Ch 11 實習 (1)
Introduction
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
所謂假設檢定，係指在尚未蒐集樣本資料、進行
推論之前，就事先對母體的某種特徵性質作一合
理的假設敘述，再利用隨機抽出的樣本及抽樣分
配，配合機率原理，以判斷此項假設是否為真

若抽出的樣本資料與所陳述的假設很不一致，檢
定的結果必然認為這個假設不對，則否定或拒絕
（reject）這個假設
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Introduction
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
若抽出的樣本資料與所陳述的假設不會很不一致，
檢定的結果就沒有充分理由斷定這個假設不對，
但也不認為這個假設是對的

假設檢定的主要精神在於尋找證據來拒絕H0，所
以假設檢定只能檢定H0是否顯著錯誤，而不能判
斷其絕對正確
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Developing Null and Alternative
Hypotheses
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
Hypothesis testing can be used to determine whether a
statement about the value of a population parameter should or
should not be rejected.

The null hypothesis, denoted by H0 , is a tentative assumption
about a population parameter.

The alternative hypothesis, denoted by H1, is the opposite of
what is stated in the null hypothesis.

Hypothesis testing is similar to a criminal trial. The hypotheses
are:
H0: The defendant is innocent
H1: The defendant is guilty
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A Summary of Forms for Null and Alternative
Hypotheses about a Population Mean
The equality part of the hypotheses always
appears in the null hypothesis.
 In general, a hypothesis test about the value
of a population mean  must take one of the
following three forms (where 0 is the
hypothesized value of the population mean).
H 0 :  = 0
H 0 :  =  0 H 0 :  = 0
H1:  < 0
H1:  > 0 H1:   0

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Concepts of Hypothesis Testing

The critical concepts of hypothesis testing.
 Example:
 An
operation manager needs to determine if the
mean demand during lead time is greater than 350.
 If so, changes in the ordering policy are needed.
 There
are two hypotheses about a population
mean:
The null hypothesis
μ = 350
 H1: The alternative hypothesis μ > 350
 H0:
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Example 1

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A spouse suspects that the average amount of
money spent on Christmas gifts for immediate
family members is above $1,200. The correct set
of hypotheses is:
a. H0: μ = 1200 vs. H1: μ < 1200
b. H0: μ > 1200 vs. H1: μ = 1200
c. H0: μ = 1200 vs. H1: μ > 1200
d. H0: μ < 1200 vs. H1: μ = 1200
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Concepts of Hypothesis Testing

Assume the null hypothesis is true (=
350).
 = 350


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Sample from the demand population, and build a
statistic related to the parameter hypothesized (the
sample mean).
Pose the question: How probable is it to obtain a
sample mean at least as extreme as the one
observed from the sample, if H0 is correct?
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Concepts of Hypothesis Testing

Assume the null hypothesis is true (=
350).
x  355
 = 350


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x  450
Since the x is much larger than 350, the mean  is
likely to be greater than 350. Reject the null
hypothesis.
In this case the mean  is not likely to be greater
than 350. Do not reject the null hypothesis.
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Types of Errors


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Two types of errors may occur when deciding
whether to reject H0 based on the statistic value.
 Type I error: Reject H0 when it is true.
 Type II error: Do not reject H0 when it is false.
Example continued
 Type I error: Reject H0 ( = 350) in favor of H1 (
> 350) when the real value of  is 350.
 Type II error: Believe that H0 is correct ( = 350)
when the real value of  is greater than 350.
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Type I and Type II Errors




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Since hypothesis tests are based on sample data,
we must allow for the possibility of errors.
The person conducting the hypothesis test
specifies the maximum allowable probability of
making a
Type I error, denoted by  and called the level of
significance.
Generally, we cannot control for the probability of
making a Type II error, denoted by .
Statistician avoids the risk of making a Type II
error by using “do not reject H0” and not “accept
H0”.
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Decision Table for Hypothesis Testing
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Null True
Null False
Fail to
reject null
Correct
Decision
Type II error
( )
Reject null
Type I error
()
Correct Decision
(Power)
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Controlling the probability of
conducting a type I error

Recall:




H0:  = 350 and H1:  > 350.
H0 is rejected if x is sufficiently large
Thus, a type I error is made if x  criticalv alue
when  = 350.
By properly selecting the critical value we can limit
the probability of conducting a type I error to an
acceptable level.
Critical value
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 = 350
x
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Steps in Hypothesis Testing
1. Establish hypotheses: state the null and
alternative hypotheses.
2. Determine the appropriate statistical test and
sampling distribution.
3. Specify the Type I error rate (
4. State the decision rule.
5. Gather sample data.
6. Calculate the value of the test statistic.
7. State the statistical conclusion.
8. Make a managerial decision.
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Testing the Population Mean When
the Population Standard Deviation is Known

Example



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A new billing system for a department store will be
cost- effective only if the mean monthly account is
more than $170.
A sample of 400 accounts has a mean of $178.
If accounts are approximately normally distributed
with
s = $65, can we conclude that the new system will be
cost effective?
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Testing the Population Mean (s is Known)

Example 11.1 – Solution


The population of interest is the credit accounts at
the store.
We want to know whether the mean account for
all customers is greater than $170.
H1 :  > 170

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The null hypothesis must specify a single value
of the parameter ,
H0 :  = 170
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Approaches to Testing

There are two approaches to test whether the
sample mean supports the alternative
hypothesis (H1)


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The rejection region method is mandatory for
manual testing (but can be used when testing is
supported by a statistical software)
The p-value method which is mostly used when a
statistical software is available.
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The Rejection Region Method
The rejection region is a range of values
such that if the test statistic falls into that
range, the null hypothesis is rejected in
favor of the alternative hypothesis.
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The Rejection Region Method
for a Right - Tail Test
Example 11.1 – solution continued
• Define a critical value xL for x that is just large enough
to reject the null hypothesis.
• Reject the null hypothesis if
x  xL
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Determining the Critical Value for the
Rejection Region


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Allow the probability of committing a Type I
error be  (also called the significance
level).
Find the value of the sample mean that is just
large enough so that the actual probability of
committing a Type I error does not exceed 
Watch…
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Determining the Critical Value –
for a Right – Tail Test
Example – solution continued
z 

 x  170
x L  170
65
400
xL
x
P(commit a Type I error) = P(reject H0 given that H0 is true)
= P( x
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 xLgiven that H0 is true)… is allowed to be 
Since P(Z  Z )   we have:
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Determining the Critical Value –
for a Right – Tail Test
Example – solution continued
 = 0.05
 x  170
xL
z 
x L  170
65
400
65
x L  170  z 
.
400
If w e select   0.05, z .05  1.645.
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65
x L  170  1.645
 175.34.
400
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Determining the Critical value
for a Right - Tail Test
Re ject the null hypothesis if
x  175.34
Conclusion
Since the sample mean (178) is greater than
the critical value of 175.34, there is sufficient
evidence to infer that the mean monthly
balance is greater than $170 at the 5%
significance level.
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The standardized test statistic

Instead of using the statistic
the standardized value z.
z

we can use
x 
s
n
Then, the rejection region becomes
z  z
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x,
One tail test
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The standardized test statistic

Example - continued



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We redo this example using the standardized
test statistic.
Recall: H0:  = 170
H1:  > 170
Test statistic:
x   178  170
z

 2.46
s n 65 400
Rejection region: z > z.05  1.645.
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The standardized test statistic

Example - continued
Re ject the null hypothesis if
Z  1.645
Conclusion
Since Z = 2.46 > 1.645, reject the null
hypothesis in favor of the alternative
hypothesis.
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P-value Method

The p-value provides information about the
amount of statistical evidence that supports the
alternative hypothesis.
– The p-value of a test is the probability of observing a
test statistic at least as extreme as the one computed,
given that the null hypothesis is true.
–虛無假設成立，獲得檢定統計量及更極端數值之機率。
–Let us demonstrate the concept on Example
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P-value Method
The probability of observing a
test statistic at least as extreme as 178,
given that  = 170 is…
P( x  178 when   170)
178  170
 P( z 
)
65 400
 P( z  2.4615)  .0069
 x  170
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x  178
The p-value
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Interpreting the p-value
Because the probability that the sample mean
will assume a value of more than 178 when 
= 170 is so small (.0069), there are reasons to
believe that
 > 170.
Note how the event
x  178 is rare under H0
when  x  170, but...
…it becomes more
probable under H1,
when  x  170
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H0 :  x  170
H1 :  x  170
x  178
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Interpreting the p-value
We can conclude that the smaller the p-value
the more statistical evidence exists to support the
alternative hypothesis.
H0 :  x  170
H1 :  x  170
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x  178
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The p-value and the Rejection Region
Methods

The p-value can be used when making decisions based
on rejection region methods as follows:
Define the hypotheses to test, and the required significance
level α.
 Perform the sampling procedure, calculate the test statistic
and the p-value associated with it.
 Compare the p-value to α. Reject the null hypothesis only if
p-value < α; otherwise, do not reject the null hypothesis.

 = 0.05
The p-value
 x  170
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x L  175.34
x  178
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Conclusions of a Test of Hypothesis
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
If we reject the null hypothesis, we conclude
that there is enough evidence to infer that the
alternative hypothesis is true.

If we do not reject the null hypothesis, we
conclude that there is not enough statistical
evidence to infer that the alternative hypothesis
is true.
The alternative hypothesis
is the more important
one. It represents what
we are investigating. Jia-Ying Chen
Example 2

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If a hypothesis is not rejected at the 0.10 level of
significance, it:
a.must be rejected at the 0.05 level.
b.may be rejected at the 0.05 level.
c.will not be rejected at the 0.05 level.
d.must be rejected at the 0.025 level.
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Example 3

Calculate the value of the test statistic, set up
the rejection region, determine the p-value,
interpret the result, and draw the sampling
distribution.
H 0 :   15
H1 :   15
s  2, n  25, x  14.3,   0.1
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Solution

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Rejection region z<z0.1=-1.28
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Example 4

A business student claims that on average an MBA students is
required to prepare more than five cases per week. To
examine the claim, a statistics professor asks a random
sample of 10 MBA students to report the number of cases
they prepare weekly. The results are exhibited here. Can the
professor conclude at the 5% significance level that the claim
is true, assuming that the number of cases is normally
distributed with a standard deviation of 1.5?
2 7 4 8 9 5 11 3 7 4
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Solution
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Example 5

A random sample of 12 second-year university students
enrolled in a business statistics course was drawn. At
the course’s completion, each student was asked how
many hours he or she spent doing homework in
statistics. The data are listed here. It is known that the
population standard deviation is σ=8. The instructor has
recommended that students devote 3 hours per week
for the duration of the 12-week semester, for a total of
36 hours. Test to determine whether there is evidence
that the average student spent less than the
recommend amount of time. Compute the p-value of
the test.
31 40 26 30 36 38 29 40 38 30 35 38
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Solution
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