Transcript 投影片 1

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

Nonstatistical Hypothesis Testing
A criminal trial is an example of hypothesis testing
without the statistics.
In a trial a jury must decide between two hypotheses.
The null hypothesis is
H0: The defendant is innocent
The alternative hypothesis or research hypothesis is
H1: The defendant is guilty
The jury does not know which hypothesis is true. They
must make a decision on the basis of evidence presented.
Nonstatistical Hypothesis Testing
In the language of statistics convicting the defendant is
called
rejecting the null hypothesis in favor of the
alternative hypothesis.
That is, the jury is saying that there is enough evidence
to conclude that the defendant is guilty (i.e., there is
enough evidence to support the alternative hypothesis).
Nonstatistical Hypothesis Testing
If the jury acquits it is stating that
there is not enough evidence to support the
alternative hypothesis.
Notice that the jury is not saying that the defendant is
innocent, only that there is not enough evidence to
support the alternative hypothesis. That is why we never
say that we accept the null hypothesis.
Nonstatistical Hypothesis Testing
There are two possible errors.
A Type I error occurs when we reject a true null
hypothesis. That is, a Type I error occurs when the jury
convicts an innocent person.
A Type II error occurs when we don’t reject a false null
hypothesis. That occurs when a guilty defendant is
acquitted.
Nonstatistical Hypothesis Testing
The probability of a Type I error is denoted as α (Greek
letter alpha). The probability of a type II error is β
(Greek letter beta).
The two probabilities are inversely related. Decreasing
one increases the other.
Nonstatistical Hypothesis Testing
5. Two possible errors can be made.
Type I error: Reject a true null hypothesis
Type II error: Do not reject a false null hypothesis.
P(Type I error) = α
P(Type II error) = β
Decision Results
H0: Innocent
Jury Trial
Actual Situation
Verdict
Innocent
Guilty
Innocent
Guilty
H0 Test
Actual Situation
Decision
Correct
Error
Accept
H0
Error
Correct
Reject
H0
H0 True
H0
False
1–
Type II
Error
()
Type I Power
Error () (1 – )
Nonstatistical Hypothesis Testing
In our judicial system Type I errors are regarded as
more serious. We try to avoid convicting innocent
people. We are more willing to acquit guilty people.
We arrange to make α small by requiring the
prosecution to prove its case and instructing the jury to
find the defendant guilty only if there is “evidence
beyond a reasonable doubt.”
Nonstatistical Hypothesis Testing
The critical concepts are theses:
1. There are two hypotheses, the null and the alternative
hypotheses.
2. The procedure begins with the assumption that the
null hypothesis is true.
3. The goal is to determine whether there is enough
evidence to infer that the alternative hypothesis is
true.
4. There are two possible decisions:
Conclude that there is enough evidence to support the
alternative hypothesis.
Conclude that there is not enough evidence to support the
alternative hypothesis.
假設檢定

假設(hypothesis) 就是我們對於母體參數的宣稱


我宣稱本校學生的平均身高為166 公分,
我宣稱本校學生的平均智商為130 分
假設檢定(hypothesis testing) 的目的就是要對這些
宣稱提供統計上的檢驗, 以統計的檢定方法來推論
假設的真偽
 決策為拒絕(reject) 該假設, 或是無法拒絕(fail to
reject) 該假設

假設檢定

對於未知的母體參數, 我們可以有各式各樣不同的
假設, 舉例來說,
[H1 ]
[H2 ]
[H3 ]
其中,
與
分別代表某固定常數
虛無假設與對立假設

a.
b.
在假設檢定中, 我們考慮兩個互斥的假設:
虛無假設(null hypothesis) 就是研究者所要檢定的假
設, 一般以H0 的符號代表。
對立假設(alternative hypothesis) 就是與虛無假設完
全相反的假設, 如果虛無假設不成立, 則對立假設
就為真, 一般以H1 或是HA 的符號代表。
虛無假設
What is tested
 Has serious outcome if incorrect decision made
 Always has equality sign: , , or 
 Designated H0 (pronounced H-oh)

對立假設

Opposite of null hypothesis

Always has inequality sign: ,, or 

Designated Ha
虛無假設與對立假設

舉例來說, 若虛無假設為

則對立假設可以是
簡單假設vs. 複合假設
如果假設中, 僅包含一個特定的假設值, 如 μ = 166,
則該假設稱作簡單假設(simple hypothesis)
 假設中, 可能的參數假設值不只一個, 則該假設稱
為複合假設(composite hypothesis), 如 μ > 166
 通常將虛無假設以簡單假設的方式呈現, 而對立假
設則為複合假設

Example

Test that the population mean is not 3

Steps:


State the question statistically (  3)
State the opposite statistically ( = 3)
—

Select the alternative hypothesis (  3)
—

Must be mutually exclusive & exhaustive
Has the , <, or > sign
State the null hypothesis ( = 3)
Example

Is the population average amount of TV viewing
different from 12 hours?

Steps

State the question statistically:  = 12

State the opposite statistically:   12

Select the alternative hypothesis: Ha:   12

State the null hypothesis: H0:  = 12
Example

Is the average cost per hat less than or equal to $20?

Steps

State the question statistically:   20

State the opposite statistically:   20

Select the alternative hypothesis: Ha:   20

State the null hypothesis: H0:   20
Example

Is the average amount spent in the bookstore greater
than $25?

Steps

State the question statistically:   25

State the opposite statistically:   25

Select the alternative hypothesis: Ha:   25

State the null hypothesis: H0:   25
假設檢定

一般來說, 透過統計上的檢定程序, 我們的決策為


拒絕H0 且接受H1 為真,
無法拒絕H0。
注意到我們不說「接受H0」的原因在於, 即使我們
找不到證據推翻H0, 並不代表H0 就是無庸置疑地為
真, 只不過是找不到充分證據來推翻H0 罷了。你或
許會在某些場合或是某些書上聽到或看到「接受
H0」的說法, 但是請記得當別人如此宣稱時, 並不代
表H0 是無庸置疑地為真(absolutely true)
Basic Idea
Sampling Distribution
It is unlikely
that we would
get a sample
mean of this
value ...
... therefore, we
reject the
hypothesis that
 = 50.
... if in fact this were
the population mean
20
 = 50
H0
Sample Means
Level of Significance
1. Probability
2. Defines unlikely values of sample statistic if null
hypothesis is true
•
Called rejection region of sampling
distribution
3. Designated (alpha)
•
Typical values are .01, .05, .10
4. Selected by researcher at start
Rejection Region (One-Tail Test)
Sampling Distribution
Level of Confidence
Rejection
Region

1–
Nonrejection
Region
Critical
Value
Ho
Value
Sample Statistic
Observed sample statistic
Rejection Region (One-Tail Test)
Sampling Distribution
Level of Confidence
Rejection
Region

1–
Nonrejection
Region
Ho
Value
Critical
Value
Observed sample statistic
Sample Statistic
Rejection Regions (Two-Tailed Test)
Sampling Distribution
Level of Confidence
Rejection
Region
Rejection
Region
1–
1/2 
1/2 
Nonrejection
Region
Critical
Value
Ho
Value
Sample Statistic
Critical
Value
Observed sample statistic
Rejection Regions (Two-Tailed Test)
Sampling Distribution
Level of Confidence
Rejection
Region
1/2 
Rejection
Region
1–
1/2 
Nonrejection
Region
Ho
Value
Critical
Critical
Value
Value
Observed sample statistic
Sample Statistic
Rejection Regions (Two-Tailed Test)
Sampling Distribution
Level of Confidence
Rejection
Region
1/2 
Rejection
Region
1–
1/2 
Nonrejection
Region
Ho
Sample Statistic
Critical Value Critical
Value
Value
Observed sample statistic
One Population Tests
One
Population
Mean
Proportion
Variance
Z Test
t Test
Z Test
c 2 Test
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
例子: 執行假設檢定

某藥廠宣稱只有5% 的人在服用過他們的新藥後,
出現嚴重副作用。食品藥物管理局對此感到懷疑,
決定應用統計方法予以檢定。在給予287 個受試者
服用此藥後, 以Xi = 1 代表第 i 個受試者有嚴重副
作用,Xi = 0 代表沒有嚴重副作用產生。顯而易見
地,
例子: 執行假設檢定

假設為
倘若我們發現, 在287 名受試者有25 位出現嚴重副
作用, 亦即, = 25/287 = 0.0871。根據這組樣本,我
們如何檢定藥廠的宣稱?
例子: 執行假設檢定
假設檢定的基本邏輯在於, 且讓我們暫時相信H0
為真
 在假設H0 : μ = 0.05 為真的情況下, 即使我們所抽
出來的每一組樣本的平均值 不會「剛好」等於
0.05, 卻應該會「相當接近」0.05
 換言之, 給定虛無假設成立的情況下, 樣本均數遠
大於0.05 的的可能性極低(very unlikely), 亦即, 在假
設H0 為真的情況下出現一個極端值的可能性將會
十分微小

例子: 執行假設檢定
因此, 如果這種「不太可能」的事件真的發生了,
我們就可以據此拒絕虛無假設
 簡而言之, 如果
的值太大, 大過於某個常數c, 我
們就拒絕虛無假設
 這樣的極端事件發生的機率要多小才算是「不太
可能」? 一般來說, 我們選取一個極小的機率, α ,
如0.10, 0.05 或是0.01, 來作為拒絕虛無假設的基礎

例子: 執行假設檢定
亦即, 根據虛無假設為真的機率分配下(稱之為虛
無分配), 的值大過於某個常數c, 且發生此極端事
件的條件機率P(
|H0為真) 非常小,我們就做
出拒絕虛無假設的決策
 此微小機率 α 通常被稱作顯著水準(significance
level)。假設檢定又被叫做顯著性檢定(test of
significance), 意指根據隨機樣本來決定是否足以顯
著地拒絕(具有充分證據拒絕)虛無假設。

例子: 執行假設檢定

因此, 統計上的「顯著」並不是指「數值」的大小,
而是指「機率」的大小。發生此極端事件的機率
小, 才稱此極端事件具顯著性
例子: 執行假設檢定

利用以上的例子說明, 則我們就是要找出一個臨界
值c 使得
亦即我們定義了一個機率微小事件: { > c}
 一旦我們找到 c 後, 且得知
, 則決策為:
{拒絕H0, 當
> c}
以上又稱為拒絕域(rejection region, RR)
例子: 執行假設檢定
值得一提的是, 我們的決策法則乃是樣本實現前所
確立下來的法則, 亦即, c 值是在樣本實現前所找出
來的臨界值, 這又是一個樣本實現前(ex ante) 的概
念
 至於拒絕與否的決策則是由樣本實現後的
與c
做比較

例子: 執行假設檢定
再者, P( > c|μ = 0.05) = 有兩大特徵
 第一, 這是一個樣本實現前(ex ante) 的機率
 第二, 這是一個條件機率, 受限於H0 : μ = 0.05 為真
的這個條件

例子: 執行假設檢定

我們回到藥廠的例子,
例子: 執行假設檢定

由於
根據CLT, 我們知道
例子: 執行假設檢定
例子: 執行假設檢定


則c 值為

若選取的 α = 0.01, 則
= 2.33, 且
故
例子: 執行假設檢定

亦即, 拒絕域為
RR ={拒絕H0, 當

≥0.08 }
在本例中, = = 0.0871 > 0.08 = c, 我們據此拒絕
H0 : μ = 0.05, 接受H1 : μ > 0.05
執行假設檢定的常用法
事實上, 我們有兩種方法執行假設檢定, 第一種方
法如上所示, 找出臨界值c。當
,則拒絕虛
無假設
 另一種方法則是呼應上一章區間估計式的建構。
我們可以建構一個與上一章相同的統計量

找出
的抽樣分配, 接著根據
的抽樣分配找出臨界值 使得
 最後, 求算
而決策法則為:
當
則拒絕H0

執行假設檢定的常用法
若
的抽樣分配已知, 稱之為實際檢定
若
的抽樣分配未知, 但符合CLT 的條件,則當
樣本夠大時,
的抽樣分配可用N(0, 1)予以近似,
我們稱之為近似檢定, 或叫做大樣本檢定
 底下我們列出假設檢定程序

假設檢定程序: θ 代表所欲檢定的母體參數
[步驟一] 設立虛無假設(H0) 與對立假設(H1)
 H0 : θ=θ0
 H1 : 三種可能
a.
b.
c.
θ > θ0 : 右尾檢定(right-tailed test, RTT)
θ < θ0 : 左尾檢定(left-tailed test, LTT)
θ ≠ θ0 : 雙尾檢定(two-tailed test, TTT)
假設檢定程序
[步驟二] 建構統計量
並找出
的
抽樣分配, (如標準常態分配, t 分配,
分配, F分配
等)。若實際抽樣分配未知, 但是可以應用CLT, 則
的抽樣分配可用標準常態分配予以近似
[步驟三] 選擇顯著水準, α
假設檢定程序
[步驟四] 根據
的抽樣分配或近似分配找出臨界值
,或是 ,並建構拒絕(rejection
region, RR)。
舉例來說, 如果 ∼ N(0, 1), 則其臨界值為
(右尾檢定), − (左尾檢定), 或是
(雙尾檢
定)。其拒絕域為
a.
b.
c.
拒絕H0, 當
拒絕H0, 當
拒絕H0, 當
[步驟五] 檢視
絕域並做出決策。
是否掉入拒
再以藥廠的例子來說明

根據CLT,

根據選取的 α = 0.01,
查表得知
, 則拒絕域為
RR ={拒絕H0, 當
}
藥廠的例子

由於μ = μ0 = 0.05,
,則
顯而易見地,
落入拒絕域,因此,
我們據此拒絕H0 : μ = 0.05
 注意到這是一個顯著水準近似於0.01 的大樣本檢
定
One Population Tests
One
Population
Mean
Proportion
Variance
Z Test
t Test
Z Test
c 2 Test
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
One-Sample Z Test for Proportion
1. Assumptions
• Random sample selected from a binomial population
• Normal approximation can be used if
npˆ 0  15 and nqˆ0  15
2.
Z-test statistic for proportion
pˆ  p0
Hypothesized population
Z
proportion
p0 q0
n
One-Proportion Z Test Example
The present packaging system
produces 10% defective
cereal boxes. Using a new
system, a random sample of
200 boxes had 11 defects.
Does the new system produce
fewer defects? Test at the .05
level of significance.
One-Proportion Z Test Solution
•
•
•
•
•
H0: p = .10
Ha: p < .10
 = .05
n = 200
Critical Value(s):
Test Statistic:
Decision:
Reject H0
.05
-1.645 0
Conclusion:
Z
One-Proportion Z Test Solution
Test Statistic:
11
 .10
pˆ  p0 200
Z

 2.12
p0 q0
.10  .90
200
n
Decision:
Reject at  = .05
Conclusion:
There is evidence new
system < 10% defective
One-Proportion Z Test Thinking Challenge
You’re an accounting manager.
A year-end audit showed 4% of
transactions had errors. You
implement new procedures. A
random sample of 500
transactions had 25 errors. Has
the proportion of incorrect
transactions changed at the .05
level of significance?
One-Proportion Z Test Solution*
•
•
•
•
•
H0: p = .04
Ha: p  .04
 = .05
n = 500
Critical Value(s):
Reject H 0
Reject H 0
.025
.025
-1.96 0 1.96 Z
Test Statistic:
Decision:
Conclusion:
One-Proportion Z Test Solution*
Test Statistic:
25
 .04
pˆ  p0 500
Z

 1.14
p0 q0
.04  .96
500
n
Decision:
Do not reject at  = .05
Conclusion:
There is evidence
proportion is not 4%
One Population Tests
One
Population
Mean
Proportion
Variance
Z Test
t Test
Z Test
 2 Test
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
Two-Tailed Z Test for Mean ( Known)
1. Assumptions
•
•
Population is normally distributed
If not normal, can be approximated by
normal distribution (n  30)
2. Alternative hypothesis has  sign
3. Z-Test Statistic
Z
X  x
x

X 

n
Two-Tailed Z Test for Mean Hypotheses
H0:= 0 Ha: ≠ 0
Reject H 0
Reject H


0
Z
Two-Tailed Z Test Finding Critical Z
What is Z given  = .05?
.500
- .025
.475

 =1

 = .025

Standardized Normal
Probability Table (Portion)
Z
.05
.06
.07
1.6 .4505 .4515 .4525
1.7 .4599 .4608 .4616
1.8 .4678 .4686 .4693
-1.96 0 1.96 Z

1.9 .4744 .4750 .4756
Two-Tailed Z Test Example
Does an average box of cereal
contain 368 grams of cereal?
A random sample of 25 boxes
showed x = 372.5. The
company has specified  to
be 15 grams. Test at the .05
level of significance.
368 gm.
Two-Tailed Z Test Solution
•
•
•
•
•
H0:  = 368
Ha:   368
  .05
n  25
Test Statistic:
Critical Value(s):
Reject H 0
Reject H 0
.025
.025
-1.96 0 1.96 Z
Decision:
Conclusion:
Two-Tailed Z Test Solution
Test Statistic:
X   372.5  368
Z

 1.50

15
n
25
Decision:
Do not reject at  = .05
Conclusion:
No evidence average
is not 368
Two-Tailed Z Test Thinking Challenge
You’re a Q/C inspector. You want to
find out if a new machine is making
electrical cords to customer specification:
average breaking strength of 70 lb. with
 = 3.5 lb. You take a sample of 36
cords & compute a sample mean of 69.7
lb. At the .05 level of significance, is
there evidence that the machine is not
meeting the average breaking strength?
Two-Tailed Z Test Solution*
•
•
•
•
•
H0:  = 70
Ha:   70
 =.05
n = 36
Critical Value(s):
Reject H 0
Reject H 0
.025
.025
-1.96 0 1.96 Z
Test Statistic:
Decision:
Conclusion:
Two-Tailed Z Test Solution*
Test Statistic:
X   69.7  70
Z

 .51

3.5
n
36
Decision:
Do not reject at  = .05
Conclusion:
No evidence average
is not 70
One-Tailed Z Test of Mean
( Known)
One-Tailed Z Test for Mean ( Known)
1. Assumptions
•
•
Population is normally distributed
If not normal, can be approximated by
normal distribution (n  30)
2. Alternative hypothesis has < or > sign
3. Z-test Statistic
Z
X  x
x

X 

n
One-Tailed Z Test for Mean Hypotheses
H0:= 0 Ha: < 0
H0:= 0 Ha: > 0
Reject H 0
Reject H0


0
Must be significantly
below 
Z
0
Z
Small values satisfy H0 .
Don’t reject!
One-Tailed Z Test Finding Critical Z
What Is Z given  = .025?
.500
- .025
.475

 =1

 = .025

Standardized Normal
Probability Table (Portion)
Z
.05
.06
.07
1.6 .4505 .4515 .4525
1.7 .4599 .4608 .4616
1.8 .4678 .4686 .4693
0 1.96 Z

1.9 .4744 .4750 .4756
One-Tailed Z Test Example
Does an average box of cereal
contain more than 368 grams
of cereal? A random sample of
25 boxes showed x = 372.5.
The company has specified  to
be 15 grams. Test at the .05
level of significance.
368 gm.
One-Tailed Z Test Solution
•
•
•
•
•
H0:  = 368
Ha:  > 368
 = .05
n = 25
Critical Value(s):
Test Statistic:
Decision:
Reject
.05
0 1.645 Z
Conclusion:
One-Tailed Z Test Solution
Test Statistic:
X   372.5  368
Z

 1.50

15
n
25
Decision:
Do not reject at  = .05
Conclusion:
No evidence average is
more than 368
One-Tailed Z Test Thinking Challenge
You’re an analyst for Ford. You
want to find out if the average miles
per gallon of Escorts is at least 32
mpg. Similar models have a
standard deviation of 3.8 mpg. You
take a sample of 60 Escorts &
compute a sample mean of 30.7
mpg. At the .01 level of
significance, is there evidence that
the miles per gallon is at least 32?
One-Tailed Z Test Solution*
•
•
•
•
•
H0:  = 32
Ha:  < 32
 = .01
n = 60
Critical Value(s):
Test Statistic:
Decision:
Reject
.01
-2.33 0
Conclusion:
Z
One-Tailed Z Test Solution*
Test Statistic:
X   30.7  32
Z

 2.65

3.8
n
60
Decision:
Reject at  = .01
Conclusion:
There is evidence average
is less than 32
三種常用假設檢定法
標準檢定統計量法： p-值(p-value)
在上一節假設檢定程序的討論中, 我們的決策法則
為: 決定一顯著水準 α 後, 找出 , 然後再比較
是否大於或小於
 這樣的作法有一個麻煩的地方, 那就是給定任何一
個不同的顯著水準 α , 我們就得查出相對應的
;
以及找出相對應的拒絕域
 一個有用的概念: p-值(p-value)
 所謂的p-值就是在H0 為真的情況下, 比觀測值至
少同樣極端之區域的機率

檢定的p-值
回到之前藥廠的例子, 我們知道 = 0.0871。
 給定H0 為真的情況下(μ = 0.05), 樣本均數
會大
於0.0871 的機率就是其p-值:
p-值

檢定的p-值
的方式表達, p-值就是

以

由於0.0019 很小, 代表在H0 為真的情況下, 我們觀
察到一件「幾乎不可能發生的事」, 因此, 這將使
我們懷疑原先的H0 為真的假設, 進而獲致拒絕H0
的決策
p-值在研究上相當便利好用, 對於研究者而言,p-值
讓我們不必再一個一個辛苦查表。許多的統計套
裝軟體都會幫我們計算出檢定的p-值
 你只需決定喜愛的顯著水準, 再比較p-值與 α 孰大
孰小即可。決策法則為:

a.
b.

拒絕H0, 當p-值≤ α
無法拒絕H0, 當p-值> α
舉例來說, 給定p-值=0.048。當 α = 0.05 時,我們拒
絕H0; 然而, 當 α = 0.01 時, 我們的決策為無法拒絕
H0
Two-Tailed Z Test p-Value Example
Does an average box of cereal
contain 368 grams of cereal? A
random sample of 25 boxes
showed x = 372.5. The
company has specified  to be
15 grams. Find the p-Value.
368 gm.
Two-Tailed Z Test p-Value Solution
X   372.5  368
Z

 1.50

15
n
25
0
1.50

Z
Z value of sample
statistic (observed)
Two-Tailed Z Test p-Value Solution
p-value is P(Z  -1.50 or Z  1.50)
1/2 p-Value
1/2 p-Value
.4332
-1.50 0

From Z table:
lookup 1.50
1.50


.5000
- .4332
.0668
Z
Z value of sample
statistic (observed)
Two-Tailed Z Test p-Value Solution
p-value is P(Z  -1.50 or Z  1.50) = .1336
1/2 p-Value
.0668
-1.50

1/2 p-Value
.0668
0
1.50
From Z table:
lookup 1.50


.5000
- .4332
.0668
Z
Z value of sample
statistic
Two-Tailed Z Test p-Value Solution
(p-Value = .1336)  ( = .05).
Do not reject H0.
1/2 p-Value = .0668
1/2 p-Value = .0668
Reject H0
Reject H0
1/2  = .025
1/2  = .025
-1.50 0
1.50
Test statistic is in ‘Do not reject’ region
Z
One-Tailed Z Test p-Value Example
Does an average box of cereal
contain more than 368 grams
of cereal? A random sample
of 25 boxes showed x = 372.5.
The company has specified 
to be 25 grams. Find the pValue.
368 gm.
One-Tailed Z Test p-Value Solution
X   372.5  368
Z

 1.50

15
n
25
0
1.50

Z
Z value of sample
statistic
One-Tailed Z Test p-Value Solution
p-Value is P(Z  1.50)

Use
alternative
hypothesis
to find
direction
p-Value
.4332

0
From Z table:
lookup 1.50
1.50


.5000
- .4332
.0668
Z
Z value of sample
statistic
One-Tailed Z Test p-Value Solution
p-Value is P(Z  1.50) = .0668

p-Value
.0668
Use
alternative
hypothesis
to find
direction
.4332

0
From Z table:
lookup 1.50
1.50


.5000
- .4332
.0668
Z
Z value of sample
statistic
One-Tailed Z Test p-Value Solution
(p-Value = .0668)  ( = .05).
Do not reject H0.
p-Value = .0668
Reject H0
 = .05
0
1.50
Test statistic is in ‘Do not reject’ region
Z
p-Value Thinking Challenge
You’re an analyst for Ford. You
want to find out if the average
miles per gallon of Escorts is at
least 32 mpg. Similar models
have a standard deviation of 3.8
mpg. You take a sample of 60
Escorts & compute a sample mean
of 30.7 mpg. What is the value of
the observed level of significance
(p-Value)?
p-Value Solution*
p-Value is P(Z  -2.65) = .004.
p-Value < ( = .01). Reject H0.

Use
alternative
hypothesis
to find
direction
p-Value
.004

.5000
- .4960
.0040
.4960
-2.65
Z value of sample
statistic

0

Z
From Z table:
lookup 2.65