chapter3 statistic

Chapter 3:Sampling Distribution

113

CHAPTER 3 : SAMPLING DISTRIBUTIONS

Sub-Topic

Sampling error.

Introduction to sampling distribution.

Sampling distribution of single mean.

Sampling distribution of the difference between two means.

Sampling distribution test : t distribution, 2 - distribution and F-distribution.

Chapter Learning Outcome

Solve the problems involve the sampling distributions for the single and two

population means.

Learning Objective

By the end of this chapter, students should be able to

Understand the concept of sampling error.

Determine the mean and standard deviation for the sampling distribution of

the sample mean.

Understand the importance of the Central Limit Theorem.

Apply the sampling distributions for mean and difference between two means.

Key Term (English to Bahasa Melayu)

English Bahasa Melayu

1. Parameter → Parameter

2. Statistic → Statistik

3. Sampling error → Ralat pensampelan

4. Central Limit Theorem → Teorem had memusat

5. Sampling distribution → Taburan pensampelan

6. Simple random sample → Sampel rawak mudah


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3.1 Sampling error

Definition 1

Sampling error of single mean is the difference between values (a statistic)

computed from a sample and the corresponding value (a parameter) computed from a

population.

Theory 1

Formula sampling error of single mean : _

xe

where

_

x sample mean

population mean.

Definition 2

A parameter is a measure computed from the entire population.

Definition 3

A statistics is a measure computed from a sample that has been selected from a

population.

Example 1

If given that the mean population is 158972 square feet and a sample size of five

shopping centre yields with sample mean 155072 square feet. Find the sampling

error.

Answer Example 1

We know that 158972 and 155072_

x .

The sampling error


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39003900158972155072_

xe Square feet.

Theory 2

Formula for population mean, N

x

where

population mean

x values in the population

N population size.

Theory 3

Fundamental statistical concepts are

The size of the sampling error depends on which sample is taken.

The sampling error may be positive or negative.

There is potentially a different value for each possible sample mean.

Definition 4

A simple random sample is a sample selected in such a manner that each possible

sample of a given size has an equal chance of being selected.

Theory 4

Formula for sample mean, is n

xx

_

where

_

x sample mean

x sample value selected from the population

n sample size.


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3.2 Introduction to sampling distribution

In the inferential statistics process, a researcher selects a random sample from the

population, computes a statistics on the sample and reaches conclusions about the

population parameter from the statistics. In this chapter, we will explore the sample

mean, __

x , as the statistic. The sample means is one of the more common statistics

used in the inferential process. To compute and assign the probability of occurrence

of a particular value of a sample mean, the researcher must know the distribution of

the sample means. One way to examine the distribution possibilities is to take a

population with a particular distribution, randomly select samples of a given size,

compute the sample means and attempt to determine how the means are distributed.

Let say 23 people selected randomly from the population of women in Ayer Hitam

Pahat between the ages of 20 and 40 years old and we computed the mean height of

the sample. We would not expect our sample mean to be equal to the mean of all

women in Ayer Hitam. It might be somewhat lower or it might be somewhat higher,

but it would not equal the population mean exactly. Similarly, if we took a second

sample of 23 people from the same population, we would not expect the mean of this

second sample to equal the mean of the first sample. Inferential statistics concerns

generalizing from sample to population. A critical part of inferential statistics

involves determining how far sample statistics are likely to vary from each other and

from the population parameter. Why we sample the population ? Why not we study

the whole population ? These all because the physical impossibility of checking all

items in the population, the cost of studying all the items in a population, the sample

results are usually adequate, contacting the whole population would often be time-

consuming and the last one is the destructive nature of certain tests such as a study of

light bulb life.

Definition 5

If samples of size n are drawn randomly from a population that has a mean of and

a standard deviation of 2 , the sample means, __

x , are approximately normally


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distributed for sufficiently large sample size ( 30n ) regardless of the shape of the

population distribution. If the population is normally distributed, the sample means

are normally distributed for any size sample. It can be shown that the mean of the

sample means is the population mean, which is __

x

and standard deviation of the

sample means (called the standard error of the mean) is the standard deviation of the

population divided by the square root of the sample size, which is nx

__ .

Definition 6

A sampling distributions is a distribution of the possible values of a statistic for a

given size sample selected from a population.

Definition 7

Sampling distribution of the mean, for random samples of n observations taken

from a population with mean, and a standard deviation, , regardless of the

population’s distribution, provided the sample size is sufficiently large, the

distribution of the sample mean, _

x will be normal with a mean equal to the

population mean, _

x

. Further, the standard deviation will equal the population

standard deviation divided by the square-root of the sample size, nx

_ . The

larger sample size is, the better an approximation to the normal distribution.

3.3 Sampling distribution of the single mean

Theory 5

Z-value for sampling distribution of _

x is n

xZ

_

where

_

x sample mean


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population mean

population standard deviation

n sample size

Theory 6 ( Calculation probability of single mean )

In order to calculate the probability of single mean, we need to follow four steps

below.

Step 1 : Write the mean of sample mean, __

x which is _

x

.

Step 2 : Write the standard deviation of sample mean, __

x which is nx

_

.

Step 3 : Write the distribution in normal distribution form which is

2

__

____ ,~xx

Nx .

Step 4 : Find the probability of sample mean,

__

____

x

x

rZPrxP

.

Example 2

What is the probability that a sample of 100 automobile insurance claim files will

yield an average claim of RM4527.77 or less if the average claim for the population is

RM4560 with standard deviation of RM600 ?

Answer Example 2

Given 100n , 77.4527_

x , 4560 and 600 .

Step 1 : 4560_ x

Step 2 : 60100

600_

nx

Step 3 : 2__

60,4560~ Nx

Step 4 :

60

456077.452777.4527

__

ZPxP


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54.0 ZP

)54.0( ZP

2946.0

Example 3

The random variable, X represent the number of box in a container, has the following

probability distribution.

X 4 5 6 7

P(x) 0.2 0.4 0.3 0.1

(a) Find the population mean and variance.

(b) Find the sample mean and variance for random samples of 36 boxes.

(c) Calculate the probability if the average number of box in 36 containers will be

less than 5.5.

Answer Example 3

(a) )(.)( xPxXE

)1.0(7)3.0(6)4.0(5)2.0(4

7.08.10.28.0

3.5

)1.0(7)3.0(6)4.0(5)2.0(4)( 22222 XE

9.48.10102.3

9.28

22 )()()(Var XEXEX

2)3.5(9.28

81.0


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(b) Mean sample, 3.5_ x

Variance sample, 0225.036

81.022_

nx

(c) Step 1 : 3.5_ x

Step 2 : 15.00225.036

81.02

_ nx

Step 3 : 2__

15.0,3.5~ Nx

Step 4 :

15.0

3.55.55.5

__

ZPxP

)33.1( ZP

33.11 ZP

09176.01

90824.0

Example 4

An electrical firm manufactures light bulbs that have a length of life that is

approximately normally distributed, with mean equal to 800 hours and a standard

deviation of 40 hours. Find the probability that a random sample of 16 bulbs will have

an average life of less than 775 hours.

Answer Example 4

Step 1 : 800_ x

Step 2 : 1016

40_

nx

Step 3 : 100,800~__

Nx

Step 4 :

10

800775775

__

ZPxP


121

)5.2( ZP

0062.0

Example 5

At a large university, the mean age of the students is 22.3 years and the standard

deviation is 4 years. A random sample of 64 students is drawn. What is the

probability that the average age of these students is greater than 23 years ?

Answer Example 5

Step 1 : 3.22_ x

Step 2 : 5.064

4_

nx

Step 3 : 2__

5.0,3.22~ Nx

Step 4 :

5.0

3.222323

__

ZPxP

)4.1( ZP

0808.0

Example 6

The breaking strength (in kg/mm) for a certain type of fabric has mean 1.86 and

standard deviation 0.27. A random sample of 80 pieces of fabric is drawn. What is the

probability that the sample mean breaking strength is less than 1.8 kg/mm ?

Answer Example 6

Given : 86.1 , 27.0 and 80n

Step 1 : 86.1_ x

Step 2 : 03018.080

27.0_

nx


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Step 3 : 2__

03018.0,80~ Nx

Step 4 :

03018.0

86.18.18.1

__

ZPxP

)99.1( ZP

0233.0

Example 7

Taking random samples of size n from an infinite population that has a standard

deviation two, show that __

x would be a more precise estimator of if sample size

were increased from four to six. Interpret the result.

Answer Example 7

The precision of __

x as an estimator of is measured by the standard deviation,

x

.

For sampling from an infinite population, nx

.

Therefore, for 2 and 4n : 142 nx

Increasing n from 4 to 16 : 5.0162 nx

Thus, with an increasing in sample size and a constant σ,

x

decreases by 50%.

Exercise 3.3

1. Bags of concrete mix labeled have a population mean weight of 100 kg and a

population standard deviation of 0.5 kg.

(a) What is the probability that the mean weight of a random sample of 50

bags is less than 99.9 kg ?

(b) If the population mean weight is increased to 100.15 kg, what is the

probability that the mean weight of a sample of size 50 will be less

than 100 kg ?


123

2. In a report stated that the average time of watching movie per week for

children with ages between two and six years is 22 hours. Assume the variable

is normally distributed and the standard deviation is five hours. A sample of

33 children with ages between two and six years is randomly selected. Find

the probability that the average time they watch movie per week will be

greater than 23.5 hours.

3. Women from aged 18 to 24, their systolic blood pressures (in mm Hg) are

normally distributed with a mean of 114.4 and a standard deviation of 13.1.

(a) If five women between the ages of 18 to 24 are randomly selected, find

the probability that her systolic blood pressure is greater than 120.

(b) If twelve women between the ages of 18 to 24 are randomly selected,

find the probability that the average of their systolic blood pressure is

less than 115.

4. A random sample of hundred is taken from a normally distributed population

with mean 20 and standard deviation equal to one. What is the probability that

__

x will take on a value 20 and 20.2 inclusive ?

5. Engineers must consider the breadths of male heads when designing

motorcycle helmets. Men have head breadths that are normally distributed

with a mean of 15.24 cm and a standard deviation of 2.54 cm.

(a) If one male is randomly selected, find the probability that his head

breadth is less than 15.75 cm.

(b) Find the probability that 100 randomly selected men have a mean head

breadths at least 16.00 cm.

6. The lifetime of a particular type of battery is normally distributed with a mean

of 1100 days and a standard deviation of 80 days. The manufacturer randomly

selects 400 batteries of this type and ships them to a departmental store.


124

(a) What is the mean and standard deviation of the sampling distribution

of __

x ?

(b) What is the probability that the average lifetime of these 400 batteries

is between 1097 and 1104 days ?

7. The time required to assemble an electronic component is normally distributed

with a mean of 25 minutes and a standard deviation of 3.5 minutes. Find the

probability that the average time required to assemble all 19 components is at

least 23 minutes.

8. The amount of sulfur in the daily emissions from a power plant has a normal

distribution with a mean of 94 and a standard deviation of 22. For a random

sample of 5 days, find the probability that the average amount of sulfur

emissions will exceed 80.

9. According to the growth chart that doctors use as a reference, the heights of

two-year-old boys are normally distributed with mean 34.5 inches and

standard deviation 1.3 inches. If six two-year-old boys are selected, what is

the probability that their average height will be between 34.1 and 35.2 inches.

10. Casual workers in a certain industry are paid on average RM5.10 per hour

which is normally distributed with standard deviation of RM2.20. A sample of

35 casual workers from the industry was selected to be respondents for the

underpaid issue questionnaires. Find the probability that the average payment

for those casual workers is

(a) at least RM6.00 per hour.

(b) greater than RM4.80 per hour.

11. Intelligent Quotients (IQ) in the general population are normally distributed

with a mean of 100 and a standard deviation of 15. A random sample of 40


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students was taken in a certain university. Find the probability that the mean

IQ of the sample is

(a) greater than 105 and less than 107.

(b) not more than 109.

12. Given a random sample of 40,321 ,,, XXXX which is drawn from

population with Poisson distribution 3.5. Find probability that the sample

mean is between 3.4 and 4.3.

13 PVC pipe is manufactured with mean diameter is 3.2 cm and standard

deviation is 1.6 cm. The distribution of diameter is normal. Find the

probability that a random sample of 64 pipes will have a sample mean

diameter is less than three centimeter.

14 Consider the PVC pipe in the previous question. How is the standard

deviation of the sample mean changed when the sample size is decreased from

64 to 9 ? Explain.

Answer Exercise 3.3

1. (a) 0.0793 (b) 0.0170

2. 0.0427

3. (a) 0.1685 (b) 0.5636

4. 0.1359

5. (a) 0.5793 (b) 0.0014

6. (a) 1100, 4 (b) 0.6147

7. 0.9936

8. 0.9222

9. 0.6799

10. (a) 0.0078 (b) 0.7910

11. (a) 0.0159 (b) 0.9826


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12. 0.6296

13. 0.1587

3.4 Sampling distribution of the difference between two means

Theory 7

Statistical analyses are very often concerned with the difference between means. A

typical example is an experiment designed to compare the mean of a control group

with the mean of an experimental group. Inferential statistics used in the analysis of

this type of experiment depend on the sampling distribution of the difference between

means.

Theory 8 ( Calculation probability of two means)

In order to calculate the probability of two means, we need to follow four steps

below.

Step 1 : Write the mean of sampling distribution which is, 212

__

1

__ xx

Step 2 : Write the standard deviation of sample mean which is,2

22

1

12

2

__

1

__

nnxx

.

Step 3 : Write the distribution in normal form,

2__

2

__

1

__

2

__

1

__ ,~xxxx

Nx .

Step 4 : Find the probability of sample mean,

2

_

1

_

1

_

1

___

2

__

1

xx

xx

r

ZPrxxP

.

Example 8

The mature citrus trees of type A have a mean height of 14.8 feet with a standard

deviation of 1.2 feet. The mature citrus trees of type B have a mean height of 12.9

feet with a standard deviation of 1.5 feet. Two samples of size 12 and 15 are

randomly selected from mature citrus tree of type A and B respectively. Find the

probability that


127

(a) the mean of type A is more than 14 feet.

(b) the mean of type B is between 12 to 14 feet.

(c) the mean of type A is two feet more than the mean of type B.

Answer Example 8

(a)

A B

Sample mean 14.8 12.9

Sample standard deviation 1.2 1.5

Sample size 12 15

Step 1 : 8.14__ AxA

Step 2 : 34641.012

2.1__

n

A

x

Step 3 : 2__

34641.0,8.14~ Nx A

Step 4 :

34641.0

8.1414)14(

__

ZPxP

)31.2( ZP

)31.2(1 ZP

01044.01

9896.0

(b) Step 1 : 9.12__ BxB

Step 2 : 38729.015

5.1__

n

B

x

Step 3 : 2__

38729.0,9.12~ Nx B

Step 4 :

38729.0

9.1214

38729.0

9.12121412

__

ZPxP

)84.232.2( ZP

)84.2()32.2(1 ZPZP


128

00226.001017.01

9878.0

(c) Step 1 : 9.19.128.14__ BAx

Step 2 : 51961.015

5.1

12

2.1 2222

__ B

B

A

A

x nn

Step 3 : 2____

51961.0,9.1~ Nxx BA

Step 4 :

51961.0

9.122

____

ZPxxP BA

)19.0( ZP

4247.0

Example 9

The result of Statistics Test 1 for two groups of management students, Section 1 and

Section 2 are normally distributed with )4,60( 2N and )2,64( 2N respectively. Two

samples of size 9 and 12 are randomly selected from Section 1 and Section 2

respectively. Find the probability that the mean of Section 2 is lower than the mean

of Section 1 ?

Answer Example 9

Section 1 Section 2

Sample mean 60 64

Sample variance 16 4

Sample size 9 12

Step 1 : 460641212

xx

Step 2 : 4529.112

4

9

16

2

2

2

1

2

1

12

nnxx


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Step 3 : 21

__

2

__

4529.1,4~ Nxx

Step 4 :

01

__

2

__

1

__

2

__

xxPxxP

4529.1

40ZP

)75.2( ZP

)75.2( ZP

0030.0

Example 10

Consider two populations of students who participate in a reading programmed prior

to taking a Japanese course. The populations are those who earn an A grade and those

who earn a B grade. Let X be the number of books read by the students who

participate in the programmed. Find the probability that the mean number of books

read by the students who earn A grade is greater than the students who earn B grade if

given the data below.

Grade A Grade B

Sample mean 37 25

Sample standard deviation 8.7014 8.5264

Sample size 8 6

Answer Example 10

Step 1 : 122537

BA

xx BA

Step 2 : 6455.46

5264.8

8

7014.8 2222

B

B

A

A

xx nnBA

Step 3 : 2____

6455.4,12~ Nxx BA

Step 4 :

0

________

BABA xxPxxP


130

6455.4

120ZP

)58.2( ZP

)58.2(1 ZP

00494.01

99506.0

Example 11

The length of computer desk is approximately normal distributed. There are two

factories produce that kind of desk. The summary statistics are given below.

Factory A Factory B

Sample mean 60.5 58.3

Sample standard deviation 3 4

Sample size 35 40

Find the probability the mean sample for the length of computer desk produced by

Factory B is greater than mean sample for the length of computer desk produced by

Factory A.

Answer Example 11

Step 1 : 2.25.603.58

AB

xx AB

Step 2 : 81064.040

4

35

3 2222

B

B

A

A

xx nnAB

Step 3 : 2____

81064.0,2.2~ Nxx AB

Step 4 :

0

________

ABAB xxPxxP


131

81064.0

)2.2(0ZP

0034.0

Exercise 3.4

1. The usage of electricity at residential area A is normally distributed with mean

of 156 kilowatt per hour and standard deviation of 43 kilowatt per hour.

Meanwhile the usage of electricity at residential area B is also normally

distributed with mean of 161 kilowatt per hour and its standard deviation is 48

kilowatt per hour. Two samples of size 20 and 25 residences are randomly

selected from residential area A and residential area B, respectively. Find the

probability that the mean of usage at residential area A is lower than the mean

of usage at residential area B.

2. A company manufactures two types of cables, brand A and brand B that have

mean breaking strengths of 4000 kg and 4500 kg and standard deviations of

300 kg and 200 kg, respectively. If 100 cables of brand A and 50 cables of

brand B are tested, what is the probability that the mean breaking strengths of

brand B will be at least 600 kg more than brand A ?

3. The effective life of a component used in a jet-turbine aircraft engine is a

random variable with mean 3465 hours and standard deviation 25 hours. The

distribution of effective life is fairly close to a normal distribution. The engine

manufacturer introduces an improvement into the manufacturing process for

this component that increases the mean life to 4050 hours and decreases the

standard deviation to 15 hours. Given a random sample of 20 components is

selected from the old process and 35 components is selected from the

improved process. What is the probability that the difference between two

sample mean improved process and old process is at most 23 hours ?


132

4. The average running times of films produced by Company A are 98.4 minutes

with standard deviation of 7.8 minutes. Companies B have a mean running

times of 110.7 minutes with standard deviation of 29.8 minutes. Assume the

populations are approximately normally distributed. What is the probability

that a random sample of 36 films from Company B will have mean running

times that at least 13 minutes more than the mean running times of a random

sample of 49 films from Company A.

5. The elasticity of polymer is affected by the concentration of a reactant. When

low concentration is used, the true mean elasticity is 55 and when high

concentration is used the mean elasticity is 60. The standard deviation of

elasticity is 4, regardless of concentration. The distribution of elasticity is

normally distributed. Two random samples of size 16 are taken. Find the

probability that the difference mean between high concentration and low

concentration is more than two.

6. If given two populations of UTHM students who participate in a debate

competition France Language. The populations are those who get a Score A

and Score B. Let X is the number of questions answered by the students who

participate in the competition. Find the probability that the mean number of

questions collect by the students who get Score B is at most than the students

who get Score A if given the data such as below.

Score A Score B

Sample mean 83 91

Sample standard deviation 12 8

Sample size 15 14

7. The mean age at death in Malaysia is 55.5 years and Singapore is 57 years.

The standard deviation is approximately 4.6 years and 5 years for each

country respectively. Samples of 130 deaths from the Malaysia Hospital and


133

120 from Singapore Hospital were selected. Find the probability that

(a) the mean age at death in Malaysia is greater than the mean age at

death in Singapore.

(b) the mean age at death in Singapore is three less than the mean age at

death in Malaysia.

8. The average life of a hand phone is 8 years for a female and 6 years for a

male, with a standard deviation of 1 and 2 years respectively. Assuming that

the lives of these hand phones follow approximately a normal distribution,

find the probability that the mean life of a random

(a) male hand phone falls between 6.6 and 7.7 years.

(b) sample of 44 females is not less than 2.5 years than the sample of 55

males hand phones.

9. A company manufactures two types of polystyrenes, type A and type B that

have mean breaking strengths of 400 g and 450 g and standard deviations of

30 g and 20 g, respectively. If 80 polystyrenes of type A and 45 polystyrenes

of type B are tested, what is the probability that the mean breaking strengths

of type B will be at most 53 g more than type A ?

10. The average running times of disks produced by Company X is 88.1 minutes

and a standard deviation of 6.1 minutes, while those of Company Y have a

mean running times of 99.3 minutes with standard deviation of 13.6 minutes.

Assume the populations are approximately normally distributed. What is the

probability that a random sample of 41 disks from Company Y will have

mean running times that at most 15 minutes more than the mean running times

of a random sample of 32 disks from Company X ?

11. A random sample of size sixteen is selected from a normal population with a

mean of 75 and a standard deviation of eight from sample A. A second sample


134

of size nine is selected from another normal population with a mean of 70 and

a standard deviation of twelve from sample B. Let AX and

BX be the two

sample means. Find

(a) the probability that mean difference between sample A and sample B

will be exceed four.

(b) the probability that mean difference between sample A and sample B

will be between 3.5 and 5.5.

12. The elasticity of polymer is affected by the concentration of a reactant. When

low concentration is used, the true mean elasticity is 55 and when high

concentration is used the mean elasticity is 60. The standard deviation of

elasticity is 4, regardless of concentration. The distribution of elasticity is

normally distributed. Two random samples of size 16 are taken. Find the

probability that the difference mean between high concentration and low

concentration is more than two.

13. The average running times of films produced by Company A is 98.4 minutes

and a standard deviation of 7.8 minutes, while those of Company B have a



probability that a random sample of 36 films from Company B will have mean

running times that at least 13 minutes more than the mean running times of a

random sample of 49 films from Company A.

14. The weight of computer chair is approximately normally distributed. There

are two company produce that kind of chair. The data in table shows as a

follows.

Company 1 Company 2

Sample mean = 20.1 Sample mean = 23.1

Sample standard deviation = 4.6 Sample standard deviation = 3.1


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Sample size = 38 Sample size = 29

Find the probability that mean weight of computer chair produced by

Company 2 is greater than weight of computer chair produced by Company 1.

15. A study was designed to estimate the difference in diastolic blood pressure

readings between men and women. The mean and standard deviation for

sixteen men are 77.37 and 8.35, while for thirteen women are 71.08 and 9.22

respectively. Assume that the readings are normally distributed, find

(a) the sampling distribution of the different between diastolic blood

pressure readings for men and women.

(b) the probability that the different between diastolic blood pressure

readings for men is greater than women.

(c) the probability that the different between diastolic blood pressure

readings from men is five less than women.

Answer Exercise 3.4

1. 0.6443 2. 0.0076

3. 0.0073 4. 0.4443

5. 0.9830 6. 0.0166

7. (a) 0070 (b) 0.9931

8. (a) 0.1844 (b) 0.0526

9. 0.7486 10. 0.9452

11. (a) 0.5871 (b) 0.1769

12. 0.983 13. 0.44433

14. 0.9993

15. (b) 0.9719 (c) 0.3483

3.5 Sampling distribution test


136

Theory 9

The t-distribution has been introduced by W. S. Gosset (1876 - 1937). He adopted the

pen name "student." Therefore, the distribution is known as 'student’s t-distribution'.

It is used to establish confidence limits and test the hypothesis when the population

variance is not known and sample size is small (less than 30). If a random sample 1x ,

2x , …, nx of n values be drawn from a normal population with mean μ and standard

deviation s, then the mean of sample nxx i

. Estimation of the variance, let 2s

be the estimate of the variance of the sample then 2s given by

1

2

2

nxxs i whereby )1( n as denominator in place ''n . The statistic '' t

is defined as nsxt 2

whereby

x sample mean, μ is actual mean of

population and n sample size and s standard deviation of sample. The formula for

1

2

nxxs i .

Note :

'' t is distributed as the student distribution with )1( n degree of freedom

(df ).

The variable '' t distribution ranges from minus infinity to plus infinity.

Such as standard normal distribution, it is also symmetrical and has mean zero.

2 of t-distribution is greater than 1, but becomes 1 as 'df' increases and thus

the sample size becomes large.

The t-distribution is lower at the mean and higher at the tails than the normal

distribution.

The t-distribution has proportionally greater area at its tails than the normal

distribution.

The t-distribution is similar in shape to the standard normal distribution,

which are symmetric about zero, uni-modal and bell-shaped.


137

The spread of a t-distribution is larger than that of a standard normal

distribution. That is, there is more probability in the tails of a t-distribution.

This makes sense because the t-statistic should have more variability that the

test statistic, Z that we use before. There is added variability in the t statistic

since it uses s , an estimate of , rather than a known, fixed value of .

Theory 10 (Finding areas under the t-distributions)

We use t-distribution table to find areas under the t-distributions. The table gives the

value of vt , which is the 100α percentage point of the t-distributions for v, degrees of

freedom. The numbers in the middle of the table are values from t-distributions. Each

row corresponds to a t-distribution with the degrees of freedom given at the beginning

of the row. The numbers in the top row are right tail areas.

Example 12

Find the value of t-distribution, if given nine degrees of freedom with alpha equal to

0.05.

Answer Example 12

Refers to t-distribution table, if we go across the row for nine degrees of freedom and

down the column for an area of 0.05, we get the t value of 1.833. That means, for t9

distribution, the area under the curve to the right of 1.833 is 0.05.

Example 13

Find the value of t-distribution, if given twenty degrees of freedom with alpha equal

to 0.001.

Answer Example 13

Refers to t-distribution table, if we go across the row for twenty degrees of freedom

and down the column for an area of 0.001, we get the t value of 3.552. That means,

for 20t distribution, the area under the curve to the right of 3.552 is 0.001.


138

Example 14

By using the statistical table, find the value of vt , .

(a) 025.0)( 14, tTP

(b) 005.0)( 24, tTP

Answer Example 14

(a) Given : 025.0)( 14, tTP

145.2025.0,14, tt v , then 025.0)145.2( TP

(b) Given : 005.0)( 24, tTP

797.2005.0,24, tt v , then 005.0)()( 24,24, tTPtTP

Theory 11

Tests like Z score, t, and F are based on the assumption that the samples were drawn

from normally distributed populations or more accurately that the sample means were

normally distributed. As these tests require assumption about the type of population

or parameters, these tests are known as 'parametric tests.' There are many situations in

which it is impossible to make any rigid assumption about the distribution of the

population from which samples are drawn. This limitation led to search for non-

parametric tests. Chi-square (Read as Ki - square) test of independence and goodness

of fit is a prominent example of a non-parametric test. The chi-square, 2 test can be

used to evaluate a relationship between two nominal or ordinal variables. The 2 is

measure of actual divergence of the observed and expected frequencies. In sampling

studies we never expect that there will be a perfect coincidence between actual and

observed frequencies and the question that we have to tackle is about the degree to

which the difference between actual and observed frequencies can be ignored as

arising due to fluctuations of sampling. If there is no difference between actual and

observed frequencies then 02 . If there is a difference, then 2 would be more


139

than 0. But the difference may also be due to sample fluctuation and thus the value of

2 should be ignored in drawing the inference. Such values of 2 under different

conditions are given in the form of tables and if the actual value is greater than the

table value, it indicates that the difference is not solely due to sample fluctuation and

that there is some other reason. On the other hand, if the calculated 2 is less than the

table value, it indicates that the difference may have arisen due to chance fluctuations

and can be ignored. Thus 2 tests enable us to find out the divergence between

theory and fact or between expected and actual frequencies are significant or not. If

the calculated value of 2 is very small, compared to table value then expected

frequencies are very little and the fit is good. If the calculated value of 2 is very

large as compared to table value then divergence between the expected and the

observed frequencies is very big and the fit is poor.

Theory 12 (Finding areas under the 2 - distributions)

We use 2 - distribution table to find areas under the 2 - distributions. The table

gives the value of v,2 which is the 100 percentage point of the 2 - distributions

for v , degrees of freedom. The numbers in the middle of the table are values from

2 - distributions. Each row corresponds to a 2 - distribution with the degrees of

freedom given at the beginning of the row. The numbers in the top row are right tail

areas.

Example 15

Find the value of 2 - distribution, if given seventeen degrees of freedom with alpha

equal to 0.95.

Answer Example 15

Refers to 2 - distribution table, if we go across the row for seventeen degrees of

freedom and down the column for an area of 0.95, we get the 2 value of 8.672. That


140

means, for 172 distribution, the area under the curve to the right of 8.672 is 0.95.

Example 16

Find the value of 2 - distribution, if given twelve degrees of freedom with alpha

equal to 0.02.

Answer Example 16

Refers to 2 - distribution table, if we go across the row for twelve degrees of

freedom and down the column for an area of 0.02, we get the 2 value of 24.054.

That means, for 122 distribution, the area under the curve to the right of 24.054 is

0.02.

Theory 13

In probability theory and statistics, the F-distribution is a continuous probability

distribution. It is also known as Snedecor's F distribution or the Fisher-Snedecor

distribution (after R.A. Fisher and George W. Snedecor). The F-distribution becomes

relevant when we try to calculate the ratios of variances of normally distributed

statistics. Suppose we have two samples with n1 and n2 observations, the ratio 2

2

2

1

s

sF

is distributed according to an F distribution (named after R.A. Fisher) with 111 nv

numerator degrees of freedom, and 122 nv denominator degrees of freedom. The

F-distribution is skewed to the right, and the F-values can be only positive.

Theory 14 (Finding areas under the F - distributions)

We use F-distribution table to find areas under the F-distributions. The table gives the

values of 21,, vvF which is the 100α percentage point of the F-distributions having 1v

degrees of freedom in the numerator and 2v degrees of freedom in the denominator.

For each pair of values of 1v and 2v , 21,, vvF is tabulated for

001.0,01.0,025.0,05.0 and the 025.0 values being bracketed. The lower

http://en.wikipedia.org/wiki/Probability_theory

http://en.wikipedia.org/wiki/Statistics

http://en.wikipedia.org/wiki/Continuous_probability_distribution

http://en.wikipedia.org/wiki/Probability_distribution

http://en.wikipedia.org/wiki/Probability_distribution

http://en.wikipedia.org/wiki/Ronald_Fisher

http://en.wikipedia.org/wiki/George_W._Snedecor


141

percentage points of the distribution may be obtained from the

relation

12

21

,,

,,1

1

vv

vvF

F

for example, 351.01

12,8,05.0

18,12,95.0 F

F .

Example 17

If 2

1s and 2

2s are the variances of independent random samples of size, 251 n and

132 n from normal population with equal variances, find

25.6

2

2

2

1

s

sP .

Answer Example 17

Variance of normal population are equal for two independent random samples,

2

2

2

1

2 .

251 n , 24125111 nv

132 n , 12113122 nv

From statistical table :

999.0

001.01

)25.6(1

)25.6(25.62

2

2

1

FP

FPs

sP

Exercise 3.4

By using the statistical table, find the probability

1. 17),898.2( vTP

2. 7),415.1( vTP

3. 30),042.2( vTP

By using the statistical table, find the value of T , if given

4. 005.0,26_____),( vTP


142

5. 995.0,21_____),( vTP

6. 9995.0,14_____),( vTP

In each of the following parts, find ,95.02 . Assume a chi- square distribution with

7. 14 degrees of freedom

8. 29 degrees of freedom

Assume a chi – square distribution with 17 degrees of freedom. Fill in the blanks.

9. 05.0)__________( 2 P

10. 005.0)__________( 2 P

11. Assume a 2 distribution with 7 degrees of freedom, find

)346.6167.2( 2 P .

12. Assume a 2 distribution with 17 degrees of freedom, find )511.19( 2 P .

13. If 2

1S and 2

2S are the variances of independent random samples of size,

91 n and 112 n from the normal population with equal variances, find

06.5

2

2

2

1

S

SP .

14. If 2

1s and 2

2s are the variances of independent random samples of size,

91 n and 132 n from normal population with equal variances, find

50.4

2

2

2

1

s

sP .

15. If 2

1s and 2

2s are the variances of independent random samples of size,

71 n and 72 n from normal population with equal variances, find


143

28.4

2

2

2

1

s

sP .

Answer Exercise 3.4

1. 0.005 2. 0.90

3. 0.975 4. 2.779

5. -2.831 6. 4.140

7. 6.571 8. 17.708

9. 27.587 10. 5.697

11. 0.45 12. 0.7

13. 0.01 14. 0.99

15. 0.95

EXERCISE CHAPTER 3

1. A simple random sample of 100 men is chosen from a population with mean

height 70 inch and standard deviation 2.5 inch. What is the probability that the

average height of the sample men is greater than 69.5 inch ?

2. A group of ball bearings have a mean weight of 5.02 grams and a standard

deviation of 0.30 grams. A random sample of 100 ball bearings chosen from

this group, find

(a) the probability that an average weight of ball bearings chosen from

this group are between 4.96 and 5.00 grams.

(b) the probability that an average weight of ball bearings chosen from

this group are more than 5.10 grams

3. Given the population 5, 5, 5, 7, 7, 8, 8, 8, 9, 9. Find mean and standard

deviation sampling distribution of

(a) If a random sample of size 40 was drawn with replacement from that

population.

(b) If a random sample of size 3 was drawn with replacement from that


144

population.

4. The mean height of 250 UPM staffs is 158 m and the standard deviation is

5 m. Find the mean and standard deviation of the sampling distribution of the

mean height for a sample size of 38 staffs.

5. A chemical engineer calculates that the populations mean yield of batch is 518

grams per milliliter with a standard deviation of 40 grams. Assume that the

distribution of yield to be approximately normal. What is the probability in a

certain month, he get yield less than 515 grams for 36 batches ?

6. The viscosity of a fluid can be measured in an experiment by dropping a small

ball into a calibrated tube containing the fluid and observing the random

variable X, the time it takes for the ball to drop the measures distance. Assume

that X is normally distributed with a mean of 20 seconds and a standard

deviation of 0.5 seconds for a particular type of liquid.

(a) What is the standard deviation of the average time of 40 experiments ?

(b) What is the probability that the average time of 40 experiments will

exceed 20.1 seconds ?

(c) Suppose the experiment is repeated only 20 times. What is the

probability that the average value of X less than 20.1 seconds ?

7. Two independent experiments are being run in which two different types of

paints are compared. Twenty specimens are painted using type A and the

drying time in hours is recorded on each. The same is done with type B.

Assume that the mean drying times of the two populations are normal,

)1,(~ NX A and )1,(~ NX B for the two types of paints.

(a) Write down the mean distribution of AX__

.

(b) Calculate

3.0

____

BA XXP .


145

8. The photo resist thickness in semiconductor manufacturing has a mean of 10

micrometers and standard deviation of 3 micrometer. Assume that the

thickness is normally distributed and that the thicknesses of different photo

resist are independent.

(a) Determine the probability that the average thickness of 10 photo resist

is either greater than 11 or less than 9 micrometers.

(b) Determine the number of photo resist that need to be measured such

that the average thickness exceeds 11 micrometers is 0.01.

9. The population of the usage per sheet of paper for old and new certain

products are distributed )60,2000(1N and )40,2500(2N respectively. Two

random samples are taken from each population of size 1n and 2n .

(a) Write down the sampling distribution of the different means of new to

old products.

(b) Find the probability of mean usage of sample of size 3021 nn for

new and old products with new products is at least 500.5 sheets more

than old products.

(c) Find the probability of mean usage of size 201 n and 252 n for

new and old products with new products is at most 502 sheets more

than old products.

10. Line Clear Manufacturing Sdn. Bhd. manufactured two type of cables A and

B that have mean breaking strengths of 2500 lb and 2400 lb with their

standard deviation 150 lb and 100 lb. If 50 cables of brand A and 25 cables of

brand B are tested, what is the probability that the mean breaking strength of

A will be

(a) at least 150 lb more than brand B ?

(b) at least 110 lb more than brand B ?

11. A light bulbs manufacturer claims that the lifetime of its light bulbs has a


146

mean of 54 months and a standard deviation of 6 months. Your consumer

advocacy group tests 50 of them. What is the probability that it finds a mean

lifetime of less than 52 months ?

12. A manufacturer of video display units is testing two microcircuit designs to

determine whether they produce equivalent mean current flow is normally

distributed with mean and standard deviation such as below. Find the

probability that the mean of design A is lower than the mean of design B.

Design A Design B

Mean 24.2 23.9

Variance 10 20

Sample size 15 10

13. The amount of time that a drive-through restaurant counter spends on a

customer is normally distributed with a mean 3.2 minutes and a standard

deviation 1.6 minutes. If a random sample of 64 customers is observed, find

the probability that the mean time at the counter is

(a) less than 2.7 minutes.

(b) more than 3 minutes.

(c) at least 3.2 minutes but less than 3.4 minutes.

14. Students may choose between a 3 semester course in physics without labs and

4 semester course with labs. The final written examination is the same for

each section. The section wit labs made an average examination grade of 84

with a standard deviation of 4 and the section without labs made an average

grade of 77 with a standard deviation of 6. Assume the populations are

approximately normally distributed. Find the probability that the sample mean

for a random sample of scores of 12 students with labs exceeds the sample

mean for a random sample of scores of 18 students without labs by at most 5.


147

15. Casual workers in a certain industry are paid on average RM5.10 per hour

which is normally distributed with standard deviation of RM2.20. A sample of

35 casual workers from the industry was selected to be respondents for the

underpaid issue questionnaires. Find the probability that the average payment

for those casual workers is

(a) at least RM6.00 per hour.

(b) greater than RM4.80 per hour.

16. The mean age at death in Malaysia is 55.5 years and Singapore is 57 years.

The standard deviation is approximately 4.6 years and 5 years for each

country respectively. Samples of 130 deaths from the Malaysia Hospital and

120 deaths from Singapore Hospital were selected. Find the probability that

(a) the mean age at death in Malaysia is greater than the mean age at death

in Singapore.

(b) the mean age at death in Singapore is three less than the mean age at

death in Malaysia.

17. Intelligent Quotients (IQ) in the general population are normally distributed

with a mean of 100 and a standard deviation of 15. A random sample of 40

students was taken in a certain university. Find the probability that

(a) the mean IQ of the sample is greater than 105 and less than 107.

(b) the mean IQ of the sample is not more than 109.

18. The average life of a hand phone is 8 years for a female and 6 years for male,

with a standard deviation of 1 and 2 years respectively. Assuming that the

lives of these hand phones follow approximately a normal distribution, find

(a) the probability that the mean life of a random male hand phone falls

between 6.6 and 7.7 years.

(b) the probability that the mean life of a random sample of 44 females is

not less than 2.5 years than the sample of 55 males hand phones.


148

19. A company manufacturers two types of polystyrenes, type A and type B that

have mean breaking strengths of 400 g and 450 g with standard deviation of

30 g and 20 g, respectively. If 80 polystyrenes of type A and 45 polystyrenes

of type B are tested, what is the probability that the mean breaking strengths

of type B will be at most 53 g more than type A ?

20. The average running times of disks produced by Company X is 88.1 minutes

and a standard deviation of 6.1 minutes, while those of Company Y have a



probability that a random sample of 41 disks from Company Y will have

mean running times that at most 15 minutes more than the mean running times

of a random sample of 32 disks from Company X ?

ANSWER EXERCISE CHAPTER 3

1. 0.97725

2. (a) 0.22867 (b) 0.00379

3. (a) 7.1, 0.2392 (b) 7.1, 0.87368

4. 158m, 0.8111 5. 0.32636

6. (a) 0.0791 (b) 0.1038 (c) 0.8133

7. (b) 0.00135

8. (a) 0.2937 (b) 49

9. (b) 0.3936 (c) 0.82381

10. (a) 0.0432 (b) 0.3658

11. 0.00914 12. 0.45620

13. (a) 0.00621 (b) 0.84134 (c) 0.34134

14. 0.13567

15. (a) 0.00776 (b) 0.791

16. (a) 0.00695 (b) 0.99305

17. (a) 0.01593 (b) 0.98257


149

18. (a) 0.18443 (b) 0.0526

19. 0.7486 20. 0.9452

SUMMARY CHAPTER 3

Sampling error of single mean : _

xe .

Population mean, N

x .

Sample mean, is n

xx

_

.

Z-value for sampling distribution of _

x is n

xZ

_

.


150

Calculate the probability of single mean :

Step 1 : Mean of sample mean, __

x which is _

x

.

Step 2 : Standard deviation of sample mean, __

x which is nx

_

.

Step 3 : Distribution in normal distribution form which is

2

__

____ ,~xx

Nx .

Step 4 : Probability of sample mean,

__

____

x

x

rZPrxP

.

Calculate the probability of two means :

Step 1 : Mean of sampling distribution which is, 212

__

1

__ xx

Step 2 : Standard deviation of sample mean which is,2

22

1

12

2

__

1

__

nnxx

.

Step 3 : Distribution in normal form,

2__

2

__

1

__

2

__

1

__ ,~xxxx

Nx .

Step 4 : Probability of sample mean,

2

_

1

_

1

_

1

___

2

__

1

xx

xx

r

ZPrxxP

.

CORRECTION PAGE CHAPTER 3


151


152

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