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clt 1
CENTRAL LIMIT THEOREM
 specifies a theoretical distribution
 formulated by the selection of all
possible random samples of a fixed
size n
 a sample mean is calculated for each
sample and the distribution of sample
means is considered
clt 2
SAMPLING DISTRIBUTION OF
THE MEAN
 The mean of the sample means is equal
to the mean of the population from
which the samples were drawn.
 The variance of the distribution is s
divided by the square root of n. (the
standard error.)
clt 3
STANDARD ERROR
Standard Deviation of the Sampling
Distribution of Means
sx = s/ /n
clt 4
How Large is Large?
 If the sample is normal, then the sampling
distribution of will also be normal, no matter
what the sample size.
 When the sample population is approximately
symmetric, the distribution becomes approximately
normal for relatively small values of n.
 When the sample population is skewed, the sample
size must be at least 30 before the sampling
distribution of becomes approximately normal.
x
x
clt 5
EXAMPLE
A certain brand of tires has a mean life of
25,000 miles with a standard deviation of
1600 miles.
What is the probability that the mean life of
64 tires is less than 24,600 miles?
clt 6
Example continued
The sampling distribution of the means
has a mean of 25,000 miles (the
population mean)
m = 25000 mi.
and a standard deviation (i.e.. standard
error) of:
1600/8 = 200
clt 7
Example continued
Convert 24,600 mi. to a z-score and use
the normal table to determine the
required probability.
z = (24600-25000)/200 = -2
P(z< -2) = 0.0228
or 2.28% of the sample means will be
less than 24,600 mi.
clt 8
ESTIMATION OF POPULATION
VALUES
 Point Estimates
 Interval Estimates
clt 9
CONFIDENCE INTERVAL
ESTIMATES for LARGE SAMPLES
 The sample has been randomly
selected
 The population standard deviation is
known or the sample size is at least
25.
clt 10
Confidence Interval Estimate of the
Population Mean
-
X: sample mean
s: sample standard deviation
n: sample size
X z
s
n
X z
s
n
 
   
m
clt 11
EXAMPLE
Estimate, with 95% confidence, the
lifetime of nine volt batteries using a
randomly selected sample where:
--
X = 49 hours
s = 4 hours
n = 36
clt 12
EXAMPLE continued
Lower Limit: 49 - (1.96)(4/6)
49 - (1.3) = 47.7 hrs
Upper Limit: 49 + (1.96)(4/6)
49 + (1.3) = 50.3 hrs
We are 95% confident that the mean
lifetime of the population of batteries is
between 47.7 and 50.3 hours.
clt 13
CONFIDENCE BOUNDS
 Provides a upper or lower bound for the
population mean.
 To find a 90% confidence bound, use the z
value for a 80% CI estimate.
clt 14
Example
 The specifications for a certain kind of
ribbon call for a mean breaking strength
of 180 lbs. If five pieces of the ribbon
have a mean breaking strength of 169.5
lbs with a standard deviation of 5.7 lbs,
test to see if the ribbon meets
specifications.
 Find a 95% confidence interval estimate
for the mean breaking strength.

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The Central Limit Theorem it specifies a theoretical distribution.

  • 1. clt 1 CENTRAL LIMIT THEOREM  specifies a theoretical distribution  formulated by the selection of all possible random samples of a fixed size n  a sample mean is calculated for each sample and the distribution of sample means is considered
  • 2. clt 2 SAMPLING DISTRIBUTION OF THE MEAN  The mean of the sample means is equal to the mean of the population from which the samples were drawn.  The variance of the distribution is s divided by the square root of n. (the standard error.)
  • 3. clt 3 STANDARD ERROR Standard Deviation of the Sampling Distribution of Means sx = s/ /n
  • 4. clt 4 How Large is Large?  If the sample is normal, then the sampling distribution of will also be normal, no matter what the sample size.  When the sample population is approximately symmetric, the distribution becomes approximately normal for relatively small values of n.  When the sample population is skewed, the sample size must be at least 30 before the sampling distribution of becomes approximately normal. x x
  • 5. clt 5 EXAMPLE A certain brand of tires has a mean life of 25,000 miles with a standard deviation of 1600 miles. What is the probability that the mean life of 64 tires is less than 24,600 miles?
  • 6. clt 6 Example continued The sampling distribution of the means has a mean of 25,000 miles (the population mean) m = 25000 mi. and a standard deviation (i.e.. standard error) of: 1600/8 = 200
  • 7. clt 7 Example continued Convert 24,600 mi. to a z-score and use the normal table to determine the required probability. z = (24600-25000)/200 = -2 P(z< -2) = 0.0228 or 2.28% of the sample means will be less than 24,600 mi.
  • 8. clt 8 ESTIMATION OF POPULATION VALUES  Point Estimates  Interval Estimates
  • 9. clt 9 CONFIDENCE INTERVAL ESTIMATES for LARGE SAMPLES  The sample has been randomly selected  The population standard deviation is known or the sample size is at least 25.
  • 10. clt 10 Confidence Interval Estimate of the Population Mean - X: sample mean s: sample standard deviation n: sample size X z s n X z s n       m
  • 11. clt 11 EXAMPLE Estimate, with 95% confidence, the lifetime of nine volt batteries using a randomly selected sample where: -- X = 49 hours s = 4 hours n = 36
  • 12. clt 12 EXAMPLE continued Lower Limit: 49 - (1.96)(4/6) 49 - (1.3) = 47.7 hrs Upper Limit: 49 + (1.96)(4/6) 49 + (1.3) = 50.3 hrs We are 95% confident that the mean lifetime of the population of batteries is between 47.7 and 50.3 hours.
  • 13. clt 13 CONFIDENCE BOUNDS  Provides a upper or lower bound for the population mean.  To find a 90% confidence bound, use the z value for a 80% CI estimate.
  • 14. clt 14 Example  The specifications for a certain kind of ribbon call for a mean breaking strength of 180 lbs. If five pieces of the ribbon have a mean breaking strength of 169.5 lbs with a standard deviation of 5.7 lbs, test to see if the ribbon meets specifications.  Find a 95% confidence interval estimate for the mean breaking strength.