chapter 2 statistical inference 2.1 estimation confidence interval estimation for mean and...

CHAPTER 2Statistical InferenceStatistical Inference

2.1 EstimationConfidence Interval Estimation for Mean and

ProportionDetermining Sample Size

2.2 Hypothesis Testing: Tests for one and two means Test for one and two proportions

What is a Hypothesis?A hypothesis is a claim (assumption) about a

population parameter:

Hypothesis testing can be used to determine whether a statement about the value of a population parameter should or should not be rejected.

population mean

population proportion

Example: The mean monthly cell phone bill in this city is μ = RM 92

Example: The proportion of adults in this city with cell phones is p = 0.68

A statistical test of hypothesis consist of :1. The Null hypothesis, 2. The Alternative hypothesis, 3. The test statistic4. The rejection region5. The conclusion

A null hypothesis is a claim (or statement) about a population parameter that is assumed to be true.

An alternative hypothesis is a statement that specifies that the population parameter has a value different in some way, from the value given in the null hypothesis.

Test Statistic is a function of the sample data on which the decision (reject or do not reject) is to be based.

Rejection region is a region for which the null hypothesis will be rejected.

Hypothesis Testing

0H

1H

The Hypothesis Testing Process

Claim: The population mean age is 50.H0: μ = 50, H1: μ ≠ 50

Sample the population and find sample mean.

Population

Sample

Suppose the sample mean age was .

This is significantly lower than the claimed mean population age of 50.

If the null hypothesis were true, the probability of getting such a different sample mean would be very small, so you reject the null hypothesis .

In other words, getting a sample mean of 20 is so unlikely if the population mean was 50, you conclude that the population mean must not be 50. (reject null hypothesis )

20X

6 Steps in 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

Alternative Hypothesis as a Research Hypothesis

Developing Null and Alternative Hypotheses

Many applications of hypothesis testing involve an attemptto gather evidence in support of a research hypothesis.

In such cases, it is often best to begin with the alternativehypothesis and make it the conclusion that the researcherhopes to support.

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

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

0H0H

Two-Tailed Test Left-Tailed Test

Right-Tailed Test

Sign in = = or = or

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

Test Statistic :

• is known and n is large

• is unknown and n is large

• is unknown and n is small

xZ

n

xZ sn

xt sn

Hypothesis Test on the Population Mean,

ExampleThe average monthly earnings for women in managerial and professional positions is RM2400. Do men in the same positionshave average monthly earnings that are higher than those for women?A random sampl 40

3600 400

 0.01

e of men in managerial and professional positions showed RM and RM . Test the appropriate hypothesis using .

n

x s

Solution

0

1

0.01

:  2400

: 2400

30

; 2.33

- 3600 - 240018.9

400

40

H

H

n

Z z z z

xZ

s

n

The hypothesis:

Use normal distribution ( )Rejection Region :

Test Statistic

0

7

HSince 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 significantly higher than those for women.

ExampleThe daily yield for a local chemical plant has averaged

880 ton for the last several years. The quality control manager would like to know whether this average has changed in recent months. She randomly selects 50 days from the computer database and computes the average and standard deviation of the n = 50 yields as = 871 tons and s = 21 tons, respectively. Test the appropriate hypothesis using α=0.05.

Solution

x

0

1

2 2

0.0252

:  880

: 880

30

1.96

H

H

n

Z z Z z

z z

The hypothesis:

Use normal distribution ( )Rejection Region : or

0

- 871-8803.03

21

50– .0 . ,

,

.

xZ

s

n

H

Test Statistic

Since 3 3 1 96 falls in the rejection region the manager can reject and conclude

that it has changed

Test statistics:

Hypothesis Test For the Difference between Two Populations Means, 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 )

Alternative hypothesis Rejection Region

1 1 2: 0H

1 1 2: 0H

1 1 2: 0H

2 2 or  ZZ z z

Z z

Z< z

•

• 1 2 1 2

2 21 2

1 2

x xt

s sn 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

and and   are unknown.

(Assume )

Example

00 .

00n 1

A university conducted an investigation to determine whether car ownership affects academic achievement was based on two random samples of 1 male students The grade point average for the 1 non

21 21

22 2

2.70 0.36 2.54

0.40 00 .

x s x

s n

owners of cars had an average and variance

equal to and as opposed to and

for the 1 car owners Do the data present sufficient evidence to indicate a differencei

? 0.05 n the mean achievements between car owners and non ownersof cars Test using

Solution

0 1 2

1 1 2

1 2 1 2

2 21 2

1 2

2 2

,

: 0

: 0

2.70 2.54.

0.36 0.40100 100

 :      ;

H

H

x xZ Z

s sn n

z z

The hypothesis to be tested are

Test statistic is:

1 84

Rejection Region Z or Z 0.05 2 0.025

0

.

. . . ,  

,

z z

H

1 96

Since 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 aca .demic achievement for the two groups

Hypothesis Test on the Population Proportion, p

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

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.

Solution

0

1

0

0 0

0.05

,

:    0.0

: 0.0

ˆ 0.07 0.042.17

0.04(0.96)200

 :     ; 1.65

. .

The hypothesis to be tested are 4

4

Test statistic is

Rejection Region Z

Since 2 17 1

H p

H p

p pZ

p q

n

z z z

0, ,  

.

65 falls in the rejection region we can reject

and conclude that the machine needs an adjustment

H

Hypothesis Test For the Difference between Two Population Proportion, 1 2p p

0 1 2

1 2 1 2 1 2

1 2

1 2

: : 0

     : 

ˆ ˆˆ

ˆ ˆ ˆ ˆ

Null Hypothesis

Test Statistics

where

H p p

p p p p x xZ p

n npq pqn n

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

ExampleA researcher want 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 a 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.

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 another toothpaste?

Solution

0 1 2

1 1 2

1 2 1 2

1 2

,

: 0

: 0

,

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

1 2

1 2

The hypothesis to be tested are

is not greater than

is greater than

Therefore test statistic is

0.01

0

01.15

1 1(0.187)(0.813)

500 400

 :  ;   2.33

,

Z z z

H

Rejection Region

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

theproportions of users of Toothpaste A who will  

.

never switch to anothertoothpaste is not greater than the proportion of users ofToothpaste B who will never switch to another toothpaste

End of Chapter 2