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Ch. 6 Confidence Intervals

6.1 Confidence Intervals for the Mean (Large Samples)

1 Find a Critical Value

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Provide an appropriate response.

1) Find the critical value zc that corresponds to a 95% confidence level.

A) ±1.96 B) ±2.575 C) ±2.33 D) ±1.645

2 Find the Margin of Error



2) Determine the sampling error if the grade point averages for 10 randomly selected students from a class of 125

students has a mean of x = 1.8. Assume the grade point average of the 125 students has a mean of

μ = 2.4.

A) 0.6 B) 2.1 C) -0.6 D) 1.5

3) A random sample of 120 students has a test score average with a standard deviation of 9.2. Find the margin oferror if c = 0.98.

A) 1.96 B) 0.18 C) 0.84 D) 0.82

4) A random sample of 150 students has a grade point average with a standard deviation of 0.78. Find the margin

of error if c = 0.98.

A) 0.15 B) 0.08 C) 0.11 D) 0.12

5) A random sample of 40 students has a mean annual earnings of $3120 and a standard deviation of $677. Find

the margin of error if c = 0.95.

A) $210 B) $77 C) $2891 D) $7

3 Construct

and

Interpret

Confidence

Intervals

for

the

Population

Mean



6) A random sample of 150 students has a grade point average with a mean of 2.86 and with a standard deviation

of 0.78. Construct the confidence interval for the population mean, μ , if c = 0.98.

A) (2.71, 3.01) B) (2.51, 3.53) C) (2.43, 3.79) D) (2.31, 3.88)

7) A random sample of 40 students has a test score with x = 81.5 and s = 10.2. Construct the confidence interval

for the population mean, μ if c = 0.90.

A) (78.8, 84.2) B) (51.8, 92.3) C) (66.3, 89.1) D) (71.8, 93.5)

8) A random sample of 40 students has a mean annual earnings of $3120 and a standard deviation of $677.

Construct the confidence interval for the population mean, μ if c = 0.95.

A) ($2910, $3330) B) ($210, $110) C) ($4812, $5342) D) ($1987, $2346)

9) A random sample of 56 fluorescent light bulbs has a mean life of 645 hours with a standard deviation of 31

hours. Construct a 95% confidence interval for the population mean.

A) (636.9, 653.1) B) (539.6, 551.2) C) (112.0, 118.9) D) (712.0, 768.0)

18

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10) A group of 49 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8.

Construct a 98% confidence interval for the population mean.

A) (21.1, 23.7) B) (20.3, 24.5) C) (19.8, 25.1) D) (18.8, 26.3)

11) A group of 40 bowlers showed that their average score was 192 with a standard deviation of 8. Find the 95%

confidence interval of the mean score of all bowlers.

A) (189.5, 194.5) B) (186.5, 197.5) C) (188.5, 195.6) D) (187.3, 196.1)

12) In a random sample of 60 computers, the mean repair cost was $150 with a standard deviation of $36.

Construct a 90% confidence interval for the population mean.A) ($142, $158) B) ($138, $162) C) ($141, $159) D) ($537, $654)

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

13) In a random sample of 60 computers, the mean repair cost was $150 with a standard deviation of $36.

a) Construct the 99% confidence interval for the population mean repair cost.

b) If the level of confidence was lowered to 95%, what will be the effect on the confidence interval?


14) In a recent study of 42 eighth graders, the mean number of hours per week that they watched television was

19.6 with a standard deviation of 5.8 hours. Find the 98% confidence interval for the population mean.

A) (17.5, 21.7) B) (14.1, 23.2) C) (18.3, 20.9) D) (19.1, 20.4)


15) In a recent study of 54 eighth graders, the mean number of hours per week that they watched television was

19.5 with a standard deviation of 5.1 hours.

a) Find the 98% confidence interval of the mean.

b) If the standard deviation is doubled to 10.2, what will be the effect on the confidence interval?


16) In a sample of 10 randomly selected women, it was found that their mean height was 63.4 inches. From

previous studies, it is assumed that the standard deviation σ is 2.4 and that the population of height

measurements is normally distributed. Construct the 95% confidence interval for the population mean.

A) (61.9, 64.9) B) (58.1, 67.3) C) (59.7, 66.5) D) (60.8, 65.4)


17) In a sample of 10 randomly selected women, it was found that their mean height was 63.4 inches. From

previous studies, it is assumed that the standard deviation, σ , is 2.4 inches and that the population of height

measurements is normally distributed.

a) Construct the 99% confidence interval for the population mean height of women.

b) If the sample size was doubled to 20 women, what will be the effect on the confidence interval?

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18) The numbers of advertisements seen or heard in one week for 30 randomly selected people in the United States

are listed below. Construct a 95% confidence interval for the true mean number of advertisements.

598 494 441 595 728 690 684 486 735 808

481 298 135 846 764 317 649 732 582 677

734 588 590 540 673 727 545 486 702 703

19) The number of wins in a season for 32 randomly selected professional football teams are listed below.

Construct a 90% confidence interval for the true mean number of wins in a season.

9 9 9 8 10 9 7 2

11 10 6 4 11 9 8 8

12 10 7 5 12 6 4 3

12 9 9 7 10 7 7 5

4 Determine the Minimum Sample Size



20) The standard IQ test has a mean of 101 and a standard deviation of 16. We want to be 98% certain that we are

within 4 IQ points of the true mean. Determine the required sample size.

A) 87 B) 10 C) 188 D) 1

21) A nurse at a local hospital is interested in estimating the birth weight of infants. How large a sample must she

select if she desires to be 99% confident that the true mean is within 2 ounces of the sample mean? The

standard deviation of the birth weights is known to be 7 ounces.

A) 82 B) 81 C) 10 D) 9

22) In order to set rates, an insurance company is trying to estimate the number of sick days that full time workers

at an auto repair shop take per year. A previous study indicated that the standard deviation was 2.8 days. How

large a sample must be selected if the company wants to be 95% confident that the true mean differs from the

sample mean by no more than 1 day?

A) 31 B) 141 C) 512 D) 1024

23) In order to efficiently bid on a contract, a contractor wants to be 95% confident that his error is less than two

hours in estimating the average time it takes to install tile flooring. Previous contracts indicate that the

standard deviation is 4.5 hours. How large a sample must be selected?

A) 20 B) 4 C) 5 D) 19

24) In order to fairly set flat rates for auto mechanics, a shop foreman needs to estimate the average time it takes to

replace a fuel pump in a car. How large a sample must he select if he wants to be 99% confident that the true

average time is within 15 minutes of the sample average? Assume the standard deviation of all times is 30

minutes.

A) 27 B) 26 C) 6 D) 5

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25) In order to set rates, an insurance company is trying to estimate the number of sick days that full time workers

at an auto repair shop take per year. A previous study indicated that the standard deviation was 2.8 days. a)

How large a sample must be selected if the company wants to be 90% confident that the true mean differs from

the sample mean by no more than 1 day? b) Repeat part (a) using a 95% confidence interval. Which level of

confidence requires a larger sample size? Explain.

5 Determine the Finite Population Correction Factor



26) There were 800 math instructors at a mathematics convention. Forty instructors were randomly selected and

given an IQ test. The scores produced a mean of 130 with a standard deviation of 10. Find a 95% confidence

interval for the mean of the 800 instructors. Use the finite population correction factor.

6 Concepts



27) A random sample of 200 high school seniors is given the SAT-V test. The mean score for this sample is x = 493.What can you say about the mean score μ of all high school seniors?

28) The grade point averages for 10 randomly selected students in a statistics class with 125 students are listed

below. What can you say about the mean score μ of all 125 students?

2.1 3.1 2.0 3.9 3.5 3.7 2.8 1.9 2.5 2.2

29) A certain confidence in interval is 9.75 < μ< 11.05. Find the sample mean x and the error of estimate E.

30) Given the same sample statistics, which level of confidence will produce the narrowest confidence interval:

75%, 85%, 90%, or 95%? Explain your reasoning.

31) The grade point averages for 10 randomly selected students in a statistics class with 125 students are listed

below.

2.0 3.2 1.8 2.9 0.9 4.0 3.3 2.9 3.6 0.8

What is the effect on the width of the confidence interval if the sample size is increased to 20? Explain your

reasoning.

6.2 Confidence Intervals for the Mean (Small Samples)

1 Find a Critical Value



1) Find the critical value, tc for c = 0.99 and n = 10.

A) 3.250 B) 3.169 C) 2.262 D) 1.833

2) Find the critical value, tc , for c = 0.95 and n = 16.

A) 2.131 B) 2.120 C) 2.602 D) 2.947

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3) Find the critical value, tc , for c = 0.90 and n = 15.

A) 1.761 B) 1.753 C) 2.145 D) 2.624

2 Construct and Interpret Confidence Intervals for the Population Mean



4) Find the value of E, the margin of error, for c = 0.90 , n = 16 and s = 2.5.

A) 1.1 B) 0.27 C) 0.84 D) 0.21

5) Find the value of E, the margin of error, for c = 0.99, n = 10 and s = 3.2.

A) 3.29 B) 2.85 C) 1.04 D) 3.21

6) Find the value of E, the margin of error, for c = 0.95, n = 15 and s = 5.2.

A) 2.88 B) 2.96 C) 2.36 D) 0.74

7) In a random sample of 28 families, the average weekly food expense was $95.60 with a standard deviation of

$22.50. Determine whether a normal distribution or a t-distribution should be used or whether neither of these

can be used to construct a confidence interval. Assume the distribution of weekly food expenses is normally

shaped.

A) Use the t-

distribution.B) Use normal distribution.

C) Cannot use normal distribution or t-distribution.

8) For a sample of 20 IQ scores the mean score is 105.8. The standard deviation, σ , is 15. Determine whether a

normal distribution or a t-distribution should be used or whether neither of these can be used to construct a

confidence interval. Assume that IQ scores are normally distributed.

A) Use normal distribution.

B) Use the t-distribution.


9) A random sample of 40 college students has a mean earnings of $3120 with a standard deviation of $677 over

the summer months. Determine whether a normal distribution or a t-distribution should be used or whether

neither of these can be used to construct a confidence interval.

A) Use normal distribution.

B) Use the t-distribution.


10) A random sample of 15 statistics textbooks has a mean price of $105 with a standard deviation of $30.25.

Determine whether a normal distribution or a t-distribution should be used or whether neither of these can be

used to construct a confidence interval. Assume the distribution of statistics textbook prices is not normally

distributed.

A) Cannot use normal distribution or t-distribution.

B) Use normal distribution.

C) Use the t-distribution.

11) Construct a 95% confidence interval for the population mean, μ. Assume the population has a normal

distribution. A sample of 20 college students had mean annual earnings of $3120 with a standard deviation of

$677.

A) ($2803, $3437) B) ($1324, $1567) C) ($2135, $2567) D) ($2657, $2891)

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distribution. A sample of 15 randomly selected students has a grade point average of 2.86 with a standard

deviation of 0.78.

A) (2.51, 3.21) B) (2.41, 3.42) C) (2.37, 3.56) D) (2.28, 3.66)


distribution. A sample of 25 randomly selected students has a mean test score of 81.5 with a standard deviation

of 10.2.

A) (77.29, 85.71) B) (56.12, 78.34) C) (66.35, 69.89) D) (87.12, 98.32)



distribution. A random sample of 20 college students has mean annual earnings of $3180 with a standard

deviation of $673.



distribution. A random sample of 16 fluorescent light bulbs has a mean life of 645 hours with a standard

deviation of 31 hours.

A) (628.5, 661.5) B) (876.2, 981.5) C) (531.2, 612.9) D) (321.7, 365.8)


distribution. A group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation

of 3.8 years.

A) (19.9, 24.9) B) (16.3, 26.9) C) (17.2, 23.6) D) (18.7, 24.1)


distribution. A study of 14 bowlers showed that their average score was 192 with a standard deviation of 8.

A) (186.3, 197.7) B) (222.3, 256.1) C) (328.3, 386.9) D) (115.4, 158.8)


18) Construct a 95% confidence interval for the population mean, μ. Assume the population has a normaldistribution. In a random sample of 26 computers, the mean repair cost was $149 with a standard deviation of

$39.



distribution. In a recent study of 22 eighth graders, the mean number of hours per week that they watched

television was 19.6 with a standard deviation of 5.8 hours.

A) (17.47, 21.73) B) (18.63, 20.89) C) (5.87, 7.98) D) (19.62, 23.12)


20) a) Construct a 95% confidence interval for the population mean, μ. Assume the population has a normal

distribution. In a random sample of 26 computers, the mean repair cost was $129 with a standard deviation of

$37.

b) Suppose you did some research on repair costs for computers and found that the standard deviation is

σ = 37. Use the normal distribution to construct a 95% confidence interval for the population mean, μ. Compare

the results.

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21) A random sample of 10 parking meters in a beach community showed the following incomes for a day.

Assume the incomes are normally distributed.

$3.60 $4.50 $2.80 $6.30 $2.60 $5.20 $6.75 $4.25 $8.00 $3.00

Find the 95% confidence interval for the true mean.

A) ($3.39, $6.01) B) ($2.11, $5.34) C) ($4.81, $6.31) D) ($1.35, $2.85)

22) The grade point averages for 10 randomly selected high school students are listed below. Assume the grade

point averages are normally distributed.

2.0 3.2 1.8 2.9 0.9 4.0 3.3 2.9 3.6 0.8

Find a 98% confidence interval for the true mean.

A) (1.55, 3.53) B) (0.67, 1.81) C) (2.12, 3.14) D) (3.11, 4.35)

23) A local bank needs information concerning the checking account balances of its customers. A random sample

of 15 accounts was checked. The mean balance was $686.75 with a standard deviation of $256.20. Find a 98%

confidence interval for the true mean. Assume that the account balances are normally distributed.

A) ($513.17, $860.33) B) ($238.23, $326.41) C) ($326.21, $437.90) D) ($487.31, $563.80)


24) A manufacturer receives an order for fluorescent light bulbs. The order requires that the bulbs have a mean life

span of 500 hours. The manufacturer selects a random sample of 25 fluorescent light bulbs and finds that they

have a mean life span of 495 hours with a standard deviation of 15 hours. Test to see if the manufacturer is

making acceptable light bulbs. Use a 95% confidence level. Assume the data are normally distributed.

25) A coffee machine is supposed to dispense 12 ounces of coffee in each cup. An inspector selects a random

sample of 40 cups of coffee and finds they have an average amount of 12.2 ounces with a standard deviation of

0.3 ounce. Use a 99% confidence interval to test whether the machine is dispensing acceptable amounts of

coffee.

6.3 Confidence Intervals for Population Proportions

1 Find a Point Estimate



1) When 435 college students were surveyed,120 said they own their car. Find a point estimate for p, the

population proportion of students who own their cars.

A) 0.276 B) 0.724 C) 0.381 D) 0.216

2) A survey of 100 fatal accidents showed that 12 were alcohol related. Find a point estimate for p, the populationproportion of accidents that were alcohol related.

A) 0.12 B) 0.88 C) 0.136 D) 0.107

3) A survey of 400 non-fatal accidents showed that 189 involved the use of a cell phone. Find a point estimate for

p, the population proportion of non-fatal accidents that involved the use of a cell phone.

A) 0.472 B) 0.527 C) 0.896 D) 0.321

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4) A survey of 250 homeless persons showed that 17 were veterans. Find a point estimate p, for the population

proportion of homeless persons who are veterans.

A) 0.068 B) 0.932 C) 0.073 D) 0.064

5) A survey of 2650 golfers showed that 379 of them are left-handed. Find a point estimate for p, the population

proportion of golfers that are left-handed.

A) 0.143 B) 0.857 C) 0.167 D) 0.125

2 Construct and Interpret Confidence Intervals for the Population Proportion



6) In a survey of 2480 golfers, 15% said they were left-handed. The surveyʹs margin of error was 3%. Construct a

confidence interval for the proportion of left-handed golfers.

A) (0.12, 0.18) B) (0.18, 0.21) C) (0.12, 0.15) D) (0.11, 0.19)


7) The Federal Bureau of Labor Statistics surveyed 50,000 people and found the unemployment rate to be 5.8%.

The margin of error was 0.2%. Construct a confidence interval for the unemployment rate.

8) When 495 college students were surveyed, 150 said they own their car. Construct a 95% confidence interval forthe proportion of college students who say they own their cars.

9) A survey of 300 fatal accidents showed that 123 were alcohol related. Construct a 98% confidence interval for

the proportion of fatal accidents that were alcohol related.

10) A survey of 400 non-fatal accidents showed that 197 involved the use of a cell phone. Construct a 99%

confidence interval for the proportion of fatal accidents that involved the use of a cell phone.


11) A survey of 280 homeless persons showed that 63 were veterans. Construct a 90% confidence interval for the

proportion of homeless persons who are veterans.A) (0.184, 0.266) B) (0.176, 0.274) C) (0.167, 0.283) D) (0.161, 0.289)

12) A survey of 2450 golfers showed that 281 of them are left-handed. Construct a 98% confidence interval for the

proportion of golfers that are left-handed.

A) (0.100, 0.130) B) (0.203, 0.293) C) (0.369, 0.451) D) (0.683, 0.712)


13) In a survey of 10 golfers, 2 were found to be left-handed. Is it practical to construct the 90% confidence interval

for the population proportion, p? Explain.

14) The USA Today claims that 44% of adults who access the Internet read the international news online. You

want to check the accuracy of their claim by surveying a random sample of 120 adults who access the Internetand asking them if they read the international news online. Fifty-two adults responded ʺyes.ʺ Use a 95%

confidence interval to test the newspaperʹs claim.

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15) A researcher at a major hospital wishes to estimate the proportion of the adult population of the United States

that has high blood pressure. How large a sample is needed in order to be 95% confident that the sample

proportion will not differ from the true proportion by more than 4%?

A) 601 B) 13 C) 1201 D) 423

16) A pollster wishes to estimate the proportion of United States voters who favor capital punishment. How large a

sample is needed in order to be 95% confident that the sample proportion will not differ from the true

proportion by more than 5%?

A) 385 B) 271 C) 10 D) 769

17) A private opinion poll is conducted for a politician to determine what proportion of the population favors

decriminalizing marijuana possession. How large a sample is needed in order to be 95% confident that the

sample proportion will not differ from the true proportion by more than 2%?

A) 2401 B) 1692 C) 4802 D) 25

18) A manufacturer of golf equipment wishes to estimate the number of left-handed golfers. How large a sample is

needed in order to be 98% confident that the sample proportion will not differ from the true proportion by

more than 2%? A previous study indicates that the proportion of left-handed golfers is 8%.

A) 999 B) 707 C) 1086 D) 17

19) A researcher wishes to estimate the number of households with two cars. How large a sample is needed in

order to be 98% confident that the sample proportion will not differ from the true proportion by more than 5%?

A previous study indicates that the proportion of households with two cars is 19%.

A) 335 B) 237 C) 413 D) 8


20) A state highway patrol official wishes to estimate the number of drivers that exceed the speed limit traveling a

certain road.

a) How large a sample is needed in order to be 95% confident that the sample proportion will not differ fromthe true proportion by more than 2%?

b) Repeat part (a) assuming previous studies found that 70% of drivers on this road exceeded the speed limit.

6.4 Confidence Intervals for Variance and Standard Deviation

1 Find Critical Values



1) Find the critical values, X 2R andX

2L , for c = 0.95 and n = 12.

A) 3.816 and 21.920 B) 3.053 and 24.725 C) 4.575 and 26.757 D) 2.603 and 19.675


2L , for c = 0.90 and n = 15.


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2L , for c = 0.98 and n = 20.



2L , for c = 0.99 and n = 10.


2 Construct

Confidence

Intervals

for

the

Population

Variance

and

Population

Standard

Deviation


Assume the sample is taken from a normally distributed population and construct the indicated confidence interval.

5) Construct a 95% confidence interval for the population standard deviation σ of a random sample of 15 men

who have a mean weight of 165.2 pounds with a standard deviation of 13.5 pounds. Assume the population is

normally distributed.

A) (9.9 , 21.3) B) (97.7, 453.3) C) (2.7, 5.8) D) (10.4, 19.7)

6) A random sample of 16 men have a mean height of 67.5 inches and a standard deviation of 1.5 inches.

Construct a 99% confidence interval for the population standard deviation, σ.

A) (1.0, 2.7) B) (1.0, 2.8) C) (0.8, 2.2) D) (1.1, 2.5)

7) A random sample of 20 women have a mean height of 62.5 inches and a standard deviation of 3.2 inches.

Construct a 98% confidence interval for the population variance, σ2.

A) (5.4, 25.5) B) (2.3, 5.0) C) (1.7, 8.0) D) (5.7, 26.8)

8) The heights (in inches) of 20 randomly selected adult males are listed below. Construct a 99% confidence

interval for the variance, σ2.

70 72 71 70 69 73 69 68 70 71

67 71 70 74 69 68 71 71 71 72

A) (1.47, 8.27) B) (21.61, 69.06) C) (1.35, 8.43) D) (2.16, 71.06)

9) The grade point averages for 10 randomly selected students are listed below. Construct a 90% confidence

interval for the population standard deviation, σ.

2.0 3.2 1.8 2.9 0.9 4.0 3.3 2.9 3.6 0.8

A) (0.81, 1.83) B) (0.32, 0.85) C) (0.53, 1.01) D) (1.10, 2.01)

10) The mean replacement time for a random sample of 12 microwave ovens is 8.6 years with a standard deviation

of 4.2 years. Construct the 98% confidence interval for the population variance, σ2.

A) (7.8, 63.6) B) (2.8, 8.0) C) (1.9, 15.1) D) (7.4, 54.3)

11) A student randomly selects 10 CDs at a store. The mean is $13.75 with a standard deviation of $1.50. Construct

a 95% confidence interval for the population standard deviation, σ.

A) ($1.03, $2.74) B) ($0.99, $2.50) C) ($1.06, $7.51) D) ($0.84, $2.24)

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12) The stem-and-leaf plot shows the test scores of 16 randomly selected students. Construct a 99% confidence

interval for the population standard deviation.

5

6

7

8

9

9

5 8 3

7 4 4 2 9

5 8 3 5

3 1 7

A) (7.61, 20.33) B) (57.97, 413.27) C) (62.18, 363.63) D) (7.89, 19.07)

13) The dotplot shows the weights (in pounds) of 15 dogs selected randomly from those adopted out by an animal

shelter last week. Construct a 98% confidence interval for the population variance.

Weights

25 30 35

Pounds

A) (3.03, 18.97) B) (1.74, 4.36) C) (3.38, 15.70) D) (2.89, 16.91)


14) A container of car oil is supposed to contain 1000 milliliters of oil. A quality control manager wants to be sure

that the standard deviation of the oil containers is less than 20 milliliters. He randomly selects 10 cans of oil

with a mean of 997 milliliters and a standard deviation of 32 milliliters. Use these sample results to construct a

95% confidence interval for the true value of σ. Does this confidence interval suggest that the variation in the oil

containers is at an acceptable level?

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Ch. 6 Confidence IntervalsAnswer Key

6.1 Confidence Intervals for the Mean (Large Samples)1 Find a Critical Value

1) A

2 Find the Margin of Error

2) A

3) A

4) A

5) A


6) A

7) A

8) A

9) A

10) A

11) A

12) A

13) a) ($138, $162)

b) A decrease in the level of confidence will decrease the width of the confidence interval.

14) A

15) a) (17.9, 21.1)

b) An increase in the standard deviation will widen the confidence interval.

16) A

17) a) (61.4, 65.4)

b) An increase in the sample size will decrease the width of the confidence interval.

18) (543.8, 658. 0)

19) (7.2, 8.8)


20) A

21) A

22) A

23) A

24) A

25) a) 22

b) 31; A 95% confidence interval requires a larger sample than a 90% confidence interval because more information is

needed from the population to be 95% confident.

5 Determine the Finite Population Correction Factor

26) (127.0, 133.0)

6 Concepts

27) The sample mean x = 493 is the best estimator of the unknown population mean μ.

28) The sample mean x = 2.77 is the best point estimate of the unknown population mean μ.

29) Sample mean x = 10.40 and the error of estimate E = 0.65.30) The 75% level of confidence will produce the narrowest confidence interval. As the level of confidence decreases, zc

decreases, causing narrower intervals.

31) The width of the interval will decrease. As n increases, E decreases because n is in the denominator of the formula

for E. So the intervals become narrower.

6.2 Confidence Intervals for the Mean (Small Samples)1 Find a Critical Value

1) A

2) A

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3) A


4) A

5) A

6) A

7) A

8) A

9) A

10) A

11) A12) A

13) A

14) ($2798, $3562)

15) A

16) A

17) A

18) ($133.24, $164.76)

19) A

20) a) ($114.05, $143.95)

b) ($114.78, $143.22); The t-confidence interval is wider.

21) A

22) A23) A

24) (488.81 , 501.19). Because the interval contains the desired life span of 500 hours, they are making good light bulbs.

25) (12.1, 12.3) Because the interval does not contain the desired amount of 12 ounces, the machine is not working

properly.

6.3 Confidence Intervals for Population Proportions1 Find a Point Estimate

1) A

2) A

3) A

4) A

5) A

2 Construct

and

Interpret

Confidence

Intervals

for

the

Population

Proportion

6) A

7) (0.058, 0.060)

8) (0.263 , 0.344)

9) (0.344, 0.476)

10) (0.428, 0.557)

11) A

12) A

13) It is not practical to find the confidence interval. It is necessary that np^ > 5 to insure that the

distribution of p^ be normal. (np

^ = 2)

14) (0.345, 0.522) Because the interval contains the reported percentage of 44%, the newspaperʹs claim is accurate.

3 Determine

the

Minimum

Sample

Size15) A

16) A

17) A

18) A

19) A

20) a) 2401

b) 2017

0


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6.4 Confidence Intervals for Variance and Standard Deviation1 Find Critical Values

1) A

2) A

3) A

4) A

2 Construct Confidence Intervals for the Population Variance and Population Standard Deviation

5) A

6) A

7) A

8) A

9) A

10) A

11) A

12) A

13) A

14) The 95% confidence interval is (22.01, 58.42). Because this interval does not contain 20, the data suggest that the

standard deviation is not at an acceptable level.

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