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Dr. Ka-fu Wong
ECON1003
Analysis of Economic Data
Ka-fu Wong © 2003
Chap 8- 1
Central Limit Theorem #1
5 balls in the bag:
0
1
2
3
4
Draw 50 ball 1000 times with replacement. Compute the sample
mean. Plot a relative frequency histogram (empirical probability
histogram) of the 1000 sample means.
The Central Limit Theorem says
1. The empirical histogram looks like a normal density.
2. Expected value (mean of the normal distribution) = 2.
3. Variance of the sample means = 2/50=0.04.
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Chap 8- 2
Confidence Interval #1
Five numbered balls in the bag:
?
?
?
?
?
Draw one sample of 50 balls with replacement. Compute the sample
mean and sample standard deviation. Suppose the sample mean is
10 and the sample standard deviation is 0.04. Can you tell us the
range of possible values the population mean may take, at 95%
confidence level?
m
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Chap 8- 3
Hypothesis testing #1
Five numbered balls in the bag:
?
?
?
?
?
Draw one sample of 50 balls with replacement. Compute the sample
mean and sample standard deviation. Suppose the sample mean is
10 and the sample standard deviation is 0.04. Do you think the balls
in this bag has a mean of 2?
2
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Chap 8- 4
Chapter Eight
One-Sample Tests of Hypothesis
GOALS
1. Define a hypothesis and hypothesis testing.
2. Describe the five step hypothesis testing
procedure.
3. Distinguish between a one-tailed and a two-tailed
test of hypothesis.
4. Conduct a test of hypothesis about a population
mean.
5. Conduct a test of hypothesis about a population
proportion.
6. Define Type I and Type II errors.
7. Compute the probability of a Type II error.
l
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Chap 8- 5
What is a Hypothesis?
 A Hypothesis is a statement about the value of
a population parameter developed for the
purpose of testing.
 Examples of hypotheses made about a
population parameter are:
 The mean monthly income for systems
analysts is $3,625.
 Twenty percent of all customers at Bovine’s
Chop House return for another meal within a
month.
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Chap 8- 6
What is Hypothesis Testing?
 Hypothesis testing is a procedure, based on
sample evidence and probability theory, used to
determine whether the hypothesis is a
reasonable statement and should not be rejected,
or is unreasonable and should be rejected.
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Chap 8- 7
Hypothesis Testing
Step 1: state null and alternative hypothesis
Step 2: select a level of significance
Step 3: identify the test statistic
Step 4: formulate a decision rule
Step 5: Take a sample, arrive at a decision
Do not reject null
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Reject null and accept alternative
Chap 8- 8
Definitions
 Null Hypothesis H0: A statement about the
value of a population parameter.
 Alternative Hypothesis H1: A statement that
is accepted if the sample data provide
evidence that the null hypothesis is false.
 Level of Significance: The probability of
rejecting the null hypothesis when it is
actually true.
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Chap 8- 9
Objectivity in formulating a hypothesis
 In court, the defendant is presumed innocent until proven
beyond reasonable doubt to be guilty of stated charges.
 The “null hypothesis”, i.e. the denial of our theory, is
presumed true until we prove beyond reasonable doubt
that it is false.
 “Beyond reasonable doubt” means that the probability
of claiming that our theory is true when it is not (null
hypothesis true) is less than an a priori set significance
level (usually 5% or 1%).
 Is the defendant guilty?
 Null: the defendant is not guilty.
 Alternative: the defendant is guilty.
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Chap 8- 10
Definitions
 Type I Error: conclude the defendant guilty when the
defendant did not commit the crime.
 Level of significance is also the maximum probability of
committing a type I error. We want to limit this Type I
Error to some small number.
 Type II Error: Conclude the defendant not guilty when the
defendant actually committed the crime.
Committed Crime
Court
Guilty
Decision
(Guilty or not) Not Guilty
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Yes
No
Correct
decision
Type I error
Type II error
Correct
decision
Chap 8- 11
Definitions
 Type I Error: Rejecting the null hypothesis when it is
actually true.
 Level of significance is also the maximum probability of
committing a type I error. We want to limit this Type I
Error to some small number.
 Type II Error: Accepting the null hypothesis when it is
actually false.
State of nature
Decision
Don’t reject
based on the null
sample
Reject null
statistic
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Null true
Null false
Correct
decision
Type II error
Type I error
Correct
decision
Chap 8- 12
Definitions
 Test statistic: A value, determined from sample
information, used to determine whether or not to
reject the null hypothesis.
 Critical value: The dividing point between the region
where the null hypothesis is rejected and the region
where it is not rejected.
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Chap 8- 13
One-Tailed Tests of Significance
 A test is one-tailed when the alternate hypothesis, H1 ,
states a direction, such as:
 H1: The mean yearly commissions earned by full-time
realtors is more than $35,000. (µ>$35,000)
 H1: The mean speed of trucks traveling on I-95 in
Georgia is less than 60 miles per hour. (µ<60)
 H1: Less than 20 percent of the customers pay cash for
their gasoline purchase. ( < .20)
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Chap 8- 14
Sampling Distribution for the Statistic Z for a
One Tailed Test, .05 Level of Significance
.95 probability
Critical
Value
z=1.65
.05 probability
0
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1
2
3
4
Rejection region
Reject the null if the test
statistic falls into this region.
Chap 8- 15
Two-Tailed Tests of Significance
 A test is two-tailed when no direction is specified in the
alternate hypothesis H1 , such as:
 H1: The mean amount spent by customers at the WalMart in Georgetown is not equal to $25. (µ  $25).
 H1: The mean price for a gallon of gasoline is not
equal to $1.54. (µ  $1.54).
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Chap 8- 16
Sampling Distribution for the Statistic Z for a
Two Tailed Test, .05 Level of Significance
.95 probability
Critical
Value
z=-1.96
Critical
Value
z=1.96
.025 probability
.025 probability
-4 -3 -2 -1
Rejection region #1
Ka-fu Wong © 2003
0
1
2
3
4
Rejection region #2
Reject the null if the test statistic falls into these two regions.Chap 8- 17
Copyright© 2002 by The McGraw-Hill Companies, Inc. All rights reserved
Testing for the Population Mean: Large Sample,
Population Standard Deviation Known
 When testing for the population mean from a large
sample and the population standard deviation is
known, the test statistic is given by:
X m
z
/ n
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Chap 8- 18
EXAMPLE 1
 The processors of Fries’ Catsup indicate on the label
that the bottle contains 16 ounces of catsup. The
standard deviation of the process is 0.5 ounces. A
sample of 36 bottles from last hour’s production
revealed a mean weight of 16.12 ounces per bottle. At
the .05 significance level is the process out of control?
 That is, can we conclude that the mean amount per
bottle is different from 16 ounces?
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Chap 8- 19
EXAMPLE 1
continued
 Step 1: State the null and the alternative
hypotheses:
H0: m = 16;
H1: m  16
 Step 2: Select the level of significance. In this case we
selected the .05 significance level.
 Step 3: Identify the test statistic. Because we know the
population standard deviation, the test statistic is z.
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Chap 8- 20
EXAMPLE 1
continued
 Step 4: State the decision rule:
Reject H0 if z > 1.96 or z < -1.96
 Step 5: Compute the value of the test statistic and arrive
at a decision.
X m
16.12  16.00
z

 1.44
 n
0.5 36
 Do not reject the null hypothesis. We cannot
conclude the mean is different from 16 ounces.
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Chap 8- 21
p-Value in Hypothesis Testing
 A p-Value is the probability, assuming that the null
hypothesis is true, of finding a value of the test statistic at
least as extreme as the computed value for the test.
 The “critical probability” for our decision to reject the
null.
 If the p-Value is smaller than the significance level, H0 is
rejected.
 If the p-Value is larger than the significance level, H0 is not
rejected.
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Chap 8- 22
Computation of the p-Value
 One-Tailed Test: p-Value = P{z ≥absolute value of
the computed test statistic value}
 Two-Tailed Test: p-Value = 2P{z ≥ absolute value of
the computed test statistic value}
 From EXAMPLE 1, z = 1.44, and because it was a
two-tailed test,
the p-Value = 2P{z ≥ 1.44} = 2(.5-.4251) = .1498.
Because .1498 > .05, do not reject H0.
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Chap 8- 23
Testing for the Population Mean: Large Sample,
Population Standard Deviation Unknown
 Here  is unknown, so we estimate it with
the sample standard deviation s.
 As long as the sample size n  30, z can be
approximated with:
X m
z
s/ n
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Chap 8- 24
EXAMPLE 2
 Roder’s Discount Store chain issues its own credit card.
Lisa, the credit manager, wants to find out if the mean
monthly unpaid balance is more than $400. The level of
significance is set at .05. A random check of 172 unpaid
balances revealed the sample mean to be $407 and the
sample standard deviation to be $38. Should Lisa
conclude that the population mean is greater than $400,
or is it reasonable to assume that the difference of $7
($407-$400) is due to chance?
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Chap 8- 25
EXAMPLE 2
continued
 Step 1: H0: m  $400, H1: m > $400
 Step 2: The significance level is .05
 Step 3: Because the sample is large we can use the z
distribution as the test statistic.
 Step 4: H0 is rejected if z>1.65
 Step 5: Perform the calculations and make a decision.
X  m $407  $400
z

 2.42
s n
$38 172
H0 is rejected. Lisa can conclude that the mean
unpaid balance is greater than $400.
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Chap 8- 26
Testing for a Population Mean: Small Sample,
Population Standard Deviation Unknown
 The test statistic is the t distribution.
 The test statistic for the one sample case is given by:
X m
t 
s/ n
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Chap 8- 27
Example 3
 The current rate for producing 5 amp fuses at Neary
Electric Co. is 250 per hour. A new machine has been
purchased and installed that, according to the supplier, will
increase the production rate. A sample of 10 randomly
selected hours from last month revealed the mean hourly
production on the new machine was 256 units, with a
sample standard deviation of 6 per hour. At the .05
significance level can Neary conclude that the new
machine is faster?
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Chap 8- 28
Example 3
continued
 Step 1: State the null and the alternate hypothesis.
H0: m  250; H1: m > 250
 Step 2: Select the level of significance. It is .05.
 Step 3: Find a test statistic. It is the t distribution because
the population standard deviation is not known and the
sample size is less than 30.
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Chap 8- 29
Example 3
continued
 Step 4: State the decision rule. There are 10 – 1 = 9
degrees of freedom. The null hypothesis is rejected if t >
1.833.
Step 5: Make a decision and interpret the results.
X  m 256  250
t

 3.162
s n
6 10
The null hypothesis is rejected. The mean number
produced is more than 250 per hour.
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Chap 8- 30
Tests Concerning Proportion
 A Proportion is the fraction or percentage that indicates
the part of the population or sample having a particular
trait of interest.
 The sample proportion is denoted by p and is found by:
Number of successes in the sample
p
Number sampled
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Chap 8- 31
Test Statistic for Testing a Single
Population Proportion
z
p 
 (1   )
n
The sample proportion is p and  is the
population proportion.
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Chap 8- 32
EXAMPLE 4
 In the past, 15% of the mail order solicitations for a
certain charity resulted in a financial contribution. A
new solicitation letter that has been drafted is sent to a
sample of 200 people and 45 responded with a
contribution. At the .05 significance level can it be
concluded that the new letter is more effective?
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Chap 8- 33
Example 4
continued
 Step 1: State the null and the alternate hypothesis.
H0:   .15 H1:  > .15
 Step 2: Select the level of significance. It is .05.
 Step 3: Find a test statistic. The z distribution is the test
statistic.
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Chap 8- 34
Example 4
continued
 Step 4: State the decision rule. The null hypothesis is
rejected if z is greater than 1.65.
 Step 5: Make a decision and interpret the results.
z
p 

 (1   )
n
45
 .15
200
 2.97
.15(1  .15)
200
The null hypothesis is rejected. More than 15 percent are
responding with a pledge. The new letter is more effective.
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Chap 8- 35
Chapter Eight
One-Sample Tests of Hypothesis
- END -
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Chap 8- 36