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CH 9

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2 views4 pages

CH 9

Uploaded by

hanandeh0791
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Chapter 9 – Sampling Distribution

 Parameters are numerical descriptive measures for populations.


 For normal distribution, the location and shape are described by µ and σ.
 Often the values of parameters are unknown. Hence, you must rely on the sample to learn
about these parameters.
 If you want the sample to provide reliable information about the population, you must
select your sample randomly.
 The numerical descriptive measures you calculate from the sample are called statistics.
These statistics vary or change for each different random sample you select; that is, they
are random variables.

Definition: The Sampling distribution of a statistic is the probability distribution for the
possible values of the statistic that result when random sample of size 𝑛 are repeatedly drawn
from the population.
i.e The probability distributions for the statistic are called Sampling Distributions. They tell
us what values of the statistics can occur and how often each value occurs.
Example: A population consists of 𝑁 = 5 numbers: 3, 6, 9, 12, 15. Assume that the
sampling distribution for 𝑥̅ is:
𝑥̅ 6 7 8 9 10 11 12
𝑃(𝑥̅ ) 0.1 0.1 0.2 0.2 0.2 0.1 0.1

1. Find 𝐸(𝑥̅ )

𝐸(𝑥̅ ) = ∑ 𝑥̅ ∗ 𝑃(𝑥̅ )
= 0.6 + 0.7 + 1.6 + 1.8 + 2 + 1.1 + 1.2 = 9
2. Find the population mean
µ= =9
 Note 𝐸(𝑥̅ ) = µ
If the population mean µ is unknown, you might choose several statistics as an estimator; the
sample mean 𝑥̅ is more widely used.
The Central Limit Theorem (CLT): If random samples of 𝑛 observations are drawn from a
nonnormal population with mean µ and standard deviation σ, then, when 𝑛 is large (𝑛 ≥
30), the sampling distribution of the sample mean 𝑥̅ is approximately normally distributed
with mean µ and standard deviation .

Note: The approximation becomes more accurate as 𝑛 becomes large.


Theorem: If random samples of 𝑛 observations are drawn from a normal population with
mean µ and standard deviation σ, the sampling distribution of the sample mean 𝑥̅ is normally
distributed with mean µ and standard deviation .

i.e The sampling distribution of the sample mean 𝒙


if the sample of 𝑛 measurements is selected from:
A population with mean µ and standard 𝑥̅ will have mean µ and standard deviation
deviation σ

population has a normal distribution 𝑥̅ exactly 𝑁(µ, )
regardless of the sample size 𝑛
population has a nonnormal distribution 𝑥̅ approximately 𝑁(µ, )
for large samples ( 𝑛 ≥ 30)

Definition: The Standard deviation of a statistic used as an estimator of a population


parameter is also called the Standard Error of the estimator ( SE ).
µ
Note: 𝑥: 𝑁(µ, 𝜎 ) → 𝑍 = ∶ 𝑁 (0, 1)

µ
𝑥̅ : 𝑁(µ, )→𝑍= ⁄√
∶ 𝑁 (0, 1)
Example: if we take a sample of size 36 from a population with mean 2 and variance 25.
Find:
1. Sampling distribution of 𝑥̅
2. 𝐸(𝑥̅ )
3. 𝑣𝑎𝑟(𝑥̅ )
4. 𝑆𝐸(𝑥̅ )
( ̅ )
5. The distribution of
6. 𝑃(𝑥̅ ≤ 3)
7. 𝑃(0 ≤ 𝑥̅ ≤ 2)

Solutions:

1. 𝑛 = 36 ≥ 30. Hence, by CLT 𝑥̅ 𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑒𝑙𝑙𝑦 𝑁(µ, ).

For this example 𝑥̅ : 𝑁(2, )

2. 𝐸(𝑥̅ ) = µ = 2
3. 𝑣𝑎𝑟(𝑥̅ ) =
4. 𝑆𝐸(𝑥̅ ) =
( ̅ ) ( ̅ )
5. Sampling distribution of ⁄
= is approximately 𝑁(0,1 )
̅ µ µ
6. 𝑃(𝑥̅ ≤ 3) = 𝑃( ⁄√
< ⁄√
)
= 𝑃( 𝑍 < ⁄
) = 𝑃(𝑍 < 1.2) = 0.8849
7. 𝑃(0 ≤ 𝑥̅ ≤ 2) = 𝑃 ( ⁄
<𝑍 < ⁄
)
= 𝑃(−2.4 < 𝑍 < 0 )
= 𝑃 (𝑍 < 0) − 𝑃(𝑍 < −2.4)

= 0.5 − 0.0082 = 0.4918


Example: suppose that 𝑥: 𝑁(20, 36). Take a random sample of size 25. Find:
1. The distribution of 𝑥̅
2. 𝑃(𝑥 ≤ 25)
3. 𝑃(𝑥̅ > 24)

Solutions:

1. 𝑥̅ : 𝑁(20, )
µ
2. 𝑃(𝑥 ≤ 25) = 𝑃( ≤ )
= 𝑃(𝑍 ≤ 0.83) = 0.7967
̅
3. 𝑃(𝑥̅ > 24) = 𝑃 ⁄√
> ⁄
= 𝑃(𝑍 > 3.33)
= 𝑃(𝑍 < −3.33) = 0.0004

Example: the distribution of the grades of Stat 107 is normal with mean 70 and variance
16. If we take a random sample of 36 students. Find:
1. The probability that the mean of grades of 36 students will be greater than 72.
2. The probability that the total of grades of 36 students will be less than 2556.

Solutions:

1. 𝑃(𝑥̅ > 72) = 𝑃 𝑍 > ⁄


= 𝑃(𝑍 > 3)
= 𝑃(𝑍 < −3) = 0.0013
∑ ∑
2. 𝑃(∑ 𝑥 < 2556) = 𝑃 < = 𝑃 <
= 𝑃(𝑥̅ < 71) = 𝑃(𝑍 < 1.5) = 0.9332

H.W: suppose a random sample of size 25 observations is selected from a population


that is normally distributed with mean 106 and standard deviation 12 find:
1. The mean and the standard deviation of the sampling distribution of the sample
mean 𝑥̅ .
2. The probability that 𝑥̅ exceed 110.
3. The probability that the sample mean deviates from the population mean by no more
than.

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