construct chapter 9 interpret confidence interval...

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Chapter 9Confidence Interval Estimation

Topics covered:Confidence Interval Estimation for the Mean (1 known)

Confidence Interval Estimation for the Mean (1 unknown)Confidence Interval Estimation for the Proportion

Determining Sample Size

Objective

Learn how to constructconstruct and

interpretinterpret confidence interval

estimatesestimates

For the mean

For the proportion

Confidence Interval Estimation

Population Mean P Population Proportion E

X Normal and/or nt30?If no stop.

np t 5 and n(1-p) t 5?If no stop.

V known? V unknown?

Limits: Limits: df=n-1

n

1zx r

Limits:

p=x/n

n

stx r

n

)p1(pzp

�r

WEEK 1WEEK 1 WEEK 2WEEK 2

Frequently asked questions

1. “1 is known” and “1 is unknown”

1 is given and 1 is not given

2. What is Z scores/values?

Go to this link to understand "z-score" :

http://www.youtube.com/watch?v=1xhCL5m4nI0&feature=BF&list=QL&index=1

http://www.youtube.com/watch?v=s0lLBcARxL4&feature=mfu_in_order&list=UL

Cut and paste the link on the URL box.

3. What are you estimating? Parameters P or S

4. How do you interpret CI?

• Go to this link http://www.youtube.com/watch?v=hP6flJdoIxc&feature=fvw

� =2

Review

� �+2��+1��r2� �r1� �+3��r3�

7 9 11 13531 X�scale�(µ�=7,� =2)

r3 r2 r1 0 +1 +2 +3 Z�scale�(µ�=0,� =1)

Convert any normal random variable X to standardized normal random variable Z.

If X is normally distributed N(�, 1), the standardized variable Z has a standard normal distribution with mean=0 (or �=0) and standard deviation 1 (or 1=1), denoted N(0,1)

V

P�

XZ

% Confidence Interval when 1 is given :

n*x z 2/

Vr

D

We needed “z value” for a % confidence Interval.

90% confidence level Confidence coefficient of 0.90 Z= ??????




% C fid I t l h i t i s% Confidence Interval when 1 is not given :

n

stx

1n,2 �DrWe needed “t value” for a % confidence Interval.

Suppose we needed to find t, when sample size (denoted as n) is 25.

90% confidence level Confidence coefficient of 0.90 t = ??????




www.notesolution.comwww.notesolution.com

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TEMPLATE : Confidence Interval for population mean

when 1 is known

:

Define : Estimate µ

95% CI :

n

1Zx

2.

r

81Note:

There are three ways to find the

50

8.19566.10.20 r

5.00.20 r

5.20�5.19 dd


Confidence interval

1. Use the formula, find z value

using the INVN function

from CASIO calculator


using SPSS function:

IDF.Normal (prob, mean, stdev)

3. Use the INTR function of CASIO

Calculator to find Confidence

Interval directly

TEMPLATE : Confidence Interval for population mean

when 1 is unknown

:

Define : Estimate µ

95% CI :

n

stx

1n,2

. �r

33064020255 r

Note:

There are two ways to find the

Confidence interval

1. Using the table to find t value and

th f l t fi d th CI

250640.20.255 r

6.130.255 r

6.268�3.241 dd

use the formula to find the CI.

2. Using the formula, find t value


IDF.T (./2, degree of freedom)

and calculator CI using SPSS

compute function.

3. Using the INTR function of CASIO


Interval directly

Confidence Interval Estimation

Population Mean P Population Proportion E

X Normal and/or nt30?If no stop.

np t 5 and n(1-p) t 5?If no stop.

V known? V unknown?

Limits: Limits: df=n-1

n

1zx r

Limits:

p=x/n

n

stx r

n

)p1(pzp

�r

WEEK 1WEEK 1 WEEK 2WEEK 2

Interpreting the CI

Read page 369, ….with 95% confidence, you conclude that the mean amount of all the sales i i i b t LI (linvoices is between LI (lower

interval) and UI (upper interval).

Referring to “population”

Confidence Interval Estimation for the proportionp p

We extends the concept of CI to categorical data

Here you are concerned with estimating the proportion of items in a population having a certain characteristic of

interest (e.g. people who have an iphone)

Confidence Interval Estimation for the proportion

• The unknown population proportion is represented by S

• The point estimate for S is the sample

proportion denoted as p where p=X/nproportion, denoted as p, where p=X/nWhere n = sample size (number of observed items/subjects in a sample)

And X=number of items in the sample having the characteristic of interest


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95% CI :)p1(p

zp�

r

Note:


Confidence interval

Check the conditions: np > 5 and n(1-p) > 5

State the n and p values

If conditions are satisfied, proceed to find CI

TEMPLATE : Confidence Interval for population

proportion, E

Define : est E = population proportion of __________________

95% CI :

n

)p(pzp r

50

)24.01(24.0575.224.0

�r

16.024.0 r

40.008.0 d�d






IDF.Normal (prob, mean,

stdev)

3. Use the INTR function of

CASIO


Interval directly

Note

n

)p1(pzp

�r

The equation contains a Z statistic since you can use the normal distribution to approximate the Binomial distribution when the sample size is sufficiently large.That is why we check the condition before we proceed to calculate the Confidence Interval for S

ExampleIn a study reported in the Wall Street Journal on

April 4, 1999. The Tupperware Corporation surveyed 1007 U.S. workers. Of the people surveyed, 665 indicated that they take their lunch to work with them. Of these 665 taking lunch, 200 reported that they take them in brown bagsbrown bags.

a) Set up a 95% confidence interval estimate of the population proportion of U.S. workers who take their lunch to work with them.

Check the conditions: np > 5 and n(1-p) > 5

State the n and p values

Solution to Question 5-36

E = population proportion of U.S. workers who take their lunch to work with them

a) Set up a 95% confidence interval estimate of the

population proportion of U.S. workers who take their lunch to work with them.

n=1007p=x/n = 665/1007 = 0.6604

np 1007(0 6604) 665 0228

Are the conditions satisfied?

np=1007(0.6604) =665.0228 n(1-p) = 1007 (1-0.6604) =341.9772

If yes, proceed to find CI

95% CI :

n

)p1(pzp

�r

Note:


Confidence interval


est.E = population proportion of U.S. workers who take their lunch to work with them



Confidence interval






IDF.Normal (prob, mean, stdev)

3. Use the INTR function of CASIO


Interval directly

95% CI :

n

)p1(pzp

�r


est. E = population proportion of U.S. workers who take their lunch to work with them



Aside:To find Z valueZ (1-0.95)/2 = Z0.05/2

Z0.025InvN(0.025,1,0)=r1.9599Round the value of Z to 4 decimals

1007

)6604.01(6604.09599.16604.0

�r

0292.06604.0 r

6896.06311.0 d�d


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Using the calculator

1-Proportion Z Interval

From the Main Menu select:STAT F4(INTR) F1(Z) F3(1-P) then enter the following items:

1-Prop ZInterval

C-Level : 0 95 EXEC-Level : 0.95 EXE

x : 665 EXE

n : 1007 EXE

Now key EXE or F1(Calc)

Calculator Result

The calculator will now show the results:

1-Prop ZInterval

Left =0.63112

Right =0.68962

p =0 66037p =0.66037

n =1007

This shows that the 95% confidence interval estimate for the population proportion is 0.6311 to 0.6896. This shows the numerical result only. You must remember to write up the complete answer as shown in class.

Using SPSS

SPSS formulas (using “compute” function in SPSS) for LL and UL

LL: 665/1007+ IDF.Normal(0.025,0,1)*SQRT(665/1007*(1-665/1007)/1007)

UL: 665/1007+IDF.Normal(0.975,0,1)*SQRT(665/1007*(1-665/1007)/1007)

n

)p1(pzp

�r

Note: The formula you input into SPSS is the same as this formula

95% CI :

n

)p1(pzp

�r

)6604.01(6604.09599166040

�r


est E = population proportion of U.S. workers who take their lunch to work with them



10079599.16604.0 r

6896.06311.0 d�d

Do Not Forget to write your interpretation of the CI you calculated above,

A point estimate for the population proportion of U.S. workers who take their lunchto work with them is approximately 0.6604 and the 95% confidence interval estimateof the population proportion of U.S. workers who take their lunch to work withthem is approximately (0.6311, 0.6896)

More examples on how to use the CASIO calculator to calculate

CI forCI for S

ExampleA bank manager wants to estimate the

proportion of all the bank’s customers who use the ATM to pay bills. A random sample of 80 customers has been selected and it was found that 43 of them use the ATM to pay bills. Determine the 99% confidence interval estimate for the proportion of all the bank’s customers who use the ATM to pay bills.


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Using the calculator

1-Proportion Z Interval

From the Main Menu select:STAT F4(INTR) F1(Z) F3(1-P)

then enter the following items:

1-Prop ZInterval

C-Level : 0 99 EXEC-Level : 0.99 EXE

x : 43 EXE

n : 80 EXE

Now key EXE or F1(Calc)

Calculator Result

The calculator will now show the results:1-Prop ZInterval

Left =0.39391Right =0.68108

0 5375p =0.5375n =80

This shows that the 99% confidence interval estimate for the population proportion is 0.394 to 0.681. This shows the numerical result only. You must remember to write up the complete answer as shown in class.

Find sample size


• In each example of confidence interval estimation, you were given the sample size.

• In a real world setting you must determine• In a real world setting, you must determine in advance how large a sampling error to allow in estimating each of the parameters.

• What is sampling error (or “margin error” as

defined in lecture 1)?

Determining sample size


Sample size determination forthe mean

Sample size determination for the proportion


Confidence Interval for population

mean when 1 is known

1Zx

.r

n

1Ze

2.

Half width of the interval“margin of error”

This quantity represents the amount of impression in the estimate that resultsfrom sampling error.

The sampling error,

n2.

Solving for n gives the sample size neededto construct the appropriate CI for estimatemean.


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“Appropriate” means that the resulting interval will have an

acceptable amount of sampling error.

221z2e

1n

To determine the sample size, you must know 3 things:1. The desired confidence level (e.g. 95% or 99%), which determines

the value of z. 2. The acceptable sampling error, e.3. The population standard deviation V

TEMPLATE For Sample size determination forthe meanDefine : Estimate µ

e

1Zn

2

222/.

2)5752( 2

121

2)575.2( 2n

n=954.81 §955

The minimum sample size is 955 people

Example 6.47

A survey is planned to determine the average annual family medical expenses of employees of a large company. The management of the company wishes to be 95% confident the sample average is located with r$50 of the truesample average is located with r$50 of the true average annual family medical expense. A pilot study indicates that the standard deviation can be estimated at $400.

a) How large a sample is necessary?

b) If management wants to be correct to within r$25, what is the sample size?

Question 6.47a

est. µ represent the population average annual family medical expenses of employees of a large company

)400 2()9599.1( 2 n

e

1Zn

2

222/.

)50( 2 n

n=245.8373 §246

The minimum sample size of 246 employees would be required.

Question 6.47b If management wants to be correct to within r$25, what is the sample size?

)400()95991(22

est. µ represent the mean annual family medical expenses in $

e2

122025.0

Z

e2

122

2.

Z

n

)25()400()9599.1(

2 n

n=983.35 §984

The minimum sample size of 984 employees would be required.

Determining sample size



Sample size determination for the proportion


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Sample size determination forthe proportion

To determine the sample size needed to estimate a population proportion (E), you use a method similar to the method for a population mean (µ).

The sampling error

)�1(� �

The sampling error,

When estimating a proportion, you replace V by

Thus the sampling error is:

n

)�1(�Ze

2.

�

Sample Size for Estimating the Proportion

e

)�1(�Zn

2

2

2.

�

Solving for n, you have the sample size necessary to developa CI for a proportion

To determine the sample size, you must know 3 things:1.The desired confidence level (e.g. 95% or 99%), which determines

the value of z. 2. The acceptable sampling error, e.3. The population proportion E

What is the value of the population proportion S?

You have two alternatives:

1. If we do not have any information about S, we use the most conservative value of S which is E=0 5S which is E=0.5

2. If we know S within some range of values, we use the value closest to 0.5

Why???

If we know E within some range of values, we use the value closest to 0.5

Why???

• Try to find a value for E that would never underestimate the sample size needed.

e

)�1(�Zn

2

2

2.

� Numerator

You need to determine the value of E that will make the quantity S(1- S)as large as possible.When E=0.5, E(1- E) value is the highest.

When S =0.9, then S(1- S)=0.9*0.1=0.09When S =0.7, then S(1- S)=0.7*0.3=0.21When S =0.5, then S(1- S)=0.5*0.5=0.25When S =0.3, then S(1- S)=0.3*0.7=0.21When S =0.1, then S(1- S)=0.1*0.9=0.09

The general rule is to round the sample size up to the next whole integer.

Example 1We wish to estimate the proportion of people who will

vote for the incumbent government party. The estimate is to be within 3% of the actual value 19 out of 20 times. What is the minimum sample size required?

Let est. S represent the population proportion of people who will vote for the incumbent government party.

106811.106703.0

)5.01)(5.0(9599.1n

2

2

| �

g p y

Since we have no information regarding value of S, we will use S =0.5

The sample size is1068 people

e

)�1(�Zn

2

2

2.

� e

)�1(�Zn

2

2025.0

�


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Example 2We wish to estimate the proportion of

students who are in favour of a laptop based business education. At similar institutions, the proportion of students

in favour is between 20% and 40%.The estimate is to be within 2% of the actualestimate is to be within 2% of the actual value at a 99% confidence level. What is the minimum sample size required?

S within some range of values, i.e. is between 20 and 40%.

we use the value closest to 0.5. That would be 0.4

SolutionLet est. E represent the population proportion of students who are in favour

of a laptop based business education.

e

)�1(�Zn

2

2

2.

�

e

)�1(�Zn

2

2025.0

�

e

398185.398002.0

)4.01)(4.0(5758.2n

2

2

| �

S within some range of values, i.e. is between 20 and 40%.

we use the value closest to 0.5. That would be 0.4

The minimum sample size is 3981 students

Finding sample size

Sample size for estimating the mean

• Given confidence level (.) and 1

e

1zn2

22

Sample size for estimating the proportion

• Given E and confidence level (.)

)�1(�2.

Z �e

e22.

n

NOTE: 1)Given the confidence level (e.g. 95% or 99%) we can compute z

value

2) Always round final answer of n (sample size) UP

For example, if n=23.2 ; you should round the value to 24

if n=34.8 ; you should round the value to 35

Homework

Topics covered:

1. Confidence Interval Estimation for the Mean (1 known)

2. Confidence Interval Estimation for the Mean (1 unknown)

3. Confidence Interval Estimation for the Proportion3. Confidence Interval Estimation for the Proportion

4. Determining Sample Size

Homework:

9.23, 9.25, 9.27, 9.29, 9.31, 9.33, 9.35, 9.37, 9.39, 9.41, 9.43, 9.45

Chapter 10

Fundamentals of Hypothesis Testing: One Sample Tests

Inferential Statistics

• In ESTIMATION (chapter 9- show how a sample could be used to develop point and interval estimates of population parameters (µ or/and S) How to use the sample mean CX or

sample proportion, p, to estimate the population mean, µ, or population proportion, S.

• In HYPOTHESIS TESTING (chapter 10) –In HYPOTHESIS TESTING (chapter 10) show how we can test whether a statement about the value of a population parameters should or should not be rejected. How the sample mean, CX, or sample

proportion, p, can be used to validate the claim that the population mean, µ, (or population proportion, S) is equal or less or more than a certain value.


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What is Hypothesis Testing?

FORESIGHT: In any experiment, researchers seek to discover or verify facts about the world around us and then put together these facts into theories or laws which clarify the relationships between the facts.

DATA COLLECTION: You make the observation andDATA COLLECTION: You make the observation and collect the data to explain the facts using the information(data) that you had collected.

TESTING: This attempted explanation is called a hypothesis. There are two hypotheses that you will state during the testing of evidence – the null hypothesis and the alternative hypothesis.

Is there any empirical evidence to support your

claim/research?

First we have to establish the statements

1. Null Hypothesis, Ho

2. Alternative Hypothesis, Hayp , a

These two hypothesis represent contradictory statements, e.g. “It’s

equal” an “It’s NOT equal”

The two hypothesis are mutually exclusive and both can’t be true at

the same time.

NULL HYPOTHESIS

• In hypothesis testing we begin with a tentative assumption about a population parameter.

• Called the NULL hypothesis• Called the NULL hypothesis

• Denoted as Ho

• The null hypothesis is written in terms of the population mean, not the sample mean

Alternative Hypothesis• After specifying Ho, we have to specify Alternative

Hypothesis

• Denoted Ha OR H1

• the alternative hypothesis states the opposite conclusion compared to the null hypothesis Ho

Example:

Ho : The average age of the population is equal to 45 years old

Ha : The average age of the population is NOT equal to 45 years old

The alternative hypothesis is also known as research hypothesis.

Developing Null and Alternative Hypothesesyp

1. Testing Research Hypotheses

2. Testing the Validity of a claim

3. Testing a Decision-Making Situations

Example: Testing Research Hypotheses

A particular automobile model that currently attains an average fuel efficiency of 24 miles per gallon. A product research groups developed a new fuel injection system specifically designed to increase the miles-per-gallon rating. To evaluate the new system, several will be

f t d i t ll d i t bil d bj t d tmanufactured, installed in automobiles, and subjected to research-controlled driving tests. The product research group is looking for evidence to conclude that the new system increases the mean-per-gallon rating.


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ExampleA particular automobile model that currently attains an

average fuel efficiency of 24 miles per gallon. A product research groups developed a new fuel injection system specifically designed to increase the miles-per-gallon rating. To evaluate the new system, several will be manufactured, installed in automobiles, and subjected to research-controlled driving tests. The product research

i l ki f id t l d th t thgroup is looking for evidence to conclude that the new system increases the miles per gallon rating.

The hypothesis statements:

Ho: µ � 24 Ha: µ > 24 : The new system increases the miles per gallon rating

As a general guideline, a research hypothesis should be stated as the

ALTERNATIVE hypothesis.

Testing the Validity of a claim

Consider the situation of a manufacturer of soft drinks who states that two-liter (2L) soft drink containers are filled with an average of at least 67.6 fluid ounces. A sample of two-liter containers will be selected, and the contents will be measured to test the manufacturer’s claim.

ExampleConsider the situation of a manufacturer of soft drinks who

states that two-liter soft drink containers are filled with an average of at least 67.6 fluid ounces. A sample of two-liter containers will be selected, and the contents will be measured to test the manufacturer’s claim.

The hypothesis statements:

Ho: µ � 67.6Ha: µ < 67.6

. Important note:

In any situation that involves testing the validity of a claim, the null hypothesisis based on the assumption that the claim is true. The alternative hypothesisis formulated so that rejection of Ho will provide statistical evidence that the stated assumption is incorrect. Action to correct the claim should be considered whenever Ho is rejected.

Example: Testing a Decision-Making Situations

On the basis of a sample of parts from a shipment just received, a quality control inspector must decide whether to accept the shipment or return the shipment to the supplier because it does not meet specifications Assume that specificationsmeet specifications. Assume that specifications for a particular part require a mean length of 2 inches per part. If the mean length is greater or

less than two-inch standard, the parts will cause quality problems in the assembly operation.

Example: Testing a Decision-Making Situations

On the basis of a sample of parts from a shipment just received, a quality control inspector must decide whether to accept the shipment or return the shipment to the supplier because it does not meet specifications. Assume that specifications for a particular part require a mean length of 2 inches per part. If the mean length is greater or less than two-inch standard, the parts will cause quality problems in the assembly operationproblems in the assembly operation.

The hypothesis statements:Ho: µ = 2 (Accept the shipment)Ha: µ � 2 (Return the shipment)

Summary Forms for Ho and Ha

For one sample

If Ho is Then Ha is And the test is called

Ho: µ = µ0 Ha: µ � µ0 Two-tailed

Ho: µ � µ0 Ha: µ > µ0 Right-tailed

Ho: µ � µ0 Ha: µ < µ0 Left-tailed

µ0 denote the hypothesized value


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Two Tailed TestHo: µ = µ0

Ha: µ � µ0

Ho: µ = µ0

Ha: µ > µ0RIGHT Tailed Test

Rejection Region Rejection Region

ONE TAIL TEST

D/2 D/2

D

Ho: µ = µ0

Ha: µ < µ0LEFT Tailed Test

Rejection Region

Rejection Region

ONE TAIL TEST

D

How do we make decision?

Statistical Decision:1. Reject Ho

2. DO NOT Reject Ho (note: do not

make the statement “Accept Ho”)

Suppose the hypothesis statements are:Ho: µ = 2Ha: µ � 2

1. If you reject the Ho, you have statistical proof thatthe alternative hypothesis is correct.

2. If you do not reject the Ho, then you have failed to prove the Ha. The failure to provethe alternative hypothesis, does not mean that you have proven the null hypothesis.

Example

You are the manager of a fast food restaurant. You want to determine whether the waiting time to place an order has changed in the last month from itschanged in the last month from its previous population mean value of 4.5 minutes. State the null and alternative hypotheses.

Solution

The null hypothesis is that the population mean has not changed from its previous values of 4.5 minutes

Ho : µ = 4.5

Th lt ti h th i i th it f thThe alternative hypothesis is the opposite of the null hypothesis. Since the null hypothesis is that the population mean is 4.5 minutes, the alternative hypothesis is that the population mean is not 4.5 minutes.

Ha: µ � 4.5

How do we make the decision?Use sample data

• To determine how likely the null hypothesis is true by considering the information gathered in a sample.

• If sample statistics,i.e. Cx , is close to µ, then Ho is trueis true.

• If the sample statistic is close to the population parameter, you have insufficient evidence to reject the Ho.

• IfCx=4.0 and µ = 4.5 , Can we said they are close?


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Continued

• Suppose Cx=3.2 and µ =4.5, Can we conclude that they are different, i.e. µ� 4.5?

• Unfortunately, the decision is not always y, yso clear – determine what is “very close” and what is “very different” is arbitrary without clear definitions.

• Hypothesis testing methodology provide very clear definitions for evaluating differences.

Regions of Rejection and Non-rejection

Note: The size of the rejectionregion is directly related to therisk involved in using only sampleevidence to make decisions about a population parameter.

Critical value

CriticalvalueRejection region

Rejection RegionNon-rejection region

Risks in Decision Making UsingHypothesis-Testing Methodology

• Since our statistical evidence is based on sample data and the corresponding sample variability, there is a risk that we may make the wrong conclusion.

There are two types of errors that can be made:

• Type 1 error and Type II errorType 1 error and Type II error

Type 1 error: Reject Ho when Ho is true. Prob of commiting Type I error = .

Type II error: Did not reject Ho when Ho is false.

Prob of commiting Type II error = �

D : Level of Significance

Prob of commiting Type I error = .

Also refer to as “Level of Significance”

You control the Type I error by deciding the i k l l th t illi t h irisk level . that you are willing to have in

rejecting the null hypothesis when it is true.

Traditionally you select . =0.01. 0.05, 0.10

Two Tailed TestHo: µ = µ0

Ha: µ � µ0

Ho: µ = µ0

Ha: µ > µ0RIGHT Tailed Test

Rejection Region Rejection Region

ONE TAIL TEST

D/2 D/2

D

Ho: µ = µ0

Ha: µ < µ0LEFT Tailed Test

Rejection Region

Rejection Region

ONE TAIL TEST

D

Regions of Rejection and Non-rejection

If . =0.01, Z=± 2.5758If . =0.05 , Z=± 1.9599If . =0.10 Z =± 1.6448

Observe: As D increases, the rejection region also

increases.

Z=-2.5758 Z=+2.5758

Rejection region Rejection RegionNon-rejection region

D/2 = 0.005 D/2 = 0.005


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� risk

• Prob of commiting Type II error

• It depends on the difference between the hypothesized and the actual value of the population parameterpopulation parameter.

• If the difference between the hypothesized and actual value of the population parameter is large, then � is small.

• 1- � = power of a statistical test.

Two kinds of errors in Hypothesis testing

Population Condition

(Truth about Population)

Decision Based on sample

(conclusion)

Ho True Ho False

(or Ha True)

Do not reject Ho Correct Conclusion Type II error1 3Do not reject Ho

(Fail to reject Ho)

Correct Conclusion

<Confidence=1-.>

Type II error

�

Reject Ho Type I error

.

Correct Conclusion

<Power= 1-�>

1. If our conclusion is “Do not Reject Ho” and in fact Ho is true, our conclusion is correct2. If our conclusion is “Reject Ho” and in fact Ho is true, we commit Type 1 error.3. If our conclusion is “Do not Reject Ho” and in fact Ha is true, we commit Type II error.4. If our conclusion is “Reject Ho” and in fact Ha is true, our conclusion is correct.

1

2

3

4

Two kinds of errors in Hypothesis testing

Population Condition

(Truth about Population)

Decision Based on sample

(conclusion)

Ho True Ho False

(or Ha True)

Do not reject Ho Correct Conclusion Type II error1 3Do not reject Ho

(Fail to reject Ho)

Correct Conclusion<Confidence=1-.>

Type II error

�

Reject Ho Type I error

.

Correct Conclusion<Power= 1-�>

1

2

3

4

Note: The error rates are dependent on each other. For a fixedsample size, n, if . is decreased then � increases and vice verse ( as . increases then � decreases)

Example

Ho : µ = 4.5

Ha: µ � 4.5

You make Type 1 error if you conclude that µ � 4.5 when µ = 4.5

You make Type II error if you conclude that µ = 4.5 when µ � 4.5

Example 7.13

In the American legal system, a defendant is presumed innocent until proven guilty. Consider a null hypothesis Ho, that the defendant is innocent, and an alternative hypothesis, Ha, that he defendant is guilty. A jury has two possible decisions: Convict the defendant (i.e. reject the null hypothesis) or do not convict the defendant (i.e. do not reject the null hypothesis).

Explain the meaning of the risks of committing either a Type I or Type II error in this example.

SolutionHo: The defendant is innocent

Ha: The defendant is guilty

Definition: Type 1 error: Reject Ho when it is true.

. = Prob of finding defendant guilty when he is innocent/

Definition: Type II error: Did not reject Ho when it is false.

� = Prob. of finding the defendant innocent when he is guilty.

Note: The error rates are dependent on each other. For a fixed

sample size, n, if . is decreased then � increases and vice

verse ( as . increases then � decreases)


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Example 7-14

Suppose the defendant is presumed guilty until proved innocent as in some other judicial systems. How do the null and alternative hypotheses differ from those inalternative hypotheses differ from those in the previous example?

What are the meanings of the risks of committing either a Type I or Type II error here?

SolutionHo: The defendant is guilty

Ha: The defendant is innocent

Definition: Type 1 error: Reject Ho when it is true.

. = Prob of finding defendant innocent when h i ilthe is guilty

Definition: Type II error: Did not reject Ho when it is false.

� = Prob. of failing to find the defendant innocent when the defendant is innocent.

Example 7:15The Food and Drug Administration (FDA) is responsible for

approving new drugs. Many consumer groups feel that the approval process is too easy, and therefore too many drugs are approved that later are found to be unsafe. On the other hand, there are a number of industry lobbyist who are pushing for a more lenient approval process so that pharmaceutical companies can get new drugs approved easier and faster Consider a null hypothesisapproved easier and faster. Consider a null hypothesis

that a new, unapproved drug is unsafe, and an

alternative hypothesis that a new, unapproved drug

is safe.

(a) Explain the risks of committing a Type I or Type II error.

SolutionThe Food and Drug Administration (FDA) is responsible for

approving new drugs. Many consumer groups feel that the approval process is too easy, and therefore too many drugs are approved that later are found to be unsafe. On the other hand, there are a number of industry lobbyist who are pushing for a more lenient approval process so that pharmaceutical companies can get new drugs approved easier and faster Consider a null hypothesisapproved easier and faster. Consider a null hypothesis

that a new, unapproved drug is unsafe, and an

alternative hypothesis that a new, unapproved drug

is safe.

(a) Explain the risks of committing a Type I or Type II error.

A Type I error is made if an unsafe drug is approved, that is concluding a

drug is safe when in fact it is not safe.

A Type II error is made if a safe drug is not approved.


approving new drugs. Many consumer groups feel that

the approval process is too easy, and therefore too

many drugs are approved that later are found to be

unsafe. On the other hand, there are a number of industry lobbyist who are pushing for a more lenient approval process so that pharmaceutical companies can get new drugs approved easier and faster Consider aget new drugs approved easier and faster. Consider a null hypothesis that a new, unapproved drug is unsafe, and an alternative hypothesis that a new, unapproved drug is safe.

(b) Which type of error are the consumer groups trying to avoid? Explain.


approving new drugs. Many consumer groups feel that

the approval process is too easy, and therefore too

many drugs are approved that later are found to be

unsafe. On the other hand, there are a number of industry lobbyist who are pushing for a more lenient approval process so that pharmaceutical companies can get new drugs approved easier and faster Consider aget new drugs approved easier and faster. Consider a null hypothesis that a new, unapproved drug is unsafe, and an alternative hypothesis that a new, unapproved drug is safe.

(b) Which type of error are the consumer groups trying to avoid?

The consumer groups are trying to avoid a Type I error.


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approving new drugs. Many consumer groups feel that the approval process is too easy, and therefore too many drugs are approved that later are found to be unsafe. On

the other hand, there are a number of industry

lobbyist who are pushing for a more lenient

approval process so that pharmaceutical companies

can get new drugs approved easier and fastercan get new drugs approved easier and faster.

Consider a null hypothesis that a new, unapproved drug is unsafe, and an alternative hypothesis that a new, unapproved drug is safe.

(c) Which type of error are the industry lobbyists trying to avoid? Explain


approving new drugs. Many consumer groups feel that the approval process is too easy, and therefore too many drugs are approved that later are found to be unsafe. On

the other hand, there are a number of industry

lobbyist who are pushing for a more lenient

approval process so that pharmaceutical companies

can get new drugs approved easier and fastercan get new drugs approved easier and faster.

Consider a null hypothesis that a new, unapproved drug is unsafe, and an alternative hypothesis that a new, unapproved drug is safe.

(c) Which type of error are the industry lobbyists trying to avoid?

The lobbyists are trying to avoid a Type II error.

How to make the statistical decision/conclusion?

Define Rejection Regions

Recall: D and E risks

• After you specify ., you know the size of the rejection region because . is the prob of rejection under the null hypothesis.

• From this fact you can then determine the• From this fact, you can then determine the critical value or values that divide the rejection and non-rejection regions.

Recall how we calculate Z value

(critical value) given D:

We needed z for a 90% confidence interval.

90% confidence Confidence coefficient(Level of Significance)

level (Level of Significance) of 0.90 o D =0.10

Z=r1.645

(2-tailed test)

SummaryRegions of Rejection and Non-rejection

Note: The size of the rejectionregion is directly related to therisk involved in using only sampleevidence to make decisions about a population parameter.

Critical value

Criticalvalue

Rejection regionRejection RegionNon-rejection region

Z=-1.645 Z= 1.645

D/2D/2


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Putting everything together for Hypothesis testing:

1. Hypothesis (Ho and Ha)

2. D2. D

3. Rejection Region(s)


construct chapter 9 interpret confidence interval...

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