Transcript Document

Welcome to MM207
Unit 7 Seminar
Dr. Bob
Hypothesis Testing and Excel
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Hypothesis Testing - Example
Criminal Trial
• Null hypothesis: Ho = defendant is not-guilty
• Alternative hypothesis: Ha = person is guilty
Procedure
• We assume Null hypothesis is true. The defendant is
not-guilty until we prove otherwise.
• Evidence is presented and we then decide whether
to:
– Reject the Null hypothesis (person is guilty)
– Do not reject the Null hypothesis (not enough evidence to
reject, but it doesn’t mean it is true)
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Hypothesis Testing Decision Errors
• Type I error
Reject a true Null hypothesis (i.e. innocent person
found guilty)
α = alpha = probability of Type I error
• Type II error
Do not reject a false Null hypothesis (i.e. guilty man
goes free)
β = beta = probability of Type II error
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Hypothesis Tests: 3 types
• Null hypothesis always in the form:
Ho: µ = k (mean equals a certain value)
• Alternative hypothesis can take 3 forms:
Ha: µ < k (left-tail, actual mean is less than stated value)
Ha: µ > k (right-tail, actual mean is greater than stated value)
Ha: µ ≠ k (two-tail, actual mean is not equal to stated value)
• The language of the problem will tell you which
alternative hypothesis you need to use. This takes
practice
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Interpreting the P-value
• The P-value is critical in determining if
Ho should be rejected.
– If P-value less than α, reject Ho
– If P-value greater α, do not reject Ho
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Section 7.2, Question 33 [page 391]
• In Illinois, a random sample of 85 eighth
grade students has a mean score of 282
with a standard deviation of 35 on a
national mathematics assessment test.
The test result prompts a state school
administrator to declare that the mean
score for the state’s eighth graders on
the examination is more than 275. At α =
0.04,is there enough evidence to support
the administrator’s claim?
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Question 33 continued
• State the Null and Alternative Hypotheses
Ho: µ ≤ 275; please note that the equality (=) is ALWAYS in
the Null)
Ha: µ > 275 (the state average greater than 275). This is the
Claim!
• Determine the Alpha level (given in this problem = 0.04)
• Calculate the Test Statistic (I am going to use Excel!)
• Interpret the Results
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Unit 7 Excel Template
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Results from Z-Test Mean
Compare these results to the Textbook
Since the one-tail p-value is less than alpha (0.04) we reject the null
and conclude that there is enough evidence to support the
administrator’s claim that it is greater than 275.
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Section 7.3, Example 4 (page 400)
• A used car dealer says that the mean price of
a 2005 Honda Pilot LX is at least $23,900. You
suspect that this claim is incorrect and find
that a random sample of 14 similar vehicles
has a mean price of $23,000 and a standard
deviation of $1113. Is there enough evidence
to reject the dealer’s claim at α = 0.05?
Assume the population is normally
distributed.
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Section 7.3, Example 4 continued
• State the Null and Alternative Hypotheses
Ho: µ ≥ $23,900; This is the Claim! [notice it can be either the
null or the alternative depending on the problem.
Ha: µ < $23,900
• Determine the Alpha level (given in this problem = 0.05)
• Calculate the Test Statistic (I am going to use Excel again!)
• Interpret the Results
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Results from the t-test Mean
Compare these results to those on page 400
We can either use the one-tail p-value (.0049 <.05) or the rejection region
method (-3.03 < -1.17709) to conclude that we reject the claim and conclude
that the mean price is less than $23,900.
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Section 7.4, Example 1 (page 408)
• A research center claims that less than 20%
of Internet users in the United States have a
wireless network in their home. In a random
sample of 100 adults, 15% say they have a
wireless network in their home. At α = 0.01, is
there enough evidence to support the
researcher’s claim?
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Section 7.4, Example 1 continued
• State the Null and Alternative Hypotheses
Ho: p ≥ 0.20;
Ha: p < 0.20; This is the Claim
• Determine the Alpha level (given in this problem = 0.01)
• Determine if np ≥ 5 and nq ≥ 5 so that you can use the z-test
• Calculate the Test Statistic (I am going to use Excel again!)
• Interpret the Results
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Results from the z-test Proportion
Compare these results to those on page 408
We can either use the one-tail p-value (.1056 >.01) or the rejection region
method (-1.25 > -2.3263) to conclude that we cannot reject the claim
that the proportion is 20%.
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