Chapter 7 Confidence Intervals and Sample Size © McGraw-Hill, Bluman, 5 th ed., Chapter 7 1.

Chapter 7

Confidence Intervals and Sample Size

Chapter 7 Overview Introduction

7-1 Confidence Intervals for the Mean When Is Known and Sample Size

7-2 Confidence Intervals for the Mean When Is Unknown

7-3 Confidence Intervals and Sample Size for Proportions

7-4 Confidence Intervals and Sample Size for Variances and Standard Deviations

Bluman, Chapter 7 2

Chapter 7 Objectives1. Find the confidence interval for the mean when

is known.

2. Determine the minimum sample size for finding a confidence interval for the mean.

3. Find the confidence interval for the mean when is unknown.

4. Find the confidence interval for a proportion.

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Chapter 7 Objectives5. Determine the minimum sample size for

finding a confidence interval for a proportion.

6. Find a confidence interval for a variance and a standard deviation.

Bluman, Chapter 7 4

7.1 Confidence Intervals for the Mean When Is Known and Sample Size

A point estimate is a specific numerical value estimate of a parameter.

The best point estimate of the population mean µ is the sample mean

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Three Properties of a Good Estimator

1. The estimator should be an unbiased estimatorunbiased estimator. That is, the expected value or the mean of the estimates obtained from samples of a given size is equal to the parameter being estimated.

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2. The estimator should be consistent. For a consistent consistent estimatorestimator, as sample size increases, the value of the estimator approaches the value of the parameter estimated.

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3. The estimator should be a relatively efficient estimatorrelatively efficient estimator; that is, of all the statistics that can be used to estimate a parameter, the relatively efficient estimator has the smallest variance.

Bluman, Chapter 7 8

Confidence Intervals for the Mean When Is Known and Sample Size

An interval estimate interval estimate of a parameter is an interval or a range of values used to estimate the parameter.

This estimate may or may not contain the value of the parameter being estimated.

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Confidence Level of an Interval Estimate

The confidence levelconfidence level of an interval estimate of a parameter is the probability that the interval estimate will contain the parameter, assuming that a large number of samples are selected and that the estimation process on the same parameter is repeated.

Bluman, Chapter 7 10

Confidence Interval

A confidence interval confidence interval is a specific interval estimate of a parameter determined by using data obtained from a sample and by using the specific confidence level of the estimate.

Formula for the Confidence Interval of the Mean for a Specific

/ 2 / 2X z X zn n

For a 90% confidence interval: / 2 1.65z

/ 2 1.96z

/ 2 2.58z For a 99% confidence interval:

For a 95% confidence interval:

95% Confidence Interval of the Mean

Maximum Error of the Estimate

/ 2E zn

The maximum error of the estimate is the maximum likely difference between the point estimate of a parameter and the actual value of the parameter.

Rounding Rule

When you are computing a confidence interval for a population mean by using raw data, round off to one more decimal place than the number of decimal places in the original data.

When you are computing a confidence interval for a population mean by using a sample mean and a standard deviation, round off to the same number of decimal places as given for the mean.

Confidence Interval for a Mean

Chapter 7Confidence Intervals and Sample Size

Section 7-1Example 7-1

Page #358

Example 7-1: Days to Sell an AveoA researcher wishes to estimate the number of days it takes an automobile dealer to sell a Chevrolet Aveo. A sample of 50 cars had a mean time on the dealer’s lot of 54 days. Assume the population standard deviation to be 6.0 days. Find the best point estimate of the population mean and the 95% confidence interval of the population mean.

The best point estimate of the mean is 54 days.

54, 6.0, 50,95% 1.96 X s n z

X z X zn n

Example 7-1: Days to Sell an Aveo

One can say with 95% confidence that the interval between 52 and 56 days contains the population mean, based on a sample of 50 automobiles.

54, 6.0, 50,95% 1.96 X s n z

6.0 6.054 1.96 54 1.96

X z X zn n

54 1.7 54 1.7 52.3 55.7

Page #358

Example 7-2: Ages of AutomobilesA survey of 30 adults found that the mean age of a person’s primary vehicle is 5.6 years. Assuming the standard deviation of the population is 0.8 year, find the best point estimate of the population mean and the 99% confidence interval of the population mean.

The best point estimate of the mean is 5.6 years.

One can be 99% confident that the mean age of all primary vehicles is between 5.2 and 6.0 years, based on a sample of 30 vehicles.

0.8 0.85.6 2.58 5.6 2.58

5.2 6.0

One can be 95% confident that an interval built around a specific sample mean would contain the population mean.

Finding for 98% CL.

2 2.33z

Example 1

A sample of the reading scores of 35 fifth-graders has a mean of 82. The standard deviation of the sample is 15.

a. Find the 95% confidence interval of the mean reading scores of all fifth-graders.

b. Find the 99% confidence interval of the mean reading scores of all fifth-graders.

c. Which interval is larger? Explain why.

15 1582 – (1.96) < < 82 + (1.96)

82 – 4.97< < 82 + 4.97

77 < < 87

X = 82 n = 35 s = 15

a. Find the 95% confidence interval of mean reading scores of all fifth-graders.

2 2– < < +s s

X z X zn n

77 < < 87

X = 82 n = 35 s = 15

Approximately 95% of the sample

means will fall within 1.96

standard errors of the population mean,

so use = 1.96.z

a. Find the 95% confidence interval of mean reading scores of all fifth-graders.

82 – (2.58)

< m < 82 + (2.58)15

82 – 6.54< < 82 + 6.54

75< < 89

X = 82 n = 35 s = 15

2 2– < < +s s

X z X zn n

75< < 89

X = 82 n = 35 s = 15

Approximately 99% of the sample

means will fall within 2.58 standard

errors of the population mean,

so use = 2.58 .z

The 99% confidence interval is larger because the confidence level is larger.

c. Which interval is larger? Explain why.

95% confidence level = 77< < 87

99% confidence level = 75< < 89

Example 2

A study of 40 English composition professors showed that they spent, on average, 12.6 minutes correcting a student’s term paper.

a. Find the 90% confidence interval of the mean time for all composition papers when = 2.5 minutes.

b. If a professor stated that he spent, on average, 30 minutes correcting a term paper, what would be your reaction?

a. Find the 90% confidence interval of the mean time for all composition papers when = 2.5 minutes.

12.6 – (1.65)

< < 12.6 + (1.65)2.5

12.6 – 0.652< < 12.6 + 0.652

11.9< < 13.3

X = 12.6 n = 40

2 2– < < +s s

X z X zn n

b. If a professor stated that he spent, on average, 30 minutes correcting a term paper, what would be your reaction?

11.9 < < 13.3

It would be highly unlikely since this is far larger than 13.3 minutes.

Page #360

Example 7-3: Credit Union AssetsThe following data represent a sample of the assets (in millions of dollars) of 30 credit unions in southwestern Pennsylvania. Find the 90% confidence interval of the mean.

12.23 16.56 4.39 2.89 1.24 2.1713.19 9.16 1.4273.25 1.91 14.6411.59 6.69 1.06 8.74 3.17 18.13 7.92 4.78 16.8540.22 2.42 21.58 5.01 1.47 12.24 2.27 12.77 2.76

Example 7-3: Credit Union Assets

Step 1: Find the mean and standard deviation. Using technology, we find = 11.091 and s = 14.405.

Step 2: Find α/2. 90% CL α/2 = 0.05.

Step 3: Find zα/2. 90% CL α/2 = 0.05 z.05 = 1.65

Table E

The Standard Normal Distribution

z .00 … .04 .05 … .09

0.00.1

1.6 0.9495 0.9505

Example 7-3: Credit Union Assets

Step 4: Substitute in the formula.

One can be 90% confident that the population mean of the assets of all credit unions is between $6.752 million and $15.430 million, based on a sample of 30 credit unions.

14.405 14.40511.091 1.65 11.091 1.65

X z X zn n

11.091 4.339 11.091 4.339 6.752 15.430

This chapter and subsequent chapters include examples using raw data. If you are using computer or calculator programs to find the solutions, the answers you get may vary somewhat from the ones given in the textbook.

This is so because computers and calculators do not round the answers in the intermediate steps and can use 12 or more decimal places for computation. Also, they use more exact values than those given in the tables in the back of this book.

These discrepancies are part and parcel of statistics.

Technology Note

where E is the maximum error of estimate. If necessary, round the answer up to obtain a whole number. That is, if there is any fraction or decimal portion in the answer, use the next whole number for sample size n.

Formula for Minimum Sample Size Needed for an Interval Estimate of the Population Mean

Page #362

Example 7-4: Depth of a RiverA scientist wishes to estimate the average depth of a river. He wants to be 99% confident that the estimate is accurate within 2 feet. From a previous study, the standard deviation of the depths measured was 4.38 feet.

Therefore, to be 99% confident that the estimate is within 2 feet of the true mean depth, the scientist needs at least a sample of 32 measurements.

99% 2.58, 2, 4.38 z E 2

2.58 4.38

31.92 32

An insurance company is trying to estimate the average number of sick days that full-time food service workers use per year. A pilot study found the standard deviation to be 2.5 days. How large a sample must be selected if the company wants to be95% confident of getting an interval that contains the true mean with a maximum error of 1 day?

s = 2.5

confidence level = 95%

maximum error = 1 day

Example 3

= (1.96)(2.5)

s = 2.5

confidence level = 95%

maximum error = 1 day

= (4.9)2

n = 24.01 n = 25 workers

7.2 Confidence Intervals for the Mean When Is Unknown

The value of , when it is not known, must be estimated by using s, the standard deviation of the sample.

When s is used, especially when the sample size is small (less than 30), critical values greater than the values for are used in confidence intervals in order to keep the interval at a given level, such as the 95%.

These values are taken from the Student t distributionStudent t distribution, most often called the t distributiont distribution.

Characteristics of the t DistributionThe t distribution is similar to the standard normal distribution in these ways:

1. It is bell-shaped.

2. It is symmetric about the mean.

3. The mean, median, and mode are equal to 0 and are located at the center of the distribution.

4. The curve never touches the x axis.

Characteristics of the t DistributionThe t distribution differs from the standard normal distribution in the following ways:

1. The variance is greater than 1.

2. The t distribution is actually a family of curves based on the concept of degrees of freedom, which is related to sample size.

3. As the sample size increases, the t distribution approaches the standard normal distribution.

Degrees of Freedom The symbol d.f. will be used for degrees of degrees of

freedomfreedom. The degrees of freedom for a confidence

interval for the mean are found by subtracting 1 from the sample size. That is, d.f. = n - 1.

Note: For some statistical tests used later in this book, the degrees of freedom are not equal to n - 1.

The degrees of freedom are n - 1.

Formula for a Specific Confidence Interval for the Mean When IsUnknown and n < 30

s sX t X t

Page #369

Find the tα/2 value for a 95% confidence interval when the sample size is 22.

Degrees of freedom are d.f. = 21.

Example 7-5: Using Table F

Page #370

Ten randomly selected people were asked how long they slept at night. The mean time was 7.1 hours, and the standard deviation was 0.78 hour. Find the 95% confidence interval of the mean time. Assume the variable is normally distributed.

Since is unknown and s must replace it, the t distribution (Table F) must be used for the confidence interval. Hence, with 9 degrees of freedom, tα/2 = 2.262.

Example 7-6: Sleeping Time

s sX t X t

0.78 0.787.1 2.262 7.1 2.262

One can be 95% confident that the population mean is between 6.54 and 7.66 inches.

Example 7-6: Sleeping Time

0.78 0.787.1 2.262 7.1 2.262

7.1 0.56 7.1 0.56

6.54 7.66

Page #370

The data represent a sample of the number of home fires started by candles for the past several years. Find the 99% confidence interval for the mean number of home fires started by candles each year.

5460 5900 6090 6310 7160 8440 9930

Step 1: Find the mean and standard deviation. The mean is = 7041.4 and standard deviation s = 1610.3.

Step 2: Find tα/2 in Table F. The confidence level is 99%, and the degrees of freedom d.f. = 6

t .005 = 3.707.

Example 7-7: Home Fires by Candles

Step 3: Substitute in the formula.

One can be 99% confident that the population mean number of home fires started by candles each year is between 4785.2 and 9297.6, based on a sample of home fires occurring over a period of 7 years.

1610.3 1610.37041.4 3.707 7041.4 3.707

s sX t X t

7041.4 2256.2 7041.4 2256.2

4785.2 9297.6

A state representative wishes to estimate the mean number of women representatives per state legislature. A random sample of 17 states is selected, and the number ofwomen representatives is shown.Based on the sample, what is the point estimate of the mean? Find the 90% confidence interval of the mean population. (Note: The population mean is actually 31.72, or about 32.) Compare this value to the point estimate and the confidence interval. There is something unusual about the data. Describe it and state how it would affect the confidence interval.

Example 4

21.2< < 45.6

X = 33.4 s = 28.7

28.7 28.733.4 – 1.746 < < 33.4 + 1.746

33.4 – 12.2< < 33.4 + 12.2

5 33 35 37 2431 16 45 19 1318 29 15 39 1858 132

– < < + 2 2s s

X Xn nt t

Example 5

A recent study of 28 employees of XYZ Company showed that the mean of the distance they traveled to work was 14.3 miles. The standard deviation of the sample mean was 2 miles. Find the 95% confidence interval of the true mean.

14.3 – 0.776< < 14.3 + 0.776

14.3 – 2.052 < < 14.3 + 2.05228 28

– < < + 2 2s s

X Xn nt t

13.5< <15.1

14.3 – 0.776< < 14.3 + 0.776

14.3 – 2.052 < < 14.3 + 2.05228 28

– < < + 2 2s s

X Xn nt t

If a manager wanted to be sure that most of his employees would not be late, how much time would he suggest they allow for the commute if the average speed is 30 miles per hour?

14.3 mi

30 mi/hr

2 hour

about 30 minutes

Example 6

The average yearly income for 28 community college instructors was $56,718. The standard deviation was $650. Find the 95% confidence interval of the

true mean.

– < < + 2 2s s

X Xn nt t

$56,466< < $56,970

– < < + 2 2s s

X Xn nt t

650 650$56, 718 – 2.052 < < $56, 718 + 2.052

If a faculty member wishes to see if he or she is being paid below average, what salary value should he or she use?

Use the lower bound of theconfidence interval: $56,466

7.3 Confidence Intervals and Sample Size for Proportions

p = population proportion

(read p “hat”) = sample proportion

For a sample proportion,

where X = number of sample units that possess the characteristics of interest and n = sample size.

ˆ ˆ ˆ ˆand 1

p q or q pn n

Page #376

In a recent survey of 150 households, 54 had central air conditioning. Find and , where is the proportion of households that have central air conditioning.

Since X = 54 and n = 150,

Example 7-8: Air Conditioned Households

54ˆ 0.36 36%

ˆ ˆ1 1 0.36 0.64 64% q p

p̂ q̂ p̂

when np 5 and nq 5.

Formula for a Specific Confidence Interval for a Proportion

ˆ ˆ ˆ ˆˆ ˆ

pq pqp z p p z

Rounding Rule: Round off to three decimal places.

Page #376

A sample of 500 nursing applications included 60 from men. Find the 90% confidence interval of the true proportion of men who applied to the nursing program.

Example 7-9: Male Nurses

60 500 0.12 p X n ˆ, 0.88q

0.12 0.88 0.12 0.880.12 1.65 0.12 1.65

500 500 p

.096 0.144 p

0.12 0.024 0.12 0.024 p

You can be 90% confident that the percentage of applicants who are men is between 9.6% and 14.4%.

ˆ ˆ ˆ ˆˆ ˆ

pq pqp z p p z

Page #377

A survey of 1721 people found that 15.9% of individuals purchase religious books at a Christian bookstore. Find the 95% confidence interval of the true proportion of people who purchase their religious books at a Christian bookstore.

Example 7-10: Religious Books

0.159 0.841 0.159 0.8410.159 1.96 0.159 1.96

1721 1721 p

0.142 0.176 p

You can say with 95% confidence that the true percentage is between 14.2% and 17.6%.

ˆ ˆ ˆ ˆˆ ˆ

pq pqp z p p z

The proportion of students in private schools is around 11%. A random sample of 450 students from a wide geographic area indicated that 55 attended private schools. Estimate the true proportion of students attending private schools with 95% confidence. How does your estimate compare to 11%?

ˆ ˆ ˆ ˆ

ˆ ˆ– < < + 2 2pq pq

p p pn n

= 55450

= 0.12 ̂q = 1– 0.12 = 0.88

Example 7

ˆ ˆ ˆ ˆ

ˆ ˆ– < < + 2 2pq pq

p p pn n

55450 = 0.12 q̂ = 1– 0.12 = 0.88

11% is contained in the confidence interval.

0.09 < p < 0.15

0.12 – 0.03 < p < 0.12 + 0.03

0.12 – 1.96

(0.12)(0.88)450

< p < 0.12 + 1.96(0.12)(0.88)

9% < p < 15% or

A survey found that out of 200 workers, 168 said they were interrupted three or more times an hour by phone messages, faxes, etc. Find the 90% confidence interval of the population proportion of workers who are interrupted three or more times an hour.

p̂ = 0.84 q̂ = 0.16

ˆ ˆ ˆ ˆ

ˆ ˆ– < < + 2 2pq pq

p p pn n

Example 8

0.797< p < 0.883

0.84 – 1.65

(0.84)(0.16)200 < p < 0.84 + 1.65

(0.84)(0.16)200

0.84 – 0.043 < p < 0.84 + 0.043

p̂ = 0.84 q̂ = 0.16

ˆ ˆ ˆ ˆ

ˆ ˆ– < < + 2 2pq pq

p p pn n

A study by the University of Michigan found that one in five 13 and 14-year-olds is a sometime smoker. To see how the smoking rate of the students at a large school district compared to the national rate, the superintendent surveyed two hundred 13 and 14-year-old students and found that 23% said they were sometime smokers. Find the 99% confidence interval of the true proportion and compare this with the University of Michigan’s study.

Example 9

Find the 99% confidence interval of the true proportion.

q̂ = 1– 0.23 = 0.77

ˆ ˆ ˆ ˆ

ˆ ˆ 2 2– < < + pq pq

p p pn n

n = 200 ̂p = 0.23

<0.23 + 2.58

(0.23)(0.77)200

0.23 – 2.58(0.23)(0.77)

200 < p

0.23 – 0.077 < p < 0.23 + 0.077

0.153 < p < 0.307

5 = 0.20

The University of Michigan study falls within this confidence interval.

A study by the University of Michigan found that one in five 13- and 14-year-olds is a sometime smoker. Find the 99% confidence interval of the true proportion and compare this with the University of Michigan’s study.

If necessary, round up to the next whole number.

Formula for Minimum Sample Size Needed for Interval Estimate of a Population Proportion

2ˆ ˆ

Page #378

A researcher wishes to estimate, with 95% confidence, the proportion of people who own a home computer. A previous study shows that 40% of those interviewed had a computer at home. The researcher wishes to be accurate within 2% of the true proportion. Find the minimum sample size necessary.

Example 7-11: Home Computers

1.960.40 0.60

2304.96

The researcher should interview a sample of at least 2305 people.

2ˆ ˆ

Page #378

The same researcher wishes to estimate the proportion of executives who own a car phone. She wants to be 90% confident and be accurate within 5% of the true proportion. Find the minimum sample size necessary.

Since there is no prior knowledge of , statisticians assign the values = 0.5 and = 0.5. The sample size obtained by using these values will be large enough to ensure the specified degree of confidence.

Example 7-12: Car Phone Ownership

1.650.50 0.50

272.25

The researcher should ask at least 273 executives.

2ˆ ˆ

p̂p̂ q̂

A medical researcher wishes to determine the percentage of females who take vitamins. He wishes to be 99% confident that the estimate is within 2 percentage points of the true proportion. A recent study of 180 females showed that 25% took vitamins.

a. How large should the sample size be?b. If no estimate of the sample proportion is available, how large should the sample be?

Example 10

p̂ = 0.25

n = 3120.187

q̂ = 0.75

n = 3121 or

n = p̂q̂

2 58 = 0.25 0 75

a. How large should the sample size be?

n = 4160.25 n = 4161 or

= 0.5 0.50.02

2/ 2ˆ ˆ =

b. If no estimate of the sample proportion is available, how large should the sample be?

p̂ = 0.5 q̂ = 0.5

Since there is no prior knowledge of

p or q, assign the values

q̂ = 0.5 and p̂ = 0.5.

7-4 Confidence Intervals for Variances and Standard Deviations

When products that fit together (such as pipes) are manufactured, it is important to keep the variations of the diameters of the products as small as possible; otherwise, they will not fit together properly and will have to be scrapped.

In the manufacture of medicines, the variance and standard deviation of the medication in the pills play an important role in making sure patients receive the proper dosage.

For these reasons, confidence intervals for variances and standard deviations are necessary.

Chi-Square Distributions The chi-square distribution must be used to calculate

confidence intervals for variances and standard deviations.

The chi-square variable is similar to the t variable in that its distribution is a family of curves based on the number of degrees of freedom.

The symbol for chi-square is (Greek letter chi, pronounced “ki”).

A chi-square variable cannot be negative, and the distributions are skewed to the right.

Chi-Square Distributions

At about 100 degrees of freedom, the chi-square distribution becomes somewhat symmetric.

The area under each chi-square distribution is equal to 1.00, or 100%.

Formula for the Confidence Interval for a Variance

2 2right left

1 1, d.f. = 1

n s n sn

Formula for the Confidence Interval for a Standard Deviation

2 2right left

1 1, d.f. = 1

n s n sn

Page #385

Find the values for and for a 90% confidence interval when n = 25.

Example 7-13: Using Table G

To find , subtract 1 - 0.90 = 0.10. Divide by 2 to get 0.05.To find , subtract 1 - 0.05 to get 0.95.

2left2

2right2left

Use the 0.95 and 0.05 columns and the row corresponding to 24 d.f. in Table G.

Example 7-13: Using Table G

2 2right leftThe value is 36.415; the value is 13.848.

Rounding Rule

When you are computing a confidence interval for a population variance or standard deviation by using raw data, round off to one more decimal places than the number of decimal places in the original data.

When you are computing a confidence interval for a population variance or standard deviation by using a sample variance or standard deviation, round off to the same number of decimal places as given for the sample variance or standard deviation.

Confidence Interval for a Variance or Standard Deviation

Page #387

Find the 95% confidence interval for the variance and standard deviation of the nicotine content of cigarettes manufactured if a sample of 20 cigarettes has a standard deviation of 1.6 milligrams.

Example 7-14: Nicotine Content

To find , subtract 1 - 0.95 = 0.05. Divide by 2 to get 0.025.

To find , subtract 1 - 0.025 to get 0.975.

In Table G, the 0.025 and 0.975 columns with the d.f. 19 row yield values of 32.852 and 8.907, respectively.

2right

Example 7-14: Nicotine Content

2 2right left

n s n s

21.5 5.5

1.5 5.5

1.2 2.3

You can be 95% confident that the true variance for the nicotine content is between 1.5 and 5.5 milligrams.

219 1.6 19 1.6

32.852 8.907

You can be 95% confident that the true standard deviation is between 1.2 and 2.3 milligrams.

Page #387

Find the 90% confidence interval for the variance and standard deviation for the price in dollars of an adult single-day ski lift ticket. The data represent a selected sample of nationwide ski resorts. Assume the variable is normally distributed.

59 54 53 52 51

39 49 46 49 48

Example 7-15: Cost of Ski Lift Tickets

Using technology, we find the variance of the data is s2=28.2.

In Table G, the 0.05 and 0.95 columns with the d.f. 9 row yield values of 16.919 and 3.325, respectively.

Example 7-15: Cost of Ski Lift Tickets

2 2right left

n s n s

215.0 76.3

15.0 76.3

3.87 8.73

You can be 95% confident that the true variance for the cost of ski lift tickets is between 15.0 and 76.3.

29 28.2 9 28.2

16.919 3.325

You can be 95% confident that the true standard deviation is between $3.87 and $8.73.

Example 11

Find the 90% confidence interval for the variance and standard deviation for the time it takes a customer to place a telephone order with a large catalog company is a sample of 23 telephone orders has a standard deviation of 3.8 minutes. Assume the variable is normally distributed.

22 3.822 3.8

12.338< <

33.924

3.06 5.07< <

29.36 < < 25.75

(n – 1)s2

right2

< 2 <(n – 1)s

Example 12

Find the 99% confidence interval for the variance and standard deviation of the weights of 25 one-gallon containers of motor oil if a sample of 14 containers has a variance of 3.2. Assume the variable is normally distributed.

Find the 99% confidence interval for the variance and standard deviation.

13(3.2)29.819 < 2 <

13(3.2)3.565

1.2 < < 3.4

1.4 < 2 < 11.7

(n – 1)s2

right2

< 2 < (n – 1)s2

n = 14 s2 = 3.2

Example 13

The number of calories in a 1-ounce serving of various kinds of regular cheese is shown. Estimate the population variance and standard deviation with 90% confidence.

110 45 100 95 110

110 100 110 95 120

130 100 80 105 105

90 110 70 125 108

15.2 < < 26.3

19(19.1913)2

30.144 < 2 <

19(19.1913)2

10.117

232.1 < 2 < 691.7

(n – 1)s2

right2

< 2 < (n – 1)s2

Example 14

A service station advertises that customers will have to wait no more than 30 minutes for an oil change. A sample of 28 oil changes has a standard deviation of 5.2 minutes. Find the 95% confidence interval of the population standard deviation of the time spent waiting for an oil change.

27(5.2)2

43.194 < 2 <

27(5.2)2

14.573

16.9 < 2 < 50.1

4.1 < < 7.1

(n – 1)s2

right2

< 2 < (n – 1)s2

Chapter 7 Confidence Intervals and Sample Size © McGraw-Hill, Bluman, 5 th ed., Chapter 7 1.

sample mean bluman

confidence intervals

specific interval

specific confidence

sample sizean interval

sample sizea point estimate

minimum sample size

sample size increases

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