Western Blot Antibody Customer Review for β-actin Monoclonal Antibody (STJ96941)
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Chapter 6, Part A [SBE 7/e, Ch. 9]Hypothesis Testing
Developing Null and Alternative HypothesesType I and Type II ErrorsPopulation Mean: σ KnownPopulation Mean: σ Unknown
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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.
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� 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.
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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.
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� 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
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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).
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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
Null and Alternative Hypotheses
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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
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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 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.
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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 makinga Type II error.
� Statisticians avoid the risk of making a Type IIerror 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 µ > 12)
Accept H0(Conclude µ < 12)
H0 True(µ < 12)
H0 False(µ > 12)Conclusion
Population Condition
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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.
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� 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.
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� 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.
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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
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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
� Critical Value Approach
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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
� 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 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.
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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
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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:σ Known
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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
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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, cumulative probability = .9932.p–value = 1 − .9932 = .0068
Because p–value = .0068 < α = .05, we reject H0.
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� p –Value Approach
p-value= .0068
0 zα =1.645
α = .05
z
z =2.47
One- Tailed Tests About a Population Mean:σ Known
Samplingdistributionof z x
n= − µσ
0/
z xn= − µ
σ0
/
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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
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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.
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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).
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Example: Glow Toothpaste
� 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.
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Example: Glow Toothpaste
� Two-Tailed Test About a Population Mean: σ Known
oz.
GlowPerform 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.
3. Compute the value of the test statistic.
α = .03
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
6
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Glow
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 = .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.
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Glow
Two- Tailed Tests About a Population Mean:σ Known
α/2 =.015
0zα/2 = 2.17
z
α/2 =.015
p-Value Approach
-zα/2 = -2.17z = 2.74z = -2.74
1/2p -value= .0031
1/2p -value= .0031
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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
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α/2 = .015
0 2.17
Reject H0Do Not Reject H0
z
Reject H0
-2.17
Glow
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
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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.)
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If the confidence interval contains the hypothesizedvalue µ0, do not reject H0. Otherwise, reject H0.
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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
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� 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.
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� 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
� Rejection Rule: Critical Value Approach
Reject H0 if p –value < α
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p - Values and the t Distribution
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.
However, we can still use the t distribution table toidentify a range for the p-value.
An advantage of computer software packages is thatthe computer output will provide the p-value for thet 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: σ Unknown
The locations where H0 is rejected are deemedthe best locations for radar traps.
H0: µ < 65
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Example: Highway Patrol
� 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.
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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
p –Value and Critical Value Approaches
H0: µ < 65Ha: µ > 65
µ− −= = =0 66.2 65 2.286
/ 4.2 / 64xts n
µ− −= = =0 66.2 65 2.286
/ 4.2 / 64xts n
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One- Tailed Test About a Population Mean:σ Unknown
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.
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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
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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
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Chapter 6, Part B [SBE 7/e, Ch. 9]Hypothesis Testing
Population ProportionHypothesis Testing and Decision MakingCalculating the Probability of Type II ErrorsDetermining the Sample Size forHypothesis Tests About a Population Mean
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� The equality part of the hypotheses always appearsin the null hypothesis.In general, a hypothesis test about the value of apopulation proportion p must take one of thefollowing three forms (where p0 is the hypothesizedvalue 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 0: H p p≥
0: aH p p> 0: aH p p>0: aH p p< 0: aH p p<0 0: H p p≤0 0: H p p≤ 0 0: H p p=0 0: H p p=
0: aH p p≠ 0: aH p p≠
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� Test Statistic
z p pp
=− 0
σz p p
p=
− 0
σ
σ pp p
n=
−0 01( )σ 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: p < p0 Reject H0 if z > zα
Reject H0 if z < -zα
Reject H0 if z < -zα/2 or z > zα/2
H0: p > p0
H0: p = p0
Tests About a Population Proportion
Reject H0 if p –value < α
� Rejection Rule: Critical Value Approach
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� Example: National Safety CouncilFor 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 withα = .05.
Two- Tailed Test About aPopulation Proportion
� Example: National Safety Council
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Two- Tailed Test About aPopulation Proportion
1. Determine the hypotheses.
2. Specify the level of significance.
3. Compute the value of the test statistic.
α = .05
p –Value and Critical Value Approaches
0: .5H p =0: .5H p =: .5aH p ≠: .5aH p ≠
σ− −
= = =0 (67 /120) .5 1.28.045644p
p pzσ− −
= = =0 (67 /120) .5 1.28.045644p
p pz
0 0(1 ) .5(1 .5) .045644120p
p pn
σ− −
= = =0 0(1 ) .5(1 .5) .045644120p
p pn
σ− −
= = =a common
error is usingin this
formula pp
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p−Value Approach
4. Compute the p -value.
5. Determine whether to reject H0.
Because p–value = .2006 > α = .05, we cannot reject H0.
Two- Tailed Test About aPopulation Proportion
For z = 1.28, cumulative probability = .8997p–value = 2(1 − .8997) = .2006
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Two- Tailed Test About aPopulation Proportion
Critical Value Approach
5. Determine whether to reject H0.
For α/2 = .05/2 = .025, z.025 = 1.96
4. 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 decisionmaker 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 thehypothesis-testing procedure be extended to includeconsideration of making a Type II error.
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Calculating the Probability of a Type II Error in Hypothesis Tests About a Population Mean
1. Formulate the null and alternative hypotheses.
3. Using the rejection rule, solve for the value of thesample mean corresponding to the critical value ofthe test statistic.
2. Using the critical value approach, use the level ofsignificance α to determine the critical value andthe rejection rule for the test.
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Calculating the Probability of a Type II Error in Hypothesis Tests About a Population Mean
4. Use the results from step 3 to state the values of thesample mean that lead to the acceptance of H0; this defines the acceptance region.
5. Using the sampling distribution of for a value of µsatisfying the alternative hypothesis, and the acceptanceregion from step 4, compute the probability that thesample mean will be in the acceptance region. (This isthe probability of making a Type II error at the chosenlevel of µ .)
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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.
Calculating the Probabilityof a Type II Error
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−= ≥
12 1.6453.2 / 40
xz −= ≥
12 1.6453.2 / 40
xz
⎛ ⎞≥ + =⎜ ⎟⎝ ⎠
3.212 1.645 12.832340
x ⎛ ⎞≥ + =⎜ ⎟⎝ ⎠
3.212 1.645 12.832340
x
4. We will accept H0 when x < 12.8323
3. Value of the sample mean that identifiesthe rejection region:
2. Rejection rule is: Reject H0 if z > 1.6451. Hypotheses are: H0: µ < 12 and Ha: µ > 12
Calculating the Probabilityof a Type II Error
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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 / 40
z µ−=
12.83233.2 / 40
z µ−=Values of µ β 1-β
5. Probabilities that the sample mean will bein the acceptance region:
Calculating the Probabilityof a Type II Error
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Calculating the Probabilityof a Type II Error
� Calculating the Probability of a Type II Error
When the true population mean µ is far abovethe null hypothesis value of 12, there is a lowprobability that we will make a Type II error.
When the true population mean µ is close tothe null hypothesis value of 12, there is a highprobability that we will make a Type II error.
Observations about the preceding table:
Example: µ = 12.0001, β = .9500
Example: µ = 14.0, β = .0104
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Power of the Test
� The probability of correctly rejecting H0 when it isfalse is called the power of the test.
� For any particular value of µ, the power is 1 – β.
� We can show graphically the power associatedwith each value of µ; such a graph is called apower curve. (See next slide.)
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Power Curve
0.00
0.10
0.20
0.30
0.400.50
0.60
0.70
0.80
0.90
1.00
11.5 12.0 12.5 13.0 13.5 14.0 14.5
Prob
abil
ity
of C
orre
ctly
Rej
ecti
ng N
ull H
ypot
hesi
s
µ
H0 False
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� The specified level of significance determines theprobability of making a Type I error.
By controlling the sample size, the probability ofmaking a Type II error is controlled.
Determining the Sample Size for a Hypothesis Test About a Population Mean
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µ0
α
µa
xx
xx
Samplingdistributionof whenH0 is trueand µ = µ0
xx
Samplingdistributionof whenH0 is false
and µa > µ0
xx
Reject H0
β
c
c
H0: µ < µ0Ha: µ > µ0
σσ =x nσσ =x n
Note:
Determining the Sample Size for a Hypothesis Test About a Population Mean
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wherezα = z value providing an area of α in the tailzβ = z value providing an area of β in the tailσ = population standard deviationµ0 = value of the population mean in H0µa = value of the population mean used for the
Type II error
nz z
a=
+
−
( )( )
α β σµ µ
2 2
02n
z z
a=
+
−
( )( )
α β σµ µ
2 2
02
Determining the Sample Size for a Hypothesis Test About a Population Mean
Note: In a two-tailed hypothesis test, use zα /2 not zα
12
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Relationship Among α, β, and n
� Once two of the three values are known, the other can be computed.
� For a given level of significance α, increasing the sample size n will reduce β.
� For a given sample size n, decreasing α will increase β, whereas increasing α will decrease b.
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Determining the Sample Size for a Hypothesis Test About a Population Mean
� Let’s assume that the director of medicalservices makes the following statements about theallowable probabilities for the Type I and Type IIerrors:•If the mean response time is µ = 12 minutes, I am willing to risk an α = .05 probability of rejecting H0.•If the mean response time is 0.75 minutes over the specification (µ = 12.75), I am willing to risk a β = .10 probability of not rejecting H0.
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Determining the Sample Size for a Hypothesis Test About a Population Mean
α = .05, β = .10zα = 1.645, zβ = 1.28µ0 = 12, µa = 12.75σ = 3.2
2 2 2 2
2 20
( ) (1.645 1.28) (3.2) 155.75 156( ) (12 12.75)a
z zn α β σ
µ µ+ +
= = = ≈− −
2 2 2 2
2 20
( ) (1.645 1.28) (3.2) 155.75 156( ) (12 12.75)a
z zn α β σ
µ µ+ +
= = = ≈− −