2007 會計資訊系統計學 ( 一 ) 上課投影片 9-1 Sampling Distributions Chapter 9

2007會計資訊系統計學 ( 一 )上課投影片

9-1

Sampling DistributionsSampling Distributions

Chapter 9


9-2

Introduction

• In real life calculating parameters of populations is prohibitive because populations are very large.

• Rather than investigating the whole population, we take a sample, calculate a statistic（統計量） related to the parameter（參數） of interest, and make an inference（推論） .

• The sampling distribution of the statistic（統計量的抽樣分配） is the tool that tells us how close is the statistic to the parameter.


9-3

Sampling Distributions（抽樣分配）• A sampling distribution is created by, as the name

suggests, sampling.

• The method we will employ on the rules of probability and the laws of expected value and variance to derive the sampling distribution.

• For example, consider the roll of one and two dice.


9-4

9.1 Sampling Distribution of the Mean

• An example– A fair die is thrown infinitely many times.– Let X represent the number of spots showing on any throw.– The probability distribution of X is

92.2)6

1()5.36()

6

1()5.32()

6

1()5.31()X(V

5.3)6

1(6)

6

1(2)

6

1(1)X(E

222

x 1 2 3 4 5 6

P(x)

1/6 1/6 1/6 1/6 1/6 1/6


9-5

• Suppose we want to estimate from the mean of a sample of size n = 2.

• What is the distribution of ?

X

Throwing a die twice – sample mean

X


9-6

Sample Mean Sample Mean Sample Mean1 1,1 1 13 3,1 2 25 5,1 32 1,2 1.5 14 3,2 2.5 26 5,2 3.53 1,3 2 15 3,3 3 27 5,3 44 1,4 2.5 16 3,4 3.5 28 5,4 4.55 1,5 3 17 3,5 4 29 5,5 56 1,6 3.5 18 3,6 4.5 30 5,6 5.57 2,1 1.5 19 4,1 2.5 31 6,1 3.58 2,2 2 20 4,2 3 32 6,2 49 2,3 2.5 21 4,3 3.5 33 6,3 4.5

10 2,4 3 22 4,4 4 34 6,4 511 2,5 3.5 23 4,5 4.5 35 6,5 5.512 2,6 4 24 4,6 5 36 6,6 6


10 2,4 3 22 4,4 4 34 6,4 511 2,5 3.5 23 4,5 4.5 35 6,5 5.512 2,6 4 24 4,6 5 36 6,6 6


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).

XX

X


9-7

Throwing a die twice – sample meanSample Mean Sample Mean Sample Mean1 1,1 1 13 3,1 2 25 5,1 32 1,2 1.5 14 3,2 2.5 26 5,2 3.53 1,3 2 15 3,3 3 27 5,3 44 1,4 2.5 16 3,4 3.5 28 5,4 4.55 1,5 3 17 3,5 4 29 5,5 56 1,6 3.5 18 3,6 4.5 30 5,6 5.57 2,1 1.5 19 4,1 2.5 31 6,1 3.58 2,2 2 20 4,2 3 32 6,2 49 2,3 2.5 21 4,3 3.5 33 6,3 4.5

10 2,4 3 22 4,4 4 34 6,4 511 2,5 3.5 23 4,5 4.5 35 6,5 5.512 2,6 4 24 4,6 5 36 6,6 6


10 2,4 3 22 4,4 4 34 6,4 511 2,5 3.5 23 4,5 4.5 35 6,5 5.512 2,6 4 24 4,6 5 36 6,6 6

1 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0

6/365/36

4/36

3/36

2/36

1/36x

1/366.0

2/365.5

3/365.0

4/364.5

5/364.0

6/363.5

5/363.0

4/362.5

3/362.0

2/361.5

1/361.0

X )(XP


9-8


• The sampling distribution of is shown below:

2:

22 XXXX andNote

2

:2

2 XXXX andNote

1/366.0

2/365.5

3/365.0

4/364.5

5/364.0

6/363.5

5/363.0

4/362.5

3/362.0

2/361.5

1/361.0

X )(XP

X

1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0

6/36

5/36

4/36

3/36

2/36

1/36

)X(P

X

46.1)36

1()5.30.6()

36

2()5.35.1()

36

1()5.30.1()X(V

5.3)36

1(0.6)

36

2(5.1)

36

1(0.1)X(E

222


9-9

Compare

• Compare the distribution of X, with the sampling distribution of .

• As well, note that:

X

1 2 3 4 5 6 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0

2

22 XX

XX


9-10

Generalize

• We can generalize the mean and variance of the sampling of two dice:

• ……to n-dice:

• The standard deviation of the sampling distribution is called the standard error:

2

22 XX

XX

nX

X

XX

22

nX


9-11

)5

(5833.

5.3

5n

2x2

x

x

)10

(2917.

5.3

10n

2x2

x

x

)25

(1167.

5.3

25n

2x2

x

x

Sampling Distribution of the Mean


9-12


)5

(5833.

5.3

5n

2x2

x

x

)10

(2917.

5.3

10n

2x2

x

x

)25

(1167.

5.3

25n

2x2

x

x

Notice that is smaller than . The larger the sample size the smaller . Therefore, tends to fall closer to , as the sample size increases.

2X

x

2X

2X


9-13


Demonstration: The variance of the sample mean is smaller than the variance of the population.

1 2 3

Mean = 1.5 Mean = 2.5Mean = 2.

Population 1.51.51.51.51.51.51.51.51.51.51.51.51.5

2.52.52.52.52.52.52.52.52.52.52.52.52.5

22222222222

Compare the variability of the populationto the variability of the sample mean.

Let us take samplesof two observations


9-14

Also,Expected value of the population =

(1 + 2 + 3)/3 = 2

Expected value of the sample mean = (1.5 + 2 + 2.5)/3 = 2



9-15

The Central Limit Theorem

• If a random sample is drawn from any population, the sampling distribution of the sample mean is approximately normal（近似常態分配） for a sufficiently large sample size.

• The larger the sample size, the more closely the sampling distribution of will resemble a normal distribution.

X


9-16

The Central Limit Theorem

• If the population is normal, then is normally distributed for all values of n.

• If the population is non-normal, then is approximately normal only for larger values of n.

• In many 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

X

X


9-17

Sampling Distribution of the Sample Mean

• 1.

• 2.

• 3. If X is normal, is normal. If X is nonnormal, is approximately normal for sufficiently large sample sizes.

• Note: the definition of “sufficiently large” depends on the extent of nonnormality of X (e.g. heavily skewed; multimodal)

X

X X

n and

n X

22X


9-18

Finite Populations• Statisticians have shown that the mean of the sampling

distribution is always equal to the mean of the population and that the standard error is equal to for infinitely large populations. However, if the population is finite the standard error is

where N is the population size and is called the finite population correction factor. （有限母體校正因子）

• If the population size is large relative to the sample size the finite population correction factor is close to 1 and can be ignored.

n/

1N

nN

nX

1N

nN


9-19

Finite Populations

• As a rule of thumb we will treat any population that is at least 20 times larger than the sample size as large. In practice most applications involve populations that qualify as large. As a consequence the finite population correction factor is usually omitted.

• There are several applications that deal with small populations. Section 12.5 introduces one of these applications.


9-20

Example 9.1(a)

• The foreman of a bottling plant has observed that the amount of soda in each “32-ounce” bottle is actually a normally distributed random variable,

• with a mean of 32.2 ounces , • and a standard deviation of .3 ounce.

• If a customer buys one bottle, what is the probability that the bottle will contain more than 32 ounces?


9-21

• The random variable X is the amount of soda in a bottle.• where X is normally distributed and =32.2 and =.3

= 32.2

0.7486

x = 327486.0)67.Z(P

)3.

2.3232X(P

)32X(P

X

Example 9.1(a)-Solution

“There is about a 75% chance that a single bottle of soda contains more than 32oz.”


9-22

Example 9.1(b)

• The foreman of a bottling plant has observed that the amount of soda in each “32-ounce” bottle is actually a normally distributed random variable,

• with a mean of 32.2 ounces,• and a standard deviation of .3 ounce.

• If a customer buys a carton of four bottles, what is the probability that the mean amount of the four bottles will be greater than 32 ounces?


9-23

= 32.2

0.7486

x = 32

• Define the random variable as the mean amount of four bottles of soda. Then:

• X is normally distributed, therefore so will .• •

9082.0)33.1Z(P

)43.

2.3232X(P)32X(P

X

32x

0.9082

2.32x

Example 9.1(b)-Solution

“There is about a 91% chance the mean of the four bottles will exceed 32oz.”

oz2.32XX

15.43.nXX

X

X


9-24

Graphically Speaking

what is the probability that one bottle will contain more than 32

ounces?

what is the probability that the mean of four bottles will exceed 32

oz?

mean=32.2


9-25

Chapter-Opening Example

• The dean of the School of Business claims that the average salary of the school’s graduates one year after graduation is $800 per week with a standard deviation of $100.

• A second-year student would like to check whether the claim about the mean is correct. He does a survey of 25 people who graduated one year ago and determines their weekly salary. He discovers the sample mean to be $750.

• To interpret his finding he 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.


9-26

Chapter-Opening Example-Solution

• Define the random variable as the mean weekly salary of 25 graduates.

• We want to compute

• Although X is likely skewed it is likely that is normally distributed. The mean of is

• The standard deviation is

)750X(P

XX

800X

2025/100n/X

X


9-27

Chapter-Opening Example-Solution

• The probability of observing a sample mean as low as $750 when the population mean is $800 is extremely small (0.0062). Because the event is quite unlikely, the claim that the mean weekly income is $800 is probably unjustified.

• With = 800 the probability of observing a sample mean as low as 750 is very small.

• It will be more reasonable to assume that is smaller than $800, because then a sample mean of $750 becomes more probable.

0062.4938.5.)5.2Z0(P5.)5.2Z(P

20

800750XP)750X(P

X

X


9-28

Using Sampling Distributions for Inference

• The sampling distribution can be used to make inferences about population parameters.

• In order to do so, the sample mean can be standardized to the standard normal distribution using the following formulation:

n

XZ


9-29

Another Way to State the Probability

• In Chapter 8 we saw that

• From the sampling distribution of the mean we have

• Substituting this definition of Z in the probability statement we produce

n/

XZ

95.)96.1n/

X96.1(P

95.)96.1Z96.1(P


9-30


• With a little algebra we rewrite the probability statement as

95.)n

96.1Xn

96.1(P

95.)n

96.1Xn

96.1(P

95.)96.1n/

X96.1(P


9-31


• Similarly

• In general

• All are probability statements about , which we’ll use in statistical inference.

90.n

645.1Xn

645.1P

1n

ZXn

ZP 2/2/

95.n

96.1Xn

96.1P

X


9-32

Return to the Chapter-Opening Example

• Substituting = 800, = 100, n = 25, and = .05, we get

• Conclusion– There is 95% chance that the sample mean falls within the interval

[760.8, 839.2] if the population mean is 800.– It is unlikely that we would observe a sample mean as low as $750

when the population mean is $800. – Since the sample mean was 750, the population mean is probably not

800.

95.2.839X8.760P

95.25

10096.1800X

25

10096.1800P

05.1n

ZXn

ZP 025.025.


9-33

Return to the Chapter-Opening Example

-1.96 -1.960n

96.1

n

96.1

.025 .025.025 .025

Standard normal distribution Z Normal distribution of x

95.)25

10096.1800x

25

10096.1800(P

.95.95

xZ

25

10096.1800(P

25

10096.1800(P


9-34

9.2 Sampling Distribution of a Proportion

• The parameter of interest for nominal data is the proportion of times a particular outcome (success) occurs.

• The estimator of a population proportion of successes “p” is the sample proportion “ ”. That is, we count the number of successes in a sample and compute:

• (read this as “p-hat”).• X is the number of successes, n is the sample size. X is binomial random variable.

p̂

n

Xpp of estimate The ˆ


9-35

9.2 Sampling Distribution of a Proportion

• Since X is binomial, probabilities about can be calculated from the binomial distribution.

• Yet, for inference about we prefer to use normal approximation to the binomial.

p̂

p̂


9-36

Normal Approximation to Binomial

• Approximate the binomial probability P(X=10) when n = 20 and p = .5 with a normal approximation superimposed ( ).?? and

176.

)5.1()5(.!10!10

!20

)10(

102010

)5.,20(

pnBinomialXP


9-37


• Binomial distribution with n=20 and p=.5 with a normal approximation superimposed ( )24.210 and

Where did these values come from?! From §7.6 we saw that:

24.25)1(

5)5.1(5.20)1(2

pnp

pnp

10.520pn


9-38


• To calculate P(X=10) using the normal distribution, we can find the area under the normal curve between 9.5 and 10.5.

where Y is a normal random variable approximating the binomial random variable X

)5.105.9()10( NormalBinomial YPXP

continuity correction factor


9-39

109.5 10.5

P(XBinomial = 10) =

~= P(9.5<Y<10.5)

= np = 20(.5) = 10; 2 = np(1 - p) = 20(.5)(1 - .5) = 5 = 51/2 = 2.24

1742.)24.2

105.10Z

24.2105.9

(P

.176

Let us build a normal distribution to approximate the binomial P(X = 10).

P(9.5<YNormal<10.5)The approximation

Normal approximation to the Binomial

The approximation is quite good!


9-40

More examples of normal approximation More examples of normal approximation to the binomialto the binomial

44.5

1413.5

P(XP(XBinomial Binomial 14)14)

P(YP(YNormalNormal< 4.5)< 4.5)

P(YP(YNormal Normal > 13.5)> 13.5)

Normal approximation to the BinomialNormal approximation to the Binomial

P(XP(XBinomialBinomial 4)4)


9-41


• Normal approximation to the binomial works best when– the number of experiments, n, (sample size) is

large,– and the probability of success, p, is close to 0.5

• For the approximation to provide good results two conditions should be met:(1) np ≥ 5(2) n(1 － p) ≥ 5


9-42

Sampling Distribution of a Sample Proportion

• Using the laws of expected value and variance, we can determine the mean, variance, and standard deviation of .

• (The standard deviation of is called the standard error of the proportion.)

p̂p̂

npp

n

pppV

ppE

p

p

)1(

)1()ˆ(

)ˆ(

ˆ

2ˆ


9-43

Sampling Distribution of a Sample Proportion

• If both np > 5 and np(1-p) > 5, then sample proportions can be standardized to a standard normal distribution using this formulation:

• Z is approximately standard normally distributed.

npp

ppZ

)1(

ˆ


9-44

• A state representative received 52% of the votes in the last election.

• One year later the representative wanted to study his popularity.

• If his popularity has not changed, what is the probability that more than half of a sample of 300 voters would vote for him?

Example 9.2


9-45

• The number of respondents who prefer the representative is binomial with n = 300 and p = .52. Thus,– np = 300(.52) = 156 and– n(1-p) = 300(1-.52) = 144 (both greater than 5)

7549.300)52.1)(52(.

52.50.

)1(

ˆ)50.ˆ(

npp

ppPpP

Example 9.2-Solution


9-46

9.3 Sampling Distribution of the Difference Between Two Means

• The final sampling distribution introduced is that of the difference between two sample means.

• Independent samples are drawn from each of two normal populations.

• We’re interested in the sampling distribution of the difference between the two sample means

21 XX


9-47

9.3 Sampling Distribution of the Difference Between Two Means

• The distribution of is normal if– The two samples are independent, and– The parent populations are normally distributed.

• If the two populations are not both normally distributed, but the sample sizes are 30 or more, the distribution of is approximately normal.

21 XX

21 XX


9-48

Sampling Distribution: Difference of two means

• The expected value and variance of the sampling distribution of are given by:

• mean:

• standard deviation:

• (also called the standard error if the difference between two means)

21 XX

2121 XX

2

22

1

21

21 nnXX


9-49

Sampling Distribution: Difference of two means

• Since the distribution of is normal and has a

• mean of

• and a standard deviation of

• We can compute Z (standard normal random variable) in this way:

21 XX

2121 XX

2

22

1

21

21 nnXX

2

22

1

21

2121 )()(

nn

XXZ


9-50

Example 9.3

• Starting salaries for MBA grads at two universities are normally distributed with the following means and standard deviations. Samples from each school are taken:

University WLU University UWO

Mean 62,000 $/yr 60,000 $/yr

Std. Dev. 14,500 $/yr 18,300 $/yr

sample size n 50 60

What is the probability that the sample mean starting salary ofUniversity WLU graduates will exceed that of the UWO grads?


9-51

Example 9.3-Solution

• “What is the probability that the sample mean starting salary of University WLU graduates will exceed that of the UWO grads?”

• We are interested in determining . Converting this to a difference of means, what is: ?

)( 21 XXP

)0( 21 XXP

7389.2389.5.

)64.()3128

20000(

)

60300,18

50500,14

)000,60000,62(0((

)0(

22

2

22

1

21

21

21

ZPZP

nn

) - XXP

XXP

21

•“There is about a 74% chance that the sample mean starting salary of U. WLU grads will exceed that of U. UWO”.

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2007 會計資訊系統計學 ( 一 ) 上課投影片 9-1 Sampling Distributions Chapter 9