Hypothesis Testing

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Chapter 9 Hypothesis Testing Section 9.1 The
Language of Hypothesis Testing

• Steps in Hypothesis Testing
• A claim is made.
• Evidence (sample data) is collected in order to
  test the claim.
• The data are analyzed in order to support or
  refute the claim.

e.g. Claim I believe the amount of corn
cultivated today is less than it was in 1990.
A hypothesis is a statement or claim regarding a
characteristic of one or more populations.
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Hypothesis testing is a procedure, based on
sample evidence and probability, used to test
claims regarding a characteristic of one or more
populations.
The null hypothesis, denoted H0, is a statement
to be tested. The null hypothesis is assumed
true until evidence indicates otherwise. In
this chapter, it will be a statement regarding
the value of a population parameter.
HA
The alternative hypothesis, denoted H1, is a
claim to be tested. We are trying to find
evidence for the alternative hypothesis. In this
chapter, it will be a claim about a population
parameter.
We have two types of alternative hypothesis,
one-sided and two-sided alternatives.
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The alternative hypothesis, denoted H1

• Equal hypothesis versus not equal hypothesis
  (two-sided test)
• H0 parameter some value
• H1 parameter ? some value
• 2a. Equal hypothesis versus less than (one-sided
  test)
• H0 parameter some value
• H1 parameter
• 2b. Equal hypothesis versus greater than
  (one-sided test)
• H0 parameter some value
• H1 parameter some value

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• Example
• Determine for the following the null and
  alternative hypotheses.
• According to the United States Department of
  Agriculture, the mean farm rent in Indiana was
  89.00 per acre in 1995. A researcher for the
  USDA claims that the mean rent has decreased
  since then.
• According to the United States Census Bureau,
  16.3 of Americans did not have health insurance
  coverage in 1998. A politician claims that this
  percentage has decreased since 1998.
• According to the United States Energy
  Information Administration, the mean expenditure
  for residential energy consumption was 1338 in
  1997. An economist claims that the mean
  expenditure for residential energy is different
  today.

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3. According to the United States Energy
Information Administration, the mean expenditure
for residential energy consumption was 1338 in
1997. An economist claims that the mean
expenditure for residential energy is different
today.
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Type I and Type II Errors Four Outcomes from
Hypothesis Testing.

• We reject the null when in fact the alternative
  is true. This decision would be correct.
• We fail to reject the null when in fact the null
  is true. This decision would be correct.
• We reject the null when in fact the null is
  true. This decision would be incorrect.
  This type of error is called a Type I error.
• We fail to reject the null when in fact the
  alternative is true. This decision would be
  incorrect. This type of error is called a
  Type II error.

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Reality
Null Hypothesis
True
False
Action
Type I
Reject
Type II
Accept
The level of significance, ?, is the probability
of making a Type I error. We refer to the
probability of making a Type II error as ?.
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• Example
• Explain for the following what it would be to
  make a Type I and a Type II error.
• According to the United States Department of
  Agriculture, the mean farm rent in Indiana was
  89.00 per acre in 1995. A researcher for the
  USDA claims that the mean rent has decreased
  since then.
• According to the United States Census Bureau,
  16.3 of Americans did not have health insurance
  coverage in 1998. A politician claims that this
  percentage has decreased since 1998.
• According to the United States Energy
  Information Administration, the mean expenditure
  for residential energy consumption was 1338 in
  1997. An economist claims that the mean
  expenditure for residential energy is different
  today.

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• Example
• Determine for the following the null and
  alternative hypotheses.
• According to the United States Department of
  Agriculture, the mean farm rent in Indiana was
  89.00 per acre in 1995. A researcher for the
  USDA claims that the mean rent has decreased
  since then.

Type I We say the mean has decreased (reject the
null), when in fact it hasnt. The truth is
µ 89. (i.e. The null is true.)
Type II We say the mean is the same or has
increased (accept null), when in fact it has
decreased. The truth is µ is false.)
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2. According to the United States Census Bureau,
16.3 of Americans did not have health insurance
coverage in 1998. A politician claims that this
percentage has decreased since 1998.
Type I We say the percentage has decreased
(reject null), when in fact it hasnt. The
truth is p 0.163. (i.e. The null is true.)
Type II We say the percentage is the same
(accept null), when in fact it has decreased.
The truth is p false.)
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3. According to the United States Energy
Information Administration, the mean expenditure
for residential energy consumption was 1338 in
1997. An economist claims that the mean
expenditure for residential energy is different
today.
Type I We say the mean has changed (reject the
null), when in fact it hasnt. The truth is
µ 1338. (i.e. The null is true.)
Type II We say the mean hasnt changed (accept
null), when in fact it has changed. The truth
is µ ? 1338. (i.e. The null is false.)
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Section 9.2 Testing a Hypothesis About ?, ?
Known
What is s?

• The Classical Method of Testing a Hypothesis.
• If a claim is made regarding the population mean
  with ? known, we use the following steps to test
  the claim provided
• The sample is obtained using simple random
  sampling
• The population from which the sample is drawn is
  normally distributed or the sample size, n, is
  large

i.e. n 30
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• Use the following steps to test the hypothesis
• Hypothesis
• Critical Value
• Test Statistic
• Compare
• Conclusion

H0 H1
Za/2 or Za
CV vs. Z
Fail to Reject or Reject the H0
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• A claim is made regarding the population mean.
  The claim is used to determine the null and
  alternative hypotheses.
• Select a level of significance ? based upon the
  seriousness of making a Type I error. The level
  of significance is used to determine the critical
  value. The critical value represents the
  maximum number of standard deviations the sample
  mean can be from ?0 before the null hypothesis is
  rejected.

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4. Compare the value of the test statistic to
that of the critical value to make a decision
regarding the null hypothesis. (Fail to reject
or Reject)
5. State the conclusion.
There is evidence at the a significance level
. There is insufficient evidence at the a
significance level .
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Example A researcher claims that the average age
of a woman before she has her first child is
greater than the 1990 mean age of 24.6 years, on
the basis of data obtained from the National
Vital Statistics Report, Vol. 48, No. 14. She
obtains a simple random sample of 40 women who
gave birth to their first child in 1999 and finds
the sample mean age to be 27.1 years. Assume
that the population standard deviation is 6.4
years. Test the researchers claim, using the
classical approach at the ? 0.05 level of
significance.

• Hypothesis
• Critical Value
• Test Statistic
• Decision
• Conclusion

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4. Decision Is this a one-sided or two-sided test?

5. Conclusion There is evidence at the 0.05
significance level that the mean age of women who
gave birth to their 1st child is greater in 1999
than in 1990.
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Testing a Hypothesis about ? with ? known, using
p-values A p-value is the probability of
observing a test statistic as extreme or more
extreme than the one observed under the
assumption that the null hypothesis is true.

• Classical Method
• Hypothesis
• Critical Value
• Test Statistic
• Decision
• Conclusion

• Use the following steps to test the hypothesis
• (p-value Method)
• Hypothesis
• Test Statistic
• p-value
• Decision
• Conclusion

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• A claim is made regarding the population mean.
  The claim is used to determine the null and
  alternative hypotheses.

3. Use the value of the test statistic to obtain
the p-value for a one-sided test from the
z-table. If the hypothesis is two-sided, double
the one-sided p-value to obtain the two-sided
p-value.
One-sided P(Z test stat) or P(Z stat) Two-sided 2One-sided
4. Make a decision if the p-value is less than
?, reject the null and if the p-value is greater
than ?, fail to reject the null.
p-value a ? fail to reject H0 p-value reject H0
5. State the conclusion.
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Example In 1990, the average farm size in Kansas
was 694 acres, according to data obtained from
the U.S. Department of Agriculture. A researcher
claims that farm sizes are larger now due to
consolidation of farms. She obtains a random
sample of 40 farms and determines the mean size
to be 731 acres. Assume that ? 212 acres.
Test the researchers claim, using the p-value
approach at the ? 0.05 level of significance.

• Hypothesis
• Test Statistic
• p-value
• Decision
• Conclusion

3. p-value P(Z 1.10) 1 P(Z 0.8643 0.1357
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4. Decision Is this a one-sided or two-sided test?
One-sided p-value 0.1357 0.05 a ? Fail
to reject H0
5. Conclusion There is insufficient evidence at
the 0.05 significance level to conclude that the
mean farm size has increased from 1990.
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Using Confidence Intervals to Test
Hypothesis When testing H0 ? ?0 vs H1 ?
? ?0, if a (1- ?)100 confidence interval
contains ?0, we do not reject the null
hypothesis. However, if the confidence interval
does not contain ?0, we have evidence that
supports the claim stated in the alternative
hypothesis and conclude ? ? ?0 at the level of
significance, ?.
Fail to reject H0
Reject H0
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Section 9.3 Testing a Hypothesis about ?, ?
Unknown
The procedures here are exactly the same except
now the test statistic will follow a
t-distribution with n 1 degrees of freedom.
Test Statistic

• p-value approach
• Hypothesis
• Test Statistic
• p-value
• Decision
• Conclusion

• Classical Method
• Hypothesis
• Critical Value
• Test Statistic
• Decision
• Conclusion

t table with n-1 df
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Example In 1989, the average age of an inmate on
death row was 36.2 years of age, according to
data obtained from the U.S. Department of
Justice. A sociologist wants to test the claim
that the average age of a death-row inmate has
changed since then. She randomly selects 32
death-row inmates and finds that their age is
38.9, with a standard deviation of 9.6. Using
the classical approach, test the sociologists
claim at the ? 0.01 level of significance.

• Hypothesis
• Critical Value
• Test Statistic
• Decision
• Conclusion

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4. Decision Is this a one-sided or two-sided test?
5. Conclusion There is insufficient evidence at
the 0.01 significance level to conclude that the
average age has changed.
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Example The mean monthly cellular telephone bill
in 1999 was 40.24, according to the Cellular
Telecommunications Industry Association. A
researcher at CTIA claims that the average
monthly billed has changed since then. He
conducts a survey of 49 cellular phone users and
determines the mean bill to be 45.15 with a
standard deviation of 21.20. Test the
researchers claim using the p-value approach at
the ? 0.10 level of significance.

• Hypothesis
• Test Statistic
• p-value
• Decision
• Conclusion

One-sided or two-sided?
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3. p-value P(t48 1.62)
2P(t48 0.10
4. Decision p-value 0.10 a Fail to Reject H0
5. Conclusion There is insufficient evidence at
the 0.10 significance level to conclude that the
monthly bill has changed.
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Section 8.3 Confidence Intervals About a
Population Proportion

• Unbiased
• Consistent
• Efficient

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Sampling Distribution of For a simple random
sample of size n such that ? 0.05, the
sampling distribution of is approximately
normal with mean p and standard deviation
, provided that np 5 and nq 5.
IF
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Constructing a (1 - ?)100 Confidence Interval
for a Population Proportion. Suppose a simple
random sample of size n is taken from a
population. A (1 - ?)100 confidence interval
for p is given by the following quantities
Upper limit
Lower limit
Best Point Estimate
Estimate C.V. SE(Estimate)
Estimate SE(Estimate)
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Example The drug Lipitor is meant to lower
cholesterol levels. In a clinical trial of 863
patients who received 10 mg doses of Lipitor
daily, 47 reported a headache as a side effect.

• Obtain a point estimate for the population
  proportion of Lipitor users who will experience
  a headache as a side effect.
• Verify that the requirements for constructing a
  confidence interval about are satisfied.
• Construct a 90 confidence interval for the
  population proportion of Lipitor users who will
  report a headache as a side effect.

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3. Construct a 90 confidence interval for the
population proportion of Lipitor users who will
report a headache as a side effect.
We are 90 confident that the true population
proportion of Lipitor users who will experience
headaches as a side effect is between 4.2 and
6.7.
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Sample Size for Estimating the Population
Proportion. The sample size required to obtain a
(1 - ?)100 confidence interval for p with a
margin of error ME is given by
where
is a prior estimate of p.
If a prior estimate of p is unavailable, the
sample size required is
Always round n up to the next integer.
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Why use 0.25 if the prior is unavailable?
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• Example
• A child psychologist wishes to estimate the
  percentage of fathers who watch their
  preschool-aged child when the mother works. What
  size sample should be obtained if she wishes the
  estimate to be within 3 percentage points with
  99 confidence if
• She uses a 1995 estimate obtained from the U.S.
  Census Bureau of 18.5
• She does not use any prior estimate.

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Section 9.4 Testing a Hypothesis About a
Population Proportion
The procedures here are exactly the same except
now the test statistic will again follow a
standard normal distribution. However, remember
that we must check that np 5 and nq 5 and
? 0.05 Test Statistic
Assuming p0 is true, thus
NOT
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Example In a survey conducted by the Gallup
Organization between August 29 and September 5,
2000, 395 of 1012 adults aged 18 years or older
said they had a gun in the house. In 1990, 47
of household had a gun. Is there significant
evidence to support the claim that the proportion
of households that have a gun has decreased since
1990 at the ? 0.01 level of significance? Use
the classical approach.
BUT, 1st check if we can assume normality

• Hypothesis
• Critical Value
• Test Statistic
• Decision
• Conclusion

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3. Test Statistic, but we need the point
estimate of p first
Under H0
5. Conclusion There is evidence at the 0.01
significance level that the true population
proportion less than 0.47.
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Example Pathological gambling is an
impulse-control disorder. The American
Psychiatry Association lists 10 characteristics
that diagnose the disorder in its DSM-IV manual.
The National Gambling Impact Study Commissions
randomly selected 2417 adults and found that 35
were pathological gamblers. Is there evidence to
support the claim that more than 1 of the adult
population are pathological gamblers at the ?
0.05 level of significance? Use p-value approach
BUT, 1st check if we can assume normality

• Hypothesis
• Test Statistic
• p-value
• Decision
• Conclusion

Again, need point estimate of p
One-sided or two-sided?
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2. Test Statistic
Under H0
3. p-value P(Z 2.214) 1 - P(Z 1 0.9866 0.0134
4. Decision p-value 0.0134 Reject H0
5. Conclusion There is evidence at the 0.05
significance level to conclude that the
proportion of pathological gamblers is greater
than 1.