Reading notes - > statistics] construction of 12-02 confidence interval -t distribution concept introduction

t Distribution

Suppose a situation ： We want to know the typical weight of sugar balls . But because only one candy store made a request , Therefore, only include 10 Representative samples , Then weigh each sugar ball . This sample is $\overline{x}=0.5Oz.,s^2=0.09$.

The same as usual ：

The first 1 Step ： Select the population statistics

We need to build a confidence interval for the mean weight of sugar balls , That is, it should be the overall average $\mu$ Build confidence intervals . Due to the need for $\mu$ The confidence interval of , So the next step is to ask $\mu$ The sampling distribution of ——$\overline{X}$ The distribution of .

The first 2 Step ： seek $\overline{X}$ Probability distribution of

Here we have a problem , We can know from the shortcut table above , When the overall distribution $X$ When it conforms to the normal distribution , We don't know the population variance ${\sigma }^2$, Need to use a point estimator $s^2$ Instead of , But on the condition that n It's big （ At least 30）, This road won't work

Another problem is , The sample is too small , The estimated value is likely to have a large error —— The error is much larger than using large samples . These potential errors mean that using a normal distribution cannot produce a sufficiently accurate $\overline{X}$ Probability , In that case, the exact confidence interval cannot be obtained .

that ,$\overline{X}$ Which distribution does it conform to ？ actually , It conforms to t Distribution .

When the sample is small ,$\overline{X}$ accord with t Distribution

When the overall distribution conforms to the normal distribution ,${\sigma }^2$ Unknown , And the sample available is very small ,$\overline{X}$ accord with t Distribution .

t The distribution is smooth 、 Symmetrical curve , The exact shape depends on the sample size . When the sample is large ,t The distribution looks like a normal distribution ; When the sample is small , The curve is flat , There are two thick tails . It has only one parameter ——v,v=n-1.n Is the size of the sample ,v go by the name of freedom .

Here's a picture , Corresponding to various v Corresponding t Distribution .

“T accord with t Distributed with degrees of freedom v” The concise expression of is ：
$$T\sim t(v)$$
（T For test statistics , The calculation method is shown below ;t(v) Express ： We are using degrees of freedom v Of t Distribution ;v=n-1）

t The use of distribution is similar to that of normal distribution —— First, the upper and lower limits of the probability interval are transformed into standard scores , Then use the probability table to find the desired result .

seek t Standard score of distribution

t The calculation method of the standard score of the distribution is the same as that of the normal distribution . Like dealing with normal distribution , Let's subtract the expectation of the sampling distribution , Then divide the difference by the standard deviation . The only difference is , We use it T instead of Z Represents the result , This is to match t Use of distribution .

We need to find out $\overline{X}$ The distribution of （ Detailed see ： Last chapter , Probability of sample mean ）, So we need $\overline{X}$ Expectations and standard deviation .$\overline{X}$ The expectation is $\mu$, The standard deviation is $\sigma /\sqrt{n}$. Because of the need to use s It is estimated that $\sigma$ The numerical , therefore t The standard score of the distribution is calculated as follows ：

We just substitute $\overline{X},\hat{\sigma }$ and n That's it .

It is known that v=n-1=9,$s^2=0.09$, be
$$\begin{gathered} T=\frac{\overline{X}-\mu }{s/\sqrt{n}} \\ =\frac{\overline{X}-\mu }{\sqrt{0.09/10}} \\ =\frac{\overline{X}-\mu }{0.0949} \end{gathered}$$
The first 3 Step ： Determine the confidence level

Confidence level means that you want to be right “ Confidence intervals contain population statistics ” How confident is this statement . Like above , Let's use it 95% Confidence level as the overall mean , So the probability that the overall mean lies in the confidence interval is 0.95.

The first 4 Step ： Find the upper and lower confidence limits

t The algorithm of upper and lower confidence limits of distribution is similar to that of normal distribution , It can be calculated by the following formula ：

We can go through t Calculate the distribution probability table t value .

Use t Distribution probability table

adopt t The distribution probability table can be found P(T>t) Medium t value . In our case ,p=0.025.

To find out t value , First, look up the... In the first column from the probability table v value , Then find the first line p value , The intersection of the two is t value . for example , lookup v=7 and p=0.05, Available t=1.895.

Find out t After value , We can find the confidence interval .

t Comparison between distribution and normal distribution

When estimating the population variance with small samples ,t More accurate distribution .

Based on small sample estimation ${\sigma }^2$ There is a question , That is, it may not accurately reflect the true value of the overall variance . in other words , We need to make the interval wider , So as to leave some error space in the confidence interval .

t The shape of the distribution follows v The value changes , Considering the size of the sample , Even if ${\sigma }^2$ There are various uncertainties in the estimation accuracy of ,t The distribution can also be ignored . When n Very hour ,t The confidence interval given by the distribution is wider than that of the normal distribution , This makes it more suitable for small samples .

Concise algorithm of confidence interval ——t Distribution

The following is about t When to use the distribution and $\mu$ A simple reminder of the confidence interval of .

To find out t(v), Need to find t Distribution probability table . So , use v=n-1 Calculate the confidence interval with the confidence level you have determined .

Example solution

Another example

ask ： If the sample size n Change , What effect will it have on the confidence interval ？

answer ： If n Reduce , Then the confidence interval becomes wider ; If n increase , Then the confidence interval narrows .

The expression of confidence interval is ：
$$statistic\pm Error\; range$$
among , Error range = c * Standard deviation of Statistics

The standard deviation of statistics depends on the size of the sample ——n The bigger it is , The smaller the standard deviation of Statistics ; That is to say ,n The larger the error range, the smaller ,n The smaller the error, the larger the error range .

As a general rule , Smaller samples form a wider confidence interval , Larger samples form a narrow confidence interval .

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summary

Now we have learned two methods of estimating population statistics .

• Last chapter , We learn to use Point estimator , The point estimator method can be used to estimate the exact value of population statistics , It is the best possible prediction based on sample data .
• This chapter , We learn to use population statistics confidence interval . This method is not an accurate estimate of the overall statistics , Instead, a numerical range with high reliability of the overall statistics is obtained .