1 Hypothesis testing can be used to determine whether Hypothesis testing can be used to determine whether a statement about the value of a population parameter

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Hypothesis testingHypothesis testing can be used to determine whether can be used to determine whether a statement about the value of a population parametera statement about the value of a population parameter

should or should not be rejected.should or should not be rejected.

The The null hypothesisnull hypothesis, , denoted by denoted by HH0 0 , , is a tentativeis a tentative assumption about a population parameter.assumption about a population parameter.

The The alternative hypothesisalternative hypothesis, denoted by , denoted by HHaa, is the, is the

opposite of what is stated in the null hypothesis.opposite of what is stated in the null hypothesis.

The hypothesis testing procedure uses data from a The hypothesis testing procedure uses data from a sample to test the two competing statementssample to test the two competing statements

indicated by indicated by HH00 and and HHaa..

© 2011 South-Western, a part of Cengage Learning© 2011 South-Western, a part of Cengage Learning 22

Alternative Hypothesis as a Research Alternative Hypothesis as a Research HypothesisHypothesis

• Example: Example: A new teaching method is developed that is A new teaching method is developed that is

believed to be better than the current method.believed to be better than the current method.

• Alternative Hypothesis: Alternative Hypothesis: The new teaching method is better. The new teaching method is better.

• Null Hypothesis: Null Hypothesis: The new method is no better than the old method.The new method is no better than the old method.

33

Null Hypothesis as an Assumption to be Challenged

• We might begin with a belief or assumption thatWe might begin with a belief or assumption that a statement about the value of a populationa statement about the value of a population

parameter is true.parameter is true.• We then use a hypothesis test to challenge theWe then use a hypothesis test to challenge the assumption and determine if there is statisticalassumption and determine if there is statistical

evidence to conclude that the assumption isevidence to conclude that the assumption is incorrect.incorrect.

• In these situations, it is helpful to develop the nullIn these situations, it is helpful to develop the null hypothesis first.hypothesis first.

44

One-tailedOne-tailed(lower-tail)(lower-tail)

One-tailedOne-tailed(upper-tail)(upper-tail)

Two-tailedTwo-tailed

0 0: H 0 0: H

0: aH 0: aH 0 0: H 0 0: H

0: aH 0: aH 0 0: H 0 0: H

0: aH 0: aH

The equality part of the hypotheses always appearsThe equality part of the hypotheses always appears

in the null hypothesis.in the null hypothesis. In general, a hypothesis test about the value of aIn general, a hypothesis test about the value of a population mean population mean must take one of the followingmust take one of the following three forms (where three forms (where 00 is the hypothesized value of is the hypothesized value of the population mean).the population mean).

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Example: Metro EMS

A major west coast city provides one of the A major west coast city provides one of the mostmostcomprehensive emergency medical services comprehensive emergency medical services in thein theworld. Operating in a multiple hospital world. Operating in a multiple hospital systemsystemwith approximately 20 mobile medical units, with approximately 20 mobile medical units, thetheservice goal is to respond to medical service goal is to respond to medical emergenciesemergencieswith a mean time of 12 minutes or less.with a mean time of 12 minutes or less.

The director of medical services wants toThe director of medical services wants toformulate a hypothesis test that could use a formulate a hypothesis test that could use a samplesampleof emergency response times to determine of emergency response times to determine whetherwhetheror not the service goal of 12 minutes or less or not the service goal of 12 minutes or less is beingis beingachieved.achieved.

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The emergency service is meetingThe emergency service is meetingthe response goal; no follow-upthe response goal; no follow-up

action is necessary.action is necessary.

The emergency service is notThe emergency service is notmeeting the response goal;meeting the response goal;

appropriate follow-up action isappropriate follow-up action isnecessary.necessary.

HH00: :

HHaa::

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

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Because hypothesis tests are based on sample data,Because hypothesis tests are based on sample data, we must allow for the possibility of errors.we must allow for the possibility of errors.

A A Type I errorType I error is rejecting is rejecting HH00 when it is true. when it is true.

The probability of making a Type I error when theThe probability of making a Type I error when the

null hypothesis is true as an equality is called thenull hypothesis is true as an equality is called the

level of significancelevel of significance..

Applications of hypothesis testing that only controlApplications of hypothesis testing that only control

the Type I error are often called the Type I error are often called significance testssignificance tests..

88

A A Type II errorType II error is accepting is accepting HH00 when it is false. when it is false.

It is difficult to control for the probability of makingIt is difficult to control for the probability of making

a Type II error.a Type II error.

Statisticians avoid the risk of making a Type IIStatisticians avoid the risk of making a Type II

error by using “do not reject error by using “do not reject HH00” and not “accept ” and not “accept HH00”.”.

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CorrectCorrectDecisionDecision

Type II ErrorType II Error

CorrectCorrectDecisionDecisionType I ErrorType I Error

RejectReject HH00

(Conclude (Conclude > 12) > 12)

AcceptAccept HH00

(Conclude(Conclude << 12) 12)

HH0 0 TrueTrue(( << 12) 12)

HH0 0 FalseFalse

(( > 12) > 12)ConclusionConclusion

Population Condition Population Condition

1010

Reject Reject HH00 if the if the pp-value -value << ..

The The pp-value-value is the probability, computed using the is the probability, computed using the test statistic, that measures the support (or lack oftest statistic, that measures the support (or lack of support) provided by the sample for the nullsupport) provided by the sample for the null hypothesis.hypothesis. If the If the pp-value is less than or equal to the level of-value is less than or equal to the level of significance significance , the value of the test statistic is in the, the value of the test statistic is in the rejection region.rejection region.

1111

pp-Value Approach-Value Approach pp-Value Approach-Value Approach

pp-value-valuepp-value-value

0000 --zz = = -1.28-1.28 --zz = = -1.28-1.28

= .10= .10 = .10= .10

zzzz

zz = =-1.46-1.46 zz = =-1.46-1.46

Sampling Sampling distribution distribution

of of

Sampling Sampling distribution distribution

of of z xn

0

/z x

n

0

/

pp-Value -Value << ,,

so reject so reject HH00..

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pp-Value Approach-Value Approach pp-Value Approach-Value Approach

pp-Value-Valuepp-Value-Value

0000 zz = = 1.751.75 zz = = 1.751.75

= .04= .04 = .04= .04

zzzz

zz = =2.292.29 zz = =2.292.29

Sampling Sampling distributiondistribution of of

Sampling Sampling distributiondistribution of of

0

/xzn

0

/xzn

pp-Value -Value << ,,so reject so reject HH00..

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The test statistic The test statistic zz has a standard normal probability has a standard normal probability distribution.distribution.

We can use the standard normal probabilityWe can use the standard normal probability distribution table to find the distribution table to find the zz-value with an -value with an areaarea of of in the lower (or upper) tail of the in the lower (or upper) tail of the distribution.distribution. The value of the test statistic that established The value of the test statistic that established thethe boundary of the rejection region is called theboundary of the rejection region is called the critical valuecritical value for the test. for the test. The rejection rule is:The rejection rule is:

• Lower tail: Reject Lower tail: Reject HH00 if if zz << - -zz

• Upper tail: Reject Upper tail: Reject HH00 if if zz >> zz

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0000 zz = = 1.281.28 zz = = 1.281.28

Reject Reject HH00Reject Reject HH00

Do Not Reject Do Not Reject HH00Do Not Reject Do Not Reject HH00

zz

SamplingSamplingdistributiondistribution

of of


of of z xn

0

/z x

n

0

/

Critical Value ApproachCritical Value Approach Critical Value ApproachCritical Value Approach

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0000 zz = 1.645 = 1.645 zz = 1.645 = 1.645



zz


of of


of of 0

/xzn

0

/xzn

Critical Value ApproachCritical Value Approach Critical Value ApproachCritical Value Approach

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Step 1Step 1. Develop the null and alternative hypotheses.. Develop the null and alternative hypotheses.

Step 2Step 2. Specify the level of significance . Specify the level of significance ..

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

pp-Value Approach-Value Approach

Step 4Step 4. Use the value of the test statistic to compute the. Use the value of the test statistic to compute the pp-value.-value.

Step 5Step 5. Reject . Reject HH00 if if pp-value -value << ..

Critical Value ApproachCritical Value Approach

Step 4Step 4. . Use the level of significance to Use the level of significance to determine the critical value and the determine the critical value and the

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

rule to determine whether to reject rule to determine whether to reject HH00..

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Example: Metro EMSExample: Metro EMS

The EMS director wants to perform a The EMS director wants to perform a hypothesishypothesistest, with a .05 level of significance, to test, with a .05 level of significance, to determinedeterminewhether the service goal of 12 minutes or less whether the service goal of 12 minutes or less is is being achieved.being achieved.

The response times for a random sample The response times for a random sample of 40of 40medical emergencies were tabulated. The medical emergencies were tabulated. The samplesamplemean is 13.25 minutes. The population mean is 13.25 minutes. The population standardstandarddeviation is believed to be 3.2 minutes.deviation is believed to be 3.2 minutes.

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1. Develop the hypotheses.1. Develop the hypotheses.

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

HH00: :

HHaa::

pp -Value and Critical Value Approaches -Value and Critical Value Approaches

3. Compute the value of the test statistic.3. Compute the value of the test statistic.

13.25 12 2.47

/ 3.2/ 40x

zn

13.25 12 2.47

/ 3.2/ 40x

zn

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5. Determine whether to reject 5. Determine whether to reject HH00..

pp –Value Approach –Value Approach

4. Compute the 4. Compute the pp –value. –value.

For For zz = 2.47, cumulative probability = .9932. = 2.47, cumulative probability = .9932.

pp–value = 1 –value = 1 .9932 = .0068 .9932 = .0068

Because Because pp–value = .0068 –value = .0068 << = .05, we reject = .05, we reject HH00..

There is sufficient statistical There is sufficient statistical evidenceevidenceto infer that Metro EMS is to infer that Metro EMS is notnot meetingmeetingthe response goal of 12 minutes.the response goal of 12 minutes.

2020

p p –Value Approach–Value Approach p p –Value Approach–Value Approach

p-valuep-value

0000 zz = =1.6451.645 zz = =1.6451.645

= .05= .05 = .05= .05

zzzz

zz = =2.472.47 zz = =2.472.47


of of


of of

z xn

0

/z x

n

0

/

2121


There is sufficient statistical There is sufficient statistical evidenceevidence

to infer that Metro EMS is to infer that Metro EMS is notnot meetingmeeting

the response goal of 12 minutes.the response goal of 12 minutes.

Because 2.47 Because 2.47 >> 1.645, we reject 1.645, we reject HH00..


For For = .05, = .05, zz.05.05 = 1.645 = 1.645

4. Determine the critical value and rejection rule.4. Determine the critical value and rejection rule.

Reject Reject HH00 if if zz >> 1.645 1.645

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The rejection rule:The rejection rule: Reject Reject HH00 if the if the pp-value -value << ..

Compute the Compute the pp-value-value using the following three steps: using the following three steps:

3. Double the tail area obtained in step 2 to obtain3. Double the tail area obtained in step 2 to obtain the the pp –value. –value.

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

If If zz is in the lower tail ( is in the lower tail (zz < 0), find the area under < 0), find the area under the standard normal curve to the left of the standard normal curve to the left of zz..

1. Compute the value of the test statistic 1. Compute the value of the test statistic zz..

2323

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

The rejection rule is:The rejection rule is:

Reject Reject HH00 if if zz << - -zz/2/2 or or zz >> zz/2/2..

Use the standard normal probability Use the standard normal probability distributiondistribution

table to find table to find zz/2/2 (the (the zz-value with an area of -value with an area of /2 in/2 in

the upper tail of the distribution).the upper tail of the distribution).

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Example: Glow Toothpaste

Quality assurance procedures call for the Quality assurance procedures call for the continuation of the filling process if the sample continuation of the filling process if the sample results are consistent with the assumption that results are consistent with the assumption that the mean filling weight for the population of the mean filling weight for the population of toothpaste tubes is 6 oz.; otherwise the process toothpaste tubes is 6 oz.; otherwise the process will be adjusted.will be adjusted.

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Perform a hypothesis test, at the .03 level of Perform a hypothesis test, at the .03 level of significance, to help determine whether the significance, to help determine whether the filling process should continue operating or be filling process should continue operating or be stopped and corrected.stopped and corrected.

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

Example: Glow Toothpaste Example: Glow Toothpaste

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1. Determine the hypotheses.1. Determine the hypotheses.

2. Specify the level of significance.2. Specify the level of significance.


= .03= .03

pp –Value and Critical Value Approaches –Value and Critical Value Approaches HH00: :

HHaa: :

6 6

0 6.1 6

2.74/ .2/ 30

xz

n

0 6.1 6

2.74/ .2/ 30

xz

n

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For For zz = 2.74, cumulative probability = .9969 = 2.74, cumulative probability = .9969

pp–value = 2(1 –value = 2(1 .9969) = .0062 .9969) = .0062

Because Because pp–value = .0062 –value = .0062 << = .03, we reject = .03, we reject HH00..

There is sufficient statistical evidence toThere is sufficient statistical evidence toinfer that the alternative hypothesis is infer that the alternative hypothesis is

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

ounces).ounces).

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/2 =/2 = .015.015/2 =/2 = .015.015

0000zz/2/2 = 2.17 = 2.17zz/2/2 = 2.17 = 2.17

zzzz

/2 =/2 = .015.015/2 =/2 = .015.015

pp-Value Approach-Value Approach

--zz/2/2 = -2.17 = -2.17--zz/2/2 = -2.17 = -2.17zz = 2.74 = 2.74zz = 2.74 = 2.74zz = -2.74 = -2.74zz = -2.74 = -2.74

1/21/2p p -value-value= .0031= .0031

1/21/2p p -value-value= .0031= .0031

1/21/2p p -value-value= .0031= .0031

1/21/2p p -value-value= .0031= .0031

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There is sufficient statistical evidence toThere is sufficient statistical evidence toinfer that the alternative hypothesis is infer that the alternative hypothesis is

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

ounces).ounces).


For For /2 = .03/2 = .015, /2 = .03/2 = .015, zz.015.015 = 2.17 = 2.17


Reject Reject HH00 if if zz << -2.17 or -2.17 or zz >> 2.17 2.17

3030

/2 = .015/2 = .015/2 = .015/2 = .015

0000 2.172.17 2.172.17

Reject Reject HH00Reject Reject HH00Do Not Reject Do Not Reject HH00Do Not Reject Do Not Reject HH00

zzzz


-2.17-2.17 -2.17-2.17



of of


of of z xn

0

/z x

n

0

/

Two-Tailed Tests About a Population Mean:Two-Tailed Tests About a Population Mean: KnownKnown

/2 = .015/2 = .015/2 = .015/2 = .015

3131

Select a simple random sample from the populationSelect a simple random sample from the population and use the value of the sample mean to developand use the value of the sample mean to develop the confidence interval for the population mean the confidence interval for the population mean .. (Confidence intervals are covered in Chapter 8.)(Confidence intervals are covered in Chapter 8.)

xx

If the confidence interval contains the hypothesizedIf the confidence interval contains the hypothesized value value 00, do not reject , do not reject HH00. Otherwise, reject . Otherwise, reject HH00..

(Actually, (Actually, HH00 should be rejected if should be rejected if 00 happens to be happens to be equal to one of the end points of the confidenceequal to one of the end points of the confidence

interval.) interval.)

3232

The 97% confidence interval for is

/ 2 6.1 2.17(.2 30) 6.1 .07924x zn

/ 2 6.1 2.17(.2 30) 6.1 .07924x zn

Because the hypothesized value for theBecause the hypothesized value for the

population mean, population mean, 00 = 6, is not in this interval, = 6, is not in this interval,the hypothesis-testing conclusion is that thethe hypothesis-testing conclusion is that the

null hypothesis, null hypothesis, HH00: : = 6, can be rejected. = 6, can be rejected.

or 6.02076 to 6.17924or 6.02076 to 6.17924

3333

Test Statistic

txs n

0/

txs n

0/

This test statistic has a This test statistic has a tt distribution distribution with with nn - 1 degrees of freedom. - 1 degrees of freedom.

3434

Rejection Rule: Rejection Rule: pp -Value Approach -Value Approach

HH00: : Reject Reject HH0 0 if if tt >> tt

Reject Reject HH0 0 if if tt << - -tt

Reject Reject HH0 0 if if tt << - - tt or or tt >> tt

HH00: :

HH00: :

Rejection Rule: Critical Value ApproachRejection Rule: Critical Value Approach

Reject Reject HH0 0 if if p p –value –value <<

3535

The format of the The format of the tt distribution table provided in most distribution table provided in most statistics textbooks does not have sufficient detailstatistics textbooks does not have sufficient detail

to determine the to determine the exactexact p p-value for a hypothesis test.-value for a hypothesis test.

However, we can still use the However, we can still use the tt distribution table to distribution table to identify a identify a rangerange for the for the pp-value.-value.

An advantage of computer software packages is thatAn advantage of computer software packages is that the computer output will provide the the computer output will provide the pp-value for the-value for the

tt distribution. distribution.

3636

A State Highway Patrol periodically samples A State Highway Patrol periodically samples vehicle speeds at various locations on a vehicle speeds at various locations on a particular roadway. The sample of vehicle particular roadway. The sample of vehicle speeds is used to test the hypothesis speeds is used to test the hypothesis HH00: : << 65.65.

One-Tailed Test About a Population Mean: Unknown

The locations where The locations where HH00 is rejected are deemed is rejected are deemed the best locations for radar traps. At Location F, the best locations for radar traps. At Location F, a sample of 64 vehicles shows a mean speed of a sample of 64 vehicles shows a mean speed of 66.2 mph with a standard deviation of 4.2 mph. 66.2 mph with a standard deviation of 4.2 mph. Use Use = .05 to test the hypothesis.= .05 to test the hypothesis.

3737

1. Determine the 1. Determine the hypotheses.hypotheses.

2. Specify the level of significance.2. Specify the level of significance.


= .05= .05

pp –Value and Critical Value Approaches –Value and Critical Value Approaches

HH00: : << 65 65

HHaa: : > 65 > 65

0 66.2 65

2.286/ 4.2/ 64

xt

s n

0 66.2 65 2.286

/ 4.2/ 64x

ts n

3838

One-Tailed Test About a Population Mean:One-Tailed Test About a Population Mean: Unknown Unknown




For For tt = 2.286, the = 2.286, the pp–value is between .025–value is between .025(for (for tt = 1.998) and .01 (for = 1.998) and .01 (for tt = 2.387). = 2.387).

.01 < .01 < pp–value < .025–value < .025

Because Because pp–value –value << = .05, we reject = .05, we reject HH00..

We are at least 95% confident that the mean We are at least 95% confident that the mean speedspeed

of vehicles at Location F is greater than 65 of vehicles at Location F is greater than 65 mph.mph.

3939



We are at least 95% confident that the mean We are at least 95% confident that the mean speed of vehicles at Location F is greater speed of vehicles at Location F is greater

than 65 mph. Location F is a good candidate than 65 mph. Location F is a good candidate for a radar trap.for a radar trap.


For For = .05 and d.f. = 64 – 1 = 63, = .05 and d.f. = 64 – 1 = 63, tt.05.05 = 1.669 = 1.669


Reject Reject HH00 if if tt >> 1.669 1.669

4040

0000 tt = =1.6691.669 tt = =1.6691.669



tt

Documents

1 Hypothesis testing can be used to determine whether Hypothesis testing can be used to determine whether a statement about the value of a population parameter