Hypothesis Testing

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Hypothesis Testing

Developing Null and Alternative Hypotheses Type I and Type II Errors Population Mean: s Known Population Mean: s Unknown

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Developing Null and Alternative Hypotheses

Hypothesis testing can be used to determine whether a statement about the value of a population parameter should or should not be rejected. The null hypothesis, denoted by H0 , is a tentative assumption about a population parameter. The alternative hypothesis, denoted by Ha, is the opposite of what is stated in the null hypothesis. The alternative hypothesis is what the test is attempting to establish.

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Testing Research Hypotheses


• The research hypothesis should be expressed as the alternative hypothesis.• The conclusion that the research hypothesis is true comes from sample data that contradict the null hypothesis.

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Testing the Validity of a Claim• Manufacturers’ claims are usually given the benefit of the doubt and stated as the null hypothesis.• The conclusion that the claim is false comes from sample data that contradict the null hypothesis.

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Testing in Decision-Making Situations


• A decision maker might have to choose between two courses of action, one associated with the null hypothesis and another associated with the alternative hypothesis.• Example: Accepting a shipment of goods from a supplier or returning the shipment of goods to the supplier

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One-tailed(lower-tail)

One-tailed(upper-tail)

Two-tailed

0 0: H

0: aH 0 0: H

0: aH 0 0: H

0: aH

Summary of Forms for Null and Alternative Hypotheses about a

Population Mean The equality part of the hypotheses always appears in the null hypothesis. In general, a hypothesis test about the value of a population mean must take one of the following three forms (where 0 is the hypothesized value of the population mean).

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Example: Metro EMS

Null and Alternative Hypotheses

Operating in a multiplehospital system with approximately 20 mobile medicalunits, the service goal is to respond to medicalemergencies with a mean time of 12 minutes or less.

A major west coast city providesone of the most comprehensiveemergency medical services inthe world.

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The director of medical serviceswants to formulate a hypothesistest that could use a sample ofemergency response times todetermine whether or not theservice goal of 12 minutes or lessis being achieved.

Example: Metro EMS


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The emergency service is meetingthe response goal; no follow-upaction is necessary.

The emergency service is notmeeting the response goal;appropriate follow-up action isnecessary.

H0:

Ha:

where: = mean response time for the population of medical emergency requests

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Type I Error

Because hypothesis tests are based on sample data, we must allow for the possibility of errors. A Type I error is rejecting H0 when it is true. The probability of making a Type I error when the null hypothesis is true as an equality is called the level of significance. Applications of hypothesis testing that only control the Type I error are often called significance tests.

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Type II Error

A Type II error is accepting H0 when it is false. It is difficult to control for the probability of making a Type II error. Statisticians avoid the risk of making a Type II error by using “do not reject H0” and not “accept H0”.

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Type I and Type II Errors

CorrectDecision Type II Error

CorrectDecisionType I ErrorReject H0

(Conclude > 12)

Accept H0(Conclude < 12)

H0 True( < 12)

H0 False( > 12)Conclusion

Population Condition

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p-Value Approach toOne-Tailed Hypothesis Testing

A p-value is a probability that provides a measure of the evidence against the null hypothesis provided by the sample.

The smaller the p-value, the more evidence there is against H0. A small p-value indicates the value of the test statistic is unusual given the assumption that H0 is true.

The p-value is used to determine if the null hypothesis should be rejected.

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p-Value Approach

p-value7

0 -za = -1.28

a = .10

z z =-1.46

Lower-Tailed Test About a Population Mean:

s Known

Samplingdistribution of z x

n s

0/

p-Value < a,so reject H0.

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p-Value Approach

p-Value

0 za = 1.75

a = .04

z z =2.29

Upper-Tailed Test About a Population Mean:

s Known


n s

0/

p-Value < a,so reject H0.

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Critical Value Approach to One-Tailed Hypothesis Testing

The test statistic z has a standard normal probability distribution.

We can use the standard normal probability distribution table to find the z-value with an area of a in the lower (or upper) tail of the distribution.

The value of the test statistic that established the boundary of the rejection region is called the critical value for the test.

The rejection rule is:• Lower tail: Reject H0 if z < -za• Upper tail: Reject H0 if z > za

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a

0 za = 1.28

Reject H0

Do Not Reject H0

z


n s

0/

Lower-Tailed Test About a Population Mean:

s Known Critical Value Approach

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a

0 za = 1.645

Reject H0

Do Not Reject H0

z


n s

0/

Upper-Tailed Test About a Population Mean:

s Known Critical Value Approach

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Steps of Hypothesis Testing

Step 1. Develop the null and alternative hypotheses.Step 2. Specify the level of significance a.

Step 3. Collect the sample data and compute the test statistic.

p-Value ApproachStep 4. Use the value of the test statistic to compute the p-value.

Step 5. Reject H0 if p-value < a.

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Critical Value ApproachStep 4. Use the level of significanceto

determine the critical value and the rejection rule.Step 5. Use the value of the test statistic and the rejection

rule to determine whether to reject H0.

Steps of Hypothesis Testing

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Example: Metro EMS

The EMS director wants toperform a hypothesis test, with a.05 level of significance, to determinewhether the service goal of 12 minutes or less is beingachieved.

The response times for a randomsample of 40 medical emergencieswere tabulated. The sample meanis 13.25 minutes. The populationstandard deviation is believed tobe 3.2 minutes.

One-Tailed Tests About a Population Mean:s Known

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1. Develop the hypotheses.

2. Specify the level of significance.a = .05

H0: Ha:

p -Value and Critical Value Approaches


3. Compute the value of the test statistic.

s

13.25 12 2.47/ 3.2/ 40

xzn

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5. Determine whether to reject H0.

p –Value Approach


4. Compute the p –value.For z = 2.47, cumulative probability = .9932.

p–value = 1 .9932 = .0068

Because p–value = .0068 < a = .05, we reject H0.There is sufficient statistical

evidenceto infer that Metro EMS is not

meetingthe response goal of 12 minutes.

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p –Value Approach

p-value

0 za =1.645

a = .05

z z =2.47



n s

0/

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There is sufficient statistical evidence

to infer that Metro EMS is not meeting

the response goal of 12 minutes.

Because 2.47 > 1.645, we reject H0.

Critical Value Approach


For a = .05, z.05 = 1.6454. Determine the critical value and rejection rule.

Reject H0 if z > 1.645

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p-Value Approach toTwo-Tailed Hypothesis Testing

The rejection rule: Reject H0 if the p-value < a .

Compute the p-value using the following three steps:

3. Double the tail area obtained in step 2 to obtain the p –value.

2. If z is in the upper tail (z > 0), find the area under the standard normal curve to the right of z. If z is in the lower tail (z < 0), find the area under the standard normal curve to the left of z.

1. Compute the value of the test statistic z.

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Critical Value Approach to Two-Tailed Hypothesis Testing

The critical values will occur in both the lower and upper tails of the standard normal curve.

The rejection rule is: Reject H0 if z < -za/2 or z > za/2.

Use the standard normal probability distribution

table to find za/2 (the z-value with an area of a/2 in the upper tail of the distribution).

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Example: Glow Toothpaste

Two-Tailed Test About a Population Mean: s Known

oz.

Glow Quality assurance procedures call forthe continuation of the filling process if thesample results are consistent with the assumption thatthe mean filling weight for the population of toothpastetubes is 6 oz.; otherwise the process will be adjusted.

The production line for Glow toothpasteis designed to fill tubes with a mean weightof 6 oz. Periodically, a sample of 30 tubeswill be selected in order to check thefilling process.

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Example: Glow Toothpaste

Two-Tailed Test About a Population Mean: s Known

oz.

Glow Perform a hypothesis test, at the .03level of significance, to help determinewhether the filling process should continueoperating or be stopped and corrected.

Assume that a sample of 30 toothpastetubes provides a sample mean of 6.1 oz.The population standard deviation is believed to be 0.2 oz.

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1. Determine the hypotheses.

2. Specify the level of significance.


a = .03

p –Value and Critical Value ApproachesGlo

w

H0: Ha: 6

Two-Tailed Tests About a Population Mean:s Known

s

0 6.1 6 2.74/ .2/ 30xz

n

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Glow



p –Value Approach

4. Compute the p –value.For z = 2.74, cumulative probability = .9969

p–value = 2(1 .9969) = .0062

Because p–value = .0062 < a = .03, we reject H0.There is sufficient statistical evidence toinfer that the alternative hypothesis is

true (i.e. the mean filling weight is not 6

ounces).

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Glow


a/2 = .015

0za/2 = 2.17

z

a/2 = .015

p-Value Approach

-za/2 = -2.17z = 2.74z = -2.74

1/2p -value= .0031

1/2p -value= .0031

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Critical Value ApproachGlo

w



There is sufficient statistical evidence toinfer that the alternative hypothesis is

true (i.e. the mean filling weight is not 6

ounces).


For a/2 = .03/2 = .015, z.015 = 2.174. Determine the critical value and rejection rule.

Reject H0 if z < -2.17 or z > 2.17

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a/2 = .015

0 2.17

Reject H0Do Not Reject H0

z

Reject H0

-2.17

Glow



n s

0/


a/2 = .015

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Confidence Interval Approach toTwo-Tailed Tests About a Population Mean Select a simple random sample from the population and use the value of the sample mean to develop the confidence interval for the population mean . (Confidence intervals are covered in Chapter 8.)

x

If the confidence interval contains the hypothesized value 0, do not reject H0. Otherwise, reject H0.

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The 97% confidence interval for is/ 2 6.1 2.17(.2 30) 6.1 .07924x znas

Confidence Interval Approach toTwo-Tailed Tests About a Population Mean

Glow

Because the hypothesized value for thepopulation mean, 0 = 6, is not in this interval,the hypothesis-testing conclusion is that thenull hypothesis, H0: = 6, can be rejected.

or 6.02076 to 6.17924

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Test Statistic

Tests About a Population Mean:s Unknown

t xs n

0

/

This test statistic has a t distribution with n - 1 degrees of freedom.

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Rejection Rule: p -Value Approach

H0: Reject H0 if t > ta

Reject H0 if t < -ta

Reject H0 if t < - ta or t > ta

H0:

H0:

Tests About a Population Mean:s Unknown

Rejection Rule: Critical Value ApproachReject H0 if p –value < a

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p -Values and the t Distribution

The format of the t distribution table provided in most statistics textbooks does not have sufficient detail to determine the exact p-value for a hypothesis test. However, we can still use the t distribution table to identify a range for the p-value. An advantage of computer software packages is that the computer output will provide the p-value for the t distribution.

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A State Highway Patrol periodically samplesvehicle speeds at various locationson a particular roadway. The sample of vehicle speedsis used to test the hypothesis

Example: Highway Patrol

One-Tailed Test About a Population Mean: s Unknown

The locations where H0 is rejected are deemedthe best locations for radar traps.

H0: < 65

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Example: Highway Patrol

One-Tailed Test About a Population Mean: s Unknown At Location F, a sample of 64 vehicles shows a

mean speed of 66.2 mph with astandard deviation of4.2 mph. Use a = .05 totest the hypothesis.

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One-Tailed Test About a Population Mean:s Unknown




a = .05

p –Value and Critical Value Approaches

H0: < 65Ha: > 65

0 66.2 65 2.286/ 4.2/ 64

xts n

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p –Value Approach


4. Compute the p –value.For t = 2.286, the p–value must be less than .025

(for t = 1.998) and greater than .01 (for t = 2.387)..01 < p–value < .025

Because p–value < a = .05, we reject H0.We are at least 95% confident that the mean speed of vehicles at Location F is greater than 65 mph.

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We are at least 95% confident that the mean speed of vehicles at Location F is greater than 65 mph. Location F is a good candidate for a radar trap.



For a = .05 and d.f. = 64 – 1 = 63, t.05 = 1.6694. Determine the critical value and rejection rule.

Reject H0 if t > 1.669

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a

0 ta =1.669

Reject H0

Do Not Reject H0

t


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Hypothesis Testing

Population Proportion Hypothesis Testing and Decision Making Calculating the Probability of Type II Errors Determining the Sample Size for a Hypothesis Test About a Population Mean

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The equality part of the hypotheses always appears in the null hypothesis. In general, a hypothesis test about the value of a population proportion p must take one of the following three forms (where p0 is the hypothesized value of the population proportion).

A Summary of Forms for Null and Alternative Hypotheses About a

Population Proportion

One-tailed(lower tail)

One-tailed(upper tail)

Two-tailed

0 0: H p p

0: aH p p0: aH p p0 0: H p p 0 0: H p p

0: aH p p

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Test Statistic

z p p

p

0

s

s pp p

n 0 01( )

Tests About a Population Proportion

where:

assuming np > 5 and n(1 – p) > 5

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Rejection Rule: p –Value Approach

H0: pp Reject H0 if z > za

Reject H0 if z < -za

Reject H0 if z < -za or z > za

H0: pp

H0: pp

Tests About a Population Proportion

Reject H0 if p –value < a Rejection Rule: Critical Value Approach

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Example: National Safety Council (NSC) For a Christmas and New Year’s week, the

National Safety Council estimated that500 people would be killed and 25,000injured on the nation’s roads. TheNSC claimed that 50% of theaccidents would be caused bydrunk driving.

Two-Tailed Test About aPopulation Proportion

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A sample of 120 accidents showed that67 were caused by drunk driving. Usethese data to test the NSC’s claim witha = .05.


Example: National Safety Council (NSC)

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a = .05

p –Value and Critical Value Approaches

0: .5H p: .5aH p

s

0 (67/ 120) .5 1.28.045644p

p pz

0 0(1 ) .5(1 .5) .045644120pp p

ns

a commonerror is using

in this formula

p

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pValue Approach

4. Compute the p -value.

5. Determine whether to reject H0.Because p–value = .2006 > a = .05, we cannot reject H0.


For z = 1.28, cumulative probability = .8997p–value = 2(1 .8997) = .2006

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For a/2 = .05/2 = .025, z.025 = 1.964. Determine the criticals value and rejection rule.

Reject H0 if z < -1.96 or z > 1.96

Because 1.278 > -1.96 and < 1.96, we cannot reject H0.

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Hypothesis Testing and Decision Making In many decision-making situations the

decision maker may want, and in some cases may be

forced, to take action with both the conclusion do not

reject H0 and the conclusion reject H0. In such situations, it is recommended that the hypothesis-testing procedure be extended to

include consideration of making a Type II error.

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Calculating the Probability of a Type II Error

in Hypothesis Tests About a Population Mean1. Formulate the null and alternative hypotheses.

3. Using the rejection rule, solve for the value of the sample mean corresponding to the critical value of the test statistic.

2. Using the critical value approach, use the level of significance a to determine the critical value and the rejection rule for the test.

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in Hypothesis Tests About a Population Mean4. Use the results from step 3 to state the values of

the sample mean that lead to the acceptance of H0;

this defines the acceptance region.5. Using the sampling distribution of for a value of satisfying the alternative hypothesis, and the acceptance region from step 4, compute the probability that the sample mean will be in the acceptance region. (This is the probability of making a Type II error at the chosen level of .)

xx

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Example: Metro EMS (revisited)

The EMS director wants toperform a hypothesis test, with a.05 level of significance, to determinewhether or not the service goal of 12 minutes or less isbeing achieved.

Recall that the response times fora random sample of 40 medicalemergencies were tabulated. Thesample mean is 13.25 minutes.The population standard deviationis believed to be 3.2 minutes.


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12 1.6453.2/ 40xz

3.212 1.645 12.832340x

4. We will accept H0 when x < 12.8323

3. Value of the sample mean that identifies the rejection region:

2. Rejection rule is: Reject H0 if z > 1.6451. Hypotheses are: H0: and Ha:


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12.0001 1.645 .9500 .050012.4 0.85 .8023 .197712.8 0.06 .5239 .476112.8323 0.00 .5000 .500013.2 -0.73 .2327 .767313.6 -1.52 .0643 .935714.0 -2.31 .0104 .9896

12.83233.2/ 40z

Values of b1-b

5. Probabilities that the sample mean will be in the acceptance region:


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When the true population mean is far above

the null hypothesis value of 12, there is a low probability that we will make a Type II error.

When the true population mean is close to the null hypothesis value of 12, there is a high probability that we will make a Type II error.

Observations about the preceding table:

Example: = 12.0001, b = .9500

Example: = 14.0, b = .0104

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Power of the Test

The probability of correctly rejecting H0 when it is false is called the power of the test. For any particular value of , the power is 1 – b.

We can show graphically the power associated with each value of ; such a graph is called a power curve. (See next slide.)

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Power Curve

0.000.100.200.300.400.500.600.700.800.901.00

11.5 12.0 12.5 13.0 13.5 14.0 14.5

Prob

abili

ty o

f Cor

rect

lyRe

ject

ing

Nul

l Hyp

othe

sis

H0 False

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The specified level of significance determines the probability of making a Type I error.

By controlling the sample size, the probability of making a Type II error is controlled.

Determining the Sample Size for a Hypothesis Test About a Population Mean

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0

a

a

x

x

Samplingdistribution of whenH0 is trueand = 0

x

Samplingdistribution of whenH0 is falseand a > 0

x

Reject H0

b

c

c

H0:

Ha:

ss x nNote:


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whereza = z value providing an area of a in the

tailzb = z value providing an area of b in the

tails= population standard deviation0 = value of the population mean in H0a = value of the population mean used

for the Type II error

nz z

a

( )( )a b s

2 2

02n

z z

a

( )( )a b s

2 2

02


Note: In a two-tailed hypothesis test, use za /2 not za

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Relationship Among a, b, and n

Once two of the three values are known, the other can be computed.

For a given level of significance a, increasing the sample size n will reduce b.

For a given sample size n, decreasing a will increase b, whereas increasing a will decrease b.

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Let’s assume that the director of medical services makes the following statements about the

allowable probabilities for the Type I and Type II errors:• If the mean response time is = 12 minutes, I

am willing to risk an a = .05 probability of rejecting H0.• If the mean response time is 0.75 minutes over the specification ( = 12.75), I am willing to risk a b = .10 probability of not rejecting H0.

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a = .05, b = .10za = 1.645, zb = 1.280 = 12, a = 12.75s= 3.2

2 2 2 2

2 20

( ) (1.645 1.28) (3.2) 155.75 156( ) (12 12.75)a

z zn a b s

Hypothesis Testing

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