Ch 4 - Estimation & Hypothesis One Sample

7/21/2019 Ch 4 - Estimation & Hypothesis One Sample

http://slidepdf.com/reader/full/ch-4-estimation-hypothesis-one-sample 1/139

Chapter a

Inferences Based on a Single Sample:

Estimation with Confidence Intervals

Business Statistics



Thinking Challenge

Suppose you’re interested

in the average amount of

money that students in

this class (the population)

have on them. How

would you find out?



Introduction

to Estimation



Statistical Methods

Statistical

Methods

EstimationHypothesis

Testing

Inferential

Statistics

Descriptive

Statistics



Estimation Process

Mean, , isunknown

Population

Sample

Mean

X = 50

Random Sample

I am 95%

confident that

is between 40 &60.



Unknown Population Parameters Are

Estimated

Estimate Population

Parameter...

with Sample

Statistic

Differences 1 - 2 x1 -x2

Variance 2 s 2

^Proportion p p

Mean x



Estimation Methods

Estimation

Interval

Estimation

Point

Estimation



Point Estimation



Estimation Methods

Estimation

Interval

Estimation

Point

Estimation



Point Estimation

1. Provides a single value

• Based on observations from one sample

2. Gives no information about how close the value isto the unknown population parameter

3. Example: Sample mean x = 3 is point

estimate of unknown population mean



Interval Estimation



Estimation Methods

Estimation

Interval

Estimation

Point

Estimation



Interval Estimation

1. Provides a range of values

• Based on observations from one sample

2. Gives information about closeness to unknown population

parameter• Stated in terms of probability

– Knowing exact closeness requires knowing unknownpopulation parameter

3. Example: Unknown population mean lies between 50 and 70with 95% confidence



Key Elements of

Interval Estimation

Sample statistic

(point estimate)Confidence interval

Confidence limit

(lower)

Confidence limit

(upper)

A probability that the population parameter falls

somewhere within the interval.



Confidence Limits

for Population Mean

x

x

x x

Z X

Z Error

Error X Z

X X Error

Error X

)5(

(4)

(3)

or(2)

)1(

Parameter =

Statistic ± Error



Many Samples Have Same Interval

x _

X

X = ± Zx

90% Samples

+1.65x-1.65x

95% Samples

+1.96x-1.96x

99% Samples

-2.58x +2.58x



Confidence Level1. Probability that the unknown population

parameter falls within interval

2. Denoted (1 –

• is probability that parameter is not within

interval

3. Typical values are 99%, 95%, 90%



Intervals & Confidence Level

x=

1 - /2 /2

X

_

x _

Sampling Distribution of Sample Mean

Large number of intervals

(1 – α)% of

intervals

contain μ

α% do not

Intervals

extend from

X – ZσX to

X + ZσX



Factors Affecting

Interval Width

1. Data dispersion

• Measured by Intervals extend from

X – ZX toX + ZX

© 1984-1994 T/Maker Co.

3. Level of confidence(1 – )

• Affects Z

2. Sample size

X n



Confidence Interval Estimates

Confidence

Intervals

Mean Proportion

σ Knownσ

Unknown



Confidence Interval Estimate

Mean ( Known)




Confidence

Intervals

Mean Proportion

σ Knownσ

Unknown



Confidence Interval

Mean ( Known)

1. Assumptions

• Population standard deviation is known

• Population is normally distributed

• If not normal, can be approximated by normaldistribution (n 30)

n Z X

n Z X

2/2/

2. Confidence interval estimate



Estimation Example

Mean ( Known)

The mean of a random sample of n = 36 is X =50. Set up a 95% confidence interval estimate for if = 10.

27.5373.46

361096.150

361096.150

2/2/

n Z X

n Z X



Thinking Challenge

You’re a Q/C inspector for

Gallo. The for 2-liter bottles

is .05 liters. A random sample

of 100 bottles showed x = 1.99

liters. What is the 90%

confidence interval estimate of

the true mean amount in 2-

liter bottles? 2 liter

© 1984-1994 T/Maker Co.

2 liter



Confidence Interval

Solution*

998.1982.1

100

05.645.199.1

100

05.645.199.1

2/2/

n Z X

n Z X



Confidence Interval Estimate

Mean ( Unknown)




Confidence

Intervals

Mean Proportion

σ Knownσ

Unknown



Confidence Interval

Mean ( Unknown)

1. Assumptions

• Population standard deviation is unknown

• Population must be normally distributed

2. Use Student’s t– distribution



Z

t

Student’s t Distribution

0

t (df = 5)

Standard

Normal

t (df = 13)Bell-ShapedSymmetric

‘Fatter’ Tails



Degrees of Freedom (df)

1. Number of observations that are free to vary aftersample statistic has been calculated

2. Example

Sum of 3 numbers is 6X 1 = 1 (or any number)

X 2 = 2 (or any number)

X 3 = 3 (cannot vary)

Sum = 6

degrees of freedom

= n - 1

= 3 - 1

= 2



v t .10 t .05 t .025

1 3.078 6.314 12.706

2 1.886 2.920 4.303

3 1.638 2.353 3.182

Student’s t Table

t values

Assume:

n = 3

df = n - 1 = 2 = .10

/2 =.05

t 0

/ 2

/ 2

t 2.920



Confidence Interval

Mean ( Unknown)

/ 2 / 2

1

S S X t X t

n n

df n



Estimation Example

Mean ( Unknown)

/ 2 / 2

8 850 2.064 50 2.06425 25

46.69 53.30

S S X t X t

n n

A random sample of n = 25 has x = 50 and s = 8.

Set up a 95% confidence interval estimate for .



Thinking Challenge

You’re a time study analyst in

manufacturing. You’ve

recorded the following task

times (min.):3.6, 4.2, 4.0, 3.5, 3.8, 3.1.

What is the 90% confidence

interval estimate of thepopulation mean task time?



Confidence Interval Solution*

x = 3.7

s = 0.38987

• n = 6, df = n - 1 = 6 - 1 = 5

• t.05 = 2.015

• 0.38987 0.38987

•3.7 - 2.015 ------------ ≤ μ ≤ 3.7 + 2.015 -------------

• √ 6 √ 6

• 3.3793 ≤ μ ≤ 4.0207



Confidence Interval Estimate of

Proportion




Confidence

Intervals

Mean Proportion

σ Knownσ

Unknown



Confidence Interval

Proportion

1. Assumptions

• Random sample selected

• Normal approximation can be used if

2 2

ˆ ˆ ˆ ˆ

ˆ ˆ

pq pq p z p p z

n n

2. Confidence interval estimate

ˆ ˆ15 and 15np nq



Estimation Example

Proportion

A random sample of 400 graduates showed 32 wentto graduate school. Set up a 95% confidence

interval estimate for p.

/ 2 / 2

ˆ ˆ ˆ ˆ

ˆ ˆ

.08 .92 .08 .92.08 1.96 .08 1.96400 400

.053 .107

pq pq p Z p p Z

n n

p

p



Thinking Challenge

You’re a production manager

for a newspaper. You want to

find the % defective. Of 200

newspapers, 35 had defects.What is the 90% confidence

interval estimate of the

population proportion defective?



Confidence Interval

Solution*

/ 2 / 2

ˆ ˆ ˆ ˆ

ˆ ˆ

.175 (.825) .175 (.825).175 1.645 .175 1.645

200 200

.1308 .2192

p q p q p z p p z

n n

p

p



Finding Sample Sizes




for Estimating

2 2

2 22

2

(1)

(2)

( )(3)

( )

x x

x

X SE Z

SE Z Z n

Z n

SE

SE = Sampling Error

I don’t want tosample too much

or too little!



Sample Size Example

What sample size is needed to be 90% confident the

mean is within 5? A pilot study suggested that thestandard deviation is 45.

2 22 2

2

22

( ) 1.645 45

219.2 220( ) 5

Z

n SE




for Estimating

ˆ ˆ

ˆ2 2

22

2

ˆ

(1)

(2)

( )(3)

( )

p p

p

p p SE Z

pqSE Z Z

n

Z pqn

SE

SE = Sampling Error

If no estimate of p is

available, use p = q = .5



Sample Size Example

What sample size is needed to estimate p with 90% confidence

and a width of .03?

(1.645) ² (0.5 0.5)

N = ----------------------- = 3006.69 ≈ 3007 (0.015)²

.03

.0152 2

width

SE



Thinking Challenge

You work in Human Resources at

Merrill Lynch. You plan to survey

employees to find their average

medical expenses. You want to be95% confident that the sample

mean is within ± $50.

A pilot study showed that was

about $400. What sample size doyou use?



Sample Size Solution*

2 2

2

2

2 2

2

( )

( )

1.96 400

50

245.86 246

Z n

SE



Hypothesis Testing Concepts



Hypothesis Testing

Population

I believe the

population meanage is 50

(hypothesis).

Mean

X = 20

Randomsample

Reject

hypothesis!

Not close.



What’s a Hypothesis?

A belief about a population

parameter

• Parameter is

population mean,

proportion, variance

• Must be stated

before analysis

I believe the mean GPA of

this class is 3.5!

© 1984-1994 T/Maker Co.



Null Hypothesis

1. What is tested

2. Has serious outcome if incorrect decision made

3. Always has equality sign: , , or 4. Designated H0 (pronounced H-oh)

5. Specified as H0: some numeric value

• Specified with = sign even if or • Example, H0: 3



Alternative Hypothesis

1. Opposite of null hypothesis

2. Always has inequality sign: , , or

3. Designated Ha

4. Specified Ha: , , or some value

• Example, Ha: < 3

Id tif i H th



Identifying Hypotheses

Steps

Example problem: Test that the population mean isnot 3

Steps:

• State the question statistically ( 3)• State the opposite statistically ( = 3)

— Must be mutually exclusive & exhaustive

• Select the alternative hypothesis ( 3) — Has the , <, or > sign

• State the null hypothesis ( = 3)



What Are the Hypotheses?

State the question statistically:

= 12 State the opposite statistically: 12

Select the alternative hypothesis: Ha: 12

State the null hypothesis: H0: = 12

Is the population average amount of TV

viewing 12 hours?





12 State the opposite statistically: = 12


State the null hypothesis: H0: = 12

Is the population average amount of TV

viewing different from 12 hours?





20 State the opposite statistically: 20


State the null hypothesis: H0: 20

Is the average cost per hat less than or equal

to $20?





25 State the opposite statistically: 25


State the null hypothesis: H0: 25

Is the average amount spent in the bookstore

greater than $25?



Basic Idea

Sample Means = 50

H0

Sampling Distribution

It is unlikely

that we would

get a samplemean of this

value ...

20

... if in fact this were

the population mean

... therefore, we

reject thehypothesis that

= 50.



Level of Significance

1. Probability

2. Defines unlikely values of sample statistic if null

hypothesis is true

• Called rejection region of samplingdistribution

3. Designated (alpha)

• Typical values are .01, .05, .10

4. Selected by researcher at start

Rejection Region



Rejection Region

(One-Tail Test)

Ho

Value Critical

Value

Sample Statistic

Rejection

Region

Nonrejection

Region


1 –

Level of Confidence

Observed sample statistic

Rejection Region



Rejection Region

(One-Tail Test)

Sampling DistributionLevel of Confidence

Ho

Value Critical

Value

Sample Statistic

Rejection

Region

Nonrejection

Region


1 –

Level of Confidence


Rejection Regions



Rejection Regions

(Two-Tailed Test)

Ho

Value Critical

Value

Critical

Value

1/2 1/2

Sample Statistic

Rejection

Region

Rejection

Region

Nonrejection

Region


1 –

Level of Confidence


Rejection Regions



Ho

Value Critical

Value

Critical

Value

1/2 1/2

Sample Statistic

Rejection

Region

Rejection

Region

Nonrejection

Region


1 –

Level of Confidence

Rejection Regions

(Two-Tailed Test)



Rejection Regions



Ho

Value Critical

Value

Critical Value

1/2 1/2

Sample Statistic

Rejection

Region

Rejection

Region

Nonrejection

Region


1 –

Level of Confidence

Rejection Regions

(Two-Tailed Test)





Decision Making Risks

Errors in



Errors in

Making Decision

1. Type I Error• Reject true null hypothesis

• Has serious consequences

• Probability of Type I Error is (alpha) — Called level of significance

2. Type II Error

• Do not reject false null hypothesis

• Probability of Type II Error is (beta)



Decision Results

H0: Innocent

Jury Trial

Actual Situation

Verdict Innocent Guilty

Innocent Correct Error

Guilty Error Correct

H0 Test

Actual Situation

Decision H0 True H0

False

Accept

H0

1 – Type II

Error

()

Reject

H0

Type I

Error ()

Power

(1 – )

& Have an



& Have an

Inverse Relationship

You can’t reduce botherrors simultaneously!



Factors Affecting 1. True value of population parameter

• Increases when difference with hypothesizedparameter decreases

2. Significance level, • Increases when decreases

3. Population standard deviation,

• Increases when increases

4. Sample size, n

• Increases when n decreases



Hypothesis Testing Steps



H0 Testing Steps

State H0

State Ha

Choose

Choose n

Choose test

• Set up critical values

• Collect data

• Compute test statistic

• Make statistical decision

• Express decision



One Population Tests

One Population

Z Test (1 & 2

tail)

t Test (1 & 2

tail)

Z Test (1 & 2

tail)

Mean Proportion Variance

2 Test (1 & 2

tail)



Two-Tailed Z Testof Mean ( Known)




One Population

Z Test (1 & 2

tail)

t Test (1 & 2

tail)

Z Test (1 & 2

tail)


2 Test (1 & 2

tail)

Two-Tailed Z Test



Two Tailed Z Test

for Mean ( Known)

1. Assumptions

• Population is normally distributed

• If not normal, can be approximated by

normal distribution (n 30)2. Alternative hypothesis has sign

x

x

X X Z

n

3. Z-Test Statistic

Two-Tailed Z Test



for Mean Hypotheses

H0: = 0 Ha: ≠ 0

Z0

Reject H0

/ 2 / 2

Reject H

Two-Tailed Z Test



.500

- .025.475

Z 0

= 1

Finding Critical Z

What is Z given = .05?

/ 2 = .025

Z .05 .07

1.6 .4505 .4515 .4525

1.7 .4599 .4608 .4616

1.8 .4678 .4686 .4693

.4744 .4756

.06

1.9 .4750

Standardized Normal

Probability Table (Portion)

1.96 -1.96



Two-Tailed Z Test Example

Does an average box of cereal

contain 368 grams of cereal? A

random sample of 64 boxes

showed x = 372.5. The companyhas specified to be 15 grams.

Test at the .05 level of

significance.

368 gm.



Two-Tailed Z Test Solution

H0:

Ha:

n

Critical Value(s):

Test Statistic:

X - μ 372.5 – 368

Z = --------- = --------------- = 2.4

σ/√ n 15/ √ 64Decision:

Conclusion:

= 368

368

.05

64

Z0 1.96-1.96

.025

Reject H0

Reject H0

.025

Reject at = .05

No evidence average

is 368

Two-Tailed Z Test Thinking



Two Tailed Z Test Thinking

Challenge

You’re a Q/C inspector. You want to find out

if a new machine is making electrical cords to

customer specification: average breaking

strength of 70 lb. with = 3.5 lb. You take asample of 36 cords & compute a sample

mean of 69.7 lb. At the .05 level of

significance, is there evidence that the

machine is not meeting the average breakingstrength?



Two-Tailed Z Test Solution*

H0:

Ha:

=

n = Critical Value(s):

Test Statistic:

Decision:

Conclusion:

= 70

70

.05

36

Z0 1.96-1.96

.025

Reject H0

Reject H0

.025

69.7 70.51

3.5

36

X Z

n

Do not reject at = .05

No evidence average

is not 70



One-Tailed Z Testof Mean ( Known)

One-Tailed Z Test



One Tailed Z Test

for Mean ( Known)

1. Assumptions• Population is normally distributed

• If not normal, can be approximated by

normal distribution (n 30)2. Alternative hypothesis has < or > sign

3. Z-test Statistic

x

x

X X Z

n

One-Tailed Z Test



for Mean Hypotheses

H0: = 0 Ha: < 0

Z0

Reject H0

Must be significantly

below

Z0

Reject H0

H0: = 0 Ha: > 0

Small values satisfy H0 .

Don’t reject!

One-Tailed Z Test



.500

- .025.475

Z 0

= 1

Finding Critical Z

What Is Z given = .025?

= .025

1.96

Z .05 .07

1.6 .4505 .4515 .4525

1.7 .4599 .4608 .4616

1.8 .4678 .4686 .4693

.4744 .4756

.06

1.9 .4750

Standardized Normal

Probability Table (Portion)

One-Tailed Z Test



Example


contain more than 368 grams of

cereal? A random sample of 64

boxes showed x = 372.5. Thecompany has specified to be 15

grams. Test at the .05 level of

significance.

368 gm.



One-Tailed Z Test Solution

H0:

Ha:

=


Test Statistic:

X - μ 372.5 – 368

Z = --------- = ---------------

σ/√ n 15/ √ 64= 2.4

Decision:

Conclusion:

= 368

> 368

.05

64

Z0 1.645

.05

Reject

Reject at = .05

No evidence average is 368

One-Tailed Z Test Thinking



g

Challenge

You’re an analyst for Ford. You want tofind out if the average miles per gallon

of Escorts is at least 32 mpg. Similar

models have a standard deviation of 3.8

mpg. You take a sample of 60 Escorts &

compute a sample mean of 30.7 mpg. At

the .01 level of significance, is there

evidence that the miles per gallon is atleast 32?



One-Tailed Z Test Solution*

H0:

Ha:

=

n =

Critical Value(s):

Test Statistic:

Decision:

Conclusion:

= 32

< 32

.01

60

Z0-2.33

.01

Reject

30.7 322.65

3.8

60

X Z

n

Reject at = .01

There is evidence average

is less than 32



Two-Tailed t Testof Mean ( Unknown)




One Population

Z Test (1 & 2

tail)

t Test (1 & 2

tail)

Z Test (1 & 2

tail)


2 Test (1 & 2

tail)

t Test for Mean



( Unknown)

1. Assumptions• Population is normally distributed

• If not normal, only slightly skewed & large

sample (n 30) taken

2. Parametric test procedure

3. t test statistic

X t

S

n

Two-Tailed t Test



t 0

Finding Critical t Values

Given: n = 3; = .10

/2 = .05

/2 = .05

df = n - 1 = 2

v t .10 t .05 t .025

1 3.078 6.314 12.706

2 1.886 2.920 4.303

3 1.638 2.353 3.182

Critical Values of t Table

(Portion)

2.920 -2.920

Two-Tailed t Test



Example

Does an average box of

cereal contain 368 grams of

cereal? A random sample

of 25 boxes had a mean of

372.5 and a standard

deviation of 12 grams. Test

at the .05 level of

significance.368 gm.

Two-Tailed t Test



Solution

H0:

Ha:

=

df = Critical Value(s):

Test Statistic:

X – μ 372.5 – 368

t = ------- = -------------- = 1.875

s/√n 12/√ 25 Decision:

Conclusion:

= 368

368

.05

25 - 1 = 24

t0 2.064-2.064

.025

Reject H0

Reject H0

.025


There is no evidence

population average is not

368

Two-Tailed t Test

Thi ki Ch ll



Thinking Challenge

You work for the FTC. A

manufacturer of detergent claims

that the mean weight of detergent is

3.25 lb. You take a random sample of16 containers. You calculate the

sample average to be 3.238 lb. with a

standard deviation of .117 lb. At the

.01 level of significance, is the

manufacturer correct?3.25 lb.

Two-Tailed t Test



Solution*

H0:

Ha:

df Critical Value(s):

Test Statistic:

X – μ 3.238 – 3.25

t = -------- = ----------------

s/√n 0.117/√16

= -0.410

Decision:

Conclusion:

= 3.25

3.25

.01

16 - 1 = 15

t0 2.947-2.947

.005

Reject H0

Reject H0

.005 Do not reject at = .01


average is not 3.25



One-Tailed t Testof Mean ( Unknown)

One-Tailed t Test



Example

Is the average capacity of batteries at least 140 ampere-

hours? A random sample of 20

batteries had a mean of 138.47 and a standard deviation of

2.66. Assume a normal

distribution. Test at the .05 level of significance.

One-Tailed t Test



Solution

H0:

Ha:

=


Test Statistic:

Decision:

Conclusion:

= 140

< 140

.05

20 - 1 = 19

t 0-1.729

.05

Reject H0

138.47 1402.57

2.66

20

X t

S

n

Reject at = .05

There is evidence population

average is less than 140

One-Tailed t Test



Thinking Challenge

You’re a marketing analyst for Wal-

Mart. Wal-Mart had teddy bears on

sale last week. The weekly sales ($

00) of bears sold in 10 stores was:8 11 0 4 7 8 10 5 8 3

At the .05 level of significance, is

there evidence that the average bear

sales per store is more than 5 ($00)?

One-Tailed t Test



Solution*

H0:

Ha:

=


Test Statistic:

Decision:

Conclusion:

= 5

> 5

.05

10 - 1 = 9

t0 1.833

.05

Reject H0

6.4 51.31

3.373

10

X t

S

n



average is more than 5



Observed Significance Levels: p-Values

V l



p-Value

1. Probability of obtaining a test statistic more extreme( or than actual sample value, given H0 is true

2. Called observed level of significance

• Smallest value of

for which H0 can be rejected3. Used to make rejection decision

• If p-value , do not reject H0

• If p-value < , reject H0

Two-Tailed Z Test

p Val e E ample



p-Value Example


contain 368 grams of cereal? A

random sample of 36 boxes

showed x = 372.5. The

company has specified to be

15 grams. Find the p-Value.

368 gm.

Two-Tailed Z Test



p-Value Solution

Z 0 1.8 Z value of sample

statistic (observed)

X - μ 372.5 - 368

Z = ---------------- = --------------------- = 1.8

σ/√ n 15/ √ 36

Two-Tailed Z Test



1/2 p-Value 1/2 p-Value

p-Value Solution

Z value of sample

statistic (observed)

p-value is P(Z -1.80 or Z 1.80)

Z 0 1.80 -1.80

From Z table:

lookup 1.50

.4641

.5000

- .4641

.0359

Two-Tailed Z Test



p-Value Solution

1/2 p-Value

.0359

1/2 p-Value

.0359

p-value is P(Z -1.80 or Z 1.80) = .0718

Z value of sample

statistic

From Z table:

lookup 1.50

.5000

- .4641

.0359

Z 0 1.80 -1.80

One-Tailed Z Test

p Value Example



p-Value Example


contain more than 368

grams of cereal? A random

sample of 36 boxes showed x = 372.5. The company has

specified to be 15 grams.

Find the p-Value.368 gm.

One-Tailed Z Test

V l S l i



p-Value Solution

Z 0 1.80

Z value of sample

statistic

X - μ 372.5 - 368Z = ---------------- = --------------------- = 1.8

σ/√ n 15/ √ 36

One-Tailed Z Test

V l S l ti



p-Value Solution

p-Value

.0359

Z value of sample

statistic

From Z table:

lookup 1.80

Usealternative

hypothesis

to find

direction

.5000

- .4641

.0359

p-Value is P(Z 1.80) = .0359

Z 0 1.80

.4641

Two-Tailed Z Test

V l S l ti



p-Value Solution

0 1.80 -1.80 Z

Reject H0 Reject H0

1/2 p-Value = .03591/2 p-Value = .0359

1/2 = .0251/2 = .025

(p-Value = .0718) ( = .05).

Do not reject H0.

Test statistic is in ‘Do not reject’ region

One-Tailed Z Test

V l S l ti



p-Value Solution

Usealternative

hypothesis

to find

direction

p-Value is P(Z 1.80)

Z value of sample

statistic

p-Value

Z 0 1.80

From Z table:

lookup 1.80

.4641

.5000

- .4641

.0359

One-Tailed Z Test

V l S l ti



= .05

p-Value Solution

0 1.50 Z

p-Value = .0359

(p-Value = .0359) < ( = .05).Do not Reject H0.

Test statistic is in ‘Do not reject’ region

p-Value

Thi ki Ch ll



Thinking Challenge

You’re an analyst for Ford. You want

to find out if the average miles per

gallon of Escorts is at least 32 mpg.

Similar models have a standarddeviation of 3.8 mpg. You take a

sample of 60 Escorts & compute a

sample mean of 30.7 mpg. What is the

value of the observed level ofsignificance (p-Value)?

p-Value

S l ti *



Usealternative

hypothesis

to find

direction

Solution*

Z 0 -2.65 Z value of sample

statistic

From Z table:

lookup 2.65

.4960

p-Value

.004 .5000- .4960

.0040

p-Value is P(Z -2.65) = .004.

p-Value < ( = .01). Reject H0.



Z Test of Proportion

Data Types



Data Types

Data

Quantitative Qualitative

ContinuousDiscrete

Qualitative Data



Qualitative Data

1. Qualitative random variables yield responses thatclassify

• e.g., Gender (male, female)

2. Measurement reflects number in category

3. Nominal or ordinal scale

4. Examples

• Do you own savings bonds?

• Do you live on-campus or off-campus?

Proportions



Proportions1. Involve qualitative variables

2. Fraction or percentage of population in a

category

3. If two qualitative outcomes, binomial

distribution

• Possess or don’t possess characteristic

4. Sample Proportion ( p)number of successes

ˆ

sample size

x p

n

^


of Proportion



of Proportion

1. Approximated by Normal

Distribution

Excludes 0 or n

2. Mean

3. Standard Error

n

p p p

)1( 00ˆ

ˆ P p


where p0 = Population Proportion

.0

.1

.2

.3

.0 .2 .4 .6 .8 1.0

P^

P(P^

) ˆ1ˆ3ˆ p pn pn

Standardizing Sampling

Distribution of Proportion



Z= 0

z= 1

Z

Distribution of Proportion

Sampling

Distribution

Standardized Normal

Distribution

P^ P

P

^

^

Z p p p

p p

n

^ p

p

^

^

( )1

^0

0 0





One Population

Z Test (1 & 2

tail)

t Test (1 & 2

tail)

Z Test (1 & 2

tail)


2 Test (1 & 2

tail)

One-Sample Z Test

for Proportion



for Proportion

1. Assumptions

• Random sample selected from a binomial population

• Normal approximation can be used if

0 0ˆ ˆ15 and 15np nq

2. Z-test statistic for proportion

0

0 0

ˆ p p Z

p q

n

Hypothesized population

proportion

One-Proportion Z Test

Example



Example

The present packaging systemproduces 10% defective cereal

boxes. Using a new system, a

random sample of 200 boxeshad 11 defects. Does the new

system produce fewer defects?

Test at the .05 level of

significance.

One Proportion Z Test Solution



One-Proportion Z Test Solution

H0:

Ha:

=


Test Statistic:

Decision:

Conclusion:

p = .10

p < .10

.05

200

Z0-1.645

.05

Reject H0

0

0 0

11.10

ˆ 200 2.12.10 .90

200

p p Z

p q

n

Reject at = .05

There is evidence new

system < 10% defective

One-Proportion Z Test Thinking

Challenge



Challenge

You’re an accounting manager. A

year-end audit showed 4% of

transactions had errors. You

implement new procedures. Arandom sample of 500 transactions

had 25 errors. Has the proportion

of incorrect transactions changed

at the .05 level of significance?

One-Proportion Z Test Solution*



One-Proportion Z Test Solution

H0: Ha:

=


Test Statistic:

Decision:

Conclusion:

p = .04

p .04

.05

500

Z0 1.96-1.96

.025

Reject H0

Reject H0

.025

0

0 0

25.04

ˆ 500 1.14.04 .96

500

p p Z

p q

n


There is evidence

proportion is not 4%



Chi-Square (2) Testof Variance





One Population

Z Test (1 & 2

tail)

t Test (1 & 2

tail)

Z Test (1 & 2

tail)


2 Test (1 & 2

tail)

Chi-Square (2) Test

for Variance



for Variance

1. Tests one population variance or standarddeviation

2. Assumes population is approximately normally

distributed3. Null hypothesis is H0: 2 = 0

2

4. Test statistic

Hypothesized pop. variance

Sample variance

2

2

2

1)

(n S

0

Chi-Square (2) Distribution



C Squa e ( ) st but o

Select simple random

sample, size n.

Compute s 2

Compute 2 = (n-1)s 2 / 2

Astronomical number

of 2 values

Population Sampling Distributions

for Different Sample

Sizes

2 1 2 3 0

Finding Critical Value Example



What is the critical 2 value given

Ha: 2 > 0.7

n = 3

=.05?


2 0

Upper Tail Area

DF .995 … .95 … .05

1 ... … 0.004 … 3.841

2 0.010 … 0.103 … 5.991

2 Table

(Portion)

df = n - 1 = 25.991

Reject

= .05





What is the critical 2 value given:

Ha: 2 < 0.7

n = 3

=.05?

What do you do

if the rejectionregion is on the

left?




What is the critical 2 value given:Ha: 2 < 0.7

n = 3

=.05?


.103 2 0

Upper Tail Area

DF .995 … .95 … .05

1 ... … 0.004 … 3.841

2 0.010 … 0.103 … 5.991

2 Table

(Portion)

Upper Tail Area

for Lower Critical

Value = 1-.05 = .95 = .05

Reject H0

df = n - 1 = 2

Chi-Square (2) Test Example



q ( ) p

Is the variation in boxes ofcereal, measured by the

variance, equal to 15 grams?

A random sample of 25 boxeshad a standard deviation of

17.7 grams. Test at the .05

level of significance.

Chi-Square (2) Test

Solution



Solution

H0: Ha:

=


Test Statistic:

Decision:

Conclusion:

2 = 152 15

.05

25 - 1 = 24

/2 = .025

= 33.42

2 22

2 2

0

( 1) (25 1) 17.7

15

n S


Ch 4 - Estimation & Hypothesis One Sample

Documents

Transcript of Ch 4 - Estimation & Hypothesis One Sample