Provide an appropriate response. Use a table of areas to ... tests/1342...Use a table of areas to obtain the shaded area under the standard normal curve. 8)-2.54 -1.27 1.27 2.54 z

Provide an appropriate response.

1) How does the standard normal distribution

differ from a nonstandard normal distribution?

Why is it necessary to standardize in order to

find percentages for nonstandard normal

variables?

2) A variable is normally distributed. 42% of the

possible observations of the variable lie

between 20 and 28. What information does this

give you about the graph of the normal curve

for this variable?

3) Two random variables are normally

distributed with the same mean. One has a

standard deviation of 10 while the other has a

standard deviation of 15. How will the graphs

of the two variables differ and how will they

be alike?

Fill in the blanks by standardizing the normally

distributed variable.

4) The amount of time that customers wait in line

during peak hours at one bank is normally

distributed with a mean of 15 minutes and a

standard deviation of 3 minutes. The

percentage of time that the waiting time

exceeds 14 minutes is equal to the area under

the standard normal curve that lies to the ___

of ___.

5) Dave drives to work each morning at about the

same time. His commute time is normally

distributed with a mean of 55 minutes and a

standard deviation of 5 minutes. The

percentage of time that his commute time

exceeds 67 minutes is equal to the area under

the standard normal curve that lies to the ___

of ___.

Use a table of areas to find the specified area under the

standard normal curve.

6) The area that lies between 0 and 3.01

7) The area that lies between -0.73 and 2.27

Use a table of areas to obtain the shaded area under the

standard normal curve.

8)

z-2.54 -1.27 1.27 2.54 z-2.54 -1.27 1.27 2.54

9)

z-2.31 2.31 z-2.31 2.31

10)

z-1.56 1.56 z-1.56 1.56

1


11) Most problems involving normally distributed

variables are one of two types.

Type A: Find a probability or percentage, e.g.

find the probability that X lies in a specified

range.

Type B: Find the observation corresponding to

a given probability or percentage.

Suppose that scores on a test are normally

distributed with a mean of 80 and a standard

deviation of 8. Which of the following

questions below are of type B?

a. Find the 80th percentile.

b. Find the cutoff for the A grade if the top 10%

get an A.

c. Find the percentage scoring more than 90.

d. Find the score that separates the bottom 30%

from the top 70%.

e. Find the probability that a randomly selected

student will score more than 80.

12) Scores on an aptitude test are normally


deviation of 60. Explain how you would find

any given percentile.

13) Suppose that scores on a test are normally


deviation of 8. You have been asked to find the

70th percentile. After sketching a standard

normal curve and shading the area of interest,

the next step in solving this problem is to use

the table of areas. Would you look for 0.7 in the

body of the table or in the left-hand column?

Explain your reasoning.

14) True or false, the mean of a normally

distributed variable can be any real number.

Find the indicated probability or percentage for the

normally distributed variable.

15) The lengths of human pregnancies are

normally distributed with a mean of 268 days

and a standard deviation of 15 days. What is

the probability that a pregnancy lasts at least

300 days?

16) Assume that the weights of quarters are

normally distributed with a mean of 5.67 g and

a standard deviation 0.070 g. A vending

machine will only accept coins weighing

between 5.48 g and 5.82 g. What percentage of

legal quarters will be rejected?

Find the specified percentile, quartile, or decile.

17) The annual precipitation for one city is

normally distributed with a mean of 25.6

inches and a standard deviation of 3.4 inches.

Find the 2nd decile.

18) Scores on an English test are normally

distributed with a mean of 31.4 and a standard

deviation of 6.3. Find the 41st percentile.

Find the requested probability.

19) The test scores of 5 students are under

consideration. The following is the dotplot for

the sampling distribution of the sample mean

for samples of size 2.

Find the probability, expressed as a percent,

that the sample mean will be within 2 points of

the population mean.






that the sample mean will be within 1 point of


2






that the sample mean will be within 3 points of


Solve the problem.

22) The ages of six members on a board of

directors of a nonprofit organization are shown

below.

Member A B C D E F

Age 32 52 43 64 41 50

Consider these board members to be a

population of interest. The table below shows

all of the possible samples of size four. For

each sample, the people in the sample, their

ages, and the sample mean are listed. Use the

table to find the mean of the variable x.

Sample Ages x

A, B, C, D

A, B, C, E

A, B, C, F

A, B, D, E

A, B, D, F

A, B, E, F

A, C, D, E

A, C, D, F

A, C, E, F

A, D, E, F

B, C, D, E

B, C, D, F

B, C, E, F

B, D, E, F

C, D, E, F

32, 52, 43, 64

32, 52, 43, 41

32, 52, 43, 50

32, 52, 64, 41

32, 52, 64, 50

32, 52, 41, 50

32, 43, 64, 41

32, 43, 64, 50

32, 43, 41, 50

32, 64, 41, 50

52, 43, 64, 41

52, 43, 64, 50

52, 43, 41, 50

52, 64, 41, 50

43, 64, 41, 50

47.75

42

44.25

47.25

49.5

43.75

45

47.25

41.5

46.75

50

52.25

46.5

51.75

49.5

23) The weights of five players on a football team

are shown below.

Player A B C D E

Weight (lb) 290 310 250 255 220

Consider these players to be a population of

interest. The table below shows all of the

possible samples of size three. For each

sample, the players in the sample, their

weights, and the sample mean are listed. Use

the table to find the mean of the variable x.

Sample Weights x

A, B, C 290, 310,250 283.3

A, B, D 290, 310, 255 285

A, B, E 290, 310 220 273.3

A, C, D 290, 250, 255 265

A, C, E 290, 250, 220 253.3

A, D, E 290, 255, 220 255

B, C, D 310, 250, 255 271.6

B, C, E 310, 250, 220 260

B, D, E 310, 255, 220 261.6

C, D, E 250, 255, 220 241.6


24) Suppose that μ represents the mean height for

a population of people. Suppose that you use a

sample mean, x, to estimate μ. Explain what is

meant by sampling error in this situation. Why

does x vary from one sample to the next? How

can you reduce the likely sampling error?

25) A population of people has a mean height of

65 inches. Andrew picks a person at random

from the population and records his or her

height. He repeats this procedure 49 times

more. Bob picks a sample of 30 people at

random from the population and records the

mean height of the sample. He repeats this

procedure 49 times more. Which set of

numbers (those recorded by Andrew or those

recorded by Bob) do you think will have more

variability? Explain your reasoning.

3

26) The mean height for a population is 65 inches

and the standard deviation is 3 inches. Let A

and B denote the events described below.

Event A: The mean height in a random sample

of 16 people is within 1 inch of the population

mean.

Event B: The mean height in a random sample

of 50 people is within 1 inch of the population

mean.

True or false, the probability of event A is

greater than the probability of event B?

27) The mean height for a population of people is

65 inches. Suppose that you pick a sample of

50 people and determine the sample mean, x.

You then repeat this procedure three more

times. You learned in class that μx

= μ. Can

you conclude that the mean of the four sample

means will be equal to the population mean of

65 inches? Why or why not?

For samples of the specified size from the population

described, find the mean and standard deviation of the

sample mean x.

28) The National Weather Service keeps records of

snowfall in mountain ranges. Records indicate

that in a certain range, the annual snowfall has

a mean of 110 inches and a standard deviation

of 14 inches. Suppose the snowfalls are

sampled during randomly picked years. For

samples of size 49, determine the mean and

standard deviation of x.

29) The mean and the standard deviation of the

sampled population are, respectively, 115.3

and 36.5.

n = 81


30) Do you agree with Tony's reasoning below?

Explain why you do or do not agree. Refer to

the Central Limit Theorem in your

explanation.

Tony: "When a balanced die is rolled, each of

the numbers 1, 2, 3, 4, 5, and 6 has an equal

chance of showing up. So, if I roll the die 50

times and find the mean of the 50 numbers, the

mean has the same chance of falling between 1

and 2 as it has of falling between 3 and 4."

31) Let x represent the number which shows up

when a balanced die is rolled. Then x is a

random variable with a uniform distribution.

Let x denote the mean of the numbers obtained

when the die is rolled 3 times. Which of the

following statements concerning the sampling

distribution of the mean, x , is true?

A) x is approximately normally distributed.

B) x is normally distributed.

C) x has a uniform distribution.

D) None of the above statements is true.

32) The typical computer random-number

generator yields numbers in a uniform

distribution between 0 and 1, with a mean of

0.500 and a standard deviation of 0.289.

Consider the following two problems, which

appear at a glance to be very similar. One can

be solved using the Central Limit Theorem.

Which one and why?

(a) Suppose a sample of size 50 is randomly

generated. Find the probability that the mean

is below 0.300. (b) Suppose a sample of size 15

is randomly generated. Find the probability

that the mean is below 0.300.

4

33) Consider the following two problems.

(a) A random-number generator yields

numbers in a uniform distribution between 0

and 1 with a mean of 0.5 and a standard

deviation of 0.289. You wish to find the

probability that the mean of a sample of 50

random numbers is greater than 0.6.

(b) Scores on an aptitude test are normally


deviation of 11. You wish to find the

probability that the score for a randomly

selected person is greater than 90.

Which of these two problems requires

application of the Central Limit Theorem?

Explain your reasoning.

Find the indicated probability or percentage for the

sampling error.

34) Scores on a chemistry final exam are normally


deviation of 50. Determine the percentage of

samples of size 4 that will have mean scores

within 35 points of the population mean score

of 280.

35) The distribution of weekly salaries at a large

company is reverse J-shaped with a mean of

$1000 and a standard deviation of $370. What

is the probability that the sampling error made

in estimating the mean weekly salary for all

employees of the company by the mean of a

random sample of weekly salaries of 80

employees will be at most $75?

36) The distribution of weekly salaries at a large

company is right skewed with a mean of $1000

and a standard deviation of $350. What is the

probability that the sampling error made in

estimating the mean weekly salary for all

employees of the company by the mean of a

random sample of weekly salaries of 50

employees will be at most $50?

37) Scores on an aptitude test are distributed with

a mean of 220 and a standard deviation of 30.

The shape of the distribution is unspecified.

What is the probability that the sampling error

made in estimating the population mean by

the mean of a random sample of 50 test scores

will be at most 5 points?

Find the requested confidence interval.

38) A college statistics professor has office hours

from 9:00 A.M. to 10:30 A.M. daily. A sample

of waiting times to see the professor (in

minutes) is 10, 12, 20, 15, 17, 10, 30, 28, 35, 28,

19, 27, 25, 22, 33, 37, 14, 21, 20, 23. Assuming

σ = 7.84, find the 99.74% confidence interval

for the population mean.

39) A college statistics professor has office hours

from 9:00 A.M. to 10:30 A.M. daily. A sample

of waiting times to see the professor (in

minutes) is 10, 12, 20, 15, 17, 10, 30, 28, 35, 28,

19, 27, 25, 22, 33, 37, 14, 21, 20, 23. Assuming

σ = 7.84, find the 95.44% confidence interval

for the population mean.

Find the requested value.

40) Physiologists often use the forced vital capacity

as a way to assess a person's ability to move air

in and out of their lungs. A researcher wishes

to estimate the forced vital capacity of people

suffering from asthma. A random sample of

15 asthmatics yields the following data on

forced vital capacity, in liters.

3.2 4.7 3.9 5.2 4.3

3.3 4.4 4.5 5.3 4.1

3.2 3.5 4.8 4.0 5.1

Use the data to obtain a point estimate of the

mean forced vital capacity for all asthmatics.

41) A researcher wishes to estimate the mean

resting heart rate for long-distance runners. A

random sample of 12 long-distance runners

yields the following heart rates, in beats per

minute.

79 64 63 77 71 81

78 58 74 70 60 63

Use the data to obtain a point estimate of the

mean resting heart rate for all long distance

runners.

5


42) Is it true that the point estimate of a population

mean must lie within the range of values

defined by the corresponding

confidence-interval estimate, regardless of the

level of confidence achieved? Explain.

43) In which of the following situations is it

reasonable to use the z-interval procedure to

obtain a confidence interval for the population

mean? Assume that the population standard

deviation is known.

A. n = 10, the data contain no outliers, the

variable under consideration is not normally

distributed.

B. n = 10, the variable under consideration is

normally distributed.

C. n = 18, the data contain no outliers, the

variable under consideration is far from being

normally distributed.

D. n = 18, the data contain outliers, the

variable under consideration is normally

distributed.

44) If the sample size is small (less than 15), under

what conditions is it reasonable to use the

z-interval procedure to obtain a confidence

interval for the population mean? If the sample

size is moderate (between 15 and 30), under

what conditions is it reasonable to use the


interval for the population mean?

45) Describe the steps involved in using the


interval for the population mean when σ is

known. Include a description of the

preliminary data analysis. Why is preliminary

data analysis needed?

46) Give an expression for the confidence level in

terms of α.

Find the confidence interval specified.

47) 31 packages are randomly selected from

packages received by a parcel service. The

sample has a mean weight of 15.6 pounds.

Assume that σ = 2.2 pounds. What is the 95%

confidence interval for the true mean weight,

μ, of all packages received by the parcel

service?

48) A random sample of 131 full-grown lobsters

had a mean weight of 15 ounces. Assume that

σ = 3.4 ounces. Construct a 95% confidence

interval for the population mean μ.

49) A sample of 32 people were randomly selected

from among the workers in a shoe factory. The

time taken for each person to polish a finished

shoe was measured. The sample mean was 3.1

minutes. Assume that σ = 0.94 minutes.

Construct a 90% confidence interval for the

true mean time, μ, to polish a shoe.


50) A confidence interval for a population mean

has a margin of error of 2.3. If the sample mean

is 69.7, obtain the confidence interval.

51) A sample mean is used to estimate a

population mean. To obtain a margin of error

of 1.5 at a confidence level of 95%, a sample

size of 120 is needed. Would the required

sample size be larger or smaller if the

researcher wished to

(a) increase the confidence level while keeping

the same margin of error?

(b) decrease the margin of error while keeping

the same confidence level?

Explain your answers.

Solve the problem.

52) A sample of 51 eggs yields a mean weight of

1.58 ounces. Assuming that σ = 0.58 ounces,

find the margin of error in estimating μ at the

95% level of confidence.

6

53) A sample of 100 women aged 18-24 yields a

mean systolic blood pressure of112.8 mm Hg.

Assuming that σ = 13.0 mm Hg, find the

margin of error in estimating μ at the 95% level

of confidence.

54) A confidence interval for a population mean,

μ, has length 184.4. Find the margin of error.

55) A confidence interval for a population mean

has a margin of error of 0.08. Determine the

length of the confidence interval.

Find the necessary sample size.

56) Scores on a certain test are normally

distributed with a variance of 100. A

researcher wishes to estimate the mean score

achieved by all adults on the test. Find the

sample size needed to assure with 95 percent

confidence that the sample mean will not differ

from the population mean by more than 4

units.

57) Weights of women in one age group are

normally distributed with a standard deviation

σ of 14 lb. A researcher wishes to estimate the

mean weight of all women in this age group.

Find how large a sample must be drawn in

order to be 90 percent confident that the

sample mean will not differ from the

population mean by more than 3.9 lb.


58) Compare the basic properties of t-curves with

the basic properties of the standard normal

curve. In what ways are t-curves similar to the

standard normal curve? In what ways are they

different?

59) Suppose that you wish to obtain a 95%

confidence interval for a population mean. The

population is normally distributed, the sample

size is 20, and the population standard

deviation is unknown. The correct procedure

to use is the t-interval procedure. If you

mistakenly use the z-interval procedure, will

the resulting confidence interval be too wide or

too narrow? Will the true confidence level

associated with this interval be greater than or

less than 95%?

60) Which of the following statements regarding

t-curves is/are true?

A. The total area under a t-curve with 10

degrees of freedom is greater than the area

under the standard normal curve.

B. The t-curve with 10 degrees of freedom is

flatter and wider than the standard normal

curve.

C. The t-curve with 10 degrees of freedom

more closely resembles the standard normal

curve than the t-curve with 20 degrees of

freedom.

61) Suppose that you wish to obtain a confidence

interval for a population mean. Under the

conditions described below, should you use

the z-interval procedure, the t-interval

procedure, or neither?

- The population standard deviation is known.

- The population is not normally distributed.

- The sample size is 12.

Find the confidence interval specified. Assume that the

population is normally distributed.

62) A laboratory tested twelve chicken eggs and

found that the mean amount of cholesterol was

226 milligrams with s = 17.9 milligrams.

Construct a 95% confidence interval for the

true mean cholesterol content of all such eggs.

63) The football coach randomly selected ten

players and timed how long each player took

to perform a certain drill. The times (in

minutes) were:

5.1 6.2 5.9 14.2 14.9

13.2 5.8 13.1 5.6 10.2

Determine a 95% confidence interval for the

mean time for all players.


64) Give an example of a situation in which you

might wish to conduct a two-tailed hypothesis

test concerning a population mean. State in

words what you wish to determine and write

the null and alternative hypotheses in words

and symbolically.

7

65) You wish to test the hypotheses shown below.

H0

: μ = 40

Ha

: μ > 40

Would you be inclined to reject the null

hypothesis if the sample mean turned out to be

much smaller than 40? Explain your thinking.

66) Suppose that you wish to conduct a hypothesis

test concerning a population mean. How

would you decide whether to conduct a

right-tailed, a left-tailed, or a two-tailed test?

Is it acceptable to decide what type of test to

conduct by examining the sample data?

67) Traditionally in hypothesis testing the null

hypothesis represents the "status quo" which

will be overturned only if there is evidence

against it. Which of the statements below

might represent a null hypothesis?

A) The defendant is guilty.

B) The treatment has no effect.

C) The teaching method raises SAT scores.

D) None of the above

68) Jenny is conducting a hypothesis test

concerning a population mean. The hypotheses

are as follows.

H0

: μ = 50

Ha

: μ > 50

She selects a sample and finds that the sample

mean is 54.2. She then does some calculations

and is able to make the following statement:

If H0

were true, the chance that the sample

mean would have come out as big ( or bigger)

than 54.2 is 0.3. Do you think that she should

reject the null hypothesis? Why or why not?

69) Give an example of a hypothesis test for which

it is important to have a small β probability.

Explain why it is important to have a small

value for β.

70) Give an example of a hypothesis test for which

it is important to have a small α probability.

Explain why it is important to have a small

value for α.

71) A bottle filling machine fills 16-ounce bottles

with juice. The amount of juice varies from

bottle to bottle, however the average amount

of juice is supposed to be 16.0 ounces. The

manufacturer performs a hypothesis test to

determine whether the machine is working

properly. The hypotheses are:

H0

: μ = 16.0 ounces

Ha

: μ < 16.0 ounces

Do you think that a consumer advocacy group

would prefer to have a small α probability or a

small β probability? Why? Do you think that

the manufacturer would prefer to have a small

α probability or a small β probability? Why?

72) A man is on trial accused of murder in the first

degree. The prosecutor presents evidence that

he hopes will convince the jury to reject the

hypothesis that the man is innocent. This

situation can be modeled as a hypothesis test

with the following hypotheses:

H0

: The defendant is innocent.

Ha

: The defendant is guilty.

If convicted, the defendant will receive the

death penalty. Do you think that a Type I error

or a Type II error would be more serious?

Why?

73) A pharmaceutical company has a new drug

which relieves headaches. However, there is

some indication that the drug may have the

side effect of increasing blood pressure.

Suppose the drug company conducts a

hypothesis test to determine whether the

medication raises blood pressure. The

hypotheses are:

H0

: The drug does not increase

blood pressure.

Ha

: The drug increases blood

pressure.

Do you think that for doctors and patients it is

more important to have a small α probability

or a small β probability? Why? Do you think

that the pharmaceutical company would prefer

to have a small α probability or a small β

probability? Why?

8

Classify the conclusion of the hypothesis test as a Type I

error, a Type II error, or a correct decision.

74) At one school, in 2005, the average amount of

time that tenth-graders spent watching

television each week was 21.6 hours. The

principal introduced a campaign to encourage

the students to watch less television. One year

later, in 2006, the principal performed a


average amount of time spent watching

television per week had decreased. The

hypotheses were:

H0

: μ = 21 hours

Ha

: μ < 21 hours

where μ is the mean amount of time, in 2006,

that tenth-graders spend watching television

each week.

Suppose that the results of the sampling lead

to nonrejection of the null hypothesis. Classify

that conclusion as a Type I error, a Type II

error, or a correct decision, if in fact the mean

amount of time, μ, spent watching television

had not decreased.

75) A manufacturer claims that the mean amount

of juice in its 16 ounce bottles is 16.1 ounces. A

consumer advocacy group wants to perform a

hypothesis test to determine whether the mean

amount is actually less than this. The

hypotheses are:

H0

: μ = 16.1 ounces

Ha

: μ < 16.1 ounces

where μ is the mean amount of juice in the

manufacturer's 16 ounce bottles.


to rejection of the null hypothesis. Classify that

conclusion as a Type I error, a Type II error, or

a correct decision, if in fact the mean amount of

juice, μ, is less than 16.1 ounces.

76) The maximum acceptable level of a certain

toxic chemical in vegetables has been set at 0.7

parts per million (ppm). A consumer health

group measured the level of the chemical in a

random sample of tomatoes obtained from one

producer to determine whether the mean level

of the chemical in these tomatoes exceeds the

recommended limit.

The hypotheses are

H0

: μ = 0.7 ppm

Ha

: μ > 0.7 ppm

where μ is the mean level of the chemical in

tomatoes from this producer. Suppose that the

results of the sampling lead to nonrejection of

the null hypothesis. Classify that conclusion as

a Type I error, a Type II error, or a correct

decision, if in fact the mean level of the

chemical in these tomatoes is greater than 0.7

ppm.

77) In the past, the mean running time for a certain

type of flashlight battery has been 8.1 hours.

The manufacturer has introduced a change in

the production method and wants to perform a


running time has increased as a result. The

hypotheses are:

H0

: μ = 8.1 hours

Ha

: μ > 8.1 hours

where μ is the mean running time of the new

batteries


to rejection of the null hypothesis. Classify that


a correct decision, if in fact the mean running

time has not increased.

78) A man is on trial accused of murder in the first

degree. The prosecutor presents evidence that

he hopes will convince the jury to reject the

hypothesis that the man is innocent. This

situation can be modeled as a hypothesis test

with the following hypotheses:

H0

: The defendant is innocent.

Ha

: The defendant is guilty.

Suppose that the null hypothesis is rejected;

i.e., the defendant is found guilty. Classify that


a correct decision, if in fact the defendant is

innocent.

9

79) The recommended dietary allowance (RDA) of

vitamin C for women is 75 milligrams per day.

A hypothesis test is to be performed to decide

whether adult women are, on average, getting

less than the RDA of 75 milligrams per day.

The hypotheses are

H0

: μ = 75 mg

Ha

: μ < 75 mg

where μ is the mean vitamin C intake (per day)

of all adult females. Suppose that the results of

the sampling lead to rejection of the null

hypothesis. Classify that conclusion as a Type I

error, a Type II error, or a correct decision, if in

fact the mean vitamin C intake of adult females

is 75 mg.

Classify the hypothesis test as two-tailed, left-tailed, or

right-tailed.








recommended limit.






running time has changed as a result.

A hypothesis test is to be performed. Determine the null

and alternative hypotheses.

82) The manufacturer of a refrigerator system for

beer kegs produces refrigerators that are

supposed to maintain a mean temperature, μ,

of 45°F, ideal for a certain type of German

pilsner. The owner of the brewery does not

agree with the refrigerator manufacturer, and

wants to conduct a hypothesis test to

determine whether the true mean temperature

differs from this value.






running time has changed as a result.

For the given hypothesis test, explain the meaning of a

Type I error, a Type II error, or a correct decision as

specified.

84) At one school, in 2005, the average amount of

time that tenth-graders spent watching

television each week was 21.6 hours. The

principal introduced a campaign to encourage

the students to watch less television. One year

later, in 2006, the principal performed a


average amount of time spent watching

television per week had decreased. The

hypotheses are:

H0

: μ = 21.6 hours

Ha

: μ < 21.6 hours

where μ is the mean amount of time, in 2006,

that tenth-graders spend watching television

each week

Explain the meaning of a correct decision.

85) In 2000, the mean math SAT score for students

at one school was 485. Five years later, in 2005,

a teacher performed a hypothesis test to

determine whether the average math SAT

score of students at the school had changed

from the 2000 mean of 485. The hypotheses

were:

H0

: μ = 485

Ha

: μ ≠ 485

where μ is the mean math SAT score, in 2005,

for students at the school

Explain the meaning of a Type I error.

10







hypotheses are:

H0

: μ = 8.0 hours

Ha

: μ > 8.0 hours


batteries . Explain the meaning of a Type II

error.







hypotheses are:

H0

: μ = 8.0 hours

Ha

: μ > 8.0 hours


batteries . Explain the meaning of a Type I

error.


of juice in its 16 ounce bottles is 16.1 ounces. A



amount is actually less than this. The

hypotheses are:

H0

: μ = 16.1 ounces

Ha

: μ < 16.1 ounces

where μ is the mean amount of juice in the

manufacturer's 16 ounce bottles. Explain the

meaning of a Type I error.

89) The recommended dietary allowance (RDA) of

vitamin C for women is 75 milligrams per day.

A hypothesis test is to be performed to decide

whether adult women are, on average, getting

less than the RDA of 75 milligrams per day.

The hypotheses are

H0

: μ = 75 mg

Ha

: μ < 75 mg

where μ is the mean vitamin C intake (per day)

of all adult females. Explain the meaning of a

Type II error.








recommended limit.

The hypotheses are

H0

: μ = 0.4 ppm

Ha

: μ > 0.4 ppm

where μ is the mean level of the chemical in

tomatoes from this producer. Explain the

meaning of a Type I error.


91) A hypothesis test is performed at the 5%

significance level to determine whether the

mean body temperature for a certain

population differs from 37.1° C. The

hypotheses are

H0

: μ = 37.1° C

Ha

: μ ≠ 37.1° C.

Explain the difference between statistical

significance and practical significance.

11

92) A hypothesis test for a population mean is to

be performed. The population standard

deviation is known. The hypotheses are

H0

: μ = 80

Ha

: μ ≠ 80.

A 99% confidence interval will also be

constructed for μ. Complete the following

statement concerning the relationship between

the hypothesis test and the confidence interval.

If 80 lies ___ (within/outside) the 99%

confidence interval for μ, then the null

hypothesis will be rejected at the ____ level of

significance.

Perform a one-sample z-test for a population mean using

the P-value approach. Be sure to state the hypotheses and

the significance level, to compute the value of the test

statistic, to obtain the P-value, and to state your

conclusion. Also, assess the strength of the evidence

against the null hypothesis.

93) In one city, the average amount of time that

tenth-graders spend watching television each

week is 21.6 hours. The principal of Birchwood

High School believes that at his school,

tenth-graders watch less television. For a

sample of 28 tenth-graders from Birchwood

High School, the mean amount of time spent

watching television per week was 19.4 hours.

Do the data provide sufficient evidence to

conclude that for tenth-graders at Birchwood

High School, the mean amount of time spent

watching television per week is less than the

city average of 21.6 hours? Perform the

appropriate hypothesis test using a

significance level of 5%. Assume that

σ = 7.2 hours.


of juice in its 16-ounce bottles is 16.1 ounces. A



amount is actually less than this. The mean

volume of juice for a random sample of

70 bottles was 15.94 ounces. Do the data

provide sufficient evidence to conclude that

the mean amount of juice for the 16-ounce

bottles, μ, is less than 16.1 ounces? Perform the

appropriate hypothesis test using a

significance level of 10%. Assume that

σ = 0.9 ounces.

95) A health insurer has determined that the

"reasonable and customary" fee for a certain

medical procedure is $1200. They suspect that

the average fee charged by one particular clinic

for this procedure is higher than $1200. The

insurer wants to perform a hypothesis test to

determine whether their suspicion is correct.

The mean fee charged by the clinic for a

random sample of 65 patients receiving this

procedure was $1280. Do the data provide

sufficient evidence to conclude that the mean

fee charged by this clinic for this procedure is

higher than $1200? Perform the appropriate

hypothesis test using a significance level of 1%.

Assume that σ = $220.


96) Suppose that you wish to perform a hypothesis

test for a population mean. Suppose that the

population standard deviation is unknown, the

population is skewed to the right, and the

sample is large. Would you perform a z-test or

a t-test? Why? Would the test be exact or

approximate?

97) Suppose that you wish to perform a hypothesis

test for a population mean using the P-value

method. Suppose that the population standard

deviation is unknown. The correct procedure

to use is the t-test. If you mistakenly use the

standard normal table to obtain the P-value,

will the value that you obtain be larger or

smaller than the correct value? Does the

mistaken use of the normal table make it more

or less likely that the null hypothesis will be

rejected?

12

Answer KeyTestname: 1342-MAY-PT-MEANS-T-Z

1) Answers will

vary. Possible

answer: The

standard

normal

distribution

has a mean of

0 and a

standard

deviation of 1

while a

nonstandard

normal

distribution

can have any

mean and

standard

deviation. It is

necessary to

standardize

nonstandard

normal

distributions

because a table

of areas is

available for

the standard

normal

distribution

but not for

other normal

distributions.Objective: (6.1)

*KnowConcepts:NormallyDistributedVariables

2) The area under

the normal

curve between

20 and 28 is

0.42.Objective: (6.1)


3) Both graphs

will have the

same shape

(they will both

be

bell-shaped)

and they will

be centered at

the same place

(the common

mean). The

graph of the

variable with

the smaller

standard

deviation will

be narrower

and taller than

the other

graph.Objective: (6.1)


4) right, -0.33Objective: (6.1)

StandardizeNormallyDistributedVariable

5) right, 2.4Objective: (6.1)

StandardizeNormallyDistributedVariable

6) 0.4987Objective: (6.2)

FindAreaUnderStandardNormalCurve

7) 0.7557Objective: (6.2)

FindAreaUnderStandardNormalCurve

8) 0.2040Objective: (6.2)

FindAreaUnderStandardNormalCurveGivenGraph

9) 0.0208Objective: (6.2)


10) 0.8812Objective: (6.2)


11) a, b, dObjective: (6.3)


12) Answers will

vary. Possible

answer: Sketch

a normal curve

and shade the

area associated

with the given

percentage.

Use a table of

areas for the

standard

normal curve

to find the

z-score

delimiting the

shaded area.

Find the

x-value

corresponding

to the z-score

by using x = μ

+ σz.Objective: (6.3)


13


13) Answers will

vary. Possible

answer: You

would look for

0.7 in the body

of the table

since it

represents a

percentage (an

area), not a

z-score.Objective: (6.3)


14) TrueObjective: (6.3)


15) 0.0166Objective: (6.3)

FindPercentage/Probability forNormalVariable

16) 1.96%Objective: (6.3)

FindPercentage/Probability forNormalVariable

17) 22.744 inchesObjective: (6.3)

FindPercentile/Quartile/Decile

18) 30.0Objective: (6.3)

FindPercentile/Quartile/Decile

19) 40%Objective: (7.1)

FindProbability ofSpecifiedSamplingError

20) 20%Objective: (7.1)


21) 60%Objective: (7.1)


22) 47Objective: (7.2)

*FindMean ofSampleMeanUsingListofSamples

23) 265 poundsObjective: (7.2)

*FindMean ofSampleMeanUsingListofSamples

24) Answers will

vary. Possible

answer:

Sampling error

is the error

involved in

using x to

estimate the

population

mean, μ. x

varies from

one sample to

the next

because there

is randomness

involved in

picking the

sample. The

sample

contains

different

people each

time and as a

result x is

different each

time. The

sampling error

can be reduced

by taking a

larger sample.Objective: (7.2)

*KnowConcepts:Mean/StandardDeviationofSampleMean

25) Answers will

vary. Possible

answer: The

numbers

recorded by

Andrew will

have greater

variability. The

standard

deviation of

sample means

(σ/ n) is

smaller than

the standard

deviation of

individual

observations

(σ).Objective: (7.2)


26) FalseObjective: (7.2)


14


27) Answers will

vary. Possible

answer: No.

The mean of

all possible

sample means

(for samples of

size 50) is

equal to 65

inches. We

cannot

conclude that

the mean of

four particular

sample means

will be 65

inches.Objective: (7.2)


28) μx= 110;

σx= 2

Objective: (7.2)FindMean/StandardDeviationofSampleMean

29) μx= 115.3;

σx= 4.1

Objective: (7.2)FindMean/StandardDeviationofSampleMean

30) Tony's

reasoning is

not correct.

Even though

the

distribution of

the individual

observations is

uniform, by

the Central

Limit

Theorem, the

distribution of

the mean of 50

numbers is

approximately

normal. So the

mean of 50

numbers is

more likely to

fall between 3

and 4 than it is

to fall between

1 and 2, since

the interval [3,

4] is closer to

the mean.Objective: (7.3)

*KnowConcepts:SamplingDistribution ofSampleMean

31) DObjective: (7.3)


32) The first may

be solved

using the

Central Limit

Theorem. The

second has too

small a sample

size.Objective: (7.3)


33) The first

problem

requires

application of

the Central

Limit Theorem

because we

need to rely on

the fact that

the sample

mean, x, has a

normal

distribution

even though x

does not. The

second

problem does

not require

application of

the Central

Limit Theorem

since it does

not concern a

sample mean.Objective: (7.3)


34) 83.84%Objective: (7.3)

FindProbability forSamplingError

15


35) 0.9298Objective: (7.3)


36) 0.6876Objective: (7.3)


37) 0.762Objective: (7.3)


38) 17.05 to 27.55

minutesObjective: (8.1)

FindConfidenceIntervalforMean

39) 18.8 to 25.8


FindConfidenceIntervalforMean

40) 4.23 litersObjective: (8.1)

FindPointEstimateofMean

41) 69.8 beats per

minuteObjective: (8.1)

FindPointEstimateofMean

42) Yes. By

definition, the

prescribed

confidence

interval

contains the

value of the

point estimate.Objective: (8.1)

KnowConcepts:Estimating aPopulationMean

43) BObjective: (8.2)

*KnowConcepts:ConfidenceIntervals(SigmaKnown)

44) When the

sample size is

small, the

z-interval

procedure

should be used

only when the

variable under

consideration

is normally

distributed or

very close to

being so.

When the

sample size is

moderate, the

z-interval

procedure can

be used unless

the data

contain

outliers or the

variable under

consideration

is far from

being

normally

distributed.Objective: (8.2)

*KnowConcepts:ConfidenceIntervals(SigmaKnown)

45) Answers will

vary. Possible

answer:

Preliminary

data analysis is

needed to

verify that the

conditions for

using the

z-interval

procedure are

satisfied. This

involves

constructing a

graph such as

a normal

probability

plot, a boxplot,

a

stem-and-leaf

plot, or a

histogram and

examining the

data. If it is

reasonable to

proceed,

calculate the

sample mean

and obtain the

confidence

interval by

using the

formula

x -

zα/2

σ

n to

x + zα/2

σ

n .

Objective: (8.2)*KnowConcepts:ConfidenceIntervals(SigmaKnown)

16


46) 100(1 - α)%Objective: (8.2)

FindAlpha orConfidenceLevelfromOneGiven

47) 14.8 to 16.4

poundsObjective: (8.2)

FindConfidenceIntervalforMean(SigmaKnown)

48) 14.4 to 15.6

ouncesObjective: (8.2)


49) 2.83 to 3.37



50) From 67.4 to

72Objective: (8.3)

*KnowConcepts:Margin ofError

51) In both cases, a

larger sample

would be

needed.

Objective: (8.3)*KnowConcepts:Margin ofError

52) 0.16 ozObjective: (8.3)

FindMargin ofError

53) 2.5 mm HgObjective: (8.3)

FindMargin ofError

54) 92.2Objective: (8.3)

FindMargin ofErrororConfidenceIntervalGivenOne

55) 0.16Objective: (8.3)

FindMargin ofErrororConfidenceIntervalGivenOne


FindSampleSizeGivenMargin ofError


FindSampleSizeGivenMargin ofError

58) For both

t-curves and

the standard

normal curve:

- The total

area under the

curve is 1.

- The curve is

bell-shaped.

- The curve is

symmetrical

about 0.

- The curve

extends

indefinitely in

both

directions,

approaching,

but never

touching, the

horizontal axis

as it does so.

The t-curves

are flatter and

wider than the

standard

normal curve,

though as the

number of

degrees of

freedom

increases, the

t-curves looks

increasingly

like the

standard

normal curve.

Objective: (8.4)*KnowConcepts:ConfidenceIntervals(SigmaUnknown)

59) The resulting

confidence

interval will be

too narrow

and the true

confidence

level will be

less than 95%.Objective: (8.4)

*KnowConcepts:ConfidenceIntervals(SigmaUnknown)


*KnowConcepts:ConfidenceIntervals(SigmaUnknown)

17


61) Neither

Objective: (8.4)*KnowConcepts:ConfidenceIntervals(SigmaUnknown)

62) 214.6 to 237.4

milligramsObjective: (8.4)

FindConfidenceIntervalforMean(SigmaUnknown)

63) 6.50 to 12.30


FindConfidenceIntervalforMean(SigmaUnknown)

64) Examples will

vary.Objective: (9.1)

*KnowConcepts:Nature ofHypothesisTesting I

65) No. The

alternative

hypothesis is

that the true

mean is

greater than

40. A sample

mean much

smaller than

40 does not

provide

evidence in

favor of this

alternative

hypothesis.

The null

hypothesis

should be

rejected only if

the sample

mean turns out

much larger

than 40.Objective: (9.1)


66) Answers will

vary. Possible

answer: The

nature of the

test depends

on what you

are trying to

determine. If

you want to

decide

whether the

population

mean is

greater than a

specified

value, the test

will be

right-tailed. If

you want to

decide

whether the

population

mean is

smaller than a

specified

value, the test

will be

left-tailed. If

you want to

decide

whether the

population

mean differs

from a

specified

value, the test

will be

two-tailed.

The nature of

the test should

be decided

before

selecting the

sample and

may not be

decided by

examining the

sample data.

Objective: (9.1)*KnowConcepts:Nature ofHypothesisTesting I



68) Answers will

vary. Possible

answer. No,

she should not

reject the null

hypothesis. If

H0

were true,

the sample

mean could

easily be as big

as 54.2 by

chance. So the

observed

sample mean

is consistent

with the null

hypothesisObjective: (9.1)


18


69) Examples will

vary. Students

should give an

example of a

test in which

the

consequences

of failing to

reject a false

null

hypothesis

would be

serious.Objective: (9.1)

*KnowConcepts:Nature ofHypothesisTesting II

70) Examples will

vary. Students

should give an

example of a

test in which

the

consequences

of rejecting a

true null

hypothesis

would be

serious.Objective: (9.1)


71) A Type II error

would mean

failing to reject

μ = 16.0

ounces when

in fact μ < 16.0

ounces. The

consumer

advocacy

group would

want to avoid

this and would

therefore want

a small β

probability.

A Type I error

would mean

concluding

that μ < 16.0

ounces when

in fact μ = 16.0

ounces. This

would

represent a

false alarm

situation

where the

manufacturer

would check

the machine

when it is

actually

working

properly. The

manufacturer

would want to

avoid this and

would

therefore want

a small α

probability.

Objective: (9.1)*KnowConcepts:Nature ofHypothesisTesting II

72) A Type II error

would mean

failing to

convict a

guilty man. A

Type I error

would mean

convicting an

innocent man,

which would

be more

serious,

especially

since this is a

death penalty

case.Objective: (9.1)


73) For doctors

and patients it

is more

important to

have a small β

probability. If

the drug really

does increase

blood

pressure, it is

important to

detect that. So

the probability

of a Type II

error should

be small. The

pharmaceutica

l company

may prefer to

have a small α

probability

since they

would lose

money if the

test concluded

that the drug

increased

blood pressure

when in fact it

didn't.Objective: (9.1)


74) Correct

decisionObjective: (9.1)

ClassifyConclusion asTypeI/TypeII/NoError

75) Correct

decisionObjective: (9.1)


76) Type II errorObjective: (9.1)


77) Type I errorObjective: (9.1)


19






80) Right-tailedObjective: (9.1)

ClassifyTestasTwo-,

Left-

, orRight-Tail

ed

81) Two-tailedObjective: (9.1)

ClassifyTestasTwo-,

Left-

, orRight-Tail

ed

82) H0

: μ = 45°F

Ha

: μ ≠ 45°F

Objective: (9.1)DetermineNullandAlternativeHypotheses

83) H0

: μ = 9.0

hours

Ha

: μ ≠ 9.0

hoursObjective: (9.1)

DetermineNullandAlternativeHypotheses

84) A correct

decision

would occur if,

in fact, μ = 21.6

hours, and the

results of the

sampling do

not lead to

rejection of

that fact; or if,

in fact, μ < 21.6

hours and the

results of the

sampling lead

to that

conclusion.Objective: (9.1)

ExplainMeaningofTypeI/TypeII/NoErrorI

85) A Type I error

would occur if,

in fact, μ = 485,

but the results

of the

sampling lead

to the

conclusion that

μ ≠ 485Objective: (9.1)


86) A Type II error

would occur if,

in fact, μ > 8.0

hours, but the

results of the

sampling fail

to lead to that



87) A Type I error

would occur if,

in fact, μ = 8.0

hours, but the

results of the

sampling lead

to the

conclusion that

μ > 8.0 hours.Objective: (9.1)


88) A Type I error

would occur if,

in fact, μ = 16.1

ounces, but the

results of the

sampling lead

to the

conclusion that

μ < 16.1

ounces.Objective: (9.1)


89) A Type II error

would occur if,

in fact, μ < 75

mg, but the

results of the

sampling fail

to lead to that


ExplainMeaningofTypeI/TypeII/NoErrorII

20


90) A Type I error

would occur if,

in fact, μ = 0.4

ppm, but the

results of the

sampling lead

to the

conclusion

that μ > 0.4

ppmObjective: (9.1)

ExplainMeaningofTypeI/TypeII/NoErrorII

91) Answers will

vary. Possible

answer: The

results are

statistically

significant at

the 5%

significance

level if the null

hypothesis is

rejected. This

means that the

data provide

evidence to

conclude that

μ ≠ 37.1° C.

However,

even if the

results are

statistically

significant, this

does not

necessarily

imply practical

significance -

the difference

between μ and

37.1° C could

be too small to

be of practical

importance.Objective: (9.4)

*KnowConcepts:HypothesisTestsforMean

92) outside, 1%Objective: (9.4)


93) H0: μ = 21.6

hours

Ha: μ < 21.6

hours

α = 0.05

z = -1.62

P-value =

0.0526

Since 0.0526 >

0.05, do not

reject H0. At

the 5%

significance

level, the data

do not provide

sufficient

evidence to

conclude that

for

tenth-graders

at Birchwood

High School

the mean

amount of

time spent

watching

television per

week is less

than the city

average of

21.6 hours. The

evidence

against the

null

hypothesis is

moderate.

Objective: (9.4)*SolveApps:PerformHypothesisTestforMean(P-V

alues)

94) H0: μ = 16.1

ounces

Ha: μ < 16.1

ounces

α = 0.10

z = -1.49

P-value =

0.0681

Since 0.0681 <

0.10, reject H0.

At the 10%

significance

level, the data

provide

sufficient

evidence to

conclude that

the mean

amount of

juice for the

16-ounce

bottles is less

than

16.1 ounces.

The evidence

against the

null

hypothesis is

moderate.

Objective: (9.4)*SolveApps:PerformHypothesisTestforMean(P-V

alues)

21


95) H0: μ = $1200

Ha: μ > $1200

α = 0.01

z = 2.93

P-value =

0.0017

Since 0.0017 <

0.01, reject H0.

At the 1%

significance

level, the data

provide

sufficient

evidence to

conclude that

the mean fee

charged by

this clinic for

this procedure

is higher than

$1200.

The evidence

against the

null

hypothesis is

very strong.Objective: (9.4)

*SolveApps:PerformHypothesisTestforMean(P-V

alues)

96) The t-test is

appropriate

since the

population

standard

deviation is

unknown. The

test would be

approximate

since the

population is

nonnormal.Objective: (9.5)


97) The P-value

obtained using

the standard

normal table

will be smaller

than the

correct

P-value which

will make it

more likely

that the null

hypothesis

will be

rejected.Objective: (9.5)


22

Documents

Provide an appropriate response. Use a table of areas to ... tests/1342...Use a table of areas to obtain the shaded area under the standard normal curve. 8)-2.54 -1.27 1.27 2.54 z