© 2010 pearson prentice hall. all rights reserved chapter sampling distributions 8

© 2010 Pearson Prentice Hall. All rights reserved

Chapter

Sampling Distributions

8

© 2010 Pearson Prentice Hall. All rights reserved 8-2

Objectives

1. Describe the distribution of the sample mean: samples from normal populations

2. Describe the distribution of the sample mean: samples from a population that is not normal

Section 8.1Distribution of the Sample Mean


Statistics such as are random variables since their value varies from sample to sample. As such, they have probability distributions associated with them. In this chapter we focus on the shape, center and spread of statistics such as .

x

x


The sampling distribution of a statistic is a probability distribution for all possible values of the statistic computed from a sample of size n.

The sampling distribution of the sample mean is the probability distribution of all possible values of the random variable computed from a sample of size n from a population with mean and standard deviation .

x

x


Illustrating Sampling Distributions

Step 1: Obtain a simple random sample of size n.

Step 2: Compute the sample mean.

Step 3: Assuming we are sampling from a finite population, repeat Steps 1 and 2 until all simple random samples of size n have been obtained. Note: once a particular sample is obtained, it cannot be

obtained a second time


The weights of pennies minted after 1982 are approximately normally distributed with mean 2.46 grams and standard deviation 0.02 grams.

Approximate the sampling distribution of the sample mean by obtaining 200 simple random samples of size n = 5 from this population.

Parallel Example 1: Sampling Distribution of the Sample Mean-Normal Population


The data on the following slide represent the sample means for the 200 simple random samples of size n = 5.

For example, the first sample of n = 5 had the following data:

2.493 2.466 2.473 2.492 2.471

Note: =2.479 for this sample

x


Sample Means for Samples of Size n =5


The mean of the 200 sample means is 2.46, the same as the mean of the population.

The standard deviation of the sample means is 0.0086, which is smaller than the standard deviation of the population (recall that we know σ = 0.02)


What role does n, the sample size, play in the standard deviation of the distribution of the sample mean?

As the size of the sample gets larger, we do not

expect as much spread in the sample means

since larger observations will offset smaller

observations.


• Approximate the sampling distribution of the sample mean by obtaining 200 simple random samples of size n = 20 from the population of weights of pennies minted after 1982 (=2.46 grams and =0.02 grams)

Parallel Example 2: The Impact of Sample Size on Sampling Variability


The mean of the 200 sample means for n =20 is still 2.46, but the standard deviation is now 0.0045 (0.0086 for n = 5, σ = 0.02).As expected, there is less variability in the distribution of the sample mean with n =20 than with n =5.

As the sample size n increases, the standard deviation of decreases


Suppose that a simple random sample of size n is

drawn from a large population with mean and

standard deviation . The sampling distribution of will have mean and standard deviation .

(The standard deviation of the sampling distribution of is called the standard error of the mean and is denoted .)

The Mean and Standard Deviation of theSampling Distribution of

x

x

x n

x

x

x

Said “mu of x bar”

Said “sigma of x bar”


The Shape of the Sampling Distribution of If X is Normal

x

If a random variable X is normally distributed, the distribution of the sample mean is normally distributed.

x


The weights of pennies minted after 1982 are approximately normally distributed with mean 2.46 grams and standard deviation 0.02 grams.

What is the probability that in a simple random sample of 10 pennies minted after 1982, we obtain a sample mean of at least 2.465 grams?

Parallel Example 3: Describing the Distribution of the Sample Mean


• is normally distributed with =2.46 and .

•

• P(Z>0.79)=1-0.7852

=0.2148.

Solution

x

x

x 0.02

100.0063

Z 2.465 2.46

0.00630.79

We are being asked to find probability that is at least 2.465 grams; that is, P ( ≥ 2.465)

Interpretation: the probability of obtaining a

sample mean weight greater than 2.465 grams from a population whose mean is 2.46 is 0.2148. If we were to take 100

simple random samples of size n = 10, we expect 21 of

the samples will have a mean weight of at least

2.456


The following table and histogram give the probability distribution for rolling a fair die:

=3.5, =1.708Note that the population distribution is NOT normal

Face on Die Relative Frequency

1 0.1667

2 0.1667

3 0.1667

4 0.1667

5 0.1667

6 0.1667

Parallel Example 4: Sampling from a Population that is Not Normal


Estimate the sampling distribution of by obtaining 200 simple random samples of size n=4 and calculating the sample mean for each of the 200 samples. Repeat for n = 10 and 30.

Below are histograms of the sampling distribution of the sample mean for each sample size.

x


Key Points from Example 4• The mean of the sampling distribution is equal to

the mean of the parent population and the standard deviation of the sampling distribution of the sample mean is regardless of the sample size.

• The Central Limit Theorem: the shape of the distribution of the sample mean becomes approximately normal as the sample size n increases, regardless of the shape of the population. We will require than the sample size is greater than or equal to 30.

n

Note: The Central Limit Theorem has only to do with the shape of the distribution of the sample means, not the center or spread


Parallel Example 5: Using the Central Limit Theorem

Suppose that the mean time for an oil change at a “10-minute oil change joint” is 11.4 minutes with a standard deviation of 3.2 minutes.

(a) If a random sample of n = 35 oil changes is selected, describe the sampling distribution of the sample mean.

(b) If a random sample of n = 35 oil changes is selected, what is the probability the mean oil change time is less than 11 minutes?

Tell me the shape, center, and spread of the distribution


Parallel Example 5: Using the Central Limit Theorem

Suppose that the mean time for an oil change at a “10-minute oil change joint” is 11.4 minutes with a standard deviation of 3.2 minutes.

(a) If a random sample of n = 35 oil changes is selected, describe the sampling distribution of the sample mean.

(b) If a random sample of n = 35 oil changes is selected, what is the probability the mean oil change time is less than 11 minutes?

Solution: is approximately normally distributed with mean=11.4 and std. dev. = .

x

3.2

350.5409

Solution: , P(Z<-0.74)=0.23.

Z 11 11.4

0.5409 0.74


Objectives

1. Describe the sampling distribution of a sample proportion

2. Compute probabilities of a sample proportion

Section 8.2Distribution of the Sample Proportion


Point Estimate of a Population Proportion

Suppose that a random sample of size n is obtained from a population in which each individual either does or does not have a certain characteristic. The sample proportion, denoted (read “p-hat”) is given by

where x is the number of individuals in the sample with the specified characteristic (connect with binomial probabilities: x can be thought of as the number of successes in

n trials.) The sample proportion is a statistic that estimates the population proportion, p.

ˆ p x

n


In a Quinnipiac University Poll conducted in May of 2008, 1,745 registered voters nationwide were asked whether they approved of the way George W. Bush is handling the economy. 349 responded “yes”. Obtain a point estimate for the proportion of registered voters who approve of the way George W. Bush is handling the economy.

Parallel Example 1: Computing a Sample Proportion

Solution:

ˆ p 349

17450.2


According to a Time poll conducted in June of 2008, 42% of registered voters believed that gay and lesbian couples should be allowed to marry.

Describe the sampling distribution of the sample proportion for samples of size n=10, 50, 100.

Parallel Example 2: Using Simulation to Describe the Distribution of the Sample Proportion

To get a sense of the shape, center, and spread of the sampling distribution of the sample proportion, we will use MINITAB to simulate the responses and construct the histograms


Key Points from Example 2

• Shape: As the size of the sample, n, increases, the shape of the sampling distribution of the sample proportion becomes approximately normal.

• Center: The mean of the sampling distribution of the sample proportion equals the population proportion, p.

• Spread: The standard deviation of the sampling distribution of the sample proportion decreases as the sample size, n, increases.


For a simple random sample of size n with population proportion p:• The shape of the sampling distribution of is

approximately normal provided np(1-p)≥10.• The mean of the sampling distribution of is

.• The standard deviation of the sampling distribution

of is

Sampling Distribution of

ˆ p

ˆ p

ˆ p

ˆ p p

ˆ p

ˆ p p(1 p)

n


Sampling Distribution of

• The model on the previous slide requires that the sampled values are independent. When sampling from finite populations, this assumption is verified by checking that the sample size n is no more than 5% of the population size N (n ≤ 0.05N).

• Regardless of whether np(1-p) ≥10 or not, the mean of the sampling distribution of is p, and the standard deviation is

ˆ p

ˆ p

ˆ p p(1 p)

n


According to a Time poll conducted in June of 2008, 42% of registered voters believed that gay and lesbian couples should be allowed to marry. Suppose that we obtain a simple random sample of 50 voters and determine which believe that gay and lesbian couples should be allowed to marry. Describe the sampling distribution of the sample proportion for registered voters who believe that gay and lesbian couples should be allowed to marry.

Parallel Example 3: Describing the Sampling Distribution of the Sample Proportion


Solution

The sample of n=50 is smaller than 5% of the population size (all registered voters in the U.S.).

Also, np(1-p)=50(0.42)(0.58)=12.18≥10.

The sampling distribution of the sample proportion is therefore approximately normal with mean=0.42 and standard deviation=

.(Note: this is very close to the standard deviation of 0.072

found using simulation in Example 2.)

0.42(1 0.42)

500.0698


According to the Centers for Disease Control and Prevention, 18.8% of school-aged children, aged 6-11 years, were overweight in 2004.

(a) In a random sample of 90 school-aged children, aged 6-11 years, what is the probability that at least 19% are overweight?

(b) Suppose a random sample of 90 school-aged children, aged 6-11 years, results in 24 overweight children. What might you conclude?

Parallel Example 4: Compute Probabilities of a Sample Proportion


• n=90 is less than 5% of the population size• np(1-p)=90(.188)(1-.188)≈13.7≥10• is approximately normal with mean=0.188 and

standard deviation =

(a) In a random sample of 90 school-aged children, aged 6-11 years, what is the probability that at least 19% are overweight?

Solution

ˆ p

(0.188)(1 0.188)

900.0412

, P(Z>0.05)=1-0.5199=0.4801

Z 0.19 0.188

0.04120.0485


• is approximately normal with mean=0.188 and standard deviation = 0.0412

(b) Suppose a random sample of 90 school-aged children, aged 6-11 years, results in 24 overweight children. What might you conclude?

Solution

ˆ p

, P(Z>1.91)=1-0.9719=0.028.

We would only expect to see about 3 samples in 100 resulting in a sample proportion of 0.2667 or more. This is an unusual sample if the true population proportion is 0.188.

ˆ p 24

900.2667

Z 0.2667 0.188

0.04121.91

© 2010 pearson prentice hall. all rights reserved chapter sampling distributions 8

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