chapter seven introduction to sampling distributions section 2 the central limit theorem
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Chapter Seven Introduction to Sampling Distributions Section 2 The Central Limit Theorem. Key Points 7.2. For a normal distribution, use mu and sigma to construct the theoretical sampling distribution for the statistic x – bar - PowerPoint PPT PresentationTRANSCRIPT
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Chapter Seven
Introduction to Sampling Distributions
Section 2
The Central Limit Theorem
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Key Points 7.2
• For a normal distribution, use mu and sigma to construct the theoretical sampling distribution for the statistic x – bar
• For large samples, use sample estimates to construct a good approximate sampling distribution for the statistic x-bar
• Learn the statement and underlying mean of the central limit theorem well enough to explain it to a friend who is intelligent, but (unfortunately) does not know much about statistics
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Let x be a random variable with a normal distribution with mean and standard deviation . Let be the sample mean
corresponding to random samples of size n taken from
the distribution.
x
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Facts about sampling distribution of the mean:
• The distribution is a normal distribution.
• The mean of the distribution is (the same mean as the original distribution).
• The standard deviation of the distribution is (the standard deviation of the original distribution, divided by the square root of the sample size).
xx
xn
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We can use this theorem to draw conclusions about means of samples taken from normal
distributions.
If the original distribution is normal, then the sampling distribution will be normal.
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The Mean of the Sampling Distribution
x
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The mean of the sampling distribution is equal to the
mean of the original distribution.
x
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The Standard Deviation of the Sampling Distribution
x
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The standard deviation of the sampling distribution is equal to the
standard deviation of the original distribution divided by the square
root of the sample size.
nx
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The time it takes to drive between cities A and B is normally distributed
with a mean of 14 minutes and a standard deviation of 2.2 minutes.
• Find the probability that a trip between the cities takes more than 15 minutes.
• Find the probability that mean time of nine trips between the cities is more than 15 minutes.
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Mean = 14 minutes, standard deviation = 2.2 minutes
• Find the probability that a trip between the cities takes more than 15 minutes.
3264.06736.000.1)45.0(
45.02.2
1415
zP
z14 15
Find this area
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Mean = 14 minutes, standard deviation = 2.2 minutes
• Find the probability that mean time of nine trips between the cities is more than 15 minutes.
73.09
2.2
n
14
x
x
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Mean = 14 minutes, standard deviation = 2.2 minutes
• Find the probability that mean time of nine trips between the cities is more than 15 minutes.
0853.04147.05.0)37.1z(P
37.173.0
1415z
14 15
Find this area
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What if the Original Distribution Is Not Normal?
Use the Central Limit Theorem!
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MOVIE!
V01081a.rm
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Central Limit Theorem
If x has any distribution with mean and standard deviation , then the sample
mean based on a random sample of size n will have a distribution that
approaches the normal distribution (with mean and standard deviation
divided by the square root of n) as n increases without bound.
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How large should the sample size be to permit the
application of the Central Limit Theorem?
In most cases a sample size of
n = 30 or more assures that the distribution will be approximately normal and the theorem will apply.
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Central Limit Theorem
• For most x distributions, if we use a sample size of 30 or larger, the distribution will be approximately normal.
x
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Central Limit Theorem
• The mean of the sampling distribution is the same as the mean of the original distribution.
• The standard deviation of the sampling distribution is equal to the standard deviation of the original distribution divided by the square root of the sample size.
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Central Limit Theorem Formula
x
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Central Limit Theorem Formula
nx
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Central Limit Theorem Formula
n
x
xz
x
x
/
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Application of the Central Limit Theorem
Records indicate that the packages shipped by a certain trucking company have a mean weight of 510 pounds and a standard deviation of 90 pounds. One hundred packages are being shipped today. What is the probability that their mean weight will be:
a. more than 530 pounds?b. less than 500 pounds?c. between 495 and 515 pounds?
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Are we authorized to use the Normal Distribution?
Yes, we are attempting to draw conclusions about means of large samples.
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Applying the Central Limit Theorem
What is the probability that their mean weight will be more than 530 pounds?Consider the distribution of sample means:
P( x > 530): z = 530 – 510 = 20 = 2.22 9 9
P(z > 2.22) = _______
9100/90,510 xx
.0132
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Applying the Central Limit Theorem
What is the probability that their mean weight will be less than 500 pounds?
P( x < 500): z = 500 – 510 = –10 = – 1.11 9 9
P(z < – 1.11) = _______.1335
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Applying the Central Limit Theorem
What is the probability that their mean weight will be between 495 and 515 pounds?
P(495 < x < 515) :
for 495: z = 495 – 510 = 15 = 1.67 9 9
for 515: z = 515 – 510 = 5 = 0.56 9 9
P( 1.67 < z < 0.56) = _______ .6648
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