时间序列图

課本41頁之公式為兩平均數差之假設檢定所需之樣本 數公式。
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Construction of the C.I.
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Xi X
S 2 i1
n 1
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Central Limit Theorem
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假設檢定(Hypothesis Testing)
“A person is innocent until proven guilty beyond a reasonable doubt.” 在沒有充分證據證明其犯罪之前, 任何人皆是清白的.
檢定結果可能為
Type I Error(a): Reject H0 while H0 is true. Type II Error(b): Fail to reject H0 while H0 is false.
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HH01
: :
無辜(Innocent) 有罪(Guilty)
Z0

x
s
/
0
n
6. Reject H0 if Z0 > 1.96 or Z0 < -1.96 (because Za/2 = Z0.025 = 1.96) 7. Computations:
Z0

51.3 50 2 / 25

3.25
8. Conclusions: Since Z0 = 3.25 > 1.96, we reject H0: = 50 at the 0.05 level of significance. We conclude that the mean burning rate differs from 50 cm/s, based on a sample of 25 measurements. In fact, there is string evidence that the mean burning rate exceeds 50 cm/s.
We therefore do not know the probability of committing a type II error (b).
Two ways of making conclusion: 1. Reject H0 2. Fail to reject H0, (Do not say accept H0) or there is not enough evidence to reject H0.
Use to performing sample size or type II error calculations.
The parameter d is defined as:
d | 0 | |d |
s
s
so that it can be used for all problems regardless of the values of 0 and s.
假設檢定 H0: = 50 cm/s H1: 50 cm/s
Null Hypothesis (H0) Vs. Alternative Hypothesis (H1) One-sided and two-sided Hypotheses A statistical hypothesis is a statement about the
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The Sample Size (I)
Given values of a and d, find the required sample size n to achieve a particular level of b..
Since
b

Za / 2

s
d
/
பைடு நூலகம்
n



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期望值與變異數之公式
E(ax1 bx2 ) aE(x1) bE(x2 ) V (ax1) a2V (x1) V (x1 x2 ) V (x1) V (x2 ) 2Cov(x1, x2 ),
其中Cov(x1, x2 ) Ex1 1x2 2
若x1, x2獨立, 則Cov(x1, x2 ) 0, 且V (x1 x2 ) V (x1) V (x2 ) E(x1 x2 ) E(x1) E(x2 ) E(x1 x2 ) E(x1) E(x2 )
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Sample and Sampling
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P-Values in Hypothesis Tests
Where Z0 is the test statistic, and (z) is the standard normal cumulative function.
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The Jury finds the person
Innocent Guilty
The Defendant is
Innocent
Guilty
Type II Error
Type I Error
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Making Conclusions
We always know the risk of rejecting H0, i.e., a, the significant level or the risk.

X
s
0
n
~
N 0,1
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Example 8-2
Aircrew escape systems are powered by a solid propellant. The burning rate of this propellant is an important product characteristic. Specifications require that the mean burning rate must be 50 cm/s. We know that the standard deviation of burning rate is 2 cm/s. The experimenter decides to specify a type I error probability or significance level of α = 0.05. He selects a random sample of n = 25 and obtains a sample average of the burning rate of x = 51.3 cm/s. What conclusions should be drawn?
these into the equation for the test statistic, and compute that value.
8. Decide whether or not H0 should be rejected and report
that in the problem context.
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1. The parameter of interest is , the meaning burning rate.
2. H0: = 50 cm/s 3. H1: 50 cm/s 4. a = 0.05
5. The test statistics is:
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General Procedure for Hypothesis Testing
1. From the problem context, identify the parameter of interest.
2. State the null hypothesis, H0. 3. Specify an appropriate alternative hypothesis, H1. 4. Choose a significance level a. 5. State an appropriate test statistic. 6. State the rejection region for the statistic. 7. Compute any necessary sample quantities, substitute
parameters of one or more populations.
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About Testing
Critical Region Acceptance Region Critical Values
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Errors in Hypothesis Testing
母體平均數(m ) = 隨機變數之期望值 E(X) 母體變異數(s 2) = 隨機變數之變異數 V(X)
E
(
X
)



All x
xf (x)dx, xp(x),
continuous discrete
V (X ) E x 2
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Significant Level (a)
a = P(type I error) = P(reject H0 while H0 is true)
n = 10, s = 2.5 s/n = 0.79
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If X1, X 2,, X n are samples drawn from a distribution
with mean and variance s 2, then X ~ N ,s 2 n .
Therefore, if H0 is true ( 0 ), then
Z0
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點估計(Point Estimation)
以抽樣得來之樣本資料, 依循某一公式計算出單一數值, 來估計母體參數, 稱為點估計.
好的點估計公式之條件:
不偏性
最小變異
常用之點估計:
母體平均數(m)
X Xi
n
母體變異數(s 2)
n
~
點圖 (Dot Diagram)
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直方圖 (Histogram)
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盒形圖 (Box Plot)
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時間序列圖 (Time Series Plot)
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期望值與變異數之公式
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Inference on the Mean of a Population -Variance Known
H0: = 0 H1: 0 , where 0 is a specified constant.
Sample mean is the unbiased point estimator for population mean.
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The Power of a Statistical Test
Power = 1 - b Power = the sensitivity of a statistical test
Za / 2

s
d
/
n


Za / 2

s
d
/
n
w hend
0
Let b Zb
d
Then, Zb

Za / 2

s
/
n
n
Za /2 Zb
d2
2s 2
whered 0
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The Operating Characteristic Curves - Normal test (z-test)