概率统计笔记 CH7

Statistical Intervals Based on a Single Sample

基于单次采样的统计区间

概率统计

• A point estimate, because it is a single number, by itself provides no information about the precision and reliability of estimation.
• An alternative to reporting a single sensible value for the parameter being estimated is to calculate and report an entire interval of plausible values—an interval estimate or confidence interval - CI(置信区间). A confidence interval is always calculated by first selecting a confidence level(置信度), which is a measure of the degree of reliability of the interval. Information about the precision of an interval estimate is conveyed by the width of the interval.

1. Basic Properties of Confidence Intervals

   • Suppose that the parameter of interest is a population mean $\mu$ and that

     1. The population distribution is normal.
     2. The value of the population standard deviation $\sigma$ is known.
        
        • Normality of the population distribution is often a reasonable assumption. However, if the value of $\mu$ is unknown, it is implausible that the value of $\sigma$ would be available.
          
          
        • If after observing $X_1=x_1,...X_n=x_n$, we compute the observed sample mean $\overline{x}$. The resulting fixed interval is called a 95% confidence interval for $\mu$. This CI can be expressed either as
          $$\lgroup\overline{x}-1.96·\frac{\sigma}{\sqrt{n}}, \overline{x}+1.96·\frac{\sigma}{\sqrt{n}}\rgroup\ \ is\ a\ 95\%\ CI\ for\ \mu$$
          or as $\overline{x}-1.96·\frac{\sigma}{\sqrt{n}} < \mu < 1.96·\frac{\sigma}{\sqrt{n}}\ \ with\ 95\%\ confidence$.
          A concise expression for the interval is $\overline{x}\pm 1.96·\sigma/\sqrt{n}$.

   • Other Levels of Confidence

     • A $100(1-\alpha)\%$ confidence interval for the mean $\mu$ of a normal population when the value of $\sigma$ is known is given by
       $$\lgroup \overline{x}-z_{\alpha/2}·\frac{\sigma}{\sqrt{n}}, \overline{x}+z_{\alpha/2}·\frac{\sigma}{\sqrt{n}}\rgroup$$
       or, equivalently, by $\overline{x}\pm z_{\alpha/2}·\sigma/\sqrt{n}$
       For example, the $99\%$ CI is $\overline{x}\pm 2.58·\sigma/\sqrt{n}$

   • Confidence Level, Precision, and Choice of Sample Size

     • Why settle for a confidence level of 95% when a level of 99% is achievable? Because the price paid for the higher confidence level is a wider interval.
       The width of the interval may be thought to specify its precision or accuracy. Then it is inversely related to the confidence level (or reliability), but positively related to the sample size n.
       An appealing strategy is to specify both the desired confidence level and interval width and then determine the necessary sample size n.
       
     • The general formula for the sample size n necessary to ensure an interval width w is obtained from $w=2·z_{\alpha/2}·\sigma/\sqrt{n}$ as
       $$n=\lgroup 2z_{\alpha/2}·\frac{\sigma}{w}\rgroup^2$$
       The smaller the desired width w, the large n must be.
       The half-width $1.96\sigma/\sqrt{n}$ of the $95\%$ CI is sometimes called the bound on the error of estimation associated with a $95\%$ confidence level.

   • Deriving a Confidence interval

     • Let $X_1,X_2,...,X_n$ denote a sample on which the CI for a parameter $\theta$ is to be based. Suppose a random variable $h(X_1,X_2,…,X_n;\theta)$ satisfying the following two properties can be found:
       
       1. The variable depends functionally on both $X_1,X_2,...,X_n$ and $\theta$.
       2. The probability distribution of the variable does not depend on θ or on any other unknown parameters.
     • In order to determine a $100(1-\alpha)%$ CI of $\theta$, we proceed as follows:
       $$P(a<h(X_1,X_2,…,X_n;\theta)<b) = 1-\alpha$$
       Because of the second property, $a$ and $b$ do not depend on $\theta$. In the normal example, we had $a=-z_{\alpha/2}$ and $b=z_{\alpha/2}$. Suppose we can isolate $\theta$ in the inequation:
       $$P(l(X_1,X_2,…,X_n)<\theta<u(X_1,X_2,…,X_n))=1-\alpha$$
       So a $100(1-\alpha)\%$ CI is $[l(X_1,X_2,…,X_n),u(X_1,X_2,…,X_n)]$. In the normal example, $l(X_1,X_2,…,X_n) = \overline{X}-z_{\alpha/2}·\sigma/\sqrt{n}$ and $u(X_1,X_2,…,X_n) = \overline{X}+z_{\alpha/2}·\sigma/\sqrt{n}$
       In general, the form of the h function is suggested by examining the distribution of an appropriate estimator $\hat{\theta}$.
       
       

2. Large-sample Confidence Intervals for a Population Mean and Proportion

   • A Large-Sample Interval for $\mu$

     • Let $X_1,X_2,…,X_n$ be a random sample from a population having a mean $\mu$ and standard deviation $\sigma$. Provided that n is large, the Central Limit Theorem (CLT) implies that X has approximately a normal distribution whatever the nature of the population distribution.
     • If n is sufficiently large, the standardized variable
       $$Z=\frac{\overline{X}-\mu}{S/\sqrt{n}}$$
       has approximately a standard normal distribution.
       This implies that
       $$\overline{x}\pm z_{\alpha/2}·\frac{s}{\sqrt{n}}$$
       is a large-sample confidence interval for $\mu$ with confidence level approximately $100(1-\alpha)\%$.
     • This formula is valid regardless of the shape of the population distribution. Generally speaking, n > 40 will be sufficient to justify the use of this interval.

   • A Large-Sample Confidence interval for a Population Proportion

     • A confidence interval for a population proportion p with confidence level approximately $100(1-\alpha)\%$ has
       $$lower\ confidence\ limit(置信下限) =\frac{\hat{p}+\frac{z_{\alpha/2}^2}{2n} - z_{\alpha/2} \sqrt{\frac{\hat{p}\hat{q}}{n}+\frac{z_{\alpha/2}^2}{4n^2}}}{1+(z_{\alpha/2}^2)/n}$$
       $$upper\ confidence\ limit(置信上限) =\frac{\hat{p}+\frac{z_{\alpha/2}^2}{2n} - z_{\alpha/2} \sqrt{\frac{\hat{p}\hat{q}}{n}+\frac{z_{\alpha/2}^2}{4n^2}}}{1+(z_{\alpha/2}^2)/n}$$
       The traditional approximate confidence limits under a large sample size:
       $$\hat{p}\pm z_{\alpha/2}\sqrt{\hat{p}\hat{q}/n}$$
     • For an interval with a desired degree of precision, equate the width of the CI for p to a prespecified width w. It gives a quadratic equation for the sample size n.
       The solution is too long...Neglecting the terms in the numerator involving $w^2$ gives
       $$n \approx \frac{4z_{\alpha/2}^2\hat{p}\hat{q}}{w^2}$$
       This expression is what results from equating the width of the traditional interval to w.

   • One-Sided Confidence Intervals (Confidence Bounds, 置信界限)
     

     • Sometimes one may want a CI with only a lower bound or an upper bound.
       For example, under the $100(1-α)\%$ confidence level and with a large sample, we have, approximately,
       $$P\left( \frac{\overline{X}-\mu}{S/\sqrt{n}} < z_{\alpha}\right) \approx 1-\alpha$$
       Rearranging the inequation in the parentheses, for a given sample, we obtain
       $$\mu > \overline{x} - z_{\alpha}·\frac{s}{\sqrt{n}}$$
       which is a one-sided CI(amounting to a lower confidence bound here).
       An upper confidence bound can be obtained similarly.
       
     • A large-sample upper confidence bound for $\mu$ is
       $$\mu < \overline{x} + z_{\alpha}·\frac{s}{\sqrt{n}}$$
       and a large-sample lower confidence bound for $\mu$ is
       $$\mu > \overline{x} - z_{\alpha}·\frac{s}{\sqrt{n}}$$
       A one-sided confidence bound for p results from replacing $z_{\alpha/2}$ by $z_{\alpha}$ and $\pm$ by either + or – in the CI formula for p.

3. Intervals Based on a Normal Population Distribution

   • Intro

     • Assumption: $X_1,X_2,…,X_n$ constitutes a random sample from a normal distribution with both $\mu$ and $\sigma$ unknown.
     • Theorem: Let $X_1,X_2,…,X_n$ be a random sample from a normal distribution with parameters $\mu$ and $\sigma^2$. Then the rv
       $$\frac{(n-1)S^2}{\sigma^2} = \frac{\sum(X_i - \overline{X})^2}{\sigma^2}$$
       has a chi-squared probability distribution with n-1df(自由度).
     • Theorem: Suppose rv's X and Y are independent, X follows a standard normal distribution, Y follows a chi-squared distribution with k degrees of freedom. Then the function of random variable
       $$T=\frac{X}{\sqrt{Y/k}}$$
       has t distribution with k degrees of freedom.
     • Theorem: When $\overline{X}$ is the mean of a random sample of size n from a normal distribution with mean $\mu$, the rv
       $$T=\frac{\overline{X} - \mu}{S/\sqrt{n}}$$
       has a t distribution with n-1 degrees of freedom(df).

   • Properties of t Ditributions

     • A t distribution is governed by only one parameter, the number of degrees of freedom of the distribution.
     • Let $t_v$ denote the density function curve for $v$ df.
       
       1. Each $t_v$ curve is bell-shaped and centered at 0.
       2. Each $t_v$ curve is more spread out than the standard normal curve.
       3. As $v$ increases, the spread of the corresponding $t_v$ curve decreases.
       4. As $v\to\infty$, the sequence of $t_v$ curves approaches the standard normal curve.
     • Notation: Let $t_{\alpha,v}$ = the number on the measurement axis for which the area under the t curve with $v$ df to the right of $t_{\alpha,v}$ is $\alpha$; $t_{\alpha,v}$ is called a t critical value(临界值).
       

   • The One-Sample t Confidence Interval

     • The standardized variable T has a t distribution with n-1 df, and the area under the corresponding t density curve between $-t_{\alpha/2, n-1}$ and $t_{\alpha/2, n-1}$ is $1-\alpha$, so
       $$P(-t_{\alpha/2, n-1} < T < t_{\alpha/2, n-1}) = 1-\alpha$$
     • Let $\bar{x}$ and s be the sample mean and sample standard deviation computed from the results of a random sample from a normal population with mean $\mu$. Then a $100(1-\alpha)\%$ confidence interval for $\mu$ is
       $$\lgroup \bar{x}-t_{\alpha/2,n-1}·\frac{s}{\sqrt{n}}, \bar{x}+t_{\alpha/2,n-1}·\frac{s}{\sqrt{n}} \rgroup$$
       or, more compactly, $\bar{x}\pm t_{\alpha/2,n-1}·\frac{s}{\sqrt{n}}$.
       An upper confidence bound with $100(1-\alpha)\%$ confidence level for $\mu$ is $\bar{x}+t_{\alpha,n-1}·s/\sqrt{n}$. Replacing + by – gives a lower confidence bound for $\mu$.
       

   • A Prediction Interval(预测区间) for a Single Future Value

     • A prediction interval - PI for a single observation to be selected from a normal population distribution is
       $$\bar{x}\pm t_{\alpha/2,n-1}·s\sqrt{1_\frac{1}{n}}$$
       The prediction level(预测水平) is $100(1-\alpha)\%$.

   • Tolerance Intervals(容许区间)

     

4. Confidence Intervals for the Variance and Standard Deviation of a Normal Population

   • A $100(1-\alpha)\%$ confidence interval for the variance $\sigma^2$ of a normal population has lower limit
     $$(n-1)s^2/\chi_{\alpha/2,n-1}^2$$
     and upper limit
     $$(n-1)s^2/\chi_{1-\alpha/2,n-1}^2$$
     A confidence interval for $\sigma$ has lower and upper limits that are the square roots of the corresponding limits in the interval for $\sigma^2$.

   

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