Chapter 9, Part A Hypothesis Tests Slide 1dscsss/teaching/mgs9920/slides/ch09a.pdf · Chapter 9,...
Transcript of Chapter 9, Part A Hypothesis Tests Slide 1dscsss/teaching/mgs9920/slides/ch09a.pdf · Chapter 9,...
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1Slide
Chapter 9, Part AHypothesis Tests
2Slide
Learning objectives
1. Understand how to develop Null and Alternative Hypotheses
2. Understand Type I and Type II Errors
3. Able to do hypothesis test about population mean when σ is known
4. Able to do hypothesis test about population mean when σ is unknown
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3Slide
Null and Alternative Hypotheses
n Hypothesis testing can be used to determine whethera statement about the value of a population parametershould or should not be rejected.
n The null hypothesis, denoted by H0 , is a tentativeassumption about a population parameter.
n The alternative hypothesis, denoted by Ha, is theopposite of what is stated in the null hypothesis.
n The alternative hypothesis is what the test isattempting to establish.
•L.O.•Hypotheses•Type I & II error•Testing –σ known•Testing –σ known
4Slide
n Testing Research Hypotheses
Null and Alternative Hypotheses:Applications
• The research hypothesis should be expressed asthe alternative hypothesis.
• The conclusion that the research hypothesis is truecomes from sample data that contradict the nullhypothesis.
•L.O.•Hypotheses•Type I & II error•Testing –σ known•Testing –σ known
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5Slide
Null and Alternative Hypotheses:Applications
n 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.
•L.O.•Hypotheses•Type I & II error•Testing –σ known•Testing –σ known
6Slide
n Testing in Decision-Making Situations
Null and Alternative Hypotheses:Applications
• 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
•L.O.•Hypotheses•Type I & II error•Testing –σ known•Testing –σ known
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7Slide
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
n The equality part of the hypotheses always appearsin the null hypothesis.
n 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).
•L.O.•Hypotheses•Type I & II error•Testing –σ known•Testing –σ known
8Slide
n 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.
•L.O.•Hypotheses•Type I & II error•Testing –σ known•Testing –σ known
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9Slide
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.
n Example: Metro EMS
Null and Alternative Hypotheses•L.O.•Hypotheses•Type I & II error•Testing –σ known•Testing –σ known
10Slide
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
•L.O.•Hypotheses•Type I & II error•Testing –σ known•Testing –σ known
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11Slide
Type I Error
n Because hypothesis tests are based on sample data,we must allow for the possibility of errors.
n A Type I error is rejecting H0 when it is true.
n The probability of making a Type I error when thenull hypothesis is true as an equality is called thelevel of significance.
n Applications of hypothesis testing that only controlthe Type I error are often called significance tests.
•L.O.•Hypotheses•Type I & II error•Testing –σ known•Testing –σ known
12Slide
Type II Error
n A Type II error is accepting H0 when it is false.
n It is difficult to control for the probability of makinga Type II error.
n Statisticians avoid the risk of making a Type IIerror by using “do not reject H0” and not “accept H0”.
•L.O.•Hypotheses•Type I & II error•Testing –σ known•Testing –σ known
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13Slide
Type I and Type II Errors
CorrectDecision Type II Error (β)
CorrectDecisionType I Error (α)
Reject H0(Conclude µ > 12)
Accept H0(Conclude µ < 12)
H0 True(µ < 12)
H0 False(µ > 12)Conclusion
Population Condition
•L.O.•Hypotheses•Type I & II error•Testing –σ known•Testing –σ known
14Slide
p-Value Approach toOne-Tailed Hypothesis Testing
n Reject H0 if the p-value < α .
n The p-value is the probability, computed using thetest statistic, that measures the support (or lack ofsupport) provided by the sample for the nullhypothesis.
n If the p-value is less than or equal to the level ofsignificance α, the value of the test statistic is in therejection region.
•L.O.•Hypotheses•Type I & II error•Testing – σ known•Testing –σ known
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15Slide
n p-Value Approach
p-value= .072
0-zα =-1.28
α = .10
z
z =-1.46
Lower-Tailed Test About a Population Mean:σ Known
Samplingdistributionof z x
n= − µσ
0/
z xn= − µ
σ0
/
p-Value < α ,so reject H0.
•L.O.•Hypotheses•Type I & II error•Testing – σ known•Testing –σ known
16Slide
n p-Value Approach
p-Value= .011
0 zα =1.75
α = .04
zz =
2.29
Upper-Tailed Test About a Population Mean:σ Known
Samplingdistributionof z x
n= − µσ
0/
z xn= − µ
σ0
/
p-Value < α ,so reject H0.
•L.O.•Hypotheses•Type I & II error•Testing – σ known•Testing –σ known
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17Slide
Critical Value Approach to One-Tailed Hypothesis Testing
n The test statistic z has a standard normal probabilitydistribution.
n 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.
n The value of the test statistic that established theboundary of the rejection region is called thecritical value for the test.
n The rejection rule is:• Lower tail: Reject H0 if z < -zα
• Upper tail: Reject H0 if z > zα
•L.O.•Hypotheses•Type I & II error•Testing – σ known•Testing –σ known
18Slide
α = .10
0−zα = −1.28
Reject H0
Do Not Reject H0
z
Samplingdistributionof z x
n= − µσ
0/
z xn= − µ
σ0
/
Lower-Tailed Test About a Population Mean:σ Known
n Critical Value Approach
•L.O.•Hypotheses•Type I & II error•Testing – σ known•Testing –σ known
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19Slide
α = .05
0 zα = 1.645
Reject H0
Do Not Reject H0
z
Samplingdistributionof z x
n= − µσ
0/
z xn= − µ
σ0
/
Upper-Tailed Test About a Population Mean:σ Known
n Critical Value Approach
•L.O.•Hypotheses•Type I & II error•Testing – σ known•Testing –σ known
20Slide
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.
•L.O.•Hypotheses•Type I & II error•Testing – σ known•Testing –σ known
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21Slide
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•L.O.•Hypotheses•Type I & II error•Testing – σ known•Testing –σ known
22Slide
n 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
•L.O.•Hypotheses•Type I & II error•Testing – σ known•Testing –σ known
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23Slide
1. Develop the hypotheses.
2. Specify the level of significance. α = .05
H0: µ < 12Ha: µ > 12
n 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
•L.O.•Hypotheses•Type I & II error•Testing – σ known•Testing –σ known
24Slide
5. Determine whether to reject H0.
We are at least 95% confident that Metro EMS is not meeting the response goal of 12 minutes.
n p –Value Approach
One-Tailed Tests About a Population Mean:σ Known
4. Compute the p –value.
For z = 2.47, cumulative probability = .9932.p–value = 1 − .9932 = .0068
Because p–value = .0068 < α = .05, we reject H0.
•L.O.•Hypotheses•Type I & II error•Testing – σ known•Testing –σ known
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25Slide
n p –Value Approach
p-value= .0068
0 zα =1.645
α = .05
zz =
2.47
One-Tailed Tests About a Population Mean:σ Known
Samplingdistributionof z x
n= − µσ
0/
z xn= − µ
σ0
/
•L.O.•Hypotheses•Type I & II error•Testing – σ known•Testing –σ known
26Slide
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.
n 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
•L.O.•Hypotheses•Type I & II error•Testing – σ known•Testing –σ known
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27Slide
p-Value Approach toTwo-Tailed Hypothesis Testing
n The rejection rule:Reject H0 if the p-value < α .
n 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.
•L.O.•Hypotheses•Type I & II error•Testing – σ known•Testing –σ known
28Slide
Critical Value Approach to Two-Tailed Hypothesis Testing
n The critical values will occur in both the lower andupper tails of the standard normal curve.
n The rejection rule is:Reject H0 if z < -zα/2 or z > zα/2.
n 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).
•L.O.•Hypotheses•Type I & II error•Testing – σ known•Testing –σ known
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29Slide
Example: Glow Toothpaste
n Two-Tailed Test About a Population Mean: σ 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.
•L.O.•Hypotheses•Type I & II error•Testing – σ known•Testing –σ known
30Slide
Example: Glow Toothpaste
n Two-Tailed Test About a Population Mean: σ Known
oz.
GlowPerform a hypothesis test, at the .03
level 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.
•L.O.•Hypotheses•Type I & II error•Testing – σ known•Testing –σ known
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31Slide
1. Determine the hypotheses.
2. Specify the level of significance.
3. Compute the value of the test statistic.
α = .03
n p –Value and Critical Value Approaches
Glow
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
•L.O.•Hypotheses•Type I & II error•Testing – σ known•Testing –σ known
32Slide
Glow
Two-Tailed Tests About a Population Mean:σ Known
5. Determine whether to reject H0.
n p –Value Approach
4. Compute the p –value.
For z = 2.74, cumulative probability = .9969p–value = 2(1 − .9969) = .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.
•L.O.•Hypotheses•Type I & II error•Testing – σ known•Testing –σ known
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33Slide
Glow
Two-Tailed Tests About a Population Mean:σ Known
α/2 =.015
0zα/2 = 2.17
z
α/2 =.015
n p-Value Approach
-zα/2 = -2.17z = 2.74z = -2.74
1/2p -value= .0031
1/2p -value= .0031
•L.O.•Hypotheses•Type I & II error•Testing – σ known•Testing –σ known
34Slide
n Critical Value Approach
Glow
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
•L.O.•Hypotheses•Type I & II error•Testing – σ known•Testing –σ known
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35Slide
α/2 = .015
0 2.17
Reject H0Do Not Reject H0
z
Reject H0
-2.17
Glow
n Critical Value Approach
Samplingdistributionof z x
n= − µσ
0/
z xn= − µ
σ0
/
Two-Tailed Tests About a Population Mean:σ Known
α/2 = .015
•L.O.•Hypotheses•Type I & II error•Testing – σ known•Testing –σ known
36Slide
Confidence Interval Approach toTwo-Tailed Tests About a Population Mean
n Select 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
n If the confidence interval contains the hypothesizedvalue µ0, do not reject H0. Otherwise, reject H0.
•L.O.•Hypotheses•Type I & II error•Testing – σ known•Testing –σ known
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37Slide
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
Glow
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
•L.O.•Hypotheses•Type I & II error•Testing – σ known•Testing –σ known
38Slide
In-class exercise
n #10 (p.355) (one-tail test)n #11 (p.355) (two-tail test)
•L.O.•Hypotheses•Type I & II error•Testing – σ known•Testing –σ known
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39Slide
n Test Statistic
Tests About a Population Mean:σ Unknown
t xs n
=− µ0
/t x
s n=
− µ0
/
This test statistic has a t distributionwith n - 1 degrees of freedom.
•L.O.•Hypotheses•Type I & II error•Testing –σ known•Testing – σ known
40Slide
n Rejection Rule: p -Value Approach
H0: µ < µ0 Reject H0 if t > tα
Reject H0 if t < -tα
Reject H0 if t < - tα/2 or t > tα/2
H0: µ > µ0
H0: µ = µ0
Tests About a Population Mean:σ Unknown
n Rejection Rule: Critical Value Approach
Reject H0 if p –value < α
•L.O.•Hypotheses•Type I & II error•Testing –σ known•Testing – σ known
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41Slide
p -Values and the t Distribution
n The format of the t distribution table provided in moststatistics textbooks does not have sufficient detailto determine the exact p-value for a hypothesis test.
n However, we can still use the t distribution table toidentify a range for the p-value.
n An advantage of computer software packages is thatthe computer output will provide the p-value for thet distribution.
•L.O.•Hypotheses•Type I & II error•Testing –σ known•Testing – σ known
42Slide
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
n One-Tailed Test About a Population Mean: σ Unknown
The locations where H0 is rejected are deemedthe best locations for radar traps.
H0: µ < 65
•L.O.•Hypotheses•Type I & II error•Testing –σ known•Testing – σ known
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43Slide
Example: Highway Patrol
n One-Tailed Test About a Population Mean: σ UnknownAt Location F, a sample of 64 vehicles shows a
mean speed of 66.2 mph with astandard deviation of4.2 mph. Use α = .05 totest the hypothesis.
•L.O.•Hypotheses•Type I & II error•Testing –σ known•Testing – σ known
44Slide
One-Tailed Test About a Population Mean:σ Unknown
1. Determine the hypotheses.
2. Specify the level of significance.
3. Compute the value of the test statistic.
α = .05
n p –Value and Critical Value Approaches
H0: µ < 65Ha: µ > 65
µ− −= = =0 66.2 65 2.286/ 4.2 / 64
xts n
µ− −= = =0 66.2 65 2.286/ 4.2 / 64
xts n
•L.O.•Hypotheses•Type I & II error•Testing –σ known•Testing – σ known
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One-Tailed Test About a Population Mean:σ Unknown
n p –Value Approach
5. Determine whether to reject H0.
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 < α = .05, we reject H0.We are at least 95% confident that the mean speed
of vehicles at Location F is greater than 65 mph.
•L.O.•Hypotheses•Type I & II error•Testing –σ known•Testing – σ known
46Slide
n Critical Value Approach
5. Determine whether to reject H0.
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.
Because 2.286 > 1.669, we reject H0.
One-Tailed Test About a Population Mean:σ Unknown
For α = .05 and d.f. = 64 – 1 = 63, t.05 = 1.669
4. Determine the critical value and rejection rule.
Reject H0 if t > 1.669
•L.O.•Hypotheses•Type I & II error•Testing –σ known•Testing – σ known
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α = .05
0 tα =1.669
Reject H0
Do Not Reject H0
t
One-Tailed Test About a Population Mean:σ Unknown
•L.O.•Hypotheses•Type I & II error•Testing –σ known•Testing – σ known
48Slide
In-class exercise
n #23 (p.362) (one-tail test)n #24 (p.362) (two-tail test)
•L.O.•Hypotheses•Type I & II error•Testing –σ known•Testing – σ known
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End of Chapter 9, Part A