Types of Hypotheses Research Hypothesis –a statement of what the researcher believes will be the...

Types of Hypotheses

• Research Hypothesis– a statement of what the researcher believes will

be the outcome of an experiment or a study.• Statistical Hypotheses

– a more formal structure derived from the research hypothesis.

• Substantive Hypotheses– a statistically significant difference does not

imply or mean a material, substantive difference.

Statistical Hypotheses• Two Parts

– a null hypothesis– an alternative hypothesis

• Null Hypothesis – nothing new is happening

• Alternative Hypothesis – something new is happening

• Notation– null: H0

– alternative: Ha

Null and Alternative Hypotheses: Example

• A manufacturer is filling 40 oz. packages with flour.

• The company wants the package contents to average 40 ounces.

ozH

ozH

a 40:

40:0

• One-tailed Tests

One-tailed and Two-tailed Tests

40:

40:0

aH

H

18.0:

18.0:0

pH

pH

a

12:

12:0

aH

H

• Two-tailed Test

Steps in Testing Hypotheses

1. Establish hypotheses: state the null and alternative hypotheses.

2. Determine the appropriate statistical test and sampling distribution.

3. Specify the Type I error rate (4. State the decision rule.5. Gather sample data.6. Calculate the value of the test statistic.7. State the statistical conclusion.8. Make a managerial decision.

Rejection and Non Rejection Regions

=40 oz

Non Rejection Region

Rejection Region

Critical Value

Rejection Region

Critical Value

Decision Table for Hypothesis Testing

(

( )

Null True Null False

Fail toreject null

CorrectDecision

Type II error)

Reject null Type I error

Correct Decision

One-tailed Tests

40:

40:0

aH

H

40:

40:0

aH

H

=40 oz

Rejection Region

Non Rejection Region

Critical Value

=40 oz

Rejection Region

Non Rejection Region

Critical Value

Two-tailed Tests

40:

40:

a

o

H

H

=12 oz

Rejection Region

Non Rejection Region

Critical Values

Rejection Region

40:

40:0

aH

H

A survey of CPAs across US found that the average net income for sole proprietor CPAs is $74914. Because this survey is now more than sever years old an accounting researcher wants to test this figure by taking a random sample of 112 sole proprietor accountants in US to determine whether the net income figure changed. Result of survey derives mean of $ 78695 and the standard deviation net incomes for sole proprietor CPAs is $14530. Test hypothesis using 5 %significance level.

914,74$:

914,74$:

a

0

H

HRejection Region

Non Rejection Region

=0

Zc 196.

Rejection Region

Zc 196.

2

025.2

025.

.Hreject not do ,96.1 If

.Hreject ,96.1 If

0

0

c

c

zz

zz

75.2

112

530,14914,74695,78

n

xz

.Hreject 1.96, = z 2.75 = z 0c

CPA Net Income Example:Critical Value Method

605,77112

530,1496.1914,74

n

zx

Upper

cc

914,74$:

914,74$:0

aH

H

223,72112

530,1496.1914,74

n

zx

Lower

cc

Rejection Region

Non Rejection Region

=0 Zc 196.

Rejection Region

Zc 196.

2

025.2

025.

72,223 77,605

96.1cz 0z 96.1cz

CPA Net Income Example:Critical Value Method

.Hreject not do ,605,7777,223 If

.Hreject ,605,77or 223,77 If

0

0

x

x x

.Hreject ,605,77695,78 Since o cxx

Rejection Region

Non Rejection Region

=0 Zc 196.

Rejection Region

Zc 196.

2

025.2

025.

72,223 77,605

96.1cz 0z 96.1cz

In an attempt to determine why customer service is important to managers in the UK, researcher surveyed managing directors of manufacturing plants in Scotland. One of the reason proposed was that customer service is a mean of retaining customer. On a scale measurement it derives mean of 4.30. suppose US researcher believe American manufacturing managers would not rate this reason as highly at 0.05 alpha value determine whether US manager rate this reason significantly lower than the 4.30. US data based on a sample of

32 and sample mean of 4.156 with standard deviation of 0.574.

30.4:

30.4:0

aH

H

Rejection Region

Non Rejection Region

0

=.05

Zc 1645.645.1cz

Rejection Region

Non Rejection Region

0

=.05

Zc 1645.

.reject not do ,6451 If

.reject ,6451 If

0

0

H.z

H.z

42.1

32

574.030.4156.4

n

sx

z

.reject not do

,645142.1

0H

.z

645.1cz

30.4:

30.4:0

aH

HRejection Region

Non Rejection Region

0

=.05

Zc 1645.

cx 4133. 4.30

133.432

574.0)645.1(30.4

n

zxc

645.1cz

133.4cx

Rejection Region

Non Rejection Region

0

=.05

Zc 1645.

cx 4133. 4.30

.reject not do ,133.4 If

.reject ,133.4 If

0

0

Hx

Hx

.reject not do ,133.4156.4 0

Hx

645.1cz

133.4cx

Rejection Region

Non Rejection Region

0

=.05

p-Value

30.4:

30.4:0

aH

H

.reject not do , value- If

.reject , < value- If

0

0

Hp

Hp

.reject not do

.05, = > .0778 = value- Since

0H

p

0778.)42.1(

42.1

32

574.030.4156.4

zpn

xz

z Test of Population Proportion

pq

p

pn

qp

ppz

-1

proportion population

proportion sampleˆ :where

ˆ

A manufacturer believes exactly 8% of its products contain at least one minor flaw. Suppose a company researcher want to test this belief. He selects a sample of 200 products and determines that 33 items have at least one minor flaw. Test the hypothesis that whether the proportion of products with at least one minor flaw is 0.08 using 0.1 significance level.

08.:

08.:

a

0

pH

pH

cZ 1645.

Critical Values

Non Rejection Region

Rejection Regions

cZ 1645.

2

05. 2

05.

cz cz

.

. .

(. )(. ).

p

Zp P

P Qn

33

200165

165 08

08 92200

4 43

If Z reject H .

If Z do not reject H .

o

o

1645

1645

. ,

. ,

Since Z reject H .o 4 43 1645. . ,

cZ 1645.

Critical Values

Non Rejection Region

Rejection Regions

cZ 1645.

2

05. 2

05.

cz cz

.0

0

reject not do 1.645, If

.reject ,645.1 If

Hz

Hz

43.4

200)92)(.08(.

08.165.ˆ

165.200

33ˆ

nqp

ppz

p

.reject ,645.143.4 Since0

Hz

A survey of the morning beverage market shows that the primary breakfast beverage for 17% of Americans is milk. A milk producer in Wisconsin, where milk is plentiful, believes the figure is higher for Wisconsin. To test this idea, she contacts a random sample of 550 Wisconsin residents and asks which primary beverage they consumed for breakfast that day. Suppose 115 replied that milk was primary beverage. Using level of significance of 0.05 test the idea that the milk figure is higher for City.

H P

H P

o

a

: .

: .

17

17

Critical Value

Non Rejection Region

Rejection Region

cZ 1645.

.0517.:

17.:0

pH

pH

a

cz

.

. .

(. )(. ).

p

Zp P

P Qn

115

550209

209 17

17 83550

2 44

If reject H .

If do not reject H .

o

o

Z

Z

1645

1645

. ,

. ,

Since Z = 2.44 reject H .o1645. ,Critical Value

Non Rejection Region

Rejection Region

cZ 1645.

.05

cz

.reject not do ,645.1 If

.reject ,645.1 If

0

0

Hz

Hz

44.2

550)83)(.17(.

17.209.ˆ

209.550

115ˆ

nqp

ppz

p

.reject ,645.144.2 Since 0Hz

A large manufacturer investigated the service it received from supplier and discovered that, in the past 32% of all material shipments were received late. However the company recently installed a JIT system in which supplier are linked more closely to the manufacturing process. A random sample of 118 deliveries since the JIT was installed reveals that 22 deliveries were late. Use the sample information to test whether the proportion of late deliveries was reduced significantly. α = 0.05

Estimating the Mean of a Normal Population: Unknown

• The population has a normal distribution.• The value of the population standard

deviation is unknown.• z distribution is not appropriate for these

conditions• t distribution is appropriate

The t Distribution

• Developed by British statistician, William Gosset

• A family of distributions -- a unique distribution for each value of its parameter, degrees of freedom (d.f.)

• Symmetric, Unimodal, Mean = 0, Flatter than a z

• t formula

n

sx

t

Comparison of Selected t Distributions to the Standard Normal

-3 -2 -1 0 1 2 3

Standard Normal

t (d.f. = 25)

t (d.f. = 1)

t (d.f. = 5)

Table of Critical Values of t

df t0.100 t0.050 t0.025 t0.010 t0.0051 3.078 6.314 12.706 31.821 63.6562 1.886 2.920 4.303 6.965 9.9253 1.638 2.353 3.182 4.541 5.8414 1.533 2.132 2.776 3.747 4.6045 1.476 2.015 2.571 3.365 4.032

23 1.319 1.714 2.069 2.500 2.80724 1.318 1.711 2.064 2.492 2.79725 1.316 1.708 2.060 2.485 2.787

29 1.311 1.699 2.045 2.462 2.75630 1.310 1.697 2.042 2.457 2.750

40 1.303 1.684 2.021 2.423 2.70460 1.296 1.671 2.000 2.390 2.660

120 1.289 1.658 1.980 2.358 2.6171.282 1.645 1.960 2.327 2.576

t

With df = 24 and = 0.05, t = 1.711.

US farmers’ production company builds large harvesters. For a harvester to be properly balanced when operation, a 25 pound plate is installed. The machine that produces these plates is set to yield plates that average 25 pound. The distribution of plates produced from the machine is normal. However, the shop supervisor is worried that the machine is out of adjustment and is producing plates that do not average 25 pnd. To test this concern, he randomly selects 20 plates and produced weights in following table. = .05

Weights in Pounds of a Sample of 20 Plates

22.622.2 23.2 27.4 24.527.026.6 28.1 26.9 24.926.225.3 23.1 24.2 26.125.830.4 28.6 23.5 23.6

Unknown, = .05

Weights in Pounds of a Sample of 20 Plates

22.622.2 23.2 27.4 24.527.026.6 28.1 26.9 24.926.225.3 23.1 24.2 26.125.830.4 28.6 23.5 23.6

20 = and 2.1933,= ,51.25 nsx

Critical Values

Non Rejection Region

Rejection Regions

ct 2 093. ct 2 093.

2

025.2

025.

H

H

o

a

:

:

25

25

df n 1 19

25:

25:0

aH

H

tX

S

n

2551 25 0

2 1933

20

104. .

. .

Since t do not reject H .o 104 2 093. . ,Critical Values

Non Rejection Region

Rejection Regions

ct 2 093. ct 2 093.

2

025.2

025.

If t reject H .

If t do not reject H .

o

o

2 093

2 093

. ,

. , .reject not do 2.093, If

.reject 2.093, If

0

0

Ht

Ht

04.1

20

1933.20.2551.25

n

sx

t

.reject not do ,093.204.1 Since0

Ht

Department of Agriculture show that the average size of farms has increased since 1940. The mean size of the farm was 174 acres, by 2000 the size was 471 acres. Suppose an researcher believes the average size of farms increased from the 2000 mean figure 471 acres. To test this he randomly sampled 23 farms and find mean of 498.78 acres and std of 46.94 acres. = .05

23 = and 46.94,= ,78.498 nsx

471:

471:0

aH

H

df n 1 22

471:

471:0

aH

H

Critical Value

Non Rejection Region

Rejection Region

ct 1717.

.05

.reject not do ,717.1 If

.reject ,717.1 If

0

0

Ht

Ht

84.2

23

94.4647178.498

n

sx

t

.reject ,717.184.2 Since0

Ht Critical Value

Non Rejection Region

Rejection Region

ct 1717.

.05

The average price per square foot for shops has been 32.28 $. A real estate firm want to determine whether that figure has changed, it sampled 19 shops finds mean price 31.67$ with std of 1.29$. = .05, Test hypothesis.