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Chapter Nine
Hypothesis
Testing
Understanding Basic Statistics
Fourth Edition
By Brase and Brase
Prepared by: Lynn Smith
Gloucester County College
Hypothesis testing is used to make
decisions concerning the value of a
parameter.
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9|2
Null Hypothesis: H0
• a working hypothesis about the
population parameter in question
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The value specified in
the null hypothesis is often:
• a historical value
• a claim
• a production specification
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Alternate Hypothesis: H1
• any hypothesis that differs from the null
hypothesis
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An alternate hypothesis is constructed in
such a way that it is the one to be accepted
when the null hypothesis must be rejected.
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A manufacturer claims that their light
bulbs burn for an average of 1000
hours. We have reason to believe
that the bulbs do not last that long.
Determine the null and alternate
hypotheses.
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A manufacturer claims that their light bulbs
burn for an average of 1000 hours. ...
• The null hypothesis (the claim) is that
the true average life is 1000 hours.
• H0: m = 1000
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… A manufacturer claims that their light
bulbs burn for an average of 1000 hours.
We have reason to believe that the bulbs
do not last that long. ...
If we reject the manufacturer’s claim,
we must accept the alternate
hypothesis that the light bulbs do
not last as long as 1000 hours.
H1: m < 1000
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9|9
Types of Statistical Tests
• Left-tailed: H1 states that the parameter is
less than the value claimed in H0.
• Right-tailed: H1 states that the parameter is
greater than the value claimed in H0.
• Two-tailed: H1 states that the parameter is
different from (  ) the value claimed in H0.
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Given the Null Hypothesis
H0: m = k
If you believe that m is less than k,
Use the left-tailed test:
H 1: m < k
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Given the Null Hypothesis
H0: m = k
If you believe that m is more than k,
Use the right-tailed test:
H 1: m > k
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Given the Null Hypothesis
H0: m = k
If you believe that m is different from k,
Use the two-tailed test:
H 1: m  k
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9 | 13
General Procedure for Hypothesis Testing
• Formulate the null and alternate
hypotheses.
• Take a simple random sample.
• Compute a test statistic corresponding
to the parameter in H0.
• Assess the compatibility of the test
statistic with H0.
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9 | 14
Hypothesis Testing about the Mean of a
Normal Distribution with a Known Standard
Deviation 
x-m
test statistic  z 
/ n
x  mean of simple random sample
m  value stated in H 0
n  sample size
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9 | 15
P-value of a Statistical Test
• Assuming H0 is true, the probability that the
test statistic (computed from sample data)
will take on values as extreme as or more
than the observed test statistic is called the
P-value of the test
• The smaller the P-value computed from
sample data, the stronger the evidence
against H0.
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P-values for Testing a Mean
Using the Standard Normal Distribution
Use the
standardiz ed sample test statistic
x-m
 zx 
/ n
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P-value for a Left-tailed Test
• P-value = probability of getting a test
statistic less than z x
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P-value for a Right-tailed Test
• P-value = probability of getting a test
statistic greater than z x
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P-value for a Two-tailed Test
• P-value = probability of getting a test
statistic lower than  z x or higher than z x
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Types of Errors in Hypothesis Testing
• Type I
• Type II
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Type I Error
• rejecting a null hypothesis which is, in
fact, true
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Type II Error
• not rejecting a null hypothesis which is,
in fact, false
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Type I and Type II Errors
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Level of Significance, Alpha (a)
• the probability of rejecting a true
hypothesis
• Alpha is the probability of a type I error
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Type II Error
• Beta = β = probability of a type II error
(failing to reject a false hypothesis)
• In hypothesis testing α and β values
should be chosen as small as possible.
• Usually α is chosen first.
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Power of the Test = 1 – β
• The probability of rejecting H0 when it is
in fact false = 1 – b.
• The power of the test increases as the
level of significance (a) increases.
• Using a larger value of alpha increases
the power of the test but also increases
the probability of rejecting a true
hypothesis.
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Probabilities Associated
with a Statistical Test
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Hypotheses and Types of Errors
A fast food restaurant indicated that the average
age of its job applicants is fifteen years. We
suspect that the true age is lower than 15.
We wish to test the claim with a level of
significance of a = 0.01.
Determine the Null and Alternate hypotheses
and describe Type I and Type II errors.
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… average age of its job applicants
is fifteen years. We suspect that the
true age is lower than 15.
H0: m = 15
H1: m < 15
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H0: m = 15
H1: m < 15
a = 0.01
A type I error would occur if we rejected
the claim that the mean age was 15,
when in fact the mean age was 15 (or
higher). The probability of committing
such an error is as much as 1%.
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H0: m = 15
H1: m < 15
a = 0.01
A type II error would occur if we failed to
reject the claim that the mean age was
15, when in fact the mean age was
lower than 15. The probability of
committing such an error is called beta.
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Concluding a Hypothesis Test Using the P-value
and Level of Significance α
• If P-value < α reject the null hypothesis
and say that the data are statistically
significant at the level α.
• If P-value > α, do not reject the null
hypothesis.
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Basic Components of a Statistical Test
• Null hypothesis, alternate hypothesis
and level of significance
• Test statistic and sampling distribution
• P-value
• Test conclusion
• Interpretation of the test results
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Null Hypothesis, Alternate Hypothesis and
Level of Significance
• If the sample data evidence against H0
is strong enough, we reject H0 and
adopt H1.
• The level of significance, α, is the
probability of rejecting H0 when it is in
fact true.
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Test Statistic and Sampling Distribution
• Mathematical tools to measure
compatibility of sample data and the
null hypothesis
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P-value
The probability of obtaining a test statistic
from the sampling distribution that is as
extreme as or more extreme than the
sample test statistic computed from the
data under the assumption that H0 is
true
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Test Conclusion
• If P-value < α reject the null hypothesis
and say that the data are statistically
significant at the level α.
• If P-value > α, do not reject the null
hypothesis.
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Interpretation of Test Results
• Give a simple explanation of conclusion
in the context of the application.
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Reject or ...
• When the sample evidence is not strong enough
to justify rejection of the null hypothesis, we fail to
reject the null hypothesis.
• Use of the term “accept the null hypothesis”
should be avoided.
• When the null hypothesis cannot be rejected, a
confidence interval is frequently used to give a
range of possible values for the parameter.
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Fail to Reject H0
• There is not enough evidence to reject
H0. The null hypothesis is retained but
not proved.
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Reject H0
• There is enough evidence to reject H0.
Choose the alternate hypothesis with
the understanding that it has not been
proven.
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Testing the Mean m When  is Known
• Let x be the appropriate random variable.
Obtain a simple random sample (of size n) of
x values and compute the sample mean x.
• State the null and alternate hypotheses and
set the level of significance α.
• If x has a normal distribution, any sample size
will work. If we cannot assume a normal
distribution, use n > 30.
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Testing the Mean m When  is Known
• Use the test statistic:
xm
z

n
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Testing the Mean m When  is Known
• Use the standard normal distribution
and the type of test (one-tailed or twotailed) to find the P-value
corresponding to the test statistic.
• If the P-value < α, then reject H0. If the
P-value > α, then do not reject H0.
• State your conclusion.
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9 | 45
Testing the Mean m
When  is Known: Example
Your college claims that the mean age
of its students is 28 years. You wish to
check the validity of this statistic with a
level of significance of a = 0.05.
Assume the standard deviation is 4.3
years.
A random sample of 49 students has a
mean age of 26 years.
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Hypothesis Test Example
H0: m = 28
H1: m  28
two
Perform a ________-tailed
test.
Level of significance = α = 0.05
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Sample Test Statistic
xm
z
 n
where
m  mean specified in H 0
σ  standard deviation of the x distributi on
n  sample size being used
x  sample test statistic
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Sample Results
x  26 , s  2.3.
Calculate the test statistic z :
xm
26  28
z

 3.26
  n 4.3  7
For a two-tailed test: P-value = 2P(z < 3.26)
= 2(0.0006) = 0.0012
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P-value and Conclusion
• P-value = 0.0012
• α = 0.05. Since the P-value < α , we
reject the null hypothesis.
• We conclude that the true average age
of students is not 28.
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Testing the Mean m When  is Unknown
• Let x be the appropriate random variable.
Obtain a simple random sample (of size n) of
x values and compute the sample mean x.
• State the null and alternate hypotheses and
set the level of significance α.
• If x has a mound shaped symmetric
distribution, any sample size will work. If we
cannot assume this, use n > 30.
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Testing the Mean m When  is Known
• Use the test statistic:
xm
t
with d.f.  n - 1
s
n
where x, s and n are sample statistics.
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Testing the Mean m When  is Unknown
• Use the Student’s t distribution and the
type of test (one-tailed or two-tailed) to
find (or estimate) the P-value
corresponding to the test statistic.
• If the P-value < α, then reject H0. If the
P-value > α, then do not reject H0.
• State your conclusion.
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Using Table 4 to Estimate P-values
Use one-tailed areas as endpoints of the interval
containing the P-value for one-tailed tests.
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P-value for One-tailed Tests
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Using Table 4 to Estimate P-values
Use two-tailed areas as endpoints of the interval
containing the P-value for one-tailed tests.
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P-value for Two-tailed Tests
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Testing the Mean m
When  is Unknown: Example
The Parks Department claims that the
mean weight of fish in a lake is 2.1 kg.
We believe that the true average
weight is lower than 2.1 kg. Assume
that the weights are mound-shaped
and symmetric and a sample of five
fish caught in the lake weighed an
average of 1.99 kg with a standard
deviation of 0.09 kg.
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Determine the P-value when testing the
claim that the mean weight of fish
caught in a lake is 2.1 kg (against the
alternate that the weight is lower). A
sample of five fish weighed an average
of 1.99 kg with a standard deviation of
0.09 kg.
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Test the Claim Using α = 10%
• Null Hypothesis:
H0: m = 2.1 kg
• Alternate Hypothesis:
H1: m < 2.1 kg
• α = 0.10
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We will complete a left-tailed test with:
x  1.99
s  0.09
n5
d.f .  4
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The Test Statistic t
x  m 1.99  2.1
t

 2.73
s
.09
n
5
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Using Table 4 with t = 2.73
and d.f. = 4
Sample t =  2.73
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The t value is between two values in the
chart.Therefore the P-value will be in a
corresponding interval.
Sample t =  2.73
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Since we are performing a one-tailed test,
we use the “one-tail area” line of the chart.
Sample t =  2.73
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Since we are performing a one-tailed test,
we use the “one-tail area” line of the chart.
Sample t =  2.73
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0.025 < P-value < 0.050
…
Sample t = 2.73
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0.025 < P-value < 0.050
• Since the range of P-values is less than
a (10%), we reject the null hypothesis.
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Interpret the results:
• At level of significance 10% we rejected
the null hypothesis that the mean
weight of fish in the lake was 2.1 kg.
• Based on our sample data, we
conclude that the true mean weight is
actually lower than 2.1 kg.
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9 | 69
Critical Region (Traditional) Method for
Hypothesis Testing
• An alternate technique to the P-value
method
• Logically equivalent to the P-value
method
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Critical Region Procedure for Testing m
When  is Known
• Let x be the appropriate random variable.
Obtain a simple random sample (of size n) of
x values and compute the sample mean x.
• State the null and alternate hypotheses and
set the level of confidence α.
• If x has a normal distribution, any sample
size will work. If we cannot assume a normal
distribution, use n > 30.
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Critical Region Method for Testing the Mean
m When  is Known
• Use the test statistic: z  x  m

n
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Critical Region Method for Testing the Mean
m When  is Known
• Using the level of significance α and
the alternate hypothesis, show the
critical region and critical values on a
graph of the sampling distribution.
• Conclude the test. If the test statistic is
in the critical region, then reject H0. If
not, do not reject H0.
• State your conclusion.
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Most Common Levels of Significance
• α = 0.05 and
• α = 0.01
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Critical Region(s)
• The values of x for which we will reject the
null hypothesis.
• The critical values are the boundaries of the
critical region(s).
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Concluding Tests Using
the Critical Region Method
• Compare the sample test statistics to
the critical value(s)
• For a left-tailed test:
• If the sample test statistic is < critical
value, reject H0.
• If the sample test statistic is > critical
value, fail to reject H0.
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Critical Region for H0: m = k
Left-tailed Test
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Concluding Tests Using the Critical
Region Method
•
•
•
•
Compare the sample test statistics to
the critical value(s)
For a right-tailed test:
If the sample test statistic is > critical
value, reject H0.
If the sample test statistic is < critical
value, fail to reject H0.
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Critical Region for H0: m = k
Right-tailed Test
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Concluding Tests Using
the Critical Region Method
• Compare the sample test statistics to
the critical value(s)
• For a two-tailed test:
• If the sample test statistic lies beyond
the critical values, reject H0.
• If the sample test statistic lies between
the critical values, fail to reject H0.
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Critical Region for H0: m = k
Two-tailed Test
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Critical Values z0 for α = 0.05
and α = 0.01: Left-tailed Test
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Critical Values z0 for α = 0.05
and α = 0.01: Right-tailed Test
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Critical Values z0 for α = 0.05
and α = 0.01: Two-tailed Test
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9 | 84
Testing the Mean m
When  is Known: Example
Your college claims that the mean age of its
students is 28 years. You wish to check the
validity of this statistic with a level of
significance of a = 0.05. Assume the
standard deviation is 4.3 years.
A random sample of 49 students has a mean
age of 26 years.
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9 | 85
Hypothesis Test Example
H0: m = 28
H1: m  28
two
Perform a ________-tailed
test.
Level of significance = α = 0.05
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9 | 86
Sample Test Statistic
xm
z
 n
where
m  mean specified in H 0
σ  standard deviation of the x distributi on
n  sample size being used
x  sample test statistic
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9 | 87
Sample Results
x  26 , s  2.3.
Calculate the test statistic z :
xm
26  28
z

 3.26
  n 4.3  7
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9 | 88
Critical Region for a
Two-tailed Test with α = 0.05
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Our z =  3.26 falls within the critical region.
z =  3.26
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Since the test statistic is
in the critical region we…
• Reject the Null Hypothesis.
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Conclusion
• We conclude that the true average age
of students is not 28.
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Tests Involving a Proportion
• We will test claims that a given
percentage of the population fits a
certain description.
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Let r be the binomial random
variable, the number of successes
out of n independent trials.
r
pˆ  is our approximat ion for p
n
q 1-p
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9 | 94
For large samples (np > 5 and nq > 5):
The distributi on of p̂ 
r
is approximat ely normal with
n
mean  m  p
and standard deviation   
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pq
n
9 | 95
Three Types of Tests of
Hypotheses for Tests of Proportions
• Left-tailed tests
• Right-tailed tests
• Two-tailed tests
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Left-Tailed Test
H 0: p = k
H 1: p < k
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Right-Tailed Test
H0: p = k
H1: p > k
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Two-Tailed Test
H0: p = k
H1: p  k
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To Convert p̂ to z :
p̂  p
z
pq
n
r
where p̂   sample statistic
n
n  number of trials
p  proportion specified in H 0
q  1 p
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9 | 100
Testing a Proportion p
• Consider a binomial experiment with n trials.
• Let p represent the population probability of
success.
• Let q = 1  p represent the population
probability of failure.
• Let r be a random variable that represents
the number of successes out of the n
binomial trials.
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Testing a Proportion p
• State the null and alternate hypotheses and
set the level of significance α.
• The number of trials should be sufficiently
large so that both np > 5 and nq > 5. (Use
p from the null hypothesis.)
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9 | 102
Testing a Proportion p
r
The p̂  distrbutio n can be approximat ed by the
n
normal distributi on using the standardiz ed
test statistic :
p̂  p
z
pq
n
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9 | 103
Testing a Proportion p
• Use the standard normal distribution
and the type of test (one-tailed or twotailed) to find the P-value
corresponding to the test statistic.
• If the P-value < α, then reject H0. If the
P-value > α, then do not reject H0.
• State your conclusion.
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In the past, college officials observed that
40% of students took advantage of early
registration. This semester, of 4830
students, 2077 took advantage of early
registration. Use a 5% level of
significance to test the claim that a higher
percentage of students now participates in
early registration.
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9 | 105
In the past, college officials observed that
40% of students took advantage of early
registration. ...
H0: m = 0.40
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… test the claim that a higher percentage of
students now participates in early
registration.
H1: m > 0.40
Use a right-tailed test.
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9 | 107
… This semester, of 4830 students, 2077
took advantage of early registration. ...
r 2077
pˆ  
 0.43
n 4830
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9 | 108
The corresponding z value:
pˆ  p 0.43  0.40
z

 4.26
pq
0.40(0.60)
n
4830
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9 | 109
Use Table 3 to Determine the P-value
Associated with z = 4.26,
• The P-value is approximately 0.000
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Since α = 0.05 and P < α
• We reject the null hypothesis.
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9 | 111
Conclusion
• Since we have rejected the hypothesis
that 40% of students participate in early
registration, we conclude that:
• A percentage higher than 40% of
students now participates in early
registration.
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9 | 112
Test the hypothesis involving a proportion:
H0: p = 0.70
H1: p  0.70
Use a = 0.01
Suppose that in 120 trials there were 80
successes.
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We find that:
pˆ  0.67
and
z  0.72
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Use Table 3 to find the P-value associated
with z = –0.72.
P(z  –0.72) = 0.2358
Since the test is a two-tailed test, double
the area in the left tail to find P.
P = 2(0.2358) = 0.4716
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When the P-value is greater than the
level of significance, we do not reject
the null hypothesis.
P = 0.4716
a = 0.01
Do not reject the null hypothesis.
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