Chapter Estimation 1 of 83 8 © 2012 Pearson Education, Inc. All rights reserved

ChapterEstimation

1 of 83

8

© 2012 Pearson Education, Inc.All rights reserved.

Chapter Outline

• 8.1 Estimating when σ is Known• 8.2 Estimating when σ is Unknown• 8.3 Confidence Intervals for Population Proportions• 8.4 Estimating - and -

© 2012 Pearson Education, Inc. All rights reserved. 2 of 83

Section 8.1

Estimating when σ is Known


Section 8.1 Objectives

• Find a point estimate and a margin of error• Construct and interpret confidence intervals for the

population mean• Determine the minimum sample size required when

estimating μ


Point Estimate for Population μ

Point Estimate• A single value estimate for a population parameter• Most unbiased point estimate of the population mean

μ is the sample mean x

Estimate Population Parameter…

with Sample Statistic

Mean: μ x


Exercise 1: Point Estimate for Population μ

A social networking website allows its users to add friends, send messages, and update their personal profiles. The following represents a random sample of the number of friends for 40 users of the website. Find a point estimate of the population mean, µ. (Source: Facebook)

140 105 130 97 80 165 232 110 214 201 122 98 65 88 154 133 121 82 130 211 153 114 58 77 51 247 236 109 126 132 125 149 122 74 59 218 192 90 117 105


Solution: Point Estimate for Population μ

The sample mean of the data is

5232130.8

40

xx

n

Your point estimate for the mean number of friends for all users of the website is 130.8 friends.


115 120 125 130 135 140 150145

Point estimate

115 120 125 130 135 140 150145

Point estimate130.8x

How confident do we want to be that the interval estimate contains the population mean μ?

Interval Estimate

Interval estimate • An interval, or range of values, used to estimate a

population parameter.


( )

Interval estimate

Right endpoint146.5

Left endpoint115.1

Confidence Levels

A confidence level, c, is any value between 0 and 1 that corresponds to the area under the standard normal curve between and .

Level of Confidence

Level of confidence c • The probability that the interval estimate contains the

population parameter.

zz = 0–zc zc

Critical values

½(1 – c) ½(1 – c)

c is the area under the standard normal curve between the critical values.

The remaining area in the tails is 1 – c .

c

Use the Standard Normal Table to find the corresponding z-scores.


Common Confidence Levels

Appendix II, Table 5: Page A23

zc

Level of Confidence

• If the level of confidence is 90%, this means that we are 90% confident that the interval contains the population mean μ.

zz = 0 zc

The corresponding z-scores are ±1.645.

c = 0.90

½(1 – c) = 0.05½(1 – c) = 0.05

–zc = –1.645 zc = 1.645


Sampling Error

Sampling error • The difference between the point estimate and the

actual population parameter value.• For μ:

the sampling error is the difference – μ μ is generally unknown varies from sample to sample

x

x


Margin of Error

• The greatest possible distance between the point estimate and the value of the parameter it is estimating for a given level of confidence, c.

• Denoted by E• | [Magnitude of sampling error]• =

• Sometimes called the maximum error of estimate or error tolerance.

When n ≥ 30, the sample standard deviation, s, can be used for σ


Distribution of Sample Means






STATEMENT REASON

) Probability that the generated confidence interval contains

= Equivalent inequality



= Shaded area in diagram

Exercise 2: Finding the Margin of Error

Use the social networking website data and a 95% confidence level to find the margin of error for the mean number of friends for all users of the website. Assume that the population standard deviation is about 53.0.


zc

Solution: Finding the Margin of Error

• Step 1: Find the critical values [Table 5 – Page A23]

zzcz = 0

0.95

0.0250.025

–zc = –1.96

95% of the area under the standard normal curve falls within 1.96 standard deviations of the mean. (You can approximate the distribution of the sample means with a normal curve by the Central Limit Theorem, because n = 40 ≥ 30.)

zc = 1.96


Solution: Finding the Margin of Error

We are 95% confident that the margin of error for the population mean is about 16.4 friends. Our estimate might be too big, or too small.


=

Step 2: Compute E

Confidence Intervals for the Population Mean

A c-confidence interval for the population mean μ

•

• The probability that the confidence interval contains μ is c.

where cx E x E E zn


Constructing Confidence Intervals for μ


Exercise 3: Construct a Confidence Interval

Construct a 95% confidence interval for the mean number of friends for all users of the website.

Solution: Recall and E ≈ 16.4130.8x

130.8 16.4

114.4

x E

130.8 16.4

147.2

x E

114.4 < μ < 147.2

Left Endpoint: Right Endpoint:


Solution: Constructing a Confidence Interval

114.4 < μ < 147.2

•

With 95% confidence, you can say that the population mean number of friends is between 114.4 and 147.2.


Exercise 4: Construct Confidence Interval: Case σ is Known

A college admissions director wishes to estimate the mean age of all students currently enrolled. In a random sample of 20 students, the mean age is found to be 22.9 years. From past studies, the standard deviation is known to be 1.5 years, and the population is normally distributed. Construct a 90% confidence interval of the population mean age.


zc

Solution: Constructing a Confidence Interval σ Known

• First find the critical values

zz = 0 zc

c = 0.90

½(1 – c) = 0.05½(1 – c) = 0.05

–zc = –1.645 zc = 1.645

zc = 1.645


• Margin of error:

• Confidence interval:


1.51.645 0.6

20cE z

n

22.9 0.6

22.3

x E

22.9 0.6

23.5

x E


22.3 < μ < 23.5



22.3 < μ < 23.5

( )• 22.922.3 23.5

With 90% confidence, we can say that the mean age of all the students is between 22.3 and 23.5 years.

Point estimate

xx E x E


Interpreting the Results

• μ is a fixed number. Once the interval is fixed, it either contains or does not contain μ .

• Incorrect: “There is a 90% probability that the actual mean μ is in the interval (22.3, 23.5).”

• Correct: “If a large number of samples is collected and a confidence interval is created for each sample, approximately 90% of these intervals will contain μ.

• Correct: “We are 90% confident that the interval (22.3,23.5) is one which contains the actual population parameter μ.”



The horizontal segments represent 90% confidence intervals for different samples of the same size.In the long run, 9 of every 10 such intervals will contain μ. μ



Since is a random variable, so are the endpoints and

After the confidence interval is numerically fixed for a specific sample, it either does or does not contain µ.

μ


TI-84: Z-Interval Function

The ZInterval (STAT / TESTS / 7: ZInterval) function computes a confidence interval for a population mean µ when the population standard deviation is known.


Input σ, OutputMargin of Error

Exercise 5: Construct Confidence Interval via TI-84: Case σ is Known

The ZInterval (STAT / TESTS / 7: ZInterval) function computes a confidence interval for a population mean µ when the population standard deviation is known.


Inputs σ = 1.5,

Output (22.3, 23.5)

Margin of Error

Minimum Sample Size

• Given a c-confidence level and a margin of error E, the minimum sample size n needed to estimate the population mean µ is given by

Round up using CELING

2

czn

E


If σ is unknown, you can estimate it using s, provided you have a preliminary sample with at least 30

Result should be rounded up by using the CELING function

Exercise 6: Minimum Sample Size

You want to estimate the mean number of friends for all users of the website. How many users must be included in the sample if you want to be 95% confident that the sample mean is within seven friends of the population mean? Assume the population standard deviation is about 53.0.


zc

Solution: Minimum Sample Size

• First find the critical values

zc = 1.96

zz = 0 zc

0.95

0.0250.025

–zc = –1.96 zc = 1.96


Solution: Minimum Sample Size

zc = 1.96 σ ≈ s ≈ 53.0 E = 7

2 21.96 53.0

220.237

czn

E

When necessary, round up to obtain a whole number.

Interpretive Statement: You should include at least 221 users in your sample in order to be 95% confident that the sample mean will be within 7 friends of the population mean.


Exercise 7: Salmon Sample Size

A sample of 50 salmon is caught and weighed. The sample standard deviation of the 50 weights is 2.15 lb. How large of a sample should be taken to be 97% confident that the sample mean is within 0.20 lb of the mean weight of the population?

Summarize your results in a complete sentence relevant to this application.


Solution: Salmon Sample Size


Critical Values

Apply Ceiling Function

545 salmon

Interpretive Statement A sample size of 545 salmon will be large enough to ensure that we are 97% confident that the sample mean is within 0.20 lb. of the population mean.

Section 8.1 Summary

• Found a point estimate and a margin of error• Constructed and interpreted confidence intervals for

the population mean• Determined the minimum sample size required when

estimating μ


Section 8.2

Estimating when σ is Unknown



• Interpret the t-distribution and use a t-distribution table

• Construct confidence intervals when n < 30, the population is normally distributed, and σ is unknown


The t-Distribution

• When the population standard deviation is unknown, the sample size is less than 30, and the random variable x is approximately normally distributed, it follows a t-distribution.

• Critical values of t are denoted by tc.

• Distribution was discovered in 1908 by Louis Gosset while he worked for the Guinness Brewing Company


𝑡=𝑥−𝜇𝑠

√𝑛

The t-Distribution

• If many samples of size n are drawn, then we get many t-scores using the equation above

• For each value of n, the t-scores can be organized into a frequency table and converted into a histogram

• They resulting graphs are known collectively as the Student’s t distribution


Properties of the t-Distribution

1. The t-distribution is bell shaped and symmetric about the mean.

2. The t-distribution is a family of curves, each determined by a parameter called the degrees of freedom. The degrees of freedom are the number of free choices left after a sample statistic such as is calculated. When you use a t-distribution to estimate a population mean, the degrees of freedom are equal to one less than the sample size. d.f. = n – 1 Degrees of freedom

x


Properties of the t-Distribution

3. The total area under a t-curve is 1 or 100%.

4. The mean, median, and mode of the t-distribution are equal to zero.

5. As the degrees of freedom increase, the t-distribution approaches the normal distribution. After 30 d.f., the t-distribution is very close to the standard normal z-distribution.

t0

Standard normal curve

The tails in the t-distribution are “thicker” than those in the standard normal distribution.d.f. = 5

d.f. = 2


Exercise 1: Critical Values of tAppendix II, Table 6– p. A24

Find the critical value tc for a 95% confidence level when the sample size is 15.

Table 5: t-Distribution

tc = 2.145

Solution: d.f. = n – 1 = 15 – 1 = 14


48

Example: Critical Values of tAppendix II, Table 5(a) – p. A23

Larson/Farber 5th ed

Rounding Convention: t-scores should be rounded to three decimal places

Solution: Critical Values of t

95% of the area under the t-distribution curve with 14 degrees of freedom lies between t = ±2.145.

t

–tc = –2.145 tc = 2.145

c = 0.95


Confidence Intervals for the Population Mean

A c-confidence interval for the population mean μ

•

• The probability that the confidence interval contains μ is c.

where c

sx E x E E t

n


Confidence Intervals and t-Distributions

1. Identify the sample statistics n, , and s.

2. Identify the degrees of freedom, the level of confidence c, and the critical value tc.

3. Find the margin of error E.

xxn

2( )1

x xsn

cE tn

s

d.f. = n – 1

x

In Words In Symbols


Confidence Intervals and t-Distributions

4. Find the left and right endpoints and form the confidence interval.

Left endpoint: Right endpoint: Interval:

x Ex E

x E x E

In Words In Symbols


Exercise 2: Construct a Confidence Interval

You randomly select 16 coffee shops and measure the temperature of the coffee sold at each. The sample mean temperature is 162.0ºF with a sample standard deviation of 10.0ºF. Find the 95% confidence interval for the population mean temperature. Assume the temperatures are approximately normally distributed.

Solution:Use the t-distribution (n < 30, σ is unknown, temperatures are approximately normally distributed).


Solution: Construct a Confidence Interval

• n =16, x = 162.0 s = 10.0 c = 0.95• df = n – 1 = 16 – 1 = 15• Critical Value Table 5: t-Distribution

tc = 2.131





102.131 5.316cE t

n s

162 5.3

156.7

x E

162 5.3

167.3

x E


156.7 < μ < 167.3© 2012 Pearson Education, Inc. All rights reserved. 55 of 83


• 156.7 < μ < 167.3

( )• 162.0156.7 167.3

With 95% confidence, you can say that the population mean temperature of coffee sold is between 156.7ºF and 167.3ºF.

Point estimate

xx E x E


No

Normal or t-Distribution?

Is n ≥ 30?

Is the population normally, or approximately normally, distributed? Cannot use the normal

distribution or the t-distribution. Yes

Is σ known?

No

Use the normal distribution with

If σ is unknown, use s instead.

cE zn

σYes

No

Use the normal distribution with .cE z

n σ

Yes

Use the t-distribution with

and n – 1 degrees of freedom.

cE tn

s


Example: Normal or t-Distribution?

You randomly select 25 newly constructed houses. The sample mean construction cost is $181,000 and the population standard deviation is $28,000. Assuming construction costs are normally distributed, should you use the normal distribution, the t-distribution, or neither to construct a 95% confidence interval for the population mean construction cost?

Solution:Use the normal distribution (the population is normally distributed and the population standard deviation is known)


Exercise 3: Construct Confidence Interval

An archeologist discovered a new, but extinct, species of miniature horse. The only seven known samples show shoulder heights (in cm) of 45.3, 47.1, 44.2, 46.8, 46.5, 45.5, and 47.6.

Find the 99% confidence interval for the mean height of the entire population of ancient horses and the margin of error E. Then summarize your results in a complete sentence relevant to this application.


Exercise 3: Construct Confidence Interval

6

14.6

1.19

= =

Left Endpoint: 46.14 – 1.67 = 44.5

Right Endpoint: 46.14 + 1.67 = 47.8

Confidence Interval


Section 8.2 Summary

• Interpreted the t-distribution and used a t-distribution table

• Constructed confidence intervals when n < 30, the population is normally distributed, and σ is unknown


Section 8.3

Confidence Intervals for Population Proportions



• Find a point estimate for the population proportion• Construct a confidence interval for a population

proportion• Determine the minimum sample size required when

estimating a population proportion


Point Estimate for Population p

Population Proportion• The probability of success in a single trial of a

binomial experiment. • Denoted by p

Point Estimate for p• The proportion of successes in a sample.

=

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Point Estimate for Population p

Point Estimate for q, the population proportion of failures

• Denoted by

Estimate Population Parameter…

with Sample Statistic

Proportion: p p̂


�̂�=1 −�̂�

Example: Point Estimate for p

In a survey of 1000 U.S. adults, 662 said that it is acceptable to check personal e-mail while at work. Find a point estimate for the population proportion of U.S. adults who say it is acceptable to check personal e-mail while at work. (Adapted from Liberty Mutual)

Solution: n = 1000 and r = 662


=

Confidence Intervals for p

A c-confidence interval for a population proportion p •

• The probability that the confidence interval contains p is c.

ˆ ˆwhereˆ ˆ cpqp E p p E E zn


Constructing Confidence Intervals for p

1. Identify the sample statistics n and r.

2. Find the point estimate

3. Verify that the sampling distribution of can be approximated by a normal distribution.

4. Find the critical value zc that corresponds to the given level of confidence c.

Use the Standard Normal Table or technology.

.̂p

5, 5ˆ ˆnp nq p̂

In Words In Symbols


�̂�=𝑟𝑛

Constructing Confidence Intervals for p

5. Find the margin of error E.

6. Find the left and right endpoints and form the confidence interval.

ˆ ˆc

pqE zn

Left endpoint: Right endpoint: Interval:

p̂ Ep̂ E

ˆ ˆp E p p E

In Words In Symbols


Exercise 1: Confidence Interval for p

In a survey of 1000 U.S. adults, 662 said that it is acceptable to check personal e-mail while at work. Construct a 95% confidence interval for the population proportion of U.S. adults who say that it is acceptable to check personal e-mail while at work.

Solution: Recall ˆ 0.662p

1 0.6ˆ ˆ1 62 0.338q p


Solution: Confidence Interval for p

• Verify the sampling distribution of can be approximated by the normal distribution

p̂

1000 0.662 2ˆ 66 5np

1000 0.338 8ˆ 33 5nq


(0.662) (0.ˆ ˆ 338)1.96 0.0291000c

pqE zn




ˆ

0.662 0.029

0.633

p E


0.633 < p < 0.691

ˆ

0.662 0.029

0.691

p E



• 0.633 < p < 0.691

With 95% confidence, you can say that the population proportion of U.S. adults who say that it is acceptable to check personal e-mail while at work is between 63.3% and 69.1%.

Point estimate

p̂p̂ E p̂ E


Exercise 2: Confidence Interval for p

Suppose that 800 students were selected at random from a student body of 20,000 and given flu shots. All 800 students were exposed to the flu, and 600 of them did not get the flu. Let p represent the probability that the shot will be successful for any single student selected at random from the entire population.

Find a 99% confidence interval for p.


Solution: Confidence Interval for pCritical Values

800 r/n 600/800 = 0.75

1 – 0.75 = 0.25 800 (0.75) = 600 [exceeds 5] 800 (0.25) = 200 [exceeds 5] =

Left Endpoint: Right Endpoint: Confidence IntervalInterpretive Statement With 99% confidence, we can say that the

probability that a flu shot will be effective for a student selected at random will be between 71% and 79%.

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75 of 83

Exercise 3: Confidence Interval for p via TI-4

Suppose that 800 students were selected at random from a student body of 20,000 and given flu shots. All 800 students were exposed to the flu, and 600 of them did not get the flu. Let p represent the probability that the shot will be successful for any single student selected at random from the entire population.

Find a 99% confidence interval for p using the 1-PropZinterval function on the TI-84.


Exercise 3: Confidence Interval for p via TI-84 1-PropZInterval Function


Inputs

Output (0.711,0.789)

Margin of Error

Minimum Sample Size

• Given a c-confidence level and a margin of error E, the minimum sample size n needed to estimate p is

• This formula assumes you have an estimate for and .

• If not, use and• Round up to nearest integer using the CEILING

function

2

ˆ ˆ czn pq

E

ˆ 0.5.qˆ 0.5p

p̂q̂


Minimum Sample Size

• Given a c-confidence level and a margin of error E, the minimum sample size n needed to estimate p is

If you have no preliminary estimate for p, use



You are running a political campaign and wish to estimate, with 95% confidence, the population proportion of registered voters who will vote for your candidate. Your estimate must be accurate within 3% of the true population proportion. Find the minimum sample size needed if

a) No preliminary estimate for p is available

b) A preliminary estimate for p is 0.31


Solution: Minimum Sample Size: No Preliminary Estimate for p is available

Critical Values

0.03

=


1068 voters

Interpretive Statement

The sample size should be at least 1068 voters in order to be 95% confident that our estimate is within 3% of the true population proportion.


Solution: Minimum Sample Size: Case: Preliminary Estimate for p is Available

You are running a political campaign and wish to estimate, with 95% confidence, the population proportion of registered voters who will vote for your candidate. Your estimate must be accurate within 3% of the true population proportion. Find the minimum sample size needed if a preliminary estimate gives ˆ 0.31p

Solution: Use the preliminary estimate

1 0.31 0. 9ˆ ˆ 61q p

ˆ 0.31p


Solution: Sample Size

• c = 0.95 zc = 1.96 E = 0.03

2 21.96

(0.31)(0.69) 913.020.

ˆ ˆ03

czn pq

E

Round up to the nearest whole number.

With a preliminary estimate of , the minimum sample size should be at least 914 voters.Need a larger sample size if no preliminary estimate is available.

ˆ 0.31p



You wish to estimate, with 90% confidence and within 2% of the true population, the proportion of adults age 18 to 29 who have high blood pressure. Find the minimum sample size needed if

a) No preliminary estimate for p is available

b) A previous survey found that 4% of adults in this age group had high blood pressure.

Solution: Because you do not have a preliminary estimate for use and ˆ 5.0.q ˆ 0.5p p,̂


Solution: Minimum Sample Size When Preliminary Estimate for p is Unavailable


Critical Values

0.02

=


1692 adults

Interpretive Statement With no preliminary estimate for p, the sample size should be at least 1692 adults in order to be 90% confident that our estimate is within 2% of the true population proportion.

Solution: Minimum Sample Size When Preliminary Estimate for p is Available


Critical Values

0.02


260 adults

Interpretive Statement With a preliminary estimate for p of 4%, the sample size should be at least 260 adults in order to be 90% confident that our estimate is within 2% of the true population proportion

Solution: Minimum Sample Size When Preliminary Estimate for p is known


c) Determine how many more adults are needed when no preliminary estimate is available compared to when a preliminary estimate for p is 4%.

Solution:

Section 8.3 Summary

• Found a point estimate for the population proportion• Constructed a confidence interval for a population

proportion• Determined the minimum sample size required when

estimating a population proportion


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Chapter Estimation 1 of 83 8 © 2012 Pearson Education, Inc. All rights reserved