Math 3680

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Math 3680 Lecture 13 Hypothesis Testing The
z Test
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• The One-Sided z Test

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• Example Before a design artist was hired to
  improve its entrance, an average of 3218 people
  entered a department store daily, with an SD of
  287.
• Since the entrance was redesigned, a simple
  random sample of 42 days has been studied. The
  results are shown on the next slide.

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• Is this statistically significant for indicating
  that the average number of people entering the
  store daily has increased?

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• Note Dont compare the difference
• 3392 - 3218 174
• with the population standard deviation 287.
• The former applies to an average, while the
  latter is for individual days.

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There are two possibilities

• The average number of people entering has not
  changed. The observed sample average of 3,392
  can be reasonably attributed to chance
  fluctuations.
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• The difference between the observed average and
  the expected average is too large to be simply
  chance. The average number of people entering
  has increased with the improved entrance.

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• Definition Null Hypothesis. The first
  hypothesis, which asserts simple chance
  fluctuations, is called the null hypothesis.
• Definition Alternative Hypothesis. The second
  hypothesis, which asserts that the average has in
  fact increased, is called the alternative
  hypothesis.

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• The null hypothesis is the default assumption.
  This is the assumption to be disproved.
• For example, if the sample average were 3219 per
  day, that would hardly be convincing evidence.
  However, if the new sample average were 5000 per
  day, we can be confident of the lure of the new
  storefront and rule out simple chance. So
  where is the cut-off value?

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• Solution.
• H0 m 3218 (The average number of customers
  entering the store has not changed due to the
  improved storefront)
• Ha m gt 3218 (The average number of customers
  entering the store has increased due to the
  improved storefront)
• We choose a 0.05
• Before continuing, why isnt Ha written as m ?
  3218?

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• Assuming H0, we have a sample of 42 days which
  are being drawn from a box with m 3218 and s
  287.
• The average has the following moments

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• Test statistic
• P-value. Assuming H0, we must find the chance of
  obtaining a test statistic at least this extreme.
  For this problem, that means

• Conclusion We reject the null hypothesis. There
  is good reason to believe that the average number
  of customers has increased after the redesign.

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• Excel Use the command
• ZTEST(A1D11, 3218, 287)

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Observations

• 1. This test of significance is called the
  z-test, named after the test statistic.
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• 2. The z-test is best used with large samples
  so that the normal approximation may be safely
  made.
• 3. Notice we have not proven beyond a shadow of
  a doubt that the new storefront was effective in
  increasing the number of patrons.

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• 4. The alternative hypothesis is that the daily
  average of patrons is greater than 3218. It is
  not that the new average is exactly equal to
  3392. In other words, the alternative hypothesis
  was a compound hypothesis, not a simple
  hypothesis.
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• 5. Small values of P are evidence against the
  null hypothesis they indicate that something
  besides chance is at work.
• 6. We are NOT saying that there is 1 chance in
  20,000 for the null hypothesis to be correct.

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Another (equivalent) procedure for hypothesis
testing In the previous problem, if the test
statistic was any number greater than 1.645, then
we would have obtained a P-value less than 0.05,
the specified a. (Why?)
We call zc 1.645 the critical value, and the
interval (1.645, ?) is called the rejection
region. Since zs lies in the rejection region,
we choose to reject H0.
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In terms of the customers, we have the critical
value
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Another (equivalent) procedure for hypothesis
testing Hypothesis testing may be correctly
conducted by using the P-value (the first
method) or by using the critical value (as we
just discussed). In scientific articles, both
are usually reported, even though the two methods
are logically equivalent. As we now discuss, the
critical value also eases computation of the
power of the test.
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Example In the previous example, suppose that
the redesign increased the average number of
patrons by 100, from 3218 to 3318. How likely is
it that a sample of only 42 days will come to the
correct conclusion (by rejecting the null
hypothesis)? Note Recall that this is called
the power of the test.
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Solution recall the critical value xc 3290.8
P( Reject H0 m 3318)
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Alternative distribution
Null distribution
73.05
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Power of the test (1-ß) as a function of the true
average
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• Example The average braking distance from 60
  mph of a Mercury Sable is 159 feet, with an SD of
  23.5 feet. Sables equipped with (hopefully)
  improved tires have just undergone early testing
  the results of the first 45 tests are shown
  below. Does this indicate that the new tires
  have decreased the braking distance? Use a
  0.05.

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• Example The Compute the probability b of
  committing a Type II error if the braking
  distance with the improved tires is now 155 feet.

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Conceptual Questions

• 1) We made a test of significance because
  (choose one)
• i) We knew what was in the box, but did not know
  how the sample would turn out or
• ii) We knew how the sample turned out, but did
  not know what was in the box.

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Conceptual Questions

• 2) The null hypothesis says that the average of
  the (sample / box) is 159 feet.

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Conceptual Questions

• 3) True or False
• a) The observed significance level of 8 depends
  on the data (i.e. sample)
• b) There are 92 chances out of 100 for the
  alternative hypothesis to be correct.

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Conceptual Questions

• 4) Suppose only 10 tests were performed instead
  of 45. Should we use the normal curve to compute
  P?

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Conceptual Questions

• 5) True or False
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• a) A highly statistically significant result
  cannot possibly be due to chance.
• b) If a sample difference is highly
  statistically significant, there is less than a
  1 chance for the null hypothesis to be correct.

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Conceptual Questions

• 6) True or False
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• a) If , then the null hypothesis
  looks plausible.
• b) If , then the null hypothesis
  looks implausible.

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P


.
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0
P
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• The Two-Sided z Test

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• Example A company claims to have designed a new
  fishing line that has a mean breaking strength of
  8 kg with an SD of 0.5 kg. Consumer Reports tests
  a random sample of 45 lines the results are
  shown below. Test the validity of the companys
  claim.

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• Notes
• To avoid data snooping, we must use a two-tailed
  test. Before the tests were actually performed,
  we had no a priori reason to think that the
  sample average would return either too high or
  too low.
• For a two-sided alternative hypothesis, the
  P-value is twice as large as for a one-sided
  alternative.

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Example. Lets take a look at the results of the
Salk vaccine trial, which we first saw back in
Lecture 2 Does it appear that the vaccine
was effective? Note If the vaccine was
ineffective, then we would expect the 199 polio
cases to be distributed with p 200745/401974
0.499398, and the 57 polio cases among the
treated was just due to a run of luck.