samples standard deviation)来替代全局方差进行预估, 更牛比了点
Without assuming the population standard deviation of the student height in survey, find the margin of error and interval estimate at 95% confidence level.
> n = length(height.response) > s = sd(height.response) # sample standard deviation > SE = s/sqrt(n); SE # standard error estimate [1] 0.68117
> E = qt(.975, df=n−1)∗SE; E # margin of error [1] 1.3429
The quality of a sample survey can be improved by increasing the sample size. The formula below provide the sample size needed under the requirement of population mean interval estimate at (1 −α) confidence level, margin of error E, and population variance σ2. Here, zα∕2 is the 100(1 − α∕2) percentile of the standard normal distribution.
样本越大当然预测就越准, 这个公式就是来算合适的样本size的
Assume the population standard deviation σ of the student height in survey is 9.48. Find the sample size needed to achieve a 1.2 centimeters margin of error at 95% confidence level.
> zstar = qnorm(.975) > sigma = 9.48 > E = 1.2 > zstar^2 ∗ sigma^2/ E^2 [1] 239.75
Multiple choice questionnaires in a survey are often used to determine the the proportion of a population with certain characteristic. For example, we can estimate the proportion of female students in the university based on the result in the sample data set survey.
Find a point estimate of the female student proportion from survey.
> library(MASS) # load the MASS package > gender.response = na.omit(survey$Sex) > n = length(gender.response) # valid responses count
> k = sum(gender.response == "Female") > pbar = k/n; pbar [1] 0.5
After we found a point sample estimate of the population proportion, we would need to estimate its confidence interval.
Let us denote the 100(1 −α∕2) percentile of the standard normal distribution as zα∕2. If the samples size n and population proportion p satisfy the condition that np ≥ 5 and n(1 − p) ≥ 5, than the end points of the interval estimate at (1 − α) confidence level is defined in terms of the sample proportion as follows.
The quality of a sample survey can be improved by increasing the sample size. The formula below provide the sample size needed under the requirement of population proportion interval estimate at (1 − α) confidence level, margin of error E, and planned proportion estimate p. Here, zα∕2 is the 100(1 − α∕2) percentile of the standard normal distribution.
本文章摘自博客园，原文发布日期：2012-02-17
相关资源：敏捷开发V1.0.pptx