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Unformatted text preview: Chapter 9 Hypothesis Testing
s
s
s s s s Developing Null and Alternative Hypotheses
Type I and Type II Errors
OneTailed Tests About a Population Mean: LargeSample Case
TwoTailed Tests About a Population Mean: LargeSample Case
Tests About a Population Mean: SmallSample Case
Tests About a Population Proportion Slide 1 Developing Null and Alternative Hypotheses
s s s s 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.
Hypothesis testing is similar to a criminal trial. The hypotheses are:
H0: The defendant is innocent
Ha: The defendant is guilty Slide 2 Developing Null and Alternative Hypotheses
s 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. Slide 3 Developing Null and Alternative Hypotheses
s 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. Slide 4 Developing Null and Alternative Hypotheses
s Testing in DecisionMaking 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. Slide 5 A Summary of Forms for Null and Alternative Hypotheses about a Population Mean
s s The equality part of the hypotheses always appears in the null hypothesis.
In general, a hypothesis test about the value of a population mean µ must take one of the following three must take one of the following three forms (where µ0 is the hypothesized value of the population mean). H0: µ > µ0
Ha: µ < µ0 H0: µ < µ0 H0: µ = µ0
≠
Ha: µ > µ0 Ha: µ µ0 Slide 6 Example: Metro EMS
s Null and Alternative Hypotheses
A major west coast city provides one of the most comprehensive emergency medical services in the world. Operating in a multiple hospital system with approximately 20 mobile medical units, the service goal is to respond to medical emergencies with a mean time of 12 minutes or less.
The director of medical services wants to formulate a hypothesis test that could use a sample of emergency response times to determine whether or not the service goal of 12 minutes or less is being achieved. Slide 7 Example: Metro EMS
s Null and Alternative Hypotheses
Hypotheses Conclusion and Action
H0: µ < 1 2 The emergency service is meeting the response goal; no followup action is necessary. Ha: µ > 1 2 The emergency service is not meeting the response goal; appropriate followup action is necessary.
Where: µ = mean response time for the population of medical emergency requests. Slide 8 Type I and Type II Errors
s
s
s
s s
s Since 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.
A Type II error is accepting H0 when it is false.
The person conducting the hypothesis test specifies the maximum allowable probability of making a
Type I error, denoted by α and called the level of significance.
Generally, we cannot control for the probability of making a Type II error, denoted by β.
Statistician avoids the risk of making a Type II error by using “do not reject H0” and not “accept H0”. Slide 9 Example: Metro EMS
s Type I and Type II Errors Population Condition Conclusion
(µ > 1 2 ) Accept H0 H0 True (µ < 1 2 ) Correct (Conclude µ < 1 2 ) Conclusion Reject H0 (Conclude µ > 1 2 ) Ha True Type II Error Type I Correct Εrror Conclusion Slide 10 The Use of pValues
s
s s The pvalue is the probability of obtaining a sample result that is at least as unlikely as what is observed.
The pvalue can be used to make the decision in a hypothesis test by noting that:
• if the pvalue is less than the level of significance α, the value of the test statistic is in the rejection region.
• if the pvalue is greater than or equal to α, the value of the test statistic is not in the rejection region.
Reject H0 if the pvalue < α. Slide 11 The Steps of Hypothesis Testing
s
s
s
s
s
s Determine the appropriate hypotheses.
Select the test statistic for deciding whether or not to reject the null hypothesis.
Specify the level of significance α for the test.
Use α to develop the rule for rejecting H0.
to develop the rule for rejecting Collect the sample data and compute the value of the test statistic.
a) Compare the test statistic to the critical value(s) in the rejection rule, or b) Compute the pvalue based on the test statistic and compare it to α, to determine whether or not to reject H0.
to determine whether or not to reject Slide 12 OneTailed Tests about a Population Mean: Large
Sample Case (n > 30)
s Hypotheses H0: µ < µ0 or
or H0: µ > µ0 Ha: µ > µ0 Ha: µ < µ0
s s Test Statistic σ Known
Known
x − µ0
z=
σ/ n σ Unknown Unknown
x − µ0
z= s/ n Rejection Rule Reject H0 if z > zα Reject H0 if z < zα
Reject Slide 13 Example: Metro EMS
s OneTailed Test about a Population Mean: Large n
Let α = P(Type I Error) = .05 Sampling distribution
x of (assuming H0 is true and µ = 12) Reject H0 Do Not Reject H0 α = .0 5
1.645σ x 12 c
(Critical value) x Slide 14 Example: Metro EMS
s OneTailed Test about a Population Mean: Large n
x Let n = 40, = 13.25 minutes, s = 3.2 minutes (The sample standard deviation s can be used to estimate the population standard deviation σ.)
x − µ 13.25 −12
z=
=
= 2.47
s/ n
3.2 / 40 Since 2.47 > 1.645, we reject H0. Conclusion: We are 95% confident that Metro EMS is not meeting the response goal of 12 minutes; appropriate action should be taken to improve service. Slide
15 Example: Metro EMS
s Using the pvalue to Test the Hypothesis
x
Recall that z = 2.47 for = 13.25. Then pvalue = .0068.
Since pvalue < α, that is .0068 < .05, we reject H0. Reject H0
Do Not Reject H0 0 pvalue= .0 0 6 8 1.645 2.47 z Slide 16 TwoTailed Tests about a Population Mean: LargeSample Case (n > 30)
s Hypotheses
Ha: µ s s µ0 Test Statistic Rejection Rule H0: µ = µ0
≠ σ Known
Known
x − µ0
z=
σ/ n σ Unknown Unknown
x − µ0
z= s/ n Reject H0 if z > zα/2 Slide 17 Example: Glow Toothpaste
s TwoTailed Tests about a Population Mean: Large n The production line for Glow toothpaste is designed to fill tubes of toothpaste with a mean weight of 6 ounces. Periodically, a sample of 30 tubes will be selected in order to check the filling process. Quality assurance procedures call for the continuation of the filling process if the sample results are consistent with the assumption that the mean filling weight for the population of toothpaste tubes is 6 ounces; otherwise the filling process will be stopped and adjusted. Slide 18 Example: Glow Toothpaste
s TwoTailed Tests about a Population Mean: Large n
A hypothesis test about the population mean can be used to help determine when the filling process should continue operating and when it should be stopped and corrected.
• Hypotheses H0: µ = 6
≠
H: µ
6 • a Rejection Rule
Α ssuming a .05 level of significance, Reject H0 if z < 1.96 or if z > 1.96 Slide 19 Example: Glow Toothpaste
s TwoTailed Test about a Population Mean: Large n
Sampling distribution
x of (assuming H0 is true and µ = 6) Reject H0 Reject H0 Do Not Reject H0 α /2 = .0 2 5 α /2 = .0 2 5 1.96 0 1.96 z Slide 20 Example: Glow Toothpaste
s TwoTailed Test about a Population Mean: Large n
Assume that a sample of 30 toothpaste tubes
provides a sample mean of 6.1 ounces and standard
deviation of 0.2 ounces.
x
Let n = 30, = 6.1 ounces, s = .2 ounces x − µ 0 6 .1 − 6
z=
=
= 2.74
s / n .2 / 30 Since 2.74 > 1.96, we reject H0. Conclusion: We are 95% confident that the mean filling weight of the toothpaste tubes is not 6 ounces. The filling process should be stopped and the filling mechanism adjusted. Slide 21 Example: Glow Toothpaste
s Using the pValue for a TwoTailed Hypothesis Test
Suppose we define the pvalue for a twotailed test as double the area found in the tail of the distribution.
With z = 2.74, the standard normal probability
table shows there is a .5000 .4969 = .0031 probability
of a difference larger than .1 in the upper tail of the
distribution.
Considering the same probability of a larger difference in the lower tail of the distribution, we have pvalue = 2(.0031) = .0062
The pvalue .0062 is less than α = .05, so H0 is rejected. Slide 22 Confidence Interval Approach to a
TwoTailed Test about a Population Mean
s s Select a simple random sample from the population and use the value of the sample mean to develop the x
confidence interval for the population mean µ.
If the confidence interval contains the hypothesized value µ0, do not reject H0. Otherwise, reject H0. Slide 23 Example: Glow Toothpaste
s Confidence Interval Approach to a TwoTailed Hypothesis Test The 95% confidence interval for µ is
σ
x ± zα / 2
= 6.1 ± 1. 96(. 2 30 ) = 6.1±. 0716
n or 6.0284 to 6.1716
Since the hypothesized value for the population mean, µ0 = 6, is not in this interval, the hypothesistesting conclusion is that the null hypothesis,
H0: µ = 6, can be rejected. Slide 24 Tests about a Population Mean:
SmallSample Case (n < 30)
s Test Statistic σ Known
Known
z = x −µ
σ/ n
0 s σ Unknown Unknown
x − µ0
t=
s/ n The t test statistic has a t distribution with n 1 degrees of freedom. (Assume population is Normal)
Rejection Rule (when σ Unknown) Unknown) OneTailed TwoTailed H0: µ < µ0 Reject H0 if t > tα H0: µ > µ0 Reject H0 if t < tα H0: µ = µ0 Reject H0 if t > tα/2 Slide 25 p Values and the t Distribution s s s The format of the t distribution table provided in most statistics textbooks does not have sufficient detail to determine the exact pvalue for a hypothesis test.
However, we can still use the t distribution table to identify a range for the pvalue.
An advantage of computer software packages is that the computer output will provide the pvalue for the
t distribution. Slide 26 Example: Highway Patrol
s OneTailed Test about a Population Mean: Small n
A State Highway Patrol periodically samples vehicle speeds at various locations on a particular roadway. The sample of vehicle speeds is used to test the hypothesis H0: µ < 65.
The locations where H0 is rejected are deemed the best locations for radar traps.
At Location F, a sample of 16 vehicles shows a mean speed of 68.2 mph with a standard deviation of 3.8 mph. Use an α = .05 to test the hypothesis. Slide 27 Example: Highway Patrol
s OneTailed Test about a Population Mean: Small n
x Let n = 16, = 68.2 mph, s = 3.8 mph α = .05, d.f. = 161 = 15, tα = 1.753 x − µ 0 68.2 − 65
t=
=
= 3.37
s / n 3.8 / 16 Since 3.37 > 1.753, we reject H0. Conclusion: We are 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. Slide 28 Summary of Test Statistics to be Used in a
Hypothesis Test about a Population Mean
Yes σknown ? Yes n > 30 ? No Yes
Use s to
estimate σ σknown ? Yes z= x −µ
σ/ n No x −µ
z=
s/ n x −µ
z=
σ/ n No
Use s to
estimate σ x −µ
t=
s/ n Popul. approx.
normal ?
No Increase n
to > 30 Slide 29 A Summary of Forms for Null and Alternative Hypotheses about a Population Proportion
s s 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). H0: p > p0 H0: p < p0 H0: p = p0 ≠ Ha: p > p0 Ha: p p0 Ha: p < p0 Slide 30 Tests about a Population Proportion:
LargeSample Case (np > 5 and n(1 p) > 5)
s Test Statistic where:
s p − p0
z=
σp σp = p0 (1 − p0 )
n Rejection Rule OneTailed TwoTailed H0: p < p0 Reject H0 if z > zα H0: p > p0 Reject H0 if z < zα H0: p = p0 Reject H0 if z > zα/2 Slide 31 Example: NSC
s TwoTailed Test about a Population Proportion: Large n
For a Christmas and New Year’s week, the National Safety Council estimated that 500 people would be killed and 25,000 injured on the nation’s roads. The NSC claimed that 50% of the accidents would be caused by drunk driving.
A sample of 120 accidents showed that 67 were caused by drunk driving. Use these data to test the NSC’s claim with α = 0.05. Slide 32 Example: NSC
s TwoTailed Test about a Population Proportion: Large n
• Hypothesis H0: p = .5 • Test Statistic σp = Ha: p .5 ≠ p0 (1 − p0 )
.5(1 − .5)
=
= .045644
n
120 p − p0 (67 / 120) − .5
z=
=
= 1.278
σp
.045644 Slide 33 Example: NSC
s TwoTailed Test about a Population Proportion: Large n
• Rejection Rule
Reject H0 if z < 1.96 or z > 1.96 • Conclusion Do not reject H0. For z = 1.278, the pvalue is .201. If we reject
H0, we exceed the maximum allowed risk of committing a Type I error (pvalue > .050). Slide 34 ...
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 Spring '08
 Priya

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