LSSG Black Belt Training. Estimation: Central Limit Theorem and Confidence Intervals. Central Limit Theorem. Assume a population with a non-normal distribution. Mean = µ Stdev = σ. If we took a sample of size 50 from this population, what would it look like?.
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LSSG Black Belt Training
Estimation: Central Limit Theorem and Confidence Intervals
Assume a population with a non-normal distribution.
Mean = µ
Stdev = σ
If we took a sample of size 50 from this population, what would it look like?
Each sample of n=50 from the same population will tend to look like the population, and the sample means will be close to the population mean.
The sample means are unbiased estimators of the population mean. They will vary randomly above and below the actual population mean.
If all such samples (n=50) were drawn, how would the sample means be distributed?
In other words, the X values will be approximately normally distributed.
How would the mean of this distribution compare to the original population mean? How about the standard deviation of this distribution?
How would sample size affect this relationship?
For sufficiently large sample sizes (typically n>30), the distribution of the sample means (X-Bar) is approximately normal, and
(Std Dev. of Population/ square root of n)
This standard deviation of the sample means is also called the standard error.
Since the X-bars are normally distributed, 95% of all samples (large enough n) from a population will yield an X-bar that is within 2 standard errors from the population mean.
We take a sample of 64 parts from a population, and want to estimate the population mean of the part length. The sample mean is 25 mm. The population standard deviation is known to be 0.2 mm.
From CLT, we know that this sample mean (25) is within 2 standard errors (actually 1.96) of the population mean, with 95% confidence.
Hence the reverse is also true.
Thus, population mean is
X-bar ± 2 * SE
Here, SE = 0.16 / √64 = 0.16/8 = 0.02
Thus 95% CI for µ is given by
25 ± 2*0.02, or 25 ± 0.04 mm
The value 0.04 is the Margin of Error (MOE)
In reality, σis generally unknown, and must be substituted with s, the sample standard deviation. In that case, the margin of error is higher, and is computed using the t-distribution rather than the standard normal (z dist). Thus instead of 1.96 standard errors for 95% confidence, we use a larger number obtained from the t-tables.
(In Excel, type =tinv(0.05,df), where df is the degrees of freedom, equal to n-1. Here in the previous problem, df is 63).