Transcript Chap010 - Ka

Dr. Ka-fu Wong
ECON1003
Analysis of Economic Data
Ka-fu Wong © 2003
Chap 10- 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 10- 2
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 10- 3
Chapter Ten
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 10- 4
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 10- 5
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 10- 6
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 10- 7
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 10- 8
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.
 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 10- 9
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 10- 10
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 10- 11
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 10- 12
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 Wal-Mart 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 10- 13
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
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0
1
2
3
4
Rejection region #2
Reject the null if the test statistic falls into these two regions.Chap 10- 14
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 
z
/ n
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Chap 10- 15
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%).
 When we are asked to test at the .05 significance level
whether a the process is out of control, we will set up the
hypothesis as:
 Null: the process is in control.
 Alternative: the process is out of control.
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Chap 10- 16
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 10- 17
EXAMPLE 1
continued
 Step 1: State the null and the alternative
hypotheses:
H0:  = 16;
H1:   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 10- 18
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 
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 10- 19
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.
 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 10- 20
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 10- 21
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 
z
s/ n
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Chap 10- 22
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 10- 23
EXAMPLE 2
continued
 Step 1: H0:   $400, H1:  > $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   $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 10- 24
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 
t 
s/ n
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Chap 10- 25
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 10- 26
Example 3
continued
 Step 1: State the null and the alternate
hypothesis.
H0:   250; H1:  > 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 10- 27
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   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 10- 28
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 10- 29
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 10- 30
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 10- 31
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 10- 32
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 10- 33
Chapter Ten
One-Sample Tests of Hypothesis
- END -
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Chap 10- 34