sfs4e ppt 09

Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc.

Chapter

Estimating the Value of a Parameter

9


Section

Estimating a Population Proportion

9.1

Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc.9-3

1. Obtain a point estimate for the population proportion

2. Construct and interpret a confidence interval for the population proportion

3. Determine the sample size necessary for estimating the population proportion within a specified margin of error

Objectives


Objective 1

• Obtain a Point Estimate for the Population Proportion


A point estimate is the value of a statistic that estimates the value of a parameter.

For example, the point estimate for the population proportion is , where x is the number of individuals in the sample with a specified characteristic and n is the sample size.

p̂ x

n


In July of 2008, a Quinnipiac University Poll asked 1783 registered voters nationwide whether they favored or opposed the death penalty for persons convicted of murder. 1123 were in favor.

Obtain a point estimate for the proportion of registered voters nationwide who are in favor of the death penalty for persons convicted of murder.

Parallel Example 1: Calculating a Point Estimate for the Population Proportion


Obtain a point estimate for the proportion of registered voters nationwide who are in favor of the death penalty for persons convicted of murder.

Solution

ˆ p 1123

17830.63


Objective 2

• Construct and Interpret a Confidence Interval for the Population Proportion


A confidence interval for an unknown parameter consists of an interval of numbers based on a point estimate.

The level of confidence represents the expected proportion of intervals that will contain the parameter if a large number of different samples is obtained. The level of confidence is denoted (1 – α)·100%.


For example, a 95% level of confidence(α = 0.05) implies that if 100 different confidence intervals are constructed, each based on a different sample from the same population, we will expect 95 of the intervals to contain the parameter and 5 not to include the parameter.


Confidence interval estimates for the population proportion are of the form

Point estimate ± margin of error.

The margin of error of a confidence interval estimate of a parameter is a measure of how accurate the point estimate is.


The margin of error depends on three factors:

Level of confidence: As the level of confidence increases, the margin of error also increases.

Sample size: As the size of the random sample increases, the margin of error decreases.

Standard deviation of the population: The more spread there is in the population, the wider our interval will be for a given level of confidence.


For a simple random sample of size n, the

sampling distribution of is approximately

normal with mean and standard deviation

, provided that

np(1 – p) ≥ 10.

NOTE: We also require that each trial be independent;

when sampling from finite populations (n ≤ 0.05N).

Sampling Distribution of

ˆ p

ˆ p p(1 p)

n

p̂

p̂ p


Suppose that a simple random sample of size n is taken from a population. A (1 – α)·100% confidence interval for p is given by the following quantities

Lower bound:

Upper bound:

Note: It must be the case that and

n ≤ 0.05N to construct this interval.

Constructing a (1 – α)·100% Confidence Interval for a Population Proportion

ˆ p z 2 ˆ p (1 ˆ p )

n

ˆ p z 2 ˆ p (1 ˆ p )

n

nˆ p (1 ˆ p ) 10


Interpretation of a Confidence Interval

A (1 – α)·100% confidence interval indicates that (1 – α)·100% of all simple random samples of size n from the population whose parameter is unknown will result in an interval that contains the parameter.


The margin of error, E, in a (1 – α) 100% confidence interval for a population proportion is given by

Margin of Error

E z2

p̂(1 p̂)

n


In July of 2008, a Quinnipiac University Poll asked 1783 registered voters nationwide whether they favored or opposed the death penalty for persons convicted of murder. 1123 were in favor.

Obtain a 90% confidence interval for the proportion of registered voters nationwide who are in favor of the death penalty for persons convicted of murder.

Parallel Example 2: Constructing a Confidence Interval for a Population Proportion


•

• and the sample size is definitely less than 5% of the population size

• α = 0.10 so zα/2 = z0.05 = 1.645

• Lower bound:

• Upper bound:

ˆ p 0.63

nˆ p (1 ˆ p ) 1783(0.63)(1 0.63) 415.6 10

Solution

0.63 1.645

0.63(1 0.63)

17830.61

0.631.645

0.63(1 0.63)

17830.65


Solution

We are 90% confident that the proportion of registered voters who are in favor of the death penalty for those convicted of murder is between 0.61and 0.65.


Objective 3

• Determine the Sample Size Necessary for Estimating a Population Proportion within a Specified Margin of Error


Sample size needed for a specified margin of error, E, and level of confidence (1 – α):

Problem: The formula uses which depends on n, the quantity we are trying to determine!

ˆ p

n ˆ p (1 ˆ p )z 2

E

2


Two possible solutions:

1. Use an estimate of based on a pilot study or an earlier study.

2. Let = 0.5 which gives the largest possible value ofn for a given level of confidence and a given margin of error.

öp

öp


The sample size required to obtain a (1 – α)·100% confidence interval for p with a margin of error E is given by

(rounded up to the next integer), where is a prior estimate of p.

ˆ p

n ˆ p (1 ˆ p )z 2

E

2

Sample Size Needed for Estimating the Population Proportion p


If a prior estimate of p is unavailable, the sample size required is

n 0.25z 2

E

2

Sample Size Needed for Estimating the Population Proportion p


A sociologist wanted to determine the percentage of residents of America that only speak English at home. What size sample should be obtained if she wishes her estimate to be within 3 percentage points with 90% confidence assuming she uses the 2000 estimate obtained from the Census 2000 Supplementary Survey of 82.4%?

Parallel Example 4: Determining Sample Size


• E = 0.03

•

•

•

We round this value up to 437. The sociologist must

survey 437 randomly selected American residents.

z 2 z0.05 1.645

ˆ p 0.824

n 0.824(1 0.824)1.645

0.03

2

436.04

Solution


SectionEstimating a Population Mean

9.2


Objectives

1. Obtain a point estimate for the population mean

2. State properties of Student’s t-distribution

3. Determine t-values

4. Construct and interpret a confidence interval for a population mean

5. Find the sample size needed to estimate the population mean within a given margin or error


Objective 1

• Obtain a Point Estimate for the Population Mean


A point estimate is the value of a statistic that estimates the value of a parameter.

For example, the sample mean, , is a point estimate of the population mean μ.

x


Pennies minted after 1982 are made from 97.5% zinc and 2.5% copper. The following data represent the weights (in grams) of 17 randomly selected pennies minted after 1982.

2.46 2.47 2.49 2.48 2.50 2.44 2.46 2.45 2.49

2.47 2.45 2.46 2.45 2.46 2.47 2.44 2.45

Treat the data as a simple random sample. Estimate the population mean weight of pennies minted after 1982.

Parallel Example 1: Computing a Point Estimate


The sample mean is

The point estimate of μ is 2.464 grams.

Solution

x 2.46 2.47 2.45

172.464


Objective 2

• State Properties of Student’s t-Distribution


Suppose that a simple random sample of size n is taken from a population. If the population from which the sample is drawn follows a normal distribution, the distribution of

follows Student’s t-distribution with n – 1 degrees of freedom where is the sample mean and s is the sample standard deviation.

Student’s t-Distribution

t x

s

n

x


a) Obtain 1,000 simple random samples of size n = 5 from a normal population with μ = 50 and σ = 10.

b) Determine the sample mean and sample standard deviation for each of the samples.

c) Compute and for each sample.

d) Draw a histogram for both z and t.

Parallel Example 1: Comparing the Standard Normal Distribution to the t-Distribution Using Simulation

z x

n

t x s

n


Histogram for z


Histogram for t


CONCLUSION:

• The histogram for z is symmetric and bell-shaped with the center of the distribution at 0 and virtually all the rectangles between –3 and 3. In other words, z follows a standard normal distribution.


CONCLUSION (continued):

• The histogram for t is also symmetric and bell-shaped with the center of the distribution at 0, but the distribution of t has longer tails (i.e., t is more dispersed), so it is unlikely that t follows a standard normal distribution. The additional spread in the distribution of t can be attributed to the fact that we use s to find t instead of σ. Because the sample standard deviation is itself a random variable (rather than a constant such as σ), we have more dispersion in the distribution of t.


1. The t-distribution is different for different degrees of freedom.

2. The t-distribution is centered at 0 and is symmetric about 0.

3. The area under the curve is 1. The area under the curve to the right of 0 equals the area under the curve to the left of 0, which equals 1/2.

4. As t increases or decreases without bound, the graph approaches, but never equals, zero.

Properties of the t-Distribution


5. The area in the tails of the t-distribution is a little greater than the area in the tails of the standard normal distribution, because we are using s as an estimate of σ, thereby introducing further variability into the t- statistic.

6. As the sample size n increases, the density curve of t gets closer to the standard normal density curve. This result occurs because, as the sample size n increases, the values of s get closer to the values of σ, by the Law of Large Numbers.

Properties of the t-Distribution


Objective 3

• Determine t-Values


Parallel Example 2: Finding t-values

Find the t-value such that the area under the t-distribution to the right of the t-value is 0.2 assuming 10 degrees of freedom. That is, find t0.20 with 10 degrees of freedom.


The figure to the left shows the graph of the t-distribution with 10 degrees of freedom.

Solution

The unknown value of t is labeled, and the area under the curve to the right of t is shaded. The value of t0.20 with 10 degrees of freedom is 0.8791.


Objective 4

• Construct and Interpret a Confidence Interval for the Population Mean


Constructing a (1 – α)100% Confidence Interval for μ

Provided• Sample data come from a simple random sample or

randomized experiment• Sample size is small relative to the population size

(n ≤ 0.05N)• The data come from a population that is normally

distributed, or the sample size is largeA (1 – α)·100% confidence interval for μ is given byLower Upper bound: bound:

where is the critical value with n – 1 df.

x t2

s

nx t

2

s

nt

2


Constructing a (1 – α)100% Confidence Interval for μ

A (1 – α)·100% confidence interval for μ is given by

Lower Upper bound: bound:

where is the critical value with n – 1 df.

x t2

s

nx t

2

s

n

t2


We will use Minitab to simulate obtaining 30 simple random samples of size n = 8 from a population that is normally distributed with μ = 50 and σ = 10. Construct a 95% confidence interval for each sample. How many of the samples result in intervals that contain μ = 50 ?

Parallel Example 2: Using Simulation to Demonstrate the Idea of a Confidence Interval


Sample Mean 95.0% CI C1 47.07 ( 40.14, 54.00)C2 49.33 ( 42.40, 56.26)C3 50.62 ( 43.69, 57.54)C4 47.91 ( 40.98, 54.84)C5 44.31 ( 37.38, 51.24)C6 51.50 ( 44.57, 58.43)C7 52.47 ( 45.54, 59.40)

C9 43.49 ( 36.56, 50.42)C10 55.45 ( 48.52, 62.38)C11 50.08 ( 43.15, 57.01)C12 56.37 ( 49.44, 63.30)C13 49.05 ( 42.12, 55.98)C14 47.34 ( 40.41, 54.27)C15 50.33 ( 43.40, 57.25)

C8 59.62 ( 52.69, 66.54)


SAMPLE MEAN 95% CIC16 44.81 ( 37.88, 51.74)C17 51.05 ( 44.12, 57.98)C18 43.91 ( 36.98, 50.84)C19 46.50 ( 39.57, 53.43)C20 49.79 ( 42.86, 56.72)C21 48.75 ( 41.82, 55.68)C22 51.27 ( 44.34, 58.20)C23 47.80 ( 40.87, 54.73)C24 56.60 ( 49.67, 63.52)C25 47.70 ( 40.77, 54.63)C26 51.58 ( 44.65, 58.51)C27 47.37 ( 40.44, 54.30)

C29 46.89 ( 39.96, 53.82)C30 51.92 ( 44.99, 58.85)

C28 61.42 ( 54.49, 68.35)


Note that 28 out of 30, or 93%, of the confidence intervals contain the population mean μ = 50.

In general, for a 95% confidence interval, any sample mean that lies within 1.96 standard errors of the population mean will result in a confidence interval that contains μ.

Whether a confidence interval contains μ depends solely on the sample mean, .

x


Construct a 99% confidence interval about the population mean weight (in grams) of pennies minted after 1982. Assume μ = 0.02 grams.

2.46 2.47 2.49 2.48 2.50 2.44 2.46 2.45 2.49

2.47 2.45 2.46 2.45 2.46 2.47 2.44 2.45

Parallel Example 3: Constructing a Confidence Interval


Weight (in grams) of Pennies


• • Lower bound:

= 2.464 – 2.575

= 2.464 – 0.012 = 2.452

• Upper bound:

= 2.464 + 2.575

= 2.464 + 0.012 = 2.476We are 99% confident that the mean weight of penniesminted after 1982 is between 2.452 and 2.476 grams.

0.0217

0.0217

t 2 2.575

x t 2 s

n

x t 2 s

n


Objective 5

• Find the Sample Size Needed to Estimate the Population Mean within a Given Margin of Error


Determining the Sample Size n

The sample size required to estimate the population mean, µ, with a level of confidence (1– α)·100% with a specified margin of error, E, is given by

where n is rounded up to the nearest whole number.

n z

2

s

E

2


Back to the pennies. How large a sample would be required to estimate the mean weight of a penny manufactured after 1982 within 0.005 grams with 99% confidence? Assume = 0.02.

Parallel Example 7: Determining the Sample Size


• • s = 0.02

• E = 0.005

Rounding up, we find n = 107.

z 2 z0.005 2.575

n z

2

s

E

2

2.575(0.02)

0.005

2

106.09


Section

Putting It Together: Which Procedure Do I Use?

9.3


Objective

1. Determine the appropriate confidence interval to construct


Objective 1

• Determine the Appropriate Confidence Interval to Construct

sfs4e ppt 09

News & Politics

pearson education

confidence interval

p1p n p p

given level of confidence

interval of numbers

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p1 p n

simple random sample