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graph on the previous page) VER. 9/11/2012. © P. KOLM 47 Example: Consistency of the sample average We saw earlier that the sample average, Yn , using a random sample with n
observations, is an unbiased estimator of the population mean, mY .
s2
Furthermore, we established that Var (Yn ) = . Therefore, Var (Yn ) 0 as
n
n ¥ , so the sample average is consistent.
Remarkably, the conclusion that the sample average is a consistent estimator of
the population mean holds even if Var (Yn ) does not exist. This result is known as
the weak law of large numbers (WLLN). VER. 9/11/2012. © P. KOLM 48 Asymptotic Property: Central Limit Theorem Intuition: The Central Limit Theorem (CLT) says that the sampling distribution
of the sample mean is approximately normally distributed, regardless of the
distribution of the underlying random sample
This is almost like magic, isn’t it? Formally, the CLT states that the standardized average of any population with mean mY and variance s 2 has an asymptotic standard normal distribution
Z= Yn  mY
sY / n N (0,1) (See: Matlab illustration) VER. 9/11/2012. © P. KOLM 49 Recall: Sampling Distribution for the Mean
2
sY
Y N (mY , ) , or equivalently
n
Y  mY N (0,1)
sY / n Assume the sample mean is drawn from the normal distribution above. From
normal distribution properties, there is a 95% chance that the population mean is
in the range:10 Y  1.96 ⋅ VER. 9/11/2012. © P. KOLM sY
n £ mY £ Y + 1.96 ⋅ sY
n
50 Confidence Interval for the Mean (With Known Standard Deviation)
2
Let us first assume that sY is known. For a particular sample, y = (y1,..., yn ) , we say that é
ù
êy  1.96 ⋅ sY , y + 1.96 ⋅ sY ú
ê
ú
n
nû
ë
is a 95% confidence interval for mY . Sometimes this is also written as y 1.96 ⋅ sY
Area = 0.95 n Area = 0.025 Area = 0.025 1.96 VER. 9/11/2012. © P. KOLM 1.96 51 Confidence Interval for the Mean (With Unknown Standard Deviation)
2 In most practical applications both mY and sY are not known
2
2 Then, we replace sY with the estimate sY With known variance, we had
Y  mY
sY / n N (0,1) It turns out that
Y  mY
sY / n tn 1 How do we construct a 95% confidence interval using this information? VER. 9/11/2012. © P. KOLM 52 We need to find the critical value, c, such that Y  c ⋅ sY
n £ mY £ Y + c ⋅ sY
n Area = 0.95 Area = 0.025 Area = 0.025 c c Then, our 95% confidence interval becomes
é
ù
êy  c ⋅ sY , y + c ⋅ sY ú
ê
ú
n
nû
ë
For the t distribution c depends on the degrees of freedom, n  1 , and the
confidence level. Its value can be calculated using statistical packages (i.e.
Matlab, Stat, R) or obtained from a statistical table VER. 9/11/2012. © P. KOLM 53 Remarks (1/2): To explicitly show that the critical value, c, depends on the confidence level , a , and the degrees of freedom, n, it is written as ca/2 (n ) There is a simple way to remember how to construct a confidence interval for the mean of a normal distribution. Recall that std (Y ) = sY / n . Thus,
sY / n is the point estimate of std (...
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This document was uploaded on 02/17/2014 for the course COURANT G63.2751.0 at NYU.
 Fall '14

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