Group assigment statistic group3

Stung Treng Campus, Cambodia

MBA Program, Promotion I

ASSIGNMENT
Subject: Business Statistics

Lectured by TepVuthy

Group Members:
1. Por Narith
2. Chor Menglon
3. Loeur Sopheaktra

Due date: 21st August 2011

Build Bright University, Stung Treng Campus, MBA, Promotion I, Group 3: Por Narith, Chor Menglon, Loeur Sopheaktra

ACADEMIC YEAR 2010– 2011
Layout of Assignment

The code numbers and page numbers of the problems are extracted from Business Static Book
authored by AMIR D. ACZEL. The problems or exercises are for group 3.

Group Student’s Name Exercise Page Contact
Number Number Number
1-Por Narith 1-6 20 Phone : 012 371 003
1-63 63 E-mail :
3. 2-10 78 narith_por01@yahoo.com
2-23 83
3-9 127
3-14 135
4-3 186
4-17 187
4-38 198
2-Chor Menglon 4-47 199 Phone : 017 920 000
4-54 201
5-1 215
5-15 228
5-22 228
6-20 261
6-28 261 Phone : 077 333 262
3- Loeur Sopheaktra 6-44 265
6-48 307
7-22 261
7-16 305


Solution

Problem 1.6
What is difference between a qualitative and a quantitative variable?

Resolution 1.6
The difference between a qualitative and a quantitative variable included

Qualitative Variable Quantitative Variable

A qualitative or categorical variable simply A quantitative variable can be described by a
record a quality. numbers of which arithmetic operations such
as averaging making sense.
If a number is used for distinguishing
members of different category of a qualitative
variable, the number assignment is arbitrary.

Problem 1.63
The following data are the numbers of tons shipped weekly across the pacific by shipping
company.

398, 412, 560 476, 544, 690, 587, 600, 613, 457, 504, 477, 530, 641, 359, 566, 452, 633, 474
499, 580, 606, 344, 455, 505, 396, 347, 441, 390, 632, 400, 582
Assume these data represent an entire population. Find the population mean and population
standard deviation.

Resolution 1.63
a. Find the population mean
X Observation X-X bar
X1 398 11382.2
X2 412 8590.97
X3 560 3059.47
X4 476 822.973
X5 544 1545.47
X6 690 34340.7
X7 587 6775.35
X8 600 9084.47
X9 613 11731.6
X10 457 2274.1
X11 504 0.47266
X12 477 766.598
X13 530 640.723
X14 641 18581.1
X15 359 21224.8
X16 566 3759.22
X17 452 2775.97


X18 633 16464.1
X19 474 941.723
X20 499 32.3477
X21 580 5671.97
X22 606 10264.2
X23 344 25820.5
X24 455 2468.85
X25 505 0.09766
X26 396 11813
X27 347 24865.3
X28 441 4056.1
X29 390 13153.2
X30 632 16208.5
X31 400 10959.5
X32 582 5977.22
Total 16150 286053
N 32

Sum(X1: X32) 16150
Mean= --------------------------------=-----------------= 504.689
N 32

Σ(X- )2 286053
Std= √---------------= √--------------= √9227.51= 94.547
N 32

Problem 2.10
The European version of roulette is difference from the US version in that the European
roulette wheel doesn’t have 00. How does this change the probability of winning when you bet
on a single numbers? European casino charge a small administration fee, which is not the case
in US casinos. Does this make sense to you based on your answer to the earlier question?

Resolution 2.10
Given that a US casino roulette wheel has 38 slots (00 and 0.36), the odds of landing on a bet
from a single number comes out to be 1 in 38 (or about 2.6%).

Percentage is found by dividing 1 (the numbers of chances you have to get your numbers) by
the numbers of outcome possible (in this case 38)

= 1/38= 0.026316

Problem 2.23
A firm has 550 employees; 380 of them have had at least some college education, and 412 of
the employees underwent a vocational training program. Future more, 357 employees both are
college-educated and have had the vocation training educated or has had the training. If
employee is chosen at random, what probability that he or she is college educated or has had
the trainings or both?


Resolution 2.23
n= 550

P(A): the probability of employees who have had at least some college education.
380
P(A)= ---------------=69.09%
550

The probability of employees who have had at least some college education is 65%.

P(B)= the probability of the employees who underwent a vocational training program.

412
P(B)= ---------------=74.91%
550

The probability of the employees who underwent a vocational training program is 75%.

The probability that he or she is college educated or has had the trainings both include

357
P(A B)=--------------------= 64.91%
550
P(AUB)=(P(A)+P(B))- P(A B)= ((69.09%+74.91%)-64.91%)= 79.09%

The probability that he or she is college educated and has had the trainings includes 79.09%.

Problem 3.9
Return on investment oversee, especially in Europe and the Pacific Rim, are expected to be
higher than those of US market in the near term, and analysis are now recommending
investment international portfolios. An investment consultant believes that the probability
distribution of return (in percent a year) on one such portfolio is as follows:

X(%) P(X)
9 0.05
10 0.15
11 0.3
12 0.2
13 0.15
14 0.1
15 0.05

a. Verify that P(x) is a probability distribution
b. What is the probability that returns will be at least 12%?
c. Find the cumulative distribution of returns

Resolution 3.9
The probability distribution of return (in percent a year) on portfolio is as follows:


X(%) 9 10 11 12 13 14 15
P(X) 0.05 0.15 0.30 0.20 0.15 0.10 0.05

a. A distribution is said to be probability distribution if it satisfies the following condition
P(X) ≥ 0 and ∑P(X) = 1
x
From the above distribution, we can see that are greater than zero and

∑P(X) = 0.05+0.15+0.30+0.20+0.15+0.10+0.05= 1
x

Since the above distribution satisfies both conditions, this is probability distribution.

b. The probability that returns will be at least 12% is
P(X≥12)= P(X=12)+ P(X=13)+ P(X=14)+ P(X=15)= 0.20+0.15+0.10+0.05= 0.5

c. The cumulative distribution of returns is given by

X(%) 9 10 11 12 13 14 15
P(X) 0.05 0.15 0.30 0.20 0.15 0.10 0.05
F(X) 0.05 0.20 0.50 0.70 0.85 0.95 1.00

Problem 3.14
An automobile dealership records the numbers of cars sold each day. The data is used in
calculating the following probabilities distribution of daily sales: Find the mean, variance,
standard deviation of the numbers sold

Resolution 3.14

x P(x) xP(x) x2P(x)
0 0.1 0 0
1 0.1 0.1 0.1
2 0.2 0.4 0.8
3 0.2 0.6 1.8
4 0.3 1.2 4.8
5 0.1 0.5 2.5
Mean of x= 2.8 Mean of x2= 10

a. The mean of numbers of the cars sold each day

E(x)= ΣxP(x)

= > E(x)= 2.8

Average numbers of the cars sole each day is 2.8.

b. Find variance

V(X) = E(x2)-[E(x)]2 = 10-(2.8)2= 2.16

= > Variance= 2.6


The square of deviation from the mean of the cars sold per day is 2.16.

c. Find standard deviation

SD(X) = [V(X)]1/2= [2.6]1/2= 1.47

= > SD(X) = 1.47

The deviation from the mean of the cars sold per day is 1.47.

Problem 4.3
Find the probability that standard normal random variable will have a value between -2.50 and
-0.89.

Resolution 4.3
P (-2.50<Z<-0.89)
P (0<X<2.50) = 0.4938
P (0<X<0.89) = 0.3133

P (-2.50<X<-0.89) = P (0<Z<2.50) - P (0<Z<0.89)
= 0.4938- 0.3133=0.1805 or 18.05% -2.50 -0.89

The probability of the Z value is between -0.89 and -2.50 is 18.05%

Problem 4.17
Find z such that P (Z>z) = 0.12

Resolution 4.17
P (Z>z) = 0.12
So p(Z <z) =0.88

z = 1.174987
z=1.174987

Problem 4.38
If X is a normally distributed random variable with mean 120 and standard deviation is 4.4,
find a value x such that the probability that X will be less than x is 0.56.

Resolution 4.38
 x= µ± Z*σ
 µ= 120
 σ= 44

P(X<x) =0.56
P(X<x)=0.50+0.06
P(X<x)= 0.06=> Z⋲0.16
P(X<x)=0.5= > Z⋲ 6
P(X<x)=0.56= > Z⋲ 6+0.16= 6.16

x = 120± (6.16)*44= 391.04
0.16


Problem 4.47
The demand for unleaded gasoline at a service station is normally distributed with mean
27,009 gallons per day and standard deviation 4,531. Find two values that will give a
symmetric 0.95 probability interval for the amount of unleaded gasoline demand daily.

Resolution 4.47
 µ= 27,009 gallon
 σ=4,530
95%
 P= 0.95%= >Z= 1.96

X= µ± Z*σ

X=27,009 ± 1.96*4530
X=27,009± 8,878.8
X= [18,130.2; 35887.8]

0.95 probability internal for the amount of unleaded gasoline demand daily is between
18,130.2 and 35887.8.

Problem 4.54
A computer system contains 45 identical microchips. The probability that any microchip will
be in working order at a given time is 0.80. A certain operation requires that at least 30 of the
chips be in working order. What is the probability that the operation will be carried out
successfully?

Resolution 4.54
 n= 45
 p=0.80
 q=1-0.80=0.20
 X=30

Binominal distributions in which n are large can be approximated using the normal
approximation. So if X is a binominal random variable with parameter p, then:

X-np
Z=-------------------
√np(1-p)

Where the mean of a binomial variable is np and variance is np(1-p), so in this problem p=0.8,
n= 45, np =36 and np(1-p)=7.2

The probability that the operation will be carried out successfully is given by

1
X- np 30+------ - 36
2
P (X≤30) = P (-------------------≤------------------------ (for the normal approximation)
√np(1-p) √7.2

The continuity correction factor of ½ is included

P(X≤30) = 0.0202


Problem 5.1
Discuss the concepts of parameter, a sample statistic, an estimator, and an estimate. What are
the relations among these entities?

Resolution 5.1
Sample statistic is a numerical measurement of sample. Population parameter is the numerical
measurement of population. Estimator of population parameter is a sample statistic used to
estimate parameter. An estimate of a parameter is a particular numerical value of estimator
obtains by sampling. When a single value is used as estimate, the estimate is called the point
estimate of the population parameter.

Problem 5.15
Under what conditions is the central limit theorem most useful in sampling to estimate the
population mean?

Resolution 5.15
The conditions that the central limit theorem is most useful in sampling to estimate the
population mean includes
 We need to know population standard deviation (σ).
 If, the population standard deviation (σ) is not known, it is replaced with estimator,
sample standard deviation (S).

Problem 5.22
An economist wishes to estimate the average family income in a certain population. The
population standard deviation is known to be $4,500, and the economist uses a random sample
of size n=225. What is the probability that the sample mean will fall within $800 of the
population mean?

Resolution 5.22
 σ= 4500
 n= 225
 µ= 800

The sample mean will fall within $800 of the population mean 81.648%

So, X1= µ-400= 800- 400= 400
X2= µ+400= 800+ 400=1200
X1- µ 400- 800
Z1= -------------------= ----------------------------= - 4/3
(σ/√N) 4500/√225

X2- µ 1200- 800
Z2=------------------= --------------------------= 4/3
(σ/√n) 4500/√225

P (-4/3) = 0.4082
P (4/3) = 0.4082

= > P (-4/3<Z<4/3) = 0.4082+ 0.4082=0.81648

The probability that the sample mean will fall within $800 of the population mean is 81.648%.


Problem 6.20
The manufacture of batteries used in small electrict appliances wants to estimate the average
life of battery. A random sample of 12 batteries yields = 34.2 hours and s=5.9 hours. Give
95% confidence of interval for the average life of batery.

Solution 6.20
 n= 12
 =34.2
95%
 s= 5.9

Confidential 95%= > Z= 1.96
σ
± Z* -------------
√s 1.96
-1.96

5.9
= 34.2 ± 1.96*-----------------------= [30.86, 37.54]
√12

95% confidence of interval for the average life of batery is between 30.86 and 37.54 hours.

Problem 6.28
To aid in planning the development of a tourist shopping area, a state agency wants to estimate
the average dollar amount spent by a tourist in an existing shopping area. A random sample of
56 tourists gives =$258 and s= $85. Give a 95% confidence interval for the average amount
spent by a tourist at the shoping area

Resolution 6.28
 n= 56
 = 258
95%
 s= 85

Confidence of 95%= > Z= 1.96
σ
±Z* -------------
√s
-1.96 1.96

85
= 258±1.96*-------------= [235.74, 293.74]
√56

95% confidence interval for the average amount spent by a tourist at the shoping area is
between 235.74$ and 293.74$.

Problem 6.44
A recent article describes the success of business schools in Europe and the demand on that
continent for MBA degree. The article reports that a survey of 280 European business position
resulted in the conclusion that only one-seventh of the position of MBAs at European business
are currently filled. Assuming that this numbers are exacted and that the sample was randomly


chosen from the entire population of interest, give a 90% confidence interval for the proportion
of filled MBA position in Europe.

Resolution 6.44

p*q
p± Zα/2*√------------
n
l 90%

90% confidence = > Z= 1.64
n= 280
1
Success case= -----*280= 40
7 1.64
-1.64
40
p= ---------=14%
280
q=1-0.14= 86%
0.14*0.86
0.14± 1.64*√--------------------= = [10.6%, 17.4%]
280

90% confidence interval for the proportion of filled MBA position in European is between
10.6% and 17.4%.

Problem 6.48
Before launching its Buyers’ Assurance Program, American Express wanted to estimate the
proportion of cardholders who would be interested in this automatic insurance coverage plan.
A random sample of 250 American Express Card holders was selected and sent questionnaires.
The result was that 121 people expressed interest in the plan. Give a 99% confidence interval
for the proportion of all interested American Express cardholders.

Resolution 6.48

 99% confidence= > Z= 2.58
 n=250

Success case=121
Success Case 121 99%
= >p=------------------------------= ---------------=48.4%
n 250
q=1-p = 1- 0.484= 51.6%

p*q
p± Zα/2*√------------
n -2.58 2.58

0.484*0.516
0.484± 2.58*---------------------------= [40.25%, 56.55%]
250


99% confidence interval for the proportion of all interested American Express cardholders is
between 40.25% and 56.55%.

Problem 7.16
An automobile manufacturer substitutes a different engine in cars that were known to have an
average miles-per-gallon ratting of 31.5 on the highway. The manufacturer wants to test
whether the new engine changes the miles-per-gallon rating of the automobile model. A
random sample of 100 trial runs gives x-bar=29.8 miles per gallon and s=6.6. Using the 0.05
level of significance, is the average miles-per-gallon ratting on the highway for cars using the
new engine different from the ratting for cars using the old engine?

Resolution 7.16
 µ =31.5
 n=100
 X-bar=29.8
 s=6.6
 α =0.05

An automobile manufacturer substitutes a different engine in cars that were known to have an
average miles-per-gallon ratting of 31.5 on the highway

We want to test whether the new engine changes the miles-per-gallon rating of the automobile
model.

Null hypothesis
H0= the average miles-per-gallon rating on the highway for cars using the new engine is not
significantly different from the ratting for cars using the old engine.
H0= µ= 31.5 mpg

Alternative Hypothesis
H1= the average miles-per-gallon rating on the highway for cars using the new engine is
significantly different from the ratting for cars using the old engine.
H1= µ≠31.5 mpg

The level of significance, α= 0.05

A random sample of 100 trial runs give X-bar = 29.8 mpg and s =6.6
α/2= α/2=
Under H0, the test statistic is given by 0.025 0.025

X-bar- µ0 29.8-31.5 Z=- Z0=- Z0=1.
Z=---------------=-------------------=- 2.58 2.58 1.96 96

σ/√n 6.6/√100

As the Z value is in the lower rejection area, H0 is rejected. The given data provides strong
evidence to conclude that the average miles-per-gallon ratting on the highway for cars using the
new engine is significantly different from the ratting for cars using the old engine.

Problem 7.22
Average total daily sales at a small food store are known to be $452.80. The store’s
management recently implemented some changes in displays of goods, order within aisles, and
other changes, and it now wants to know whether average sales volume has changed. A


random sample of 12 days show X-bar= $501.9 and s= $65.00. Using α = 0.05, is the sampling
result significant? Explain

Resolution 7.22

 µ0=452.80
 n=12
 X-bar=501.90
 s=65
 α=0.05

It was supposed to test the average total daily sales has changed from $452.8 or not.
H0= the average total daily sales has not changed from 452.8
µ0= $452.8
H1= the average total daily sales have changed from $452.8.
µ0≠ $452.8

The level of significance, α= 0.05

The test statistic is

X-bar - µ
t= -------------------------
s√n

501.90-452.80
t =--------------------------------=2.62
65/√12

P-value of the test = 2P (T>2.62) = 0.024

Since the P-value is less than the significance level, α=0.05, we reject the null hypothesis.
Hence, we may conclude that the average total daily sales have changed from $452.8.


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