Question

1

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Fifty-three percent of the population

possesses a certain characteristic N. If 15 people are randomly selected from

the population, what is the probability that exactly 9 possess characteristic

N?

.1653

.1780

.1102

.2009

Question

2

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.johnwiley.net.au/highered/bus_stats/demo_probs/5-4.html”>Click

here to access demo problem

Solve the following problems by using

the binomial tables.

a.Ifn= 20 andp= 0.4, find .0/msohtmlclip1/01/clip_image003.gif”>=.0/msohtmlclip1/01/clip_image001.gif” alt=”Description: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif”>.

b.Ifn= 20 andp= 0.4, find .0/msohtmlclip1/01/clip_image003.gif”>=.0/msohtmlclip1/01/clip_image001.gif” alt=”Description: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif”>.

c.Ifn= 20 andp= 0.6, find .0/msohtmlclip1/01/clip_image003.gif”>=.0/msohtmlclip1/01/clip_image001.gif” alt=”Description: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif”>.

d.Ifn= 20 andp= 0.8, find .0/msohtmlclip1/01/clip_image003.gif”>=.0/msohtmlclip1/01/clip_image001.gif” alt=”Description: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif”>.

e.Ifn= 15 andp= 0.4, find .0/msohtmlclip1/01/clip_image003.gif”>=.0/msohtmlclip1/01/clip_image001.gif” alt=”Description: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif”>.

f.Ifn= 10 andp= 0.7, find .0/msohtmlclip1/01/clip_image003.gif”>=.0/msohtmlclip1/01/clip_image001.gif” alt=”Description: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif”>.

Round your answer to 3 decimal places.

The tolerance is +/- 0.005

Question

3

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In

a binomial experiment, p is .29 and n = 18. What is the standard deviation of

this binomial distribution?

3.425

1.046

1.925

2.129

Question 4

Suppose a researcher wants to work a

binomial problem by using a normal curve approximation. The researcher should

use the:

binomial correction factor

adjustment for discrete distributions

correction for continuity

finite correction factor

Question 5

A

researcher is working a binomial problem using a normal curve approximation.

In the binomial problem, the researcher is trying to determine the

probability of 51 < x=””>< 56.=”” in=”” working=”” the=”” problem=”” by=”” the=”” normal=”” curve,=”” the=”” solution=”” will=”” be=”” found=”” at:=”” 50.5=””>< x=””>< 55.5=”” 50.5=””>< x=””>< 56.5=”” 50.5=”” ≤=”” x=”” ≤=”” 56.5=”” 51.5=””>< x=””>< 55.5=”” question=”” 6=”” .0/msohtmlclip1/01/clip_image002.gif”=”” alt=”Description: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif”>

Suppose 57% of all shoppers use credit cards for their purchase

in department stores. If 80 such shoppers are randomly selected, what

is the probability that more than 49 use a credit card?

.2578

.2946

.1894

.3106

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.0/msohtmlclip1/01/clip_image005.gif” alt=”Description: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif”>

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Question

7

Suppose 29% of all commuter cars

leaving the city at 5 P.M. are going somewhere other than home. If 45 commuter

cars leaving downtown at 5 P.M. are tracked, what is the probability that more

than 11 are going somewhere other than home?

.1950

.6950

.7486

.2486

Question

8

A company believes that it controls

more than 40% of the total market share for one of its products. To prove this

belief, a random sample of 165 purchases of this product are contacted. It is

found that 70 of the 165 purchased this company’s brand of the product. If a

researcher wants to conduct a statistical test for this problem, the

alternative hypothesis would be:

the population proportion is greater

than 0.40.

the population proportion is not

equal to 0.40.

the population mean is equal to 0.40.

the population proportion is less

than 0.40.

Question

9

A bank inspector monitors the default

rate on personal loans at Victorian banks. One standard that she examines is

that no more than 5% of personal loans should be in default. On each Friday,

the default rate is calculated for a sample of 500 personal loans. Last

Friday’s sample contained 30 defaulted loans. The bank inspector’s alternate hypothesis

is:

p ≠ .05

p = .05

p < 0.05=”” p=””> 0.05

Question

10

A

business researcher is testing the following hypotheses using a 10% level of

significance.

H0: p = .70

Ha: p < .70=”” a=”” sample=”” of=”” 415=”” is=”” taken=”” and=”” the=”” sample=”” proportion=”” is=”” .66.=”” the=”” business=”” researcher's=”” decision=”” from=”” this=”” test=”” is:=”” not=”” enough=”” information=”” to=”” conduct=”” the=”” hypothesis=”” test.=”” reject=”” the=”” null=”” hypothesis.=”” take=”” a=”” larger=”” sample.=”” fail=”” to=”” reject=”” the=”” null=”” hypothesis.=”” question=”” 11=”” .0/msohtmlclip1/01/clip_image002.gif”=”” alt=”Description: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif”>

A

study by Hewitt Associates showed that 79% of companies offer employees

flexible scheduling. Suppose a researcher believes that in accounting firms

this figure is lower. The researcher randomly selects 415 accounting firms and

through interviews determines that 303 of these firms have flexible scheduling.

With a 1% level of significance, does the test show enough evidence to conclude

that a significantly lower proportion of accounting firms offer employees

flexible scheduling?

The

value of the test statistic rounded to 2 decimal places isz=.0/msohtmlclip1/01/clip_image001.gif” alt=”Description: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif”>and we.0/msohtmlclip1/01/clip_image001.gif” alt=”Description: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif”>.

The

tolerance is +/- 0.05.

Question

12

A business researcher wants to compare

her observed distribution of frequency data to an expected distribution of data

using the chi-square goodness-of-fit test. The data are given below. The

degrees of freedom for this test are:

observed

expected

12

9

20

7

38

32

24

38

18

20

11

7

5

4

10

6

Question

13

A business researcher wants to compare

her observed distribution of frequency data to an expected distribution of data

using the chi-square goodness-of-fit test. The data are given below. The

observed chi-square value for this test is:

observed

expected

15

9

20

7

38

32

24

38

18

20

11

10

34.73

33.9

47.63

41.46

Question 14

A researcher wants to test the

following observed distribution of values to determine if the values are

uniformly distributed. The researcher is using the chi-square goodness-of-fit

test for this analysis.

213

219

209

210

216

199

217

213

For α = .10, the researcher’s

decision is to:

fail to reject the null hypothesis that the observed distribution is

not uniform.

reject the null hypothesis that the observed distribution is

uniform.

reject the null hypothesis that the observed distribution is not

uniform.

fail to reject the null hypothesis that the observed distribution is

uniform.

Question

15

The following percentages come from a

national survey of the ages of prerecorded-music shoppers. A local survey

produced the observed values. Does the evidence in the observed data indicate

that we should reject the national survey distribution for local prerecorded-music

shoppers? Use α = 0.01(Alpha

= 0.01)

Find the observed value of chi-square.

Round the answer to 2 decimal places.

Age

Precent

from Survey

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10-14

9

22

15-19

23

50

20-24

22

43

25-29

14

29

30-34

10

19

.0/msohtmlclip1/01/clip_image003.gif”>35

22

49

The observed x2 =.0/msohtmlclip1/01/clip_image001.gif” alt=”Description: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif”>.

There is.0/msohtmlclip1/01/clip_image001.gif” alt=”Description: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif”>evidence to declare that the distribution of

observed frequencies is different from the distribution of expected

frequencies.

The tolerance is +/- 0.05.

Question

16

.johnwiley.net.au/highered/bus_stats/demo_probs/12.4/index.html”>Click

here to access demo problem

A group of 30-year-olds is interviewed

to determine whether the type of music most listened to by people in their age

category is independent of the geographic location of their residence. Use the

chi-square test of independence, α = 0.01 (Alpha = 0.01), and the following contingency table to determine

whether music preference is independent of geographic location.

Find the observed value of chi-square.

Round the answer to 2 decimal places.

Type

of Music Preferred

Geographic

Region

Rock

R & B

Country

Classical

Victoria

137

28

6

18

Western Australia

130

36

47

6

Tasmania

154

25

10

12

The observed x2 =.0/msohtmlclip1/01/clip_image001.gif” alt=”Description: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif”>.

Type of music preferred is.0/msohtmlclip1/01/clip_image001.gif” alt=”Description: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif”>of region of the country.

The tolerance is +/- 0.05.

Question

17

A researcher interviewed 2,067 people

and asked whether they were the primary decision makers in the household when

buying a new car last year. Two hundred seven were men and had bought a new car

last year. Sixty-five were women and had bought a new car last year. Eight

hundred eleven of the responses were from men who did not buy a car last year.

Nine hundred eighty-four were from women who did not buy a car last year. Use

these data to determine whether gender is independent of being a major decision

maker in purchasing a car last year. Let α = 0.05 (Alpha = 0.05)

Find the observed value of chi-square.

Round the answer to 2 decimal places.

The observed x2 =.0/msohtmlclip1/01/clip_image001.gif” alt=”Description: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif”>.

Purchasing a car or not is.0/msohtmlclip1/01/clip_image001.gif” alt=”Description: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif”>of gender.

The tolerance is +/- 0.05.

Question

18

Is a manufacturerâs geographic location

independent of type of customer? Use the following data for companies with

primarily industrial customers and companies with primarily retail customers to

test this question. Let α= 0.10 (Alpha = 0.10)

Find the observed value of chi-square.

Round the answer to 2 decimal places.

Geographic Location

Customer

Type

Northern Territory

Western Australia

South Australia

Industrial

Customer

230

115

68

Retail

Customer

185

143

89

The observed x2 =.0/msohtmlclip1/01/clip_image001.gif” alt=”Description: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif”>.

Type of customer is.0/msohtmlclip1/01/clip_image001.gif” alt=”Description: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif”>of geographic region.

The tolerance is +/- 0.05.

Question

19

The

game showDeal or No Dealinvolves a series of opportunities for the

contestant to either accept an amount of money from the show’sbankeror to decline it and open a specific number of

briefcases in the hope of exposing and, thereby eliminating, low amounts of

money from the game, which would lead the banker to increase the amount of the

next offer. Suppose that 700 people aged 21 years and older were selected at

random. Each of them watched an episode of the show until exactly four

briefcases were left unopened. The money amounts in these four briefcases were

$750, $5000, $50000, and $400000, respectively. The banker’s offer to the

contestant was $81600 if the contestant would stop the game and accept the

offer. If the contestant were to decline the offer, he or she would choose one

briefcase out of these four to open, and then there would be a new offer. All

700 persons were asked whether they would accept the offer (Deal) for $81600 or

turn it down (No Deal), as well as their ages. The responses of these 700

persons are listed in the following table.

Age

Group (years)

21-29

30-39

40-49

50-59

60

and over

Deal

76

84

89

92

63

No

Deal

56

70

60

63

47

Test

at the 5% significance level whether the decision to accept or not to accept

the offer (Deal or No Deal) and age group are dependent.

The

decision to accept/not accept the offer and age group are.0/msohtmlclip1/01/clip_image001.gif” alt=”Description: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif”>