Question 1 .0/msohtmlclip1/01/clip_image002.gif” alt=”Description: http://edugen.wileyplus.com/eduge

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

.0/msohtmlclip1/01/clip_image002.gif” alt=”Description: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif”>

.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

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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

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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”>

.0/msohtmlclip1/01/clip_image004.gif” alt=”Description: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif”>

Question
7

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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

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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

.0/msohtmlclip1/01/clip_image002.gif” alt=”Description: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif”>

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

.0/msohtmlclip1/01/clip_image002.gif” alt=”Description: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif”>

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&#39;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

.0/msohtmlclip1/01/clip_image002.gif” alt=”Description: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif”>

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

.0/msohtmlclip1/01/clip_image002.gif” alt=”Description: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif”>

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

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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

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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

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

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

.0/msohtmlclip1/01/clip_image002.gif” alt=”Description: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif”>

.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

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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

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Is a manufacturer’s 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

.0/msohtmlclip1/01/clip_image002.gif” alt=”Description: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif”>

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”>