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統計學: 應用與進階
第13 章: 假設檢定
假設檢定的基本觀念
 如何執行假設檢定?
 假設檢定程序
 檢定的p-值
 誤差機率與檢定力
 檢定力函數

One Population Tests
One
Population
Mean
Proportion
Variance
Z Test
t Test
Z Test
c2 Test
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
t Test for Mean ( Unknown)
1. Assumptions
• Population is normally distributed
• If not normal, only slightly skewed & large sample (n
 30) taken
2. Parametric test procedure
3. t test statistic
X 
t
S
n
Two-Tailed t Test Finding Critical t Values
Given: n = 3;  = .10
df = n - 1 = 2

 /2 = .05


 /2 = .05
Critical Values of t Table
(Portion)
v
t .10
t .05
t .025
1 3.078 6.314 12.706
2 1.886 2.920 4.303
-2.920 0 2.920 t
3 1.638 2.353 3.182

例子: 實際檢定
設
且
令
 試在顯著水準為 α 下, 檢定

未知
例子: 實際檢定

在
未知的情況下, 我們所選的
根據
 接下來, 我們求算

若
我們知道
,則拒絕H0
為
Two-Tailed t Test Example
Does an average box of
cereal contain 368 grams of
cereal? A random sample
of 36 boxes had a mean of
372.5 and a standard
deviation of 12 grams. Test
at the .05 level of
significance.
368 gm.
Two-Tailed t Test Solution
•
•
•
•
•
H0:  = 368
Ha:   368
 = .05
df = 36 - 1 = 35
Critical Value(s):
Reject H0
Reject H0
.025
.025
-2.030
0 2.030
t
Test Statistic:
Decision:
Conclusion:
Two-Tailed t Test Solution
Test Statistic:
X   372.5  368
t

  2.25
S
12
n
36
Decision:
Reject at  = .05
Conclusion:
There is evidence population
average is not 368
Two-Tailed t TestThinking Challenge
You work for the FTC. A
manufacturer of detergent claims
that the mean weight of detergent
is 3.25 lb. You take a random
sample of 64 containers. You
calculate the sample average to
be 3.238 lb. with a standard
deviation of .117 lb. At the .01
level of significance, is the
manufacturer correct?
3.25 lb.
Two-Tailed t Test Solution*
•
•
•
•
•
H0:  = 3.25
Ha:   3.25
  .01
df  64 - 1 = 63
Critical Value(s):
Reject H 0
Reject H0
.005
.005
-2.656
0 2.656
t
Test Statistic:
Decision:
Conclusion:
Two-Tailed t Test Solution*
Test Statistic:
X   3.238  3.25
t

  .82
S
.117
n
64
Decision:
Do not reject at  = .01
Conclusion:
There is no evidence
average is not 3.25
One-Tailed t Test
of Mean ( Unknown)
One-Tailed t Test Example
Is the average capacity of
batteries at least 140 amperehours? A random sample of 20
batteries had a mean of 138.47
and a standard deviation of 2.66.
Assume a normal distribution.
Test at the .05 level of
significance.
One-Tailed t Test Solution
•
•
•
•
•
H0:  = 140
Ha:  < 140
 = .05
df = 20 - 1 = 19
Critical Value(s):
Test Statistic:
Decision:
Reject H0
Conclusion:
.05
-1.729
0
t
One-Tailed t Test Solution
Test Statistic:
X   138.47  140
t

  2.57
S
2.66
n
20
Decision:
Reject at  = .05
Conclusion:
There is evidence population
average is less than 140
One-Tailed t Test Thinking Challenge
You’re a marketing analyst for
Wal-Mart. Wal-Mart had teddy
bears on sale last week. The
weekly sales ($ 00) of bears sold
in 10 stores was:
8 11 0 4 7 8 10 5 8 3
At the .05 level of significance, is
there evidence that the average
bear sales per store is more than
5 ($ 00)?
One-Tailed t Test Solution*
•
•
•
•
•
H0:  = 5
Ha:  > 5
 = .05
df = 10 - 1 = 9
Critical Value(s):
Test Statistic:
Decision:
Reject H0
.05
0 1.833
Conclusion:
t
One-Tailed t Test Solution*
Test Statistic:
X   6.4  5
t

  1.31
S
3.373
n
10
Decision:
Do not reject at  = .05
Conclusion:
There is no evidence
average is more than 5
檢定誤差
在檢定的過程中, 我們被迫要在虛無假設與對立假
設之間做出選擇, 一如法官必須在有罪與無罪的判
決上做出決策
 法官可能會冤枉好人, 將一名無罪的人送入監獄;
法官亦可能會錯縱罪人, 將一名有罪之人當庭開釋
 同理, 在假設檢定的過程中, 我們可能會在H0為真
的情況下拒絕H0, 亦有可能在H0 為假的情況下, 作
出無法拒絕H0 的決策

檢定誤差
無法拒絕H0
拒絕H0
H0 為真
正確決策
型一誤差
H0 為假
型二誤差
正確決策
型一誤差

犯下型一誤差的機率我們以 α 表示
α = P(型一誤差) = P(拒絕H0 | H0為真)

我們亦稱顯著水準 α 為該檢定的「檢定範圍」或
是「檢定大小」(size of the test)
型二誤差
犯下型二誤差的機率我們以 β 表示
β = P(型二誤差) = P(無法拒絕H0 | H0為假).
 β 取決於H0 為假, 也就是取決於母體參數的真值
 如果我們以 θ 代表母體參數的真值, 則 β 就是 θ 的
函數 β = β (θ)
 令 π(θ) = 1 − β (θ) = P(拒絕H0 | H0為假) 代表了我
們的檢定正確地拒絕了不為真的H0,
 我們稱 π(θ) 為檢定的檢定力(power)

型一誤差的機率(α) 與型二誤差的機率(β)
檢定力

再以藥廠為例, 依照之前提及, 在 α = 0.01 的顯著
水準下, 拒絕域為
RR ={拒絕H0, 當
≥0.08 }
檢定力: 如果母體參數μ 的真值為0.09

β = P(型二誤差) = P(無法拒絕H0 | H0為假).

而檢定力 π = 1 − β = 1 − 0.2776 = 0.7224
Finding Power Step 1

Hypothesis:
H0: 0  368
Ha: 0 < 368
n

15
25

Reject H0
 = .05
Do Not
Draw
Reject H0
0 = 368
X
Finding Power Steps 2 & 3

Hypothesis:
H0: 0  368
Ha: 0 < 368
n


Reject H0
Do Not
Draw
Reject H0
15
25
 = .05
0 = 368
‘True’
Situation:
a = 360 (Ha)

Specify

Draw
1-
a = 360

X
X
Finding Power Step 4

Hypothesis:
H0: 0  368
Ha: 0 < 368
n


Reject H0
Do Not
Draw
Reject H0
15
25
 = .05
0 = 368
‘True’
Situation:
a = 360 (Ha)

Specify
X

15
 368  1.64
n
25
 363.065

X L  0  Z
Draw

1-
a = 360
363.065
X
Finding Power Step 5

n
Hypothesis:
H0: 0  368
Ha: 0 < 368


Reject H0
Do Not
Draw
Reject H0
15
25
 = .05
0 = 368
‘True’
Situation:
a = 360 (Ha)

Draw


Specify
Z Table
X

15
 368  1.64
n
25
 363.065
X L  0  Z
 = .154

1- =.846
a = 360
363.065
X
檢定力函數
根據以上的討論, 我們知道檢定力是母體參數真值
的函數, 故又稱檢定力函數
 檢定力函數的定義為:
π(m) = P(拒絕H0|θ = m)
 因此, π(θ0) = P(拒絕H0| θ = θ0) = , 亦即,在H0 為真
的情況下, π(θ0) 就是顯著水準, 就是犯型一誤差的
機率 α
 如果H1 為真, 則 π(m) 就是該檢定的檢定力,就是不
犯型二誤差的機率1 − β(m)

α 與 β 間的抵換關係
檢定力函數
Power Curves
Power
H0:  0
Power H0:  0
Possible True Values for a
Power
Possible True Values for a
H0:  =0
Possible True Values for a
 = 368 in
Example
誤差機率與檢定力
α + β 通常被當作衡量檢定好壞的指標。我們希望
α + β 越小越好。此外, 注意到 α + β 的機率值可以
大於1
 雖然我們希望 α + β 越小越好, 但是在樣本數 n 固
定的情況下, α 越小則 β 越大; 反之, β 越小則 α 越
大。亦即 α 與 β 之間存在著抵換關係(trade-off)
 如果我們可以增加樣本數n, 則 α 與 β 將會隨著n
增加而同時減少。

 &  Have an Inverse Relationship
You can’t reduce both
errors simultaneously!


Factors Affecting 
1. True value of population parameter
•
Increases when difference with hypothesized
parameter decreases
2. Significance level, 
•
Increases when decreases
3. Population standard deviation, 
•
Increases when  increases
4. Sample size, n
•
Increases when n decreases
One Population Tests
One
Population
Mean
Proportion
Variance
Z Test
t Test
Z Test
c2 Test
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
Chi-Square (c2) Test
for Variance
1. Tests one population variance or standard deviation
2. Assumes population is approximately normally
distributed
3. Null hypothesis is H0: 2 = 02
4. Test statistic
c 
2
(n 1)  S
2
0
2
Sample variance
Hypothesized pop. variance
Chi-Square (c2) Distribution
Population
Select simple random
sample, size n.
Compute s 2


Sampling Distributions
for Different Sample
Sizes
Compute c2 = (n-1)s 2 /2
0
Astronomical number
of c2 values
1
2
3 c2
Finding Critical Value Example
What is the critical c2 value given:
Ha: 2 > 0.7
Reject
n=3
 =.05?
 = .05
df = n - 1 = 2
c2 Table
(Portion)
0
DF .995
1
...
2 0.010
5.991
c2
Upper Tail Area
…
.95
…
… 0.004
…
… 0.103
…
.05
3.841
5.991
Finding Critical Value Example
What is the critical c2 value given:
Ha: 2 < 0.7
n=3
What do you do
 =.05?
if the rejection
region is on the
left?
Finding Critical Value Example
What is the critical c2 value given:
Ha: 2 < 0.7
Upper Tail Area
Reject H0
for Lower Critical
n=3
Value = 1-.05 = .95

=
.05
 =.05?
df = n - 1 = 2
c2 Table
(Portion)
0 .103
DF .995
1
...
2 0.010
c2
Upper Tail Area
…
.95
…
… 0.004
…
… 0.103
…
.05
3.841
5.991
Chi-Square (c2) Test Example
Is the variation in boxes of
cereal, measured by the
variance, equal to 15 grams?
A random sample of 25
boxes had a standard
deviation of 17.7 grams.
Test at the .05 level of
significance.
Chi-Square (c2) Test Solution
•
•
•
•
•
H0: 2 = 15
Ha: 2  15
 = .05
df = 25 - 1 = 24
Critical Value(s):
Test Statistic:
Decision:
 /2 = .025
Conclusion:
0 12.401
39.364
c2
Chi-Square (c2) Test Solution
Test Statistic:
c 
2
(n  1) S

2
0
2
(25  1) 17.7

2
15
= 33.42
Decision:
Do not reject at  = .05
Conclusion:
There is no evidence
2 is not 15
2
例子: 實際檢定

某工廠製作某型的塑膠管，經長期測定其直徑之
標準差為0.09毫米，且設母體為常態分布。最近僱
用一批新操作員，抽取10個產品測定其直徑之標
準差為0.12毫米。在α=0.05下，此批新的操作員
是否可以勝任此工作？
Chi-Square (c2) Test Solution
•
•
•
•
•
H0: 2 = 0.092
Ha: 2 > 0.092
 = .05
df = 10 - 1 = 9
Critical Value(s):
Test Statistic:
Decision:
 = .005
Conclusion:
0 12.401
16.916
c2
Chi-Square (c2) Test Solution
Test Statistic:
c 
2
(n  1) S

2
0
2
(10  1)(0.12)

2
(0.09)
 16
Decision:
Do not reject at  = .05
Conclusion:
There is no evidence
2 is not 0.081
2
Thinking Challenge
How would you try to answer these questions?
• Who gets higher grades:
males or females?
• Which program is faster to
learn: Word or Excel?
Target Parameters
Difference between Means
1– 2
Difference between
Proportions
p1– p2
Ratio of Variances
( 1 )2
2
( 2 )
Possible Estimator
•
( X 1  X 2 ) for ( 1  2 )
( p1  p2 ) for ( p1  p2 )
2
1
2
2
S
S

for

2
1
2
2
Test Statistics
• What are the possible test statistics?
• Do we know the sampling distribution?
Two Population Inference
Two
Populations
Mean
Paired
Proportion
Variance
Z
F
Indep.
Z
t
t
(Large
sample)
(Small
sample)
(Paired
sample)
Comparing
Two Independent Means
Two Population Inference
Two
Populations
Mean
Paired
Proportion
Variance
Z
F
Indep.
Z
t
t
(Large
sample)
(Small
sample)
(Paired
sample)
Sampling Distribution
1
Population
1
2
2
1
Select simple random
sample, n1. Compute
X1
Compute X1 – X2
for every pair of
samples
Astronomical number
of X1 – X2 values
Population
2
Select simple random
sample, n2. Compute
X2
Sampling
Distribution
1 - 2
One Population Case
Base on CLT：
2
σ1
X 1～N(1 ,
)
n1
2
σ2
X 2～N( 2 ,
)
n2
Two Population Case (independent)
•
So far we do not know the sampling distribution
of X 1  X 2 , If these two populations are
independent, then we have
X 1  X 2 should be normal i. i. d
E ( X )  E ( X 1  X 2 )  1  2
V ( X )  V ( X1  X 2 )  V ( X1 )  V ( X 2 ) 
X1  X 2
N ( 1  2 ,
 12
n1

 22
n2
)
 12
n1

 22
n2
Large-Sample Inference for
Two Independent Means
Two Population Inference
Two
Populations
Mean
Paired
Proportion
Variance
Z
F
Indep.
Z
t
t
(Large
sample)
(Small
sample)
(Paired
sample)
Conditions Required for Valid LargeSample Inferences about μ1 – μ2
Assumptions
• Independent, random samples
• Can be approximated by the normal distribution when n1 
30 and n2  30
Large-Sample Confidence Interval for
μ1 – μ2 (Independent Samples)
Confidence Interval
X
1
 X 2   Z 2

2
1
n1


2
2
n2
Hypotheses for Means of Two
Independent Populations
Research Questions
Hypothesis
No Difference
Any Difference
Pop 1 Pop 2 Pop 1  Pop 2
Pop 1 < Pop 2 Pop 1 > Pop 2
H0
1  2  0 1  2  0
1  2  0
Ha
1  2  0 1  2  0
1  2  0
Large-Sample Test for μ1 – μ2
(Independent Samples)
Two Independent Sample Z-Test Statistic
z
( x1  x2 )  ( 1  2 )
 12
n1

 22
n2
Hypothesized
difference
Large-Sample Confidence Interval
Example
You’re a financial analyst for Charles Schwab.
You want to estimate the difference in dividend
yield between stocks listed on NYSE and
NASDAQ. You collect the following data:
NYSE
NASDAQ
Number
121
125
Mean
3.27
2.53
Std Dev
1.30
1.16
What is the 95% confidence interval
for the difference between the mean
dividend yields?
© 1984-1994 T/Maker Co.
Large-Sample Confidence Interval
Solution
X
1
 X 2   Z 2
 12
n1

 22
n2
(1.3) 2 (1.16) 2

(3.27  2.53)  1.96
125
121
.43  1  2  1.05
Hypotheses for Means of Two
Independent Populations
Research Questions
Hypothesis
No Difference
Any Difference
Pop 1 Pop 2 Pop 1  Pop 2
Pop 1 < Pop 2 Pop 1 > Pop 2
H0
1  2  0 1  2  0
1  2  0
Ha
1  2  0 1  2  0
1  2  0
Large-Sample Test for μ1 – μ2
(Independent Samples)
Two Independent Sample Z-Test Statistic
z
( x1  x2 )  ( 1  2 )
 12
n1

 22
n2
Hypothesized
difference
Large-Sample Test
Example
You’re a financial analyst for Charles Schwab.
You want to find out if there is a difference in
dividend yield between stocks listed on NYSE
and NASDAQ. You collect the following data:
NYSE
NASDAQ
Number
121
125
Mean
3.27
2.53
Std Dev
1.30
1.16
Is there a difference in average
yield ( = .05)?
© 1984-1994 T/Maker Co.
Large-Sample Test
Solution
•
•
•
•
•
H0: 1 - 2 = 0 (1 = 2)
Ha: 1 - 2  0 (1  2)
  .05
n1= 121 , n2 = 125
Critical Value(s):
Reject H0
Reject H0
.025
-1.96 0 1.96
.025
z
Large-Sample Test
Solution
• Test Statistic:
(3.27  2.53)  0
z
 4.69
1.698 1.353

121
125
• Decision:
Reject at  = .05
• Conclusion:
There is evidence of a difference in means
Large-Sample Test
Thinking Challenge
You’re an economist for the Department of Education.
You want to find out if there is a difference in
spending per pupil between urban and rural high
schools. You collect the following:
Urban
Rural
Number
35
35
Mean
$ 6,012
$ 5,832
Std Dev
$ 602
$ 497
Is there any difference in population
means ( = .10)?
Large-Sample Test
Solution*
•
•
•
•
•
H0: 1 - 2 = 0 (1 = 2)
Ha: 1 - 2  0 (1  2)
  .10
n1 = 35 , n2 = 35
Critical Value(s):
Reject H 0
.05
-1.645
Reject H0
.05
0 1.645
z
Large-Sample Test
Solution*
• Test Statistic:
z
(6012  5832)  0
2
602 497

35
35
2
 1.36
• Decision:
Do not reject at  = .10
• Conclusion:
There is no evidence of a difference in means
兩獨立樣本之區間估計
若有兩母體, 其均數分別為 與 , 變異數為
與
 我們有興趣的參數分別為兩均數的差
或是兩母體變異之比值
 我們可以由每一個母體抽出大小分別為n1 與n2 的
兩獨立隨機樣本, 以求得
或是
的區
間估計式
 譬如說, 不同統計班的平均成績是否有差異

例一: 當
與
已知, 且來自常態母體時,
的區間估計式
為兩獨立

隨機樣本


則
的100 · (1 − α )% 區間估計式為