Chapter 2: Statistical Inferences

4.1 IntroductionThe field of statistical inference consist of those methods used to make decisions or to draw conclusions about a population. These methods utilize the information contained in a sample from the population in drawing conclusions. Statistical Inference may be divided into two major areas: parameter estimation and hypothesis testing.

Definition 2.1: An Interval EstimateIn interval estimation, an interval is constructed around the point estimate and it is stated that this interval is likely to contain the corresponding population parameter. Definition 2.2: Confidence Level and Confidence Interval Each interval is constructed with regard to a given confidence level and is called a confidence interval. The confidence level associated with a confidence interval states how much confidence we have that this interval contains the true population parameter. The confidence level is denoted by .

2.1 Estimation

1 100%

2

EQT 373

Confidence Intervals

Population Mean

σ Unknown

Confidence Intervals

PopulationProportion

σ Known

2.1.1 Confidence Interval Estimates for Population Mean

2

2 2

2

The (1 )100% Confidence Interval of Population Mean,

(i) if   is known and normally distributed population

or

(ii) if    is unknown, large (

x zn

x z x zn n

sx z n n

n

2 2

30)

or s s

x z x zn n

4

1, 2

1, 1,2 2

(iii) if    is unknown, normally distributed population

and small  sample size 30     

or

n

n n

sx t

n

n

s sx t x t

n n

5

Example 2.1

2

If a random sample of size 20 from a normal population

with the variance 225 has the mean 64.3, construct

a 95% confidence interval for the population mean, .

n

x

6

solution

7

0.025

2

It is known that, 20,  64.3 and 15

For 95% CI,

95% 100(1– )%

1–  0.95

  0.05   

0.0252

1.96

                    

n x

z z

8

2

Hence, 95% CI   

15 64.3 1.96

20

64.3 6.57

[57.73,70.87]

@

x zn

57.73 70.87

Thus, we are 95% confident that the mean of random variable

is between 57.73 and 70.87

Example 2.2 :

A publishing company has just published a new textbook. Before the

company decides the price at which to sell this textbook, it wants to know the

average price of all such textbooks in the market. The research department at

the company took a sample of 36 comparable textbooks and collected the

information on their prices. This information produced a mean price RM 70.50

for this sample. It is known that the standard deviation of the prices of

all such textbooks is RM4.50. Construct a 90% confidence interval for the mean

price of all such college textbooks.

9

solution

10

1 2

2.1.2 The (1 )100% Confidence Interval Estimates

on the Differences between Two Population Means,

2 21 2

1 2 1 22

1 2

2 21 2

1 2 1 22

1 2

1 2

2 21 2

1 22

1

(i) if   and is known and normally

distributed population

(ii) if   and is unknown, large

30, 30

or

x x zn n

s sx x z n

n n

n n

s sx x z

n n

2 2

1 21 2

22 1 2

,s s

x x zn n

11

2 21 2

1 2 , 2 1 21 2

1 2

22 21 2

1 222 2

1 2

1 1 2 2

(iii) if  ( , ) is unknown, normally distributed

population and small sample size   30, 30  

with 1 1

1 1

v

s sx x t

n n

n n

s sn n

vs s

n n n n

2

12

1 2 1 2

1 2

1 22

for (ii) and (iii) cases, if ( , ) is unknown but we assume = ,

the pooled estimator is used to estimate , then the formula is given as follows:

for large sample size   30, 30 ,

pS

n n

x x z

1 2

1 2

1 2

1 2 2, 21 2

2 21 1 2 2

1 2

1 1

for small sample size   30, 30

1 1

1 1with

2

p

n n p

p

Sn n

n n

x x t Sn n

n s n sS

n n

13

Example 2.3 :

The scientist wondered whether there was a difference in the average daily

intakes of dairy products between men and women. He took a sample of n =50

adult women and recorded their daily intakes of dairy products in grams per day.

He did the same for adult men. A summary of his sample results is listed below.

Construct a 95% confidence interval for the difference in the average daily intakes

of daily products for men and women. Can you conclude that there is a difference

in the average daily intakes of daily products for men and women?

14

Men Women

Sample size 50 50

Sample mean 780 grams per day 762 grams per day

Sample standard deviation

35 30

Solution

15

2 22 21 2

1 2

1 22

Hence, 95% CI   :

35 30780 762 1.96

50 50

18 12.78

[5.22,30.78]

Thus, we should be willin

s sx x z

n n

1 2

g to conclude that there is a difference in the

average daily intakes of daily products for men and women as

0 there is no difference between two population means

is not one of the possible value

1 2s for ( ).

2.1.3 Confidence Interval Estimates for Population Proportion

2

2 2

The (1 )100% Confidence Interval for for Large Samples ( 30)

ˆ ˆ1ˆ

or

ˆ ˆ ˆ ˆ1 1ˆ ˆ

p n

p pp z

n

p p p pp z p p z

n n

16

Example 2.4 According to the analysis of Women Magazine in June 2005, “Stress has become a common part of everyday life among working women in Malaysia. The demands of work, family and home place an increasing burden on average Malaysian women”. According to this poll, 40% of working women included in the survey indicated that they had a little amount of time to relax. The poll was based on a randomly selected of 1502 working women aged 30 and above. Construct a 95% confidence interval for the corresponding population proportion.

17

SolutionLet be the proportion of all working women age 30 and above,

ˆwho have a limited amount of time to relax, and let  be the

corresponding sample proportion. From the given information,

ˆ 1502 , 0.

p

p

n p

2

ˆ ˆ40,   1 1 – 0.40 0.60

ˆ ˆˆHence, 95% CI    

0.40(0.60) 0.40 1.96

1502

0.40 0.02478

[0.375,0.425

q p

pqp z

n

] or 37.5% to 42.5%

Thus, we can state with 95% confidence that the proportion of all

working women aged 30 and above who have a limited amount of

time to relax is between 37.5% and 42.5%. 

18

1 2 1 2

2.1.4 The (1 )100% Confidence Interval Estimates for the

Differences between Two Proportions, , ( 30, 30)P P n n

1 1 2 21 2

21 2

ˆ ˆ ˆ ˆ1 1ˆ ˆ

p p p pp p z

n n

19

Example 2.5:A researcher wanted to estimate the difference between the percentages of users of two toothpastes who will never switch to another toothpaste. In a sample of 500 users of Toothpaste A taken by this researcher, 100 said that the will never switch to another toothpaste. In another sample of 400 users of Toothpaste B taken by the same researcher, 68 said that they will never switch to another toothpaste. Construct a 97% confidence interval for the difference between the proportions of all users of the two toothpastes who will never switch.

20

Solutions

Toothpaste A : n1 = 500 and x1 = 100

Toothpaste B : n2 = 400 and x2 = 68

The sample proportions are calculated;

Thus, with 97% confidence we can state that the difference between the two

population proportions is between -0.026 and 0.086.

21

1 2

1 1 2 21 2

1 22

100 68ˆ ˆ0.20; 0.17

500 400

A 97% confidence interval ;

ˆ ˆ ˆ ˆ(1 ) (1 )ˆ ˆ

0.20(0.80) 0.17(0.83)0.20 0.17 2.17

500 400

0.03 0.05628

[ 0.026,0.086]

p p

p p p pp p Z

n n

EQT 373

2.1.5 Determining Sample Size

For the Mean

Determining Sample Size

For theProportion

The required sample size can be found to reach a desired margin of error (e) with a specified level of confidence (1 - ). The margin of error is also called error of estimation

Definition 2.3 (Estimating the Population Mean):

Error of Estimation and Determining the Sample size

23

2

/2

If is used as an estimate of , we can be 100(1- )% confident

that the error | | will not exceed a specified amount when the

sample size is

x

x E

zn

E

Determining Sample Size for population mean problems

For the Mean

DeterminingSample Size

Sampling error (margin of error)

/2X Zn

/2e Zn

Example 2.6

If = 45, what sample size is needed to estimate the mean within ± 5 with 90% confidence

So the required sample size is n = 220(Always round up)

2 2 2 2

2 2

σ (1.645) (45)219.19

5

Zn

e

Example 2.7:A team of efficiency experts intends to use the mean of a

random sample of size n=150 to estimate the average mechanical aptitude of assembly-line workers in a large industry (as measured by a certain standardized test). If, based on experience, the efficiency experts can assume thatfor such data, what can they assert with probability 0.99 about the maximum error of their estimate?

26

6.2

Solutions

27

0.005

/2

Substituting 150, 6.2, and 2.575 into the expression

for the maximum error, we get

2.575(6.2) 1.30

150

Thus,

n z

zE

n

the efficiency experts can assert with probability 0.99 that their

error will be less than 1.30.

Definition 2.4 (Estimating the Population Proportion):

28

/2

/2

ˆIf is used as an estimate of , we can assert with

ˆ ˆ(1 )(1- )100% confidence that the error is less than .

ˆ ˆ(1 )If we set and solve for , the appropriate

sample size is

xp p

n

p pz

n

p pE z n

n

2

/2

ˆ ˆ (1 )z

n p pE

EQT 373

Determining

Sample Size

For theProportion

2

2

ˆ ˆ(1 )Z p pn

e

Now solve for n to get

Determining Sample Size for population proportion problems

ˆ ˆ(1- )p pe Z

n

How large a sample would be necessary to estimate the true proportion defective in a large population within ±3%, with 95% confidence? (Assume a sample yields p = 0.12)

For 95% confidence, we have Zα/2 = 1.96

e = 0.03; p = 0.12

Example 2.8

Solution:

So use n = 451

2 2/2

2 2

ˆ ˆ(1 ) (1.96) (0.12)(1 0.12)450.74

(0.03)

Z p pn

e

Example 2.9:A study is made to determine the proportion of voters in a sizable community who favor the construction of a nuclear power plant. If 140 of 400 voters selected at random favor the project and we use as an estimate of the actual proportion of all voters in the community who favor the project, what can we say with 99% confidence about the maximum error?

31

140ˆ 0.35

400p

Solution

32

Example 2.10:How large a sample required if we want to be 95% confident that the error in using to estimate p is less than 0.05? If , find the required sample size.

33

p̂ˆ 0.12p

Solution

34

Hypothesis and Test Procedures  A statistical test of hypothesis consist of :1. The Null hypothesis, 2. The Alternative hypothesis, 3. The test statistic and its p-value4. The rejection region5. The conclusion

2.2 Hypothesis Testing

35

0H

1H

Definition 2.5:Hypothesis testing can be used to determine whether a statement about the value of a population parameter should or should not be rejected.Null hypothesis, H0 : A null hypothesis is a claim (or statement)

about a population parameter that is assumed to be true. (the null hypothesis is either rejected or fails to be rejected.)Alternative hypothesis, H1 : An alternative hypothesis is a claim

about a population parameter that will be true if the null hypothesis is false.Test Statistic is a function of the sample data on which the decision is to be based.p-value is the probability calculated using the test statistic. The smaller the p-value, the more contradictory is the data to .

36

0H

Definition 2.6: p-value The p-value is the smallest significance level at which the null hypothesis is rejected.

37

0

0

Using the value approach, we reject the null hypothesis,  if

value for one tailed test

value for two tailed test2

and we do not reject the null hypothesis,  if

value    for one tailed

p H

p

p

H

p

test

value    for two tailed test2

p

6 Steps to Perform a Hypothesis Testing

1. State the null hypothesis, H0 and the alternative hypothesis, H1

2. Choose the level of significance, and the sample size, n

3. Determine the appropriate test statistic

4. Determine the critical values that divide the rejection and non rejection regions

5. Collect data and compute the value of the test statistic

6. Make the statistical decision and state the managerial conclusion. If the test statistic falls into the non rejection region, do not reject the null hypothesis H0. If the test statistic falls into the rejection region, reject the null hypothesis. Express the managerial conclusion in the context of the problem

It is not always obvious how the null and alternative hypothesis should be formulated.

When formulating the null and alternative hypothesis, the nature or purpose of the test must also be taken into account. We will examine:

1) The claim or assertion leading to the test.2) The null hypothesis to be evaluated.3) The alternative hypothesis.4) Whether the test will be two-tail or one-tail.5) A visual representation of the test itself.

In some cases it is easier to identify the alternative hypothesis first. In other cases the null is easier.

Developing Null and Alternative Hypothesis

39

Alternative Hypothesis as a Research Hypothesis

• Many applications of hypothesis testing involve an attempt to gather evidence in support of a research hypothesis.

• In such cases, it is often best to begin with the alternative hypothesis and make it the conclusion that the researcher hopes to support.

• The conclusion that the research hypothesis is true is made if the sample data provide sufficient evidence to show that the null hypothesis can be rejected.

40

Example: A new drug is developed with the goalof lowering blood pressure more than the existing drug.

• Alternative Hypothesis: The new drug lowers blood pressure more than the existing drug. • Null Hypothesis: The new drug does not lower blood pressure more than the existing drug.

41

Null Hypothesis as an Assumption to be Challenged

• We might begin with a belief or assumption that a statement about the value of a population parameter is true.

• We then using a hypothesis test to challenge the assumption and determine if there is statistical evidence to conclude that the assumption is incorrect.

• In these situations, it is helpful to develop the null hypothesis first.

42

Example: The label on a soft drink bottle states that it contains at least 67.6 fluid ounces.

• Null Hypothesis: The label is correct. µ > 67.6 ounces.• Alternative Hypothesis: The label is incorrect. µ < 67.6 ounces.

43

Example: Average tire life is 35000 miles.

• Null Hypothesis: µ = 35000 miles

• Alternative Hypothesis: µ 35000 miles

44

How to decide whether to reject or accept ?The entire set of values that the test statistic may assume is divided into two regions. One set, consisting of values that support the and lead to reject , is called the rejection region. The other, consisting of values that support the is called the acceptance region.

Tails of a Test

0H

1H

0H

0H

Two-Tailed Test Left-Tailed Test

Right-Tailed Test

Sign in =

Sign in < >

Rejection Region In both tail In the left tail In the right tail

0H

1H

Rejection Region2 2 or  ZZ z z Z z Z< z

2.2.1 a) Testing Hypothesis on the Population Mean,

Null Hypothesis :

Test Statistic :

46

0 0 :H

• any population, is known and n is largeor• normal population, is known and n is small

• any population, is unknown and n is large

• normal population, is unknown and n is small

xZ

n

xZ s

n

1

xt sn

v n

Example 2.11

47

The average monthly earnings for women in managerial and

professional positions is 2400. Do men in the same positions

have average monthly earnings that are higher than those for women ?

A random sampl

RM

e of 40 men in managerial and professional

positions showed 3600 and 400. Test the appropriate

hypothesis using  0.01

n

x RM s RM

Solution

48

0

1

0.01

1.The hypothesis to be tested are,

:  2400

: 2400

2.We use normal distribution 30

3. Rejection Region : Z ; 2.33

 4. Test Statistic     

H

H

n

z z z

0

3600 2400 18.97

400

40

Since 18.97 2.33, falls in the rejection region, we reject 

and conclude that average monthly earnings for men in managerial

and professional positions are significant

xZ

s

n

H

ly higher than those for women.

Test statistics:

b) Hypothesis Test For the Difference between Two Populations Means,

49

1 2

0 1 2Null Hypothesis :         : 0H

•

•

•

1 2

For two large and independent samples

and and   are known.

1 2 1 2

2 21 2

1 2

x xZ

n n

1 2

1 2

For two large and independent samples

and and   are unknown.

(Assume )

1 2 1 2

2 21 2

1 2

x xZ

s sn n

1 2 1 2

1 2

2 21 1 2 2

1 2

1 1

1 1with

2

p

p

x xZ

Sn n

n s n sS

n n

1 2

1 2

For two large and independent samples

and and   are unknown.

(Assume )

50

•

1 2 1 2

2 21 2

1 2

22 21 2

1 22 22 2

1 2

1 1 2 2

1 11 1

x xt

s sn n

s sn n

vs s

n n n n

1 2 1 2

1 2

1 2

1 1

2

p

x xt

Sn n

v n n

1 2

1 2

For two small and independent samples

and and   are unknown.

(Assume )

1 2

1 2

• For two small and independent samples   

taken from two normally distributed  

populations. and   are unknown.

(Assume )

Alternative hypothesis Rejection Region

51

1 1 2: 0H

1 1 2: 0H

1 1 2: 0H

2 2 or  ZZ z z

Z z

Z< z

Example 2.12

52

A university conducted an investigation to determine whether

car ownership affects academic achievement was based on two

random samples of 100 male students, each drawn from the

student body. The grad 1

1

2 221 2

2

e point average for the 100 non

owners of cars had an average and variance equal to 2.70

and 0.36 as opposed to 2.54 and 0.40 for the

100 car owners. Do the data present sufficient

n

x

s x s

n

evidence to

indicate a difference in the mean achievements between car owners

and non owners of cars? Test using 0.05

Solution

53

0 1 2

1 1 2

1 2 1 2

2 21 2

1 2

The hypothesis to be tested are ,

: 0

: 0

Therefore, test statistic is

2.70 2.54 1.84

0.36 0.40

100 100

H

H

x xZ Z

s s

n n

z=-1.96 z=+1.96

Do NotReject H 0

00 Reject HReject H

2 2 0.05 2 0.025Rejection Region : Z or  Z   ; 1.96z z z z

0Since 1.84 does not exceed 1.96 and not less than 1.96, we fail to reject 

and that is, there is not sufficient evidence to declare that there is a difference

in the average academic achievement for

H

the two groups.

54

2.2.2 a) Testing Hypothesis on the Population Proportion, p

55

0 0

0

0 0

Null Hypothesis : :

ˆTest Statistic     :          

H p p

p pZ

p q

n

Alternative hypothesis Rejection Region

1 0:H p p

1 0:H p p

1 0:H p p

2 2 or  ZZ z z

Z z

Z< z

Example 2.13

When working properly, a machine that is used to make chips for calculators

does not produce more than 4% defective chips. Whenever the machine

produces more than 4% defective chips it needs an adjustment. To check if

the machine is working properly, the quality control department at the

company often takes sample of chips and inspects them to determine if the

chips are good or defective. One such random sample of 200 chips taken

recently from the production line contained 14 defective chips. Test at the 5%

significance level whether or not the machine needs an adjustment.

56

Solution

57

0

1

0

0 0

0.05

The hypothesis to be tested are ,

:    0.04

: 0.04

Test statistic is

ˆ 0.07 0.042.17

0.04(0.96)200

Rejection Region : Z     ; 1.65

Since 2.17 1.

H p

H p

p pZ

p q

n

z z z

065, falls in the rejection region, we can reject  

and conclude that the machine needs an adjustment.

H

b) Hypothesis Test For the Difference between Two Population Proportion,

58

1 2p p

0 1 2

1 2 1 2 1 2

1 2

1 2

1 2 1 2

1 2

Null Hypothesis : : 0

Test Statistics     : 

ˆ ˆˆ where

ˆ ˆ ˆ ˆ(1 ) (1 )

ˆ ˆ

1 1ˆ ˆ

H p p

p p p p x xZ p

n np p p pn n

p p p pZ

pqn n

59

Alternative hypothesis Rejection Region

1 1 2: 0H p p

1 1 2: 0H p p

1 1 2: 0H p p

2 2 or  ZZ z z

Z z

Z< z

Example 2.14:Reconsider Example 2.5, At the significance level 1%, can we conclude that the proportion of users of Toothpaste A who will never switch to another toothpaste is higher than the proportion of users of Toothpaste B who will never switch to anothertoothpaste?

60

Solution

61

0 1 2 1 2

1 1 2 1 2

1 2 1 2

1 2

The hypothesis to be tested are ,

: 0 is not greater than

: 0 is greater than

Therefore, test statistic is

ˆ ˆ 0.20 0.17

1 1ˆ ˆ

H p p p p

H p p p p

p p p pZ Z

pqn n

0.01

0

0 1.15

1 1(0.187)(0.813)

500 400

Rejection Region : Z ;   2.33

Since 1.15 2.33, we fail to reject and therefore, we conclude that the

proportions of users of  Toothpaste A who will

z z

H

never switch to another toothpaste

is not greater than the proportion of users of  Toothpaste B who will never switch

to another toothpaste.