1. Sampling Distributions

2. Parameter & Statistic Parameter Sample Statistic
Summary measure about population
Sample Statistic
Summary measure about sample
P in Population & Parameter
S in Sample & Statistic

3. Common Statistics & Parameters
Sample Statistic
Population Parameter
Mean  X 
Standard Deviation S 
Variance S2 2
Binomial Proportion p ^

4. Sampling Distribution
Theoretical probability distribution
Random variable is sample statistic
Sample mean, sample proportion, etc.
Results from drawing all possible samples of a fixed size
List of all possible [x, p(x)] pairs
Sampling distribution of the sample mean

5. Sampling from Normal Populations

6. 定理1
平均數
變異數

7. 定理2

8. 定理：Let Y1,Y2,…,Yn be a random sample of size n from a normal distribution with mean μand varianceσ2. Then
is normally distribution with mean
And variance

9. Proof:

12. Properties of the Sampling Distribution of x

13. Standard Error of the Mean
1. Standard deviation of all possible sample means, x ● Measures scatter in all sample means, x
Less than population standard deviation
3. Formula (sampling with replacement)

14. Sampling from Normal Populations
Central Tendency
Dispersion
Sampling with replacement
Population Distribution m
= 50 s
= 10 X
Sampling Distribution
n =16 X = 2.5
n = 4 X = 5 m X
= 50 -

15. Standardizing the Sampling Distribution of x
Standardized Normal Distribution m
= 0 s
= 1 Z

16. Thinking Challenge
You’re an operations analyst for AT&T. Long-distance telephone calls are normally distribution with  = 8 min. and  = 2 min. If you select random samples of 25 calls, what percentage of the sample means would be between 7.8 & 8.2 minutes?
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17. Sampling Distribution Solution*8 s ` X
= .4
7.8
8.2 s
= 1
–.50 Z
.50
.3830
Standardized Normal Distribution
.1915

18. Sampling from Non-Normal Populations

19. Developing Sampling Distributions
Suppose There’s a Population ...
Population size, N = 4
Random variable, x
Values of x: 1, 2, 3, 4
Uniform distribution
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20. Population Characteristics
Summary Measures
Population Distribution
P(x) .3 .2
Have students verify these numbers. .1 .0 x 1 2 3 4

21. All Possible Samples of Size n = 2
2nd Observation 1 2 3 4
1st
Obs
16 Sample Means
2nd Observation 1 2 3 4
1st
Obs
1,1
1,2
1,3
1,4
1.0
1.5
2.0
2.5
2,1
2,2
2,3
2,4
1.5
2.0
2.5
3.0
3,1
3,2
3,3
3,4
2.0
2.5
3.0
3.5
4,1
4,2
4,3
4,4
2.5
3.0
3.5
4.0
Sample with replacement

22. Sampling Distribution of All Sample Means
2nd Observation 1 2 3 4
1st
Obs
16 Sample Means
Sampling Distribution of the Sample Mean .0 .1 .2 .3
1.0
1.5
2.0
2.5
3.0
3.5
4.0
P(x) x
1.0
1.5
2.0
2.5
1.5
2.0
2.5
3.0
2.0
2.5
3.0
3.5
2.5
3.0
3.5
4.0

23. Summary Measures of All Sample Means
Have students verify these numbers.

24. Sampling Distribution
Comparison
Population
Sampling Distribution
P(x) .0 .1 .2 .3 1 2 3 4 .0 .1 .2 .3
1.0
1.5
2.0
2.5
3.0
3.5
4.0
P(x) x x

25. Sampling Distribution of the Mean…
A fair die is thrown infinitely many times, with the random variable X = # of spots on any throw. The probability distribution of X is: …and the mean and variance are calculated as well: x 1 2 3 4 5 6
P(x)
1/6
9.25

26. Sampling Distribution of Two Dice
A sampling distribution is created by looking at all samples of size n=2 (i.e. two dice) and their means… While there are 36 possible samples of size 2, there are only 11 values for , and some (e.g. =3.5) occur more frequently than others (e.g. =1).
9.26

27. Sampling Distribution of Two Dice…
6/36
5/36
4/36
3/36
2/36
1/36
P( )
9.27
9.27

28. Compare… Compare the distribution of X…
…with the sampling distribution of .
As well, note that:
9.28

29. Sampling from Non-Normal Populations

30. Law of Large Numbers
The law of large numbers states that, under general conditions, will be near with very high probability when n is large.
The conditions for the law of large numbers are
Yi , i=1, …, n, are i.i.d.
The variance of Yi , , is finite.

32. Central Limit Theorem…
The sampling distribution of the mean of a random sample drawn from any population is approximately normal for a sufficiently large sample size.
The larger the sample size, the more closely the sampling distribution of X will resemble a normal distribution.
9.32

33. Central Limit Theorem…
If the population is normal, then X is normally distributed for all values of n.
If the population is non-normal, then X is approximately normal only for larger values of n.
In most practical situations, a sample size of 30 may be sufficiently large to allow us to use the normal distribution as an approximation for the sampling distribution of X.
9.33

36. Sampling from Non-Normal Populations
Central Tendency
Dispersion
Sampling with replacement
Population Distribution s
= 10 m
= 50 X
Sampling Distribution
n = 4 X = 5
n =30 X = 1.8 m -
= 50 X X

37. X Central Limit Theorem As sample size gets large enough (n  30) ...
sampling distribution becomes almost normal. X

38. Central Limit Theorem Example
The amount of soda in cans of a particular brand has a mean of 12 oz and a standard deviation of .2 oz. If you select random samples of 50 cans, what percentage of the sample means would be less than oz?
SODA

39. Central Limit Theorem Solution*
Shaded area exaggerated
Sampling Distribution 12 s ` X
= .03
11.95 s
= 1
–1.77 Z
.0384
Standardized Normal Distribution
.4616

40. Example
One survey interviewed 25 people who graduated one year ago and determines their weekly salary. The sample mean to be $750. To interpret the finding one needs to calculate the probability that a sample of 25 graduates would have a mean of $750 or less when the population mean is $800 and the standard deviation is $100. After calculating the probability, he needs to draw some conclusions.
9.40

41. Example
We want to find the probability that the sample mean is less than $750. Thus, we seek
The distribution of X, the weekly income, is likely to be positively skewed, but not sufficiently so to make the distribution of nonnormal. As a result, we may assume that is normal with mean
and standard deviation
9.41

42. Example
Thus,
The probability of observing a sample mean as low as $750 when the population mean is $800 is extremely small. Because this event is quite unlikely, we would have to conclude that the dean's claim is not justified.
9.42

43. Using the Sampling Distribution for Inference
Here’s another way of expressing the probability calculated from a sampling distribution.
P(-1.96 < Z < 1.96) = .95
Substituting the formula for the sampling distribution
With a little algebra
9.43

44. Using the Sampling Distribution for Inference
Returning to the chapter-opening example where µ = 800, σ = 100, and n = 25, we compute or
This tells us that there is a 95% probability that a sample mean will fall between and Because the sample mean was computed to be $750, we would have to conclude that the dean's claim is not supported by the statistic.
9.44

45. Using the Sampling Distribution for Inference
For example, with µ = 800, σ = 100, n = 25 and α= .01, we produce
9.45

46. Sampling Distributions The Proportion
The proportion of the population having some characteristic is denoted π.

47. Sampling Distributions The Proportion
Standard error for the proportion:
Z value for the proportion:

48. Sampling Distributions The Proportion: Example
If the true proportion of voters who support Proposition A is π = .4, what is the probability that a sample of size 200 yields a sample proportion between .40 and .45?
In other words, if π = .4 and n = 200, what is
P(.40 ≤ p ≤ .45) ?

49. Sampling Distributions The Proportion: Example
Find :
Convert to standardized normal:

50. Sampling Distributions The Proportion: Example
Use cumulative normal table:
P(0 ≤ Z ≤ 1.44) = P(Z ≤ 1.44) – 0.5 = .4251
Standardized Normal Distribution
Sampling Distribution
.4251
Standardize
.40
.45
1.44 p Z