chapter confidence intervals 1 of 83 6 © 2012 pearson education, inc. all rights reserved

ChapterConfidence Intervals

1 of 83

6

© 2012 Pearson Education, Inc.All rights reserved.

Chapter Outline

• 6.1 Confidence Intervals for the Mean (Large Samples)

• 6.2 Confidence Intervals for the Mean (Small Samples)

• 6.3 Confidence Intervals for Population Proportions

• 6.4 Confidence Intervals for Variance and Standard Deviation

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

Section 6.1

Confidence Intervals for the Mean (Large Samples)


Section 6.1 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 meanx

Estimate Population Parameter…

with Sample Statistic

Mean: μ x


Example: 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)

Enter this in list 1. We will use this for future examples in this section

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.


The problem with a Point Estimate is that the real world probability of hitting that point is virtually zero. -In other words, who has 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

Therefore, we estimate µ using an Interval estimate

• An interval, or range of values, used to estimate a population parameter.


( )

Interval estimate

Right endpoint146.5

Left endpoint115.1

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.


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


Previously- we have calculated this as InvNorm(.05)

Levels of Confidence

• There are 5 standard levels of confidence in statistics:

• 80% z = +- 1.28 (InvNorm(.10)

• 90% z = +- 1.645 (InvNorm(.05)

• 95% z = +- 1.96 (InvNorm(.025)

• 98% z = +- 2.33 (InvNorm(.01)

• 99% z = +- 2.575 (InvNorm(.005)•NOTE: To have a 90% confidence level, we must have an area under the curve = .90•Therefore, our left limit is -1.645, and our right limit is 1.645•When using this Z value as a Z critical value (and not as a left limit/right limit) we only use the POSITIVE Z value

KNOW THESEWrite them downIt will save you time

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

• Therefore the sampling error is hard to calculate –if at all

x

x


Margin of Error

Margin of error • The greatest possible distance between the point estimate

(sample mean) and the value of the parameter (Population mean) it is estimating for a given level of confidence, c.

• Denoted by E.

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

c x cE z zn

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

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Critical Z Score

NOTE: If we had entered the long list of numbers into list 1, we could then calculate S using STAT: 1 –VAR Stats

Example: 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 the sample standard deviation is about 53.0.


zc

Solution: Finding the Margin of Error

• First find the critical values

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


This slide is showing how they got a Z-scoreWe have this on our cheat sheet of scores

Solution: Finding the Margin of Error

53.01.96

4016.4

c c

sE z z

n n

You don’t know σ, but since n ≥ 30, you can use s in place of σ.

You are 95% confident that the margin of error for the population mean is about 16.4 friends.


On a Calculator:Stat TestZintervalChoose StatsEnter ơEnter “0” for x-bar (we don’t know it)Enter nEnter c-LevelCalculate

On the Calculator

• The problem with calculating “The maximum error of Estimate” E is calculating the sample Standard Deviation S (they may not give you a population standard deviation to use)

• This is easily overcome, however, if you make a list, then run 1-var Stats

• You need this to see S the sample deviation• If you make a list, you can then go to Stats TestsZinterval• Here, it asks for input• You can either type in the mean, SD etc. or choose “Data” if you

made a list.• You then choose the confidence level for example .95• It will then tell you the bottom and top value for your interval

estimate. It will also give you the sample means (which is also the point estimate) and the sample standard deviation S

• Nice…

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


Example: Constructing 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.


Example: Constructing a Confidence Interval σ 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


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


Sometimes this is easier than using Z interval…


22.3 < μ < 23.5

( )• 22.922.3 23.5

With 90% confidence, you 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. It is either in the confidence interval or not.

• Incorrect: “There is a 90% probability that the actual mean 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.


Interpreting the Results

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


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

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

• You will need to do this manually. Make a note of this equation.

2

czn

E


Example: 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 sample standard deviation is about 53.0.

In this example, we are setting our margin of error –how close do we want to be- and still maintaining our confidence level


zc

Solution: Sample Size


zc = 1.96

zz = 0 zc

0.95

0.0250.025

–zc = –1.96 zc = 1.96



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. Always round up.

You should include at least 221 users in your sample.


Section 6.1 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


Assignment

• Page 311 5-32 skip 17-20

• Page 312 35-50 skip 47,48

Larson/Farber 5th ed 34

Chapter 6 Quiz 1 ( 5 points each)

A college admissions director wishes to estimate the mean age of all students currently enrolled. In a random sample of 25 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. 1.What is the point estimate of the mean age?

2.What is the Z critical value for 95%?

3.What is the Margin of Error (E) using this data (at 95%)

4.Construct a 95% confidence interval of the population mean age.

35 of 83

Section 6.2

Confidence Intervals for the Mean (Small Samples)



• 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.

t x s

n

38 of 83

Question: When do we use the T-Distribution instead of the Normal Distribution? Answer: When the population σ is unknown and the sample is less than 30. KNOW THIS

How would we figure out if the variable x is approximately normally distributed?

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


the number of degrees of freedom is the number of values in the final calculation of a statistic that are free to vary.

Degrees of Freedom

• In statistics, the number of degrees of freedom is the number of values in the final calculation of a statistic that are free to vary

• Remember, this is calculated as n-1• For every n-value, there is the possibility of variance,

so we account for that• Why n-1 then?• What if you had 25 chairs, and 25 students. As each

student walked in the door, they would have a choice (freedom) of where to sit. Except for the last student. His choice is made. So you have 25-1, or 24, choices

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


T-Distribution Curve

• The tails in a t-distribution are “thicker” than those in the standard normal distribution

T-Distribution Curves

• William S. Gosset developed this curve

• He published it under the pseudo-name “Student” thus –if you Google a T-curve, you may find “Student’s T-Distribution”

Example: Critical Values of t

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


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


Example (using a Calculator)

• Finding the Critical Values of T 2ndVarsinvT(area,D.F.) Area is almost the same as the confidence level –but

you have to add one tail to get (in essence) the right limit, or critical T-value.

For example, find the critical value of T for a 95% confidence when the sample size is 15

Area = .95 + one tail (1+.95)/2 = .975 D.F. = 15-1 = 14 InvT(.975,14) = 2.1447 which is the T-Critical value Therefore, 95% of the area under the t-distribution curve

with 14 degrees of freedom is between -2.145 and 2.145

Chapter 6 Quiz 2

Instructions• Using data from problem 64 on

page 315, do the following:

1. Enter data into list 1

2. Calculate the sample standard deviation (s), the population standard deviation (Ơ) and sample mean (xbar) calc/1-Var Stats

3. Calculate a 95% confidence interval test/Zinterval (be sure to use Ơ since you have it, and input data, not “stats”)

Provide the answers on a piece of paper

1. n=

2. xbar=

3. Ơ= S=_______

4. Z critical score=

5. Margin of Error (E) =

6. Confidence interval=

__________< µ < ________

• Each question worth 5 points


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.

• This should look familiar, it is the SAME EQUATION we used to calculate the margin of Error in a normal distribution, except instead of a Zcritical score, we have a Tcritical score and we use the sample standard deviation

where c

sx E x E E t

n


Same equation except use T instead of Z, and S instead of Ợ

On a Calculator


• On a Calculator:

• Stat TestTinterval

• Choose Stats

• Enter sx

• Enter “0” for x-bar (we don’t know it)

• Enter n

• Enter c-Level

• Calculate

• Chose the positive answer

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



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).



• 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


InvT(.975,15)



• Confidence interval: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. 54 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


Example

• You randomly select 16 restaurants and measure the temperature of the coffee

• The sample mean is 162 degrees (F)

• The sample Standard Deviation (S) = 10 deg.

• Find the 95% confidence interval for the mean temperature. Assume they are normally distributed

• Go to: StatTestTinterval. Choose “Stats” and enter information

• You should get: 156.57---167.33, with a mean of 162

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)


Section 6.2 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


Assignment

• Page 323 1-34 Skip 13-16


Chapter 6 Quiz 3• Using data from problem 24 on page 324, and assuming a

98% confidence level, answer the following:

1. n=

2. s=

3. D.F.=

4. xbar=

5. Tcritical=

6. Confidence interval (based on 98% confidence level)

7. Hint: Use InvT for question 5

8. Hint: Use Tinterval for question 6


5 points each question

Chapter 6 Quiz 4

• Solve question #35 on page 325

1. Calculate the confidence interval indicated in the problem

2. Is the company making acceptable tennis balls?

3. Why or why not?


Section 6.3

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.

• Denoted by read as “p hat”

number of successes in sampleˆ sample sizexpn


Point Estimate for Population p

Point Estimate for q, the population proportion of failures

• Denoted by

• Read as “q hat”

1ˆ ˆq p



Proportion: p p̂


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 x = 662

662 0.662 66.2%1000ˆ xp

n


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 x.

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.

ˆ xpn

Use the Standard Normal Table or technology.

.̂p

5, 5ˆ ˆnp nq p̂

In Words In Symbols


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


Example: 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



• Confidence interval:

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


On a Calculator

• You can calculate a p confidence interval on a calculator

• Stats Tests1PropZInt

• Enter X (the number of successes in the sample)

• Enter n (the size of the sample)

• Enter the confidence level

• This will give you the confidence level, and the p-hat

• It will not give you the Margin of Error (E) should you need that, you gotta go old school


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

2

ˆ ˆ czn pq

E

ˆ 0.5.qˆ 0.5p

p̂q̂


Example: Sample Size

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

1. no preliminary estimate is available.

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



• c = 0.95 zc = 1.96 E = 0.03

2 21.96

(0.5)(0.5) 1067.110.

ˆ03

ˆ czn pq

E

Round up to the nearest whole number.

With no preliminary estimate, the minimum sample size should be at least 1068 voters.


Example: Sample SizeYou 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

2. a preliminary estimate gives .

ˆ 0.31p

Solution: Use the preliminary estimate

1 0.31 0. 9ˆ ˆ 61q p

ˆ 0.31p



• c = 0.95 zc = 1.96 E = 0.03

2 21.96

(0.31)(0.69) 913.020.

ˆ ˆ03

czn pq

E


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


Section 6.3 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


Assignment

• Page 332 11-24


Chapter 6 Quiz 5You are running a political campaign and wish to estimate, with 98% 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.

(hint: solve for n)

You do not have a preliminary point estimate

to the population proportion.

Show equation 5 points for equation, 5 points for solution



• c = 0.98 zc = 2.33 E = 0.03

2 21.96

(0.31)(0.69) 913.020.

ˆ ˆ03

czn pq

E


Using the standard value of .50 for p-hat, and .50 for q-hat, the minimum sample size should be at least 1509 voters.


2.33.50 .50 1508.02

Section 6.4

Confidence Intervals for Variance and Standard Deviation



• Interpret the chi-square distribution and use a chi-square distribution table

• Use the chi-square distribution to construct a confidence interval for the variance and standard deviation


The Chi-Square Distribution

• The point estimate for σ2 is s2

• The point estimate for σ is s

• s2 is the most unbiased estimate for σ2



Variance: σ2 s2

Standard deviation: σ s


The Chi-Square Distribution

• You can use the chi-square distribution to construct a confidence interval for the variance and standard deviation.

• If the random variable x has a normal distribution, then the distribution of

forms a chi-square distribution for samples of any size n > 1.

22

2( 1)n s

σ


Properties of The Chi-Square Distribution

1. All chi-square values χ2 are greater than or equal to zero.

2. The chi-square distribution is a family of curves, each determined by the degrees of freedom. To form a confidence interval for σ2, use the χ2-distribution with degrees of freedom equal to one less than the sample size.

• d.f. = n – 1 Degrees of freedom

3. The area under each curve of the chi-square distribution equals one.


Properties of The Chi-Square Distribution

4. Chi-square distributions are positively skewed.

Chi-square Distributions


• There are two critical values for each level of confidence.

• The value χ2R represents the right-tail critical value

• The value χ2L represents the left-tail critical value.

Critical Values for χ2

The area between the left and right critical values is c.

χ2

c

12

c

12

c

2L 2

R


Example: Finding Critical Values for χ2

Find the critical values and for a 95% confidence interval when the sample size is 18.

Solution:• d.f. = n – 1 = 18 – 1 = 17 d.f.

• Area to the right of χ2R =

1 0.95 0.0252

12

c

• Area to the right of χ2L =

1 0.95 0.9752

12

c

2L2

R

• Each area in the table represents the region under the chi-square curve to the right of the critical value.


Use Table 6 on A-19

Solution: Finding Critical Values for χ2

Table 6: χ2-Distribution

2R 2

L

95% of the area under the curve lies between 7.564 and 30.191.

30.191 7.564


Confidence Interval for σ:

•

Confidence Intervals for σ2 and σ

2 2

2 2( 1) ( 1)

R L

n s n s 2σ

• The probability that the confidence intervals contain σ2 or σ is c.

Confidence Interval for σ2:

•

2 2

2 2( 1) ( 1)

R L

n s n s σ


Confidence Intervals for σ2 and σ

1. Verify that the population has a normal distribution.

2. Identify the sample statistic n and the degrees of freedom.

3. Find the point estimate s2.

4. Find the critical values χ2R and χ2

L that correspond to the given level of confidence c.

Use Table 6 in Appendix B.

22 )

1x xsn

(

d.f. = n – 1

In Words In Symbols


Confidence Intervals for 2 and

5. Find the left and right endpoints and form the confidence interval for the population variance.

6. Find the confidence interval for the population standard deviation by taking the square root of each endpoint.

2 2

2 2( 1) ( 1)

R L

n s n s 2σ

2 2

2 2( 1) ( 1)

R L

n s n s σ

In Words In Symbols



You randomly select and weigh 30 samples of an allergy medicine. The sample standard deviation is 1.20 milligrams. Assuming the weights are normally distributed, construct 99% confidence intervals for the population variance and standard deviation.

Solution:• d.f. = n – 1 = 30 – 1 = 29 d.f.



• The critical values are χ2

R = 52.336 and χ2L = 13.121

• Area to the right of χ2R =

1 0.99 0.0052

12

c

• Area to the right of χ2L =

1 0.99 0.9952

12

c



2

22 (30 1)(1.20) 0.8052.336

( 1)

R

n s

Confidence Interval for σ2:

2

22 (30 1)(1.20) 3.1813.121

( 1)

L

n s

Left endpoint:

Right endpoint:

0.80 < σ2 < 3.18With 99% confidence, you can say that the population variance is between 0.80 and 3.18.



2 2(30 1)(1.20) (30 1)(1.20)52.336 13.121

Confidence Interval for σ :

0.89 < σ < 1.78With 99% confidence, you can say that the population standard deviation is between 0.89 and 1.78 milligrams.

2 2

2 2( 1) ( 1)

R L

n s n s σ


Section 6.4 Summary

• Interpreted the chi-square distribution and used a chi-square distribution table

• Used the chi-square distribution to construct a confidence interval for the variance and standard deviation


Assignment

• Page 341 3-12


chapter confidence intervals 1 of 83 6 © 2012 pearson education, inc. all rights reserved

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