Copyright © 2009 Cengage Learning Ch 9 實習. Copyright © 2009 Cengage Learning 9.2 Sampling...

Copyright © 2009 Cengage Learning

Ch 9 實習Ch 9 實習

Copyright © 2009 Cengage Learning 9.2

Sampling Distributions…

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…


x 1 2 3 4 5 6

P(x) 1/6 1/6 1/6 1/6 1/6 1/6

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:


Sampling Distribution of Two DiceA sampling distribution is created by looking atall 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).


Sampling Distribution of Two Dice…

The sampling distribution of is shown below:

1.0 1/361.5 2/362.0 3/362.5 4/363.0 5/363.5 6/364.0 5/364.5 4/365.0 3/365.5 2/366.0 1/36

P( )

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

P( )


Compare…

Compare the distribution of X…

…with the sampling distribution of .

As well, note that:

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


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:


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.


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.


Sampling Distribution of the Sample Mean1.

2.

3. If X is normal, X is normal. If X is nonnormal, X 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)


Standard Deviation of sample meanStandard Deviation of sample mean

Finite PopulationFinite Population Infinite Infinite Population Population

• A finite population is treated as being A finite population is treated as being infinite if infinite if nn//NN << .05. .05.

• is the finite correction is the finite correction factor.factor.

• is referred to as the is referred to as the standard error of the standard error of the meanmean..

x n

N nN

( )1

x n

N nN

( )1

x n

x n

( ) / ( )N n N 1( ) / ( )N n N 1

xx

Sampling Distribution of sample mean


Example 1

• The amount of time the university professors devote to their jobs per week is normally distributed with a mean of 52 hours and a standard deviation of 6 hours

• a. What is the probability that a professor works for more than 60 hours per week?

• b. Find the probability that the mean amount of work per week for three randomly selected professors is more than 60 hours.

• c. Find the probability that if three professors are randomly selected, all three work for more than 60 hours per week.


Solution• a.

• b.

• c.


Using the Sampling Distribution for InferenceHere’s another way of expressing the probability calculated from a sampling distribution.

P(-1.96 < Z < 1.96) = .95Substituting the formula for the sampling distribution

With a little algebra

95.)96.1n/

X96.1(P

95.)n

96.1Xn

96.1(P


Using the Sampling Distribution for InferenceReturning 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 760.8 and 839.2. 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.

95.)25

10096.1800X

25

10096.1800(P

95.)2.839X8.760(P


The estimate of p = The estimate of p =

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

• To estimate the population proportion ‘p’ we use the sample proportion.

Sampling Distribution of a Proportion

pp̂̂ ==XXnn

The number of successes


• 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.

pp̂̂

pp̂̂

Sampling Distribution of a Proportion


Normal approximation to the Binomial–Normal approximation to the binomial works best when

•the number of experiments (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:

np 5; n(1 - p) 5


Normal approximation to the Binomial

ExampleApproximate the binomial probability P(x=10) when n = 20 and p = .5

The parameters of the normal distribution used to approximate the binomial are:

= np; 2 = np(1 - p)


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


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

44.5

1413.5

P(X P(X 14) 14)

P(Y< 4.5)P(Y< 4.5)

P(Y > 13.5)P(Y > 13.5)

Normal approximation to the BinomialNormal approximation to the Binomial

P(X P(X 4) 4)


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.)

Sample proportions can be standardized to a standard normal distribution using this formulation:


Sampling Distribution of pppp

pp pn

N nN

( )1

1 p

p pn

N nN

( )1

1 p

p pn

( )1 p

p pn

( )1

p p

• Standard Deviation of

Finite Population Infinite Population

– is referred to as the standard error of the proportion.


Example 2

• A psychologist believes that 80% of male drivers when lost continue to drive, hoping to find the location they seek rather than ask directions. To examine this belief, he took a random sample of 350 male drivers and asked each what they did when lost. If the belief is true, determine the probability that less than 75% said they continue driving.


Solution


Sampling Distribution: Difference of two meansThe final sampling distribution introduced is that of the difference between two sample means. This requires:

independent random samples be drawn from each of two normal populations

If this condition is met, then the sampling distribution of the difference between the two sample means, i.e.will be normally distributed.(note: if the two populations are not both normally distributed, but the sample sizes are “large” (>30), the distribution of is approximately normal)


Sampling Distribution: Difference of two meansThe 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)


Example 9.3…

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:


Example 3

• The manager of a restaurant believes that waiters and waitresses who introduce themselves by telling customers their names will get larger tips than those who don’t. In fact, she claims that the average tip for the former group is 18% whereas that of the latter is only 15%. If tips are normally distributed with a standard deviation of 3 %, what is the probability that in a random sample of 10 tips recorded from waiters and waitresses who introduce themselves and 10 tips from waiters and waitresses who don’t, the mean of the former will exceed that of the latter?


Solution

Copyright © 2009 Cengage Learning Ch 9 實習. Copyright © 2009 Cengage Learning 9.2 Sampling...

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