Chapter 9Hypothesis Testing

Developing Null and Alternative HypothesesType I and Type II ErrorsPopulation Mean: σ Known

Developing Null and Alternative Hypotheses

Hypothesis testing can be used to determine whethera statement about the value of a population parametershould or should not be rejected.The null hypothesis, denoted by H0 , is a tentativeassumption about a population parameter.The alternative hypothesis, denoted by Ha, is theopposite of what is stated in the null hypothesis.

The alternative hypothesis is what the test isattempting to establish.

� Testing Research Hypotheses

Developing Null and Alternative Hypotheses

• The research hypothesis should be expressed asthe alternative hypothesis.

• The conclusion that the research hypothesis is truecomes from sample data that contradict the nullhypothesis.

Developing Null and Alternative Hypotheses

� Testing the Validity of a Claim

• Manufacturers’ claims are usually given the benefitof the doubt and stated as the null hypothesis.

• The conclusion that the claim is false comes fromsample data that contradict the null hypothesis.

� Testing in Decision-Making Situations

Developing Null and Alternative Hypotheses

• A decision maker might have to choose betweentwo courses of action, one associated with the nullhypothesis and another associated with thealternative hypothesis.

• Example: Accepting a shipment of goods from asupplier or returning the shipment of goods to thesupplier

One-tailed(lower-tail)

One-tailed(upper-tail)

Two-tailed

0 0: H μ μ≥0 0: H μ μ≥

0: aH μ μ< 0: aH μ μ<0 0: H μ μ≤0 0: H μ μ≤

0: aH μ μ> 0: aH μ μ>0 0: H μ μ=0 0: H μ μ=

0: aH μ μ≠ 0: aH μ μ≠

Summary of Forms for Null and Alternative Hypotheses about a Population Mean

� The equality part of the hypotheses always appearsin the null hypothesis.In general, a hypothesis test about the value of apopulation mean μ must take one of the followingthree forms (where μ0 is the hypothesized value ofthe population mean).

� 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.

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

Null and Alternative Hypotheses

Null and Alternative Hypotheses

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: μ < 12

Ha: μ > 12

where: μ = mean response time for the populationof medical emergency requests

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 thenull hypothesis is true as an equality is called thelevel of significance.

� Applications of hypothesis testing that only controlthe Type I error are often called significance tests.

Type II Error

� A Type II error is accepting H0 when it is false.

� It is difficult to control for the probability of makinga Type II error.

� Statisticians avoid the risk of making a Type IIerror by using “do not reject H0” and not “accept H0”.

Type I and Type II Errors

CorrectDecision Type II Error

CorrectDecisionType I ErrorReject H0

(Conclude μ > 12)

Accept H0(Conclude μ < 12)

H0 True(μ < 12)

H0 False(μ > 12)Conclusion

Population Condition

p-Value Approach toOne-Tailed Hypothesis Testing

Reject H0 if the p-value < α .

The p-value is the probability, computed using thetest statistic, that measures the support (or lack ofsupport) provided by the sample for the nullhypothesis.

If the p-value is less than or equal to the level ofsignificance α, the value of the test statistic is in therejection region.

� p-Value Approach� p-Value Approach

p-value= .072p-value= .072

00-zα =-1.28-zα =-1.28

α = .10α = .10

zz

z =-1.46z =-1.46

Lower-Tailed Test About a Population Mean:σ Known

Samplingdistributionof

Samplingdistributionof z x

n= − μσ

0/z x

n= − μσ

0/

p-Value < α ,so reject H0.

� p-Value Approach� p-Value Approach

p-Value= .011p-Value= .011

00 zα =1.75zα =1.75

α = .04α = .04

zzz =2.29z =2.29

Upper-Tailed Test About a Population Mean:σ Known

Samplingdistributionof

Samplingdistributionof z x

n= − μσ

0/z x

n= − μσ

0/

p-Value < α ,so reject H0.

Critical Value Approach to One-Tailed Hypothesis Testing

The test statistic z has a standard normal probabilitydistribution.We can use the standard normal probabilitydistribution table to find the z-value with an areaof α in the lower (or upper) tail of the distribution.The value of the test statistic that established theboundary of the rejection region is called thecritical value for the test.

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

α = .10α = .10

00−zα = −1.28−zα = −1.28

Reject H0Reject H0

Do Not Reject H0Do Not Reject H0

z

Samplingdistributionof

Samplingdistributionof z x

n= − μσ

0/z x

n= − μσ

0/

Lower-Tailed Test About a Population Mean:σ Known

� Critical Value Approach� Critical Value Approach

α = .05α = .05

00 zα = 1.645zα = 1.645

Reject H0Reject H0

Do Not Reject H0Do Not Reject H0

z

Samplingdistributionof

Samplingdistributionof z x

n= − μσ

0/z x

n= − μσ

0/

Upper-Tailed Test About a Population Mean:σ Known

� Critical Value Approach� Critical Value Approach

Steps of Hypothesis Testing

Step 1. Develop the null and alternative hypotheses.Step 2. Specify the level of significance α.Step 3. Collect the sample data and compute the test

statistic.

p-Value Approach

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

Step 5. Reject H0 if p-value < α.

Critical Value ApproachStep 4. Use the level of significance to determine the

critical value and the rejection rule.

Step 5. Use the value of the test statistic and the rejectionrule to determine whether to reject H0.

Steps of Hypothesis Testing

� 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:σ Known

1. Develop the hypotheses.

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

H0: μ < 12Ha: μ > 12

p -Value and Critical Value Approaches

One-Tailed Tests About a Population Mean:σ Known

3. Compute the value of the test statistic.

μσ

− −= = =

13.25 12 2.47/ 3.2/ 40

xzn

μσ

− −= = =

13.25 12 2.47/ 3.2/ 40

xzn

5. Determine whether to reject H0.

We are at least 95% confident that Metro EMS is not meeting the response goal of 12 minutes.

p –Value Approach

One-Tailed Tests About a Population Mean:σ Known

4. Compute the p –value.

For z = 2.47, probability = .4932.p–value = .5 − .4932 = .0068

Because p–value = .0068 < α = .05, we reject H0.

� p –Value Approach� p –Value Approach

p-value= .0068p-value= .0068

00 zα =1.645zα =1.645

α = .05α = .05

zz

z =2.47z =2.47

One-Tailed Tests About a Population Mean:σ Known

Samplingdistributionof

Samplingdistributionof z x

n= − μσ

0/z x

n= − μσ

0/

5. Determine whether to reject H0.

We are at least 95% confident that Metro EMS is not meeting the response goal of 12 minutes.

Because 2.47 > 1.645, we reject H0.

Critical Value Approach

One-Tailed Tests About a Population Mean:σ Known

For α = .05, z.05 = 1.645

4. Determine the critical value and rejection rule.

Reject H0 if z > 1.645

p-Value Approach toTwo-Tailed Hypothesis Testing

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

Compute the p-value using the following three steps:

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

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

1. Compute the value of the test statistic z.

Critical Value Approach to Two-Tailed Hypothesis Testing

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

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

Use the standard normal probability distributiontable to find zα/2 (the z-value with an area of α/2 inthe upper tail of the distribution).

Example: Glow Toothpaste

� Two-Tailed Test About a Population Mean: σ Known

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.

Example: Glow Toothpaste

� Two-Tailed Test About a Population Mean: σ Known

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.

1. Determine the hypotheses.

2. Specify the level of significance.

3. Compute the value of the test statistic.

α = .03

p –Value and Critical Value Approaches

H0: μ = 6Ha: 6μ ≠ 6μ ≠

Two-Tailed Tests About a Population Mean:σ Known

μσ

− −= = =0 6.1 6 2.74

/ .2/ 30xz

nμ

σ− −

= = =0 6.1 6 2.74/ .2/ 30

xzn

Two-Tailed Tests About a Population Mean:σ Known

5. Determine whether to reject H0.

p –Value Approach

4. Compute the p –value.

For z = 2.74, cumulative probability = .4969p–value = 2(.5 − .4969) = .0062

Because p–value = .0062 < α = .03, we reject H0.We are at least 97% confident that the mean

filling weight of the toothpaste tubes is not 6 oz.

Two-Tailed Tests About a Population Mean:σ Known

α/2 =.015

α/2 =.015

00zα/2 = 2.17zα/2 = 2.17

zz

α/2 =.015

α/2 =.015

p-Value Approach

-zα/2 = -2.17-zα/2 = -2.17z = 2.74z = 2.74z = -2.74z = -2.74

1/2p -value= .0031

1/2p -value= .0031

1/2p -value= .0031

1/2p -value= .0031

Critical Value Approach

Two-Tailed Tests About a Population Mean:σ Known

5. Determine whether to reject H0.

We are at least 97% confident that the mean filling weight of the toothpaste tubes is not 6 oz.

Because 2.47 > 2.17, we reject H0.

For α/2 = .03/2 = .015, z.015 = 2.17

4. Determine the critical value and rejection rule.

Reject H0 if z < -2.17 or z > 2.17

α/2 = .015α/2 = .015

00 2.172.17

Reject H0Reject H0Do Not Reject H0Do Not Reject H0

zz

Reject H0Reject H0

-2.17-2.17

Critical Value Approach

Samplingdistributionof

Samplingdistributionof z x

n= − μσ

0/z x

n= − μσ

0/

Two-Tailed Tests About a Population Mean:σ Known

α/2 = .015α/2 = .015

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

xx

If the confidence interval contains the hypothesizedvalue μ0, do not reject H0. Otherwise, reject H0.

The 97% confidence interval for μ is

/ 2 6.1 2.17(.2 30) 6.1 .07924x znα

σ± = ± = ±/ 2 6.1 2.17(.2 30) 6.1 .07924x z

nασ

± = ± = ±

Confidence Interval Approach toTwo-Tailed Tests About a Population Mean

Because the hypothesized value for thepopulation mean, μ0 = 6, is not in this interval,the hypothesis-testing conclusion is that thenull hypothesis, H0: μ = 6, can be rejected.

or 6.02076 to 6.17924