Download Lecture 30 - Sampling Distribution Mean 2

Sampling Distribution
of a Sample Mean
Lecture 30
Section 8.4
Tue, Mar 15, 2005
The Central Limit Theorem


Begin with a population that has mean  and
standard deviation .
For sample size n, the sampling distribution of
the sample mean is approximately normal with
Mean  
Variance 
2
n
Standard deviation 

n
The Central Limit Theorem
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The approximation gets better and better as the
sample size gets larger and larger.
For many populations, the distribution is almost
exactly normal when n  10.
For almost all populations, if n  30, then the
distribution is almost exactly normal.
The Central Limit Theorem
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
Special Case: If the original population is exactly
normal, then the sampling distribution of the
sample mean is exactly normal for any sample
size.
This is all summarized on page 500.
Example

The population {1, 2, 3} has
Mean 2.
 Variance 2/3.
 Standard deviation (2/3) = 0.8165.

Example

When n = 3, the sample mean is (very)
approximately normal with
Mean 2.
 Standard deviation 0.8165/3 = 0.4714.

Example

When n = 30, the sample mean is approximately
(almost exactly) normal with
Mean 2.
 Standard deviation 0.8165/30 = 0.1491.

Example

If I collect, with replacement, a sample of 30
values from this population, what is the
probability that my sample mean will be at least
2.2?
Let’s Do It!

Let’s do it! 8.9, p. 502 – Probability of Accepting
the Shipment.
Find P(X > 250).
 Note that P(X < 250) is.

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Let’s do it! 8.11, p. 504 – Testing Hypotheses
about the Mean Weight of Nuts.


Find P(X < 15.8), i.e., the p-value.
Let’s do it! 8.10, p. 503 –Mean Grocery
Expenditure.
Estimating the Population Mean


Example 8.12, p. 504 – Estimating the
Population Mean Grocery Expenditure.
The sampling distribution ofx is approximately
normal with
x = $60.
 x = $35/100 = $3.50.


Based on the Empirical Rule, 95% of all samples
have a mean within $7.00 of $60, that is,
between $53 and $67.
Estimating a Population Proportion

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See the article “Water on airlines often
unacceptable, finds EPA.”
They found that the water in 20 out of 158
airliners contained coliform.
So p^ = 20/158 = 0.1266 = 12.66%.
What is a good estimate of p, the planes whose
water contains coliform as a proportion of all
planes?
Estimating a Population Proportion



Based on our theory, there is a 95% chance that
p^ is within 2 standard deviations of p.
Therefore, there is a 95% chance that p is within
2 standard deviations of p^.
That is, there’s a 95% chance that p is between p^
– 2p^ and p^ + 2p^.
Estimating a Population Proportion
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Compute p^ = 0.0265 = 2.65%.
Therefore, we are 95% sure that the true
proportion is between 7.36% and 17.96%.
This is called a 95% confidence interval.
How clean are municipal water systems?
Based on the data, is it reasonable to believe that
the water on airliners is in fact cleaner than the
water in municipal water systems?