the sampling distribution of the sample mean

The Sampling Distribution of the The Sampling Distribution of the Sample Mean Sample Mean

The Sampling Distribution of the The Sampling Distribution of the Sample Proportion Sample Proportion

Chapter 6 Sampling Distributions(样本分布 )

Chapter 6 Sampling Distributions

Sample Mean

Partner Hours

Dunn 22

Hardy 26

Kiers 30

Malory 26

Tillman 22

Example: The law firm of Hoya and Associates has five partners. At their weekly partners meeting each reported the number of hours they billed clients for their services last week.

If two partners are selected randomly, how many different samples are possible?

Partners Total Mean 1,2 48 24 1,3 52 26 1,4 48 24 1,5 44 22 2,3 56 28 2,4 52 26 2,5 48 24 3,4 56 28 3,5 52 26 4,5 48 24

A total of 10 different samples

23/4/24 Chapter 6 Sampling Distributions

Sample Mean Frequency Relative Frequency probability

22 1 1/10

24 4 4/10

26 3 3/10

28 2 2/10

As a sampling distribution

Different samples of the same size from the same population

will yield different sample means


Sample Mean Let there be a population of units of size Let there be a population of units of size NN Consider all its samples of a fixed size Consider all its samples of a fixed size n n ((n<Nn<N)) For all possible samples of size For all possible samples of size nn, we obtain a population , we obtain a population

of sample means. That is, of sample means. That is, is a random variable which is a random variable which may have all these means as its valuesmay have all these means as its values

Before we draw the sample, the sample mean Before we draw the sample, the sample mean is a random variable.

We consider the probability distribution of the random We consider the probability distribution of the random variable variable , i.e., the probability distribution for the , i.e., the probability distribution for the population of sample meanspopulation of sample means

x

x

x


Section 6.1 The Sampling Distribution of Section 6.1 The Sampling Distribution of the Sample Meanthe Sample Mean

The The sampling distribution of the samplesampling distribution of the sample meanmean is the probability distribution of the is the probability distribution of the population of the sample means obtainable population of the sample means obtainable from all possible samples of size from all possible samples of size n n from a from a population of size population of size N.N.


Example 6.1Example 6.1 The Stock Return Case We have a population of the percent returns from six stocks

In order, the values of % return are:10%, 20%, 30%, 40%, 50%, and 60%

Label each stock A, B, C, …, F in order of increasing % return

The mean rate of return is 35% with a standard deviation of 17.078%

Any one stock of these stocks is as likely to be picked as any other of the six Uniform distribution with N = 6 Each stock has a probability of being picked of 1/6


Stock

% Return

Frequency

Relative Frequency

Stock A 10 1 1/6 Stock B 20 1 1/6 Stock C 30 1 1/6 Stock D 40 1 1/6 Stock E 50 1 1/6 Stock F 60 1 1/6 Total 6 1

The Stock Return Case 2 ﹟


Now, select all possible samples of size Now, select all possible samples of size nn = = 2 from this population of stocks of size 2 from this population of stocks of size N N = = 66 That is, select all possible pairs of stocksThat is, select all possible pairs of stocks

How to select?How to select? Sample randomlySample randomly Sample without replacementSample without replacement Sample without regard to orderSample without regard to order



Result: There are 15 possible Result: There are 15 possible samples of size samples of size nn = 2 = 2

Calculate the sample mean of each and Calculate the sample mean of each and every sampleevery sample

For example, if we choose the two stocks For example, if we choose the two stocks with returns 10% and 20%, then the sample with returns 10% and 20%, then the sample mean is 15%mean is 15%

26C



SampleMean Frequency

RelativeFrequency

15 1 1/1520 1 1/1525 2 2/1530 2 2/1535 3 3/1540 2 2/1545 2 2/1550 1 1/1555 1 1/15



The population of N = 6 stock returns has a uniform distribution.

But the histogram of n = 15 sample mean returns:

1. Seems to be centered over the same mean return of 35%, and

2. Appears to be bell-shaped and less spread out than the histogram of individual returns

Observations


Example 6.2Example 6.2 Sampling all the stocks Consider the population of returns of all 1,815

stocks listed on NYSE for 1987 See Figure 6.2(a) on next slide The mean rate of return was –3.5% with a

standard deviation of 26% Draw all possible random samples of size n = 5

and calculate the sample mean return of each Sample with a computer See Figure 6.2(b) on next slide


Observations Both histograms appear to be bell-shaped and

centered over the same mean of –3.5% The histogram of the sample mean returns looks

less spread out than that of the individual returns Statistics

Mean of all sample means: = = -3.5% Standard deviation of all possible means:

%63.115

26

nx

x


If the population of individual items is normal, then the population of all sample means is also normal

Even if the population of individual items is not normal, there are circumstances that the population of all sample means is normal (see Central Limit Theorem(中心极限定理 ) later)

General Conclusions


The mean of all possible sample means equals the The mean of all possible sample means equals the population meanpopulation mean That is,That is, =

The standard deviation sx of all sample means is less than the standard deviation of the population That is, <

Each sample mean averages out the high and the low measurements, and so are closer to m than many of the individual population measurements

General Conclusions

x

x


The empirical rule holdsThe empirical rule holds for the sampling distribution of for the sampling distribution of the sample meanthe sample mean 68.26%68.26% of all possible sample means are within (plus or of all possible sample means are within (plus or

minus) one standard deviationminus) one standard deviation ofof 95.44% of all possible observed values of x are within

(plus or minus) two ofof In the example., 95.44% of all possible sample

mean returns are in the interval [-3.5 ± (211.63)] = [-3.5 ± 23.26]

That is, 95.44% of all possible sample means are between -26.76% and 19.76%

99.73% of all possible observed values of x are within (plus or minus) three of

x

x

x


Properties of the SamplingDistribution of the Sample Mean #1

• If the population being sampled is normal, If the population being sampled is normal, then so is the sampling distribution of the then so is the sampling distribution of the sample meansample mean,

• The mean of the sampling distribution of is

=

That is, the mean of all possible sample means is the same as the population mean

x

x

x

x


• The variance The variance 22 of the sampling distribution of the sampling distribution

of of is is

That is, the variance of the sampling That is, the variance of the sampling distribution of distribution of is is directly proportional to the variance of directly proportional to the variance of

the population, and the population, and inversely proportional to the sample inversely proportional to the sample

sizesize


nx

22

x

x

x



• The standard deviation of the sampling distribution of is

That is, the standard deviation of the sampling distribution of is directly proportional to the standard

deviation of the population, and inversely proportional to the square root

of the sample size

nx

xx

x


is the point estimate of , and the larger the sample size n, the more accurate the estimate, because when n increases, decreases, is more clustered to the is more clustered to the populationpopulation

In order to reduceIn order to reduce , take bigger samples!

Notesx

xx

x


Example 6.3Example 6.3 Car Mileage Case Population of all midsize cars of a particular make and

model Population is normal with mean and standard

deviation Draw all possible samples of size Draw all possible samples of size nn Then the sampling distribution of the sample mean is Then the sampling distribution of the sample mean is

normal with meannormal with mean == and standard deviation In particular, draw samples of size:

n = 5 n = 49

nx x


2.23615

x

749

x

So, all possible sample means for n=49 will be more closely clustered around than the case of n =5


Central Limit Theorem(中心极限定理 ) #1

If the population is non-normal, what is the shape of If the population is non-normal, what is the shape of the sampling distribution of the sample means?the sampling distribution of the sample means?

In fact the sampling distribution is approximately In fact the sampling distribution is approximately normal if the sample is large enough, even if the normal if the sample is large enough, even if the population is non-normalpopulation is non-normal

by the “Central Limit Theorem”by the “Central Limit Theorem”


No matter what is the probability distribution that describes the population, if the sample size n is large enough, then the population of all possible sample means is approximately normal with mean and standard deviation

Further, the larger the sample size n, the closer the sampling distribution of the sample means is to being normal In other words, the larger n, the better the

approximation

xnx


Random Sample (x1, x2, …, xn)

Population Distribution

(, )(right-skewed)

X

as n large

n, xx

Sampling Distribution of Sample Means

(nearly normal)

x


Example 6.4Example 6.4 Effect of the Sample SizeEffect of the Sample SizeThe larger the sample The larger the sample size, the more nearly size, the more nearly normally distributed is normally distributed is the population of all the population of all possible sample meanspossible sample means

Also, as the sample size Also, as the sample size increases, the spread of increases, the spread of the sampling distribution the sampling distribution decreasesdecreases


How Large?How Large? How large is “large enough?”How large is “large enough?” If the sample size n is at least 30, then for most sampled If the sample size n is at least 30, then for most sampled

populations, the sampling distribution of sample means is populations, the sampling distribution of sample means is approximately normalapproximately normal

Refer to Figure 6.6 on next slideRefer to Figure 6.6 on next slide Shown in Fig 6.6(a) is an exponential (right Shown in Fig 6.6(a) is an exponential (right

skewed) distributionskewed) distribution In Figure 6.6(b), 1,000 samples of size In Figure 6.6(b), 1,000 samples of size nn = 5 = 5

Slightly skewed to rightSlightly skewed to right In Figure 6.6(c), 1,000 samples with In Figure 6.6(c), 1,000 samples with nn = 30 = 30

Approximately bell-shaped and normalApproximately bell-shaped and normal If the population is normal, the sampling distribution of If the population is normal, the sampling distribution of

is normal regardless of the sample sizeis normal regardless of the sample sizex

x


Example: Central Limit Theorem Simulation


Recall from Chapter 2 mileage example, = 31.5531 mpg for a sample of size n=49 With With s = 0.7992s = 0.7992

Does this give statistical evidence that the population Does this give statistical evidence that the population meanmean is greater than 31 mpg?is greater than 31 mpg? That is, does the sample mean give evidence thatThat is, does the sample mean give evidence that is is

at least at least 31 mpg31 mpg?? Calculate the probability of observing a sample mean

that is greater than or equal to 31.5531 mpg if = = 31 31 mpgmpg WantWant P(> 31.5531 if = 31)

Reasoning from Sample Distribution

x

x


Use Use ss as the point estimate for as the point estimate for so that so that

0.114349

79920

.n

x

ThenThen

84.41143.0

315531.31

5531.3131 if 5531.31

zP

zP

zPxPx

x

But But z = 4.84z = 4.84 is off the standard normal table is off the standard normal table The largest The largest zz value in the table is value in the table is 3.093.09, which has a , which has a

right hand tail area of right hand tail area of 0.0010.001


Probability that > 31.5531 when = 31= 31x


z = 4.84 > 3.09z = 4.84 > 3.09, so , so P(z ≥ 4.84) < 0.001P(z ≥ 4.84) < 0.001 That is, ifThat is, if = 31 mpg, then fewer than 1 in 1,000 of all

possible samples have a mean at least as large as samples have a mean at least as large as observedobserved

Have either of the following explanations:Have either of the following explanations: If is actually is actually 3131 mpg, then picking this sample is an mpg, then picking this sample is an

almost unbelievable thingalmost unbelievable thingOROR is not 31 mpgis not 31 mpg

Difficult to believe such a small chance would occur, so conclude that there is strong evidence thatevidence that does not equal 31 mpg.. is in fact larger than 31 mpgis in fact larger than 31 mpg


Section 6.2 The Sampling Distribution Section 6.2 The Sampling Distribution of the Sample Proportion(of the Sample Proportion( 样本比例样本比例 ))

For a population of units, we select samples of size n, and For a population of units, we select samples of size n, and calculate its proportion for the units of the sample to be calculate its proportion for the units of the sample to be fall into a particular category. fall into a particular category. is a random variable and has its probability distribution.is a random variable and has its probability distribution.

The probability distribution of all possible sample The probability distribution of all possible sample proportions is called the proportions is called the sampling distribution of the sampling distribution of the sample proportionsample proportion

p̂

p̂


the sampling distribution of isp̂

approximately normal, if n is large (meet the conditions that np≥5 and n(1-p)≥5)

has mean pp ˆ

has standard deviation

npp

p̂

1

where p is the population proportion for the category


Example 6.5Example 6.5 The Cheese Spread CaseThe Cheese Spread Case A food processing company developed a new cheese

spread spout which may save production cost. If only less than 10% of current purchasers do not accept the design, the company would adopt and use the new spout.

1000 current purchasers are randomly selected and inquired, and 63 of them say they would stop buying the cheese spread if the new spout were used. So, the sample proportion =0.063.

To evaluate the strength of this evidence, we ask: if p=0.1, what is the probability of observing a sample of size 1000 with sample proportion 0.063?≦

p̂

p̂


If p=0.10, since n=1000, np≥5 and n(1-p)≥5, is approximately normal with p̂

,1.0ˆ pp ,094868.01

ˆ

n

ppp

.001.090.3094868.0

10.0063.0

063.01.0 if 063.0

zP

zP

zPppPp

p


So, if p=0.1, the chance of observing at most 63 out of 1000 randomly selected customers do not accept the new design is less than 0.001

But such observation does occur. This means that we have extremely strong evidence that p≠0.1, and p is in fact less than 0.1

Therefore, the company can adopt the new design

the sampling distribution of the sample mean

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