Sample Means(耶鲁大学教材)
Sample Means
The sample mean![](http://www.stat.yale.edu/Courses/1997-98/101/xbar.gif)
![](http://www.stat.yale.edu/Courses/1997-98/101/mu.gif)
![](http://www.stat.yale.edu/Courses/1997-98/101/mu.gif)
![](http://www.stat.yale.edu/Courses/1997-98/101/sigma.gif)
![](http://www.stat.yale.edu/Courses/1997-98/101/xbardef.gif)
By the properties of means and variances of random variables, the mean and variance of the sample mean are the following:
![](http://www.stat.yale.edu/Courses/1997-98/101/smmnvar.gif)
![](http://www.stat.yale.edu/Courses/1997-98/101/xbar.gif)
For example, suppose the random variable X records a randomly selected student's score on a national test, where the population distribution for the score is normal with mean 70 and standard deviation 5 (N(70,5)). Given asimple
random sample (SRS) of 200 students, the distribution of the sample mean score has mean 70 and standard deviation 5/sqrt(200) = 5/14.14 = 0.35.
Distribution of the Sample Mean
When the distribution of the population is normal, then the distribution of the sample mean is also normal. For a normal population distribution with mean![](http://www.stat.yale.edu/Courses/1997-98/101/mu.gif)
![](http://www.stat.yale.edu/Courses/1997-98/101/sigma.gif)
![](http://www.stat.yale.edu/Courses/1997-98/101/mu.gif)
![](http://www.stat.yale.edu/Courses/1997-98/101/smvar.gif)
This result follows from the fact that any linear combination of independent normal random variables is also normally distributed. This means that for two independent normal random variablesX and
Y and any constants a and b, aX + bY will be normally distributed. In the case of the sample mean, the linear combination is
=(1/n)*(X1 + X2 + ... Xn).
For example, consider the distributions of yearly average test scores on a national test in two areas of the country. In the first area, the test scoreX is normally distributed with mean 70 and standard deviation 5. In the second area, the yearly
average test scoreY is normally distributed with mean 65 and standard deviation 8. The differenceX - Y between the two areas is normally distributed, with mean 70-65 = 5 and variance 5² + 8² = 25 + 64 = 89. The standard deviation is the square
root of the variance, 9.43. The probability that areaX will have a higher score than area
Y may be calculated as follows:
P(X > Y) = P(X - Y > 0)
= P(((X - Y) - 5)/9.43 > (0 - 5)/9.43)
= P(Z > -0.53) = 1 - P(Z < -0.53) = 1 - 0.2981 = 0.7019.
Area X will have a higher average score than area Y about 70% of the time.
The Central Limit Theorem
The most important result about sample means is the Central Limit Theorem. Simply stated, this theorem says that for a large enough sample sizen, the distribution of the sample mean![](http://www.stat.yale.edu/Courses/1997-98/101/xbar.gif)
![](http://www.stat.yale.edu/Courses/1997-98/101/sigma.gif)
A formal statement of the Central Limit Theorem is the following:
If is the mean of a random sampleX1, X2, ... , Xn of size
n from a distribution with a finite mean
and a finite positive variance
²,
then the distribution ofW =
isN(0,1) in the limit as n approaches infinity.
This means that the variable is distributedN(
,
).
One well-known application of this theorem is the normal approximation to the binomial distribution.
Example
Using the MINITAB "RANDOM" command with the "UNIFORM" subcommand, I generated 100 samples of size 50 each from the Uniform(0,1) distribution. The mean of this distribution is 0.5, and its standard deviation is approximately 0.3. I then applied the "RMEAN" command to calculate the sample mean across the rows of my sample, resulting in 50 sample mean values (each of which represents the mean of 100 observations). The MINITAB "DESCRIBE" command gave the following information about the sample mean data:Descriptive Statistics Variable N Mean Median Tr Mean StDev SE Mean C101 50 0.49478 0.49436 0.49450 0.02548 0.00360 Variable Min Max Q1 Q3 C101 0.43233 0.55343 0.47443 0.51216The mean 0.49 is nearly equal to the population mean 0.5. The desired value for the standard deviation is the population standard deviation divided by the square root of the size of the sample (which is 10 in this case), approximately 0.3/10 = 0.03. The calculated value for this sample is 0.025. To evaluate the normality of the sample mean data, I used the "NSCORES" and "PLOT" commands to create a normal quantile plot of the data, shown below.