Population proportion and sample proportion

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© 蘇蘇蘇 2007 蘇蘇蘇蘇 蘇蘇 () Page 1 Population proportion and Population proportion and sample proportion sample proportion • 蘇蘇蘇蘇蘇蘇蘇蘇蘇蘇蘇蘇蘇蘇蘇… 蘇蘇蘇蘇… 蘇蘇蘇 蘇蘇蘇蘇蘇蘇蘇蘇蘇蘇蘇蘇蘇 「」「」 count) 蘇蘇蘇蘇蘇 (proportion) • 蘇蘇蘇蘇蘇蘇蘇蘇蘇蘇蘇蘇蘇蘇蘇蘇蘇 蘇 蘇蘇蘇 「」 (a si ngle proportion) 蘇蘇 蘇蘇蘇蘇蘇蘇蘇 。一 蘇蘇蘇 蘇蘇蘇蘇蘇蘇蘇 一。

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Population proportion and sample proportion. 生活中很多的調查都僅問是否贊成 … 、是否支持 … ,然後計算「贊成」與「反對」的人數( count) 所佔之比例 (proportion) 。 本章要介紹如何用統計方法來推論單一的「比例」 (a single proportion) 。下一章將會介紹如何來推論一組比例的分配。. Population proportion and sample proportion. - PowerPoint PPT Presentation

Transcript of Population proportion and sample proportion

Page 1: Population proportion and sample proportion

© 蘇國賢 2007社會統計(上) Page 1

Population proportion and sample Population proportion and sample proportionproportion

• 生活中很多的調查都僅問是否贊成…、是否支持…,然後計算「贊成」與「反對」的人數( count) 所佔之比例(proportion) 。

• 本章要介紹如何用統計方法來推論單一的「比例」 (a single proportion) 。下一章將會介紹如何來推論一組比例的分配。

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Population proportion and sample Population proportion and sample proportionproportion

• 想要估計總統大選阿扁的得票率,即投票給阿扁的人佔所有投票者的比例,我們可以利用適當的抽樣方法取處樣本數為 n 的樣本,然後觀察樣本中支持阿扁的人數佔整個樣本n 的比例,即可得到樣本中的阿扁支持率,稱之為樣本比例。

• 如果我們知道樣本比例的抽樣分配,即樣本比例的期望值,變異數,及分配形狀,則可以用樣本比例來推估母體比例。

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Sampling Distribution of the Sampling Distribution of the Sample ProportionSample Proportion

• Let p denote the proportion of items in a population that possess a certain characteristic (unemployed, income below poverty level).

• To estimate p, we take a random sample of n observation from the population and count the number X of items in the sample that possess the characteristic.

• The sample proportion p^ = X/n is used to estimate the population proportion p.

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Sampling Distribution of the Sample Sampling Distribution of the Sample ProportionProportion

• 若一隨機試驗只有兩種課能的結果( X=1 支持阿扁 , X=0 不支持阿扁),若母體數總共為 N( 所有投票人),若母體中有 K 個人會投票給阿扁,則支持阿扁的母體比例 (population proportion) 為

• p = K/N (N= 母體個數, K= 支持阿扁總人數)

定義定義

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Sampling Distribution of the Sampling Distribution of the Sample ProportionSample Proportion

• 上次總統大選的有效投票數 12,664,393 (N)• 其中阿扁得 4,977,697 (K)• 母體比例為 39.30%

定義定義

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Sampling Distribution of the Sampling Distribution of the Sample ProportionSample Proportion

• 若母體 N 中隨機抽取 n 個元素為樣本,表為(X1, X2, …Xn) ,且 n 個樣本中有 k 個人支持阿扁,支持阿扁所佔的比例稱為樣本比例 (sample proportion) :

• (n= 樣本個數, k= 樣本個數)• k 為樣本中,支持阿扁 (X=1) 的個數總和。

定義定義

n

X

n

kp

n

ii

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Sampling Distribution of the Sampling Distribution of the Sample ProportionSample Proportion

• 在大選前,民調中心調查 1500 個樣本 (n=1500) ,其中有 573 人支持阿扁 (k=573) ,樣本支持比例為 38.2%

• 抽樣誤差為

定義定義

%1.12.383.39ˆ pp

隨著每一次樣本所抽取的對象不同,所計算出的樣本比例也會有差異,因此樣本比例本身為一隨機變數。

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The Bernoulli DistributionThe Bernoulli Distribution

• P(X=1) = p

• P(X=0) = (1-p)

• If we let q = 1- p, then the p.f of X can be written as follows:

定義定義

otherwise 0

1,0for )(

1 XqpXf

XX

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The Bernoulli DistributionThe Bernoulli Distribution

• E(X) = 1·p +0·q = p ( X的期望值等於母體比例)

• E(X2) =X2 f(x)=12·p+02·q = p

• Var(X) = E(X2) –[E(X)]2 =p-p2

=p(1-p) = p·q

定義定義

otherwise 0

1,0for )(

1 XqpXf

XX

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Sampling Distribution of the Sampling Distribution of the Sample ProportionSample Proportion

• The Normal Approximation Rule for Proportion: Let p denote the proportion of a population possessing some characteristics of interest. Take a random sample of n observations from the population. Let X denote the number of items in the sample possessing the characteristic. We estimate the population proportion p by the sample proportion p^=X/n. If np5, and nq 5, the random variable p^ has approximately a normal distribution with:

pq/n pq/n)p̂Var( )ˆ( p̂ ppE

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Sampling Distribution of the Sampling Distribution of the Sample ProportionSample Proportion

• 證明 )ˆ( ppE

)(1

)()ˆ( XEnn

XEpE

)](...)()([1

)...(1

21

21

n

n

XEXEXEn

XXXEn

n

npppp

n

pXE

]...[1

)(

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Sampling Distribution of the Sampling Distribution of the Sample ProportionSample Proportion

• 證明

)var(1

)var()ˆvar(2

Xnn

Xp

)]var(...)var()[var(1

)...(1

212

212

n

n

XXXn

XXXVarn

22]...[

1

)var(

n

npqpqpqpq

n

pqX

pq/n)p̂Var(

assume X1, X2…Xn independent

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Sampling Distribution of the Sampling Distribution of the Sample ProportionSample Proportion

• If the distribution of p^ is approximately normal, and

)1,0(~pq/n

ˆN

ppZ

pq/n pq/n)p̂Var( )ˆ( p̂ ppE

),(~ˆn

pqpNp

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例題例題• 假設這一次的大選會有 55% 的選民支持阿扁,假設我們任取 n=400 人的隨機樣本來預測阿扁的當選率,我們預測阿扁會輸的的機率為?

45.55.1,55.,400 qpn

5)45(.400,5)55(.400 nqnp

)00062.400

)45)(.55(.,55(.~ˆ

n

pqNp

2025.

05.

00062.

)55(.)50(.

Z

0228.)2()5.ˆ( ZPpP

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例題例題• Of your first 15 grandchildren, what is the chance there will be

more than 10 boys? (assume equal probability of male/female)• “more than 10 boys””the proportion of boys is more than

10/15”• Use the Normal Approximation Rule:

)15/)5)(.5(./,50(.~ˆ npqNp

29.1129.

5.1510

ˆ

SE

ppZ

099.)29.1()15/10ˆ( ZPpP

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Confidence intervals for proportions Confidence intervals for proportions (large samples)(large samples)

we know that p^ ~N(p, pq/n) , where q = 1-p and np5 and nq 5) ≧ ≧

)1,0(~/

ˆN

npq

ppZ

)(1 2/2/ zZzP

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Value of ZValue of Zαα

• P(Z z≧ α/2) =α/2 • P(Z -z≦ α/2) =α/2 • P(-zα/2 Z z≦ ≦ α/2) =(1-α)

0.399

1.338104

f x( )

44 x4 3 2 1 0 1 2 3 4

0

0.2

0.4

α/2

1-α/2-α/2

=1-α

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Confidence intervals for proportions Confidence intervals for proportions (large samples)(large samples)

)1,0(~/

ˆN

npq

ppZ

)/

ˆ(1 2/2/

z

npq

ppzP

)(1 2/2/ zZzP

)/ˆ/ˆ(1 2/2/ npqzppnpqzpP

)/ˆ,/ˆ( 2/2/ npqzpnpqzp

上面的公式必須要有母體比例 p 才能估計標準誤

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Confidence intervals for proportions Confidence intervals for proportions (large samples)(large samples)

)/ˆ,/ˆ( 2/2/ npqzpnpqzp

因為沒有 p 與 q 的資訊,在樣本數夠大時,我們通常以樣本的比例 p^ 來估計母體的標準誤:

)/ˆˆˆ,/ˆˆˆ( 2/2/ nqpzpnqpzp

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Confidence interval for the population Confidence interval for the population proportion proportion pp

Let p denote the population proportion. Suppose we take a large random sample of n observations and obtain the sample proportion p^. A confidence interval for the population proportion having level of confidence 100(1-α)% is given by

)/ˆˆˆ,/ˆˆˆ( 2/2/ nqpZpnqpZp

定義定義

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Wilson estimateWilson estimate

• 用樣本比例取代母體比例來估計標準誤並不一定正確。

• 例如:丟一個銅板三次得到三次都得正面,則

• p^=3/3=1

• p^(1-p^)/n=1(1-0)/3=0

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Wilson estimateWilson estimate

We must know the s.d. of the population to get a CI for p.

• Unfortunately, modern computer studies reveal the confidence intervals based on this approach can be quite inaccurate, even for large samples.

-- When the sample is not a SRS.

-- When the sample size is small

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Wilson estimateWilson estimate

• The Wilson estimate ~

Add 2 successes and 2 failures (so that the sample proportion is slightly moved away from 0 and 1.)

-- Because this estimate was first suggested by Edwin Bidwell Wilson in 1927, we call it the Wilson estimate.

4

2~ˆ

n

xp

n

xp

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Wilson estimateWilson estimate

• 的抽樣分配趨近於平均數為 p 、標準差為 的常態分配。

• An approximate level C confidence interval for p is

• The margin of error is

p~

4

)1(

n

pp

pSEzp ~~

4

)~1(~~

n

ppzSEzm p

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Confidence interval for the population Confidence interval for the population proportion proportion pp

政府想要估計月收入低於 $25,000NT 的家庭。 500 個家庭接受訪問,其中有 200 戶人家年收入少於 25000. 求 p 的 95% 信賴區間?

4.504/202/~ nxp

例題例題

6.~1~ pq

)504/)6)(.4(.96.14,.504/)6)(.4(.96.14(.

(.3572, .4428)

))4/(~~~,)4/(~~~( 2/2/ nqpZpnqpZp

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例題例題• 從台北市隨機抽取 500 個人,詢問是否贊成公投,結果有 312 名贊成。試求台北市贊成公投比率 95% 信賴區間。

623.0504

314~ p

504

)623.01(623.0623.0

504

)623.01(623.0623.0 025.0025.0

ZpZ

042.0623.0042.0623.0 p

665.0581.0 p

, p的信賴區間為:

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One-sided confidence intervals for the One-sided confidence intervals for the population proportionpopulation proportion

Suppose that we take a random sample of n observations from some population having unknown proportion p. Suppose we wish to find the lower confidence limit LCL such that the probability is (1-) that p exceeds LCL.

The one-sided interval (LCL, 1.00) is a left-sided confidence interval.

The LCL is given by: nqpZpLCL /ˆˆˆ

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One-sided confidence intervals for the One-sided confidence intervals for the population proportionpopulation proportion

Construct a right-sided 95% CI for the proportion of defective items produced by a machine if 16 items are found to be defective in a random sample of 100 items.

)4/(~~~ nqpZpUCL

17.104/18~ p 05.95.1

2306.104/)83)(.17(.645.117. UCL

The 95% right-sided CI for p is (0, .2306) This mean that we can be 95% confident that the population proportion is less than .2306

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Determining the sample sizeDetermining the sample size 決定樣本大小決定樣本大小

Margin of Error

Suppose that we take a random sample from some population. Then a 100(1-)% confidence interval for the population proportion extends at most a distance m on each side of the sample proportion if the number of observations is ?

))4/(~~~,)4/(~~~( 2/2/ nqpzpnqpzp

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Determining the sample sizeDetermining the sample size 決定樣本大小決定樣本大小

問題是我們還不知道 (因為樣本數都還沒決定),所以上述公式無法使用,除非我們有 p 的推估值。

(1) 我們可以用 pilot study 來得到 p 的估計值。

(2) 在不知道的樣本比例情形下,我們可以採用最保守的估計,也就是最大的變異 .5*.5=.25 來估計 n 。

m

qpzn

n

qpzm

~~4

4

~~2/2/

2

22/

~~)(4

m

qpzn

p~

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Sample size and confidence interval for the Sample size and confidence interval for the proportionproportion

如果母體比率無法推估,則樣本數:

2

22/ )(25.

4m

zn

如果母體比率 p 可以推估,則樣本數:

2

22/

~~)(4

m

qpzn

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Sample size and confidence interval for the Sample size and confidence interval for the proportionproportion

民意調查機構想知道某總統候選人得票的比率,請問至少要多大的樣本數才可以使此機構在 95% 的信賴度下,估計的誤差界不會超過 .03 ?

.5q̂p̂ 03.0m 96.1 95.1 2/ z

10681.067,103.

)96.1(25.)(25.4

2

2

2

22/

m

zn

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Sample size and confidence interval for the Sample size and confidence interval for the proportionproportion

民意調查機構想知道某總統候選人得票的比率。假設該公司要求樣本比例與母體之誤差不能超過 0.01 ,且有 95 %的信賴度,則樣本數應為何?

2/1p

9604)01.0(

4/1)96.1(

)01.0(

)1()(4

2

2

2

22/

PPZn 可解得

p未知,故以 代入,

故至少應選取 9,600個樣本點。

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Tests of the population proportionTests of the population proportion

樣本比例的抽樣分配 f(p^) :如果母體的比例為 p, 且 np5 and nq 5 , 則樣本比例 p^ 為一常態分配 ~N(p, pq/n)

The Normal Approximation Rule for Proportion: If np5, and nq 5, the random variable p^ has approximately a normal distribution with:

pq/n pq/n)p̂Var( )ˆ( p̂ ppE

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Sampling Distribution of the Sampling Distribution of the Sample ProportionSample Proportion

• If the distribution of p^ is approximately normal, then random variable

)1,0(~pq/n

ˆN

ppZ

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Tests of the population proportionTests of the population proportion

設 np5 and nq 5 檢證下列假說:H0: p = p0 or H0: pp0 H1: p < p0

如果 H0為真,則樣本比率 ~N(p0, p0q0/n)

)1,0(~/nqp

ˆ

00

0 Npp

Z

Reject H0 if Z < -z or p^ < p^* ( critical value approach)

nqpzpp /ˆ 000*

假設為真時的母體比例

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Page 614, Procedure 12.2B (cont.)Page 614, Procedure 12.2B (cont.)

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例:例: Testing a population ProportionTesting a population Proportion

藍營立法委員宣稱民調顯示 60% 的民眾支持連戰出訪中國,綠營團體宣稱支持的民眾不會超過 60% ,妳用 100 的樣本來驗證:H0: p = .6 v.s. H1: p < .6

假設 55 個樣本支持連戰出訪,以 5% 的顯著水準,我們可以推翻藍營立委的說法嗎?

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例:例: Testing a population ProportionTesting a population Proportion

Solution:

If H0 is true, then p^ has a normal distribution with mean p =.6 and variance pq/n = (.6)(.4)/100 = .0024If we use a one-tailed test at the 5% level of significance, the critical region consists of all values of Z less than –z = -z.05 = -1.645從樣本中得知 p^=x/n = 55/100 =.55

02.1100/)4)(.6(.

60.55.

/nqp

ˆ

00

0

pp

z

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例:例: Testing a population ProportionTesting a population Proportion

We do not reject H0

0

1

645.1-1.02

519.100/)4)(.6(.645.16./ˆ 000* nqpzpp

1539.)02.1,( ZPvaluep

實際上觀察到的樣本比例為 .55>.519 因此無法推翻虛擬假設

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Sampling distribution of the difference Sampling distribution of the difference between sample proportionsbetween sample proportions

• Suppose we take independent sample of size n1 and n2 from two population. Let p1 and p2 be the proportion of items in each population that possess a certain characteristics, and let q1=(1-p1), q2=(1-p2). If n1p1>5, n1q1>5, n2p2>5, n2q2>5, then the random variable (p1^-p2^) is approximately normally distributed with

2121 )ˆˆ( ppppE

2

22

1

1121 )ˆˆ(

n

qp

n

qpppVar

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例題例題

• 假設某行銷公司想要知道某電視節目在高、低收入人口中受歡迎的程度。假設高收入的人中有 40%喜歡看此節目,在低收入人口中喜歡此節目的佔 50% 。這家行銷公司從高收入的人口中抽取 100 人的樣本,從低收入中抽 200 人的樣本。請問兩樣本比率差距小於 .05 的機率?

?)05.ˆˆ5.0( 21 ppP

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例題例題

200,5.,5.

100,6.,4.

122

111

nqp

nqp

5,5

5,5

2222

1111

qnpn

qnpn

10.5.4.21 pp

00365.200

)5)(.5(.

100

)6)(.4(.

2

22

1

11 n

qp

n

qp

)00365.

)1.0(5.0ˆˆ

00365.

)1.0(05.( 2211 zppzP

1967.)48.2ˆˆ83(. 21 ppP

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Confidence intervals for the difference of Two Confidence intervals for the difference of Two population proportionpopulation proportion

2

22

1

11,2

212

22

1

11,2

21

ˆˆˆˆ)ˆˆ(,

ˆˆˆˆ)ˆˆ(

n

qp

n

qpZpp

n

qp

n

qpZpp

Let p1 denote the observed proportion of successes in a random sample of n1 observation from a population with proportion p1 successes, and let p2 denote the observed proportion of successes in an independent random sample of n2 observations from a population with proportion p2 successes. A 100(1- α) % confidence interval for (p1 – p2) is given by the interval

This result holds provided n1p1 5≧ n1q1 5 ≧ n2p2 5 and ≧ n2q2 5≧

Page 49: Population proportion and sample proportion

© 蘇國賢 2007社會統計(上) Page 49

Tests concerning differences of Tests concerning differences of proportionsproportions

• 欲檢定兩母體的比率是否等於某特定值(相等),假設母體 1 的比率為 p1,母體 2 的比率為 p2 :

• H0: p1 –p2 = D0

• 分別從兩母體中抽取樣本 n1, n2並計算樣本比率為 p^1 p^2。

Page 50: Population proportion and sample proportion

© 蘇國賢 2007社會統計(上) Page 50

Tests concerning differences of Tests concerning differences of proportionsproportions

• 若虛擬假設為真 H0: p1 –p2 = D0,且 n1p1≥5, n1q1

≥5, n2p2≥5, n2q2≥5

),(~)ˆˆ(2

22

1

112121 n

qp

n

qpppNpp

• 通常我們想要檢驗虛擬假設 H0: p1 –p2 =0 的情形,即 H0: p1 = p2

Page 51: Population proportion and sample proportion

© 蘇國賢 2007社會統計(上) Page 51

Tests concerning differences of Tests concerning differences of proportionsproportions

),(~)ˆˆ(2

22

1

112121 n

qp

n

qpppNpp

• 由於 p1和 p2為未知,我們無法計算變異數。• 由於我們假設 p1 = p2,一個合理估計母體變異數的方法為同時利用兩樣本的資料來估計母體比率p = p1 = p2,稱為 pooled sample proportion 。

21

2211

21

21 ˆˆˆ

nn

pnpn

nn

xxp

Page 52: Population proportion and sample proportion

© 蘇國賢 2007社會統計(上) Page 52

Tests concerning differences of Tests concerning differences of proportionsproportions

21

2211

21

21 ˆˆˆ

nn

pnpn

nn

xxp

)(ˆˆ

0)ˆˆ(

/ˆˆ/ˆˆ

0)ˆˆ(

21

21

21

21

21

nnnn

qp

pp

nqpnqp

ppz

檢定 H0: p1 – p2 =0 v.s. H1: p1 – p2 ≠0

Page 53: Population proportion and sample proportion

© 蘇國賢 2007社會統計(上) Page 53

例題例題• 第一台機器生產的 400產品中,有 23 個瑕疵品,第二台機器生產的 400 樣本種,有 17 個瑕疵品,請用 5% 的顯著水準測驗這兩台機器的品質是否相當。 96.1 valuecritical,05. 025. z

05.800

40

400400

1723ˆ

21

21

nn

xxp

2121 /ˆˆ/ˆˆ的變異數為 )ˆˆ( nqpnqppp

0002375.400

)95)(.05(.

400

)95)(.05(.

Page 54: Population proportion and sample proportion

© 蘇國賢 2007社會統計(上) Page 54

例題例題• 第一台機器生產的 400產品中,有 23 個瑕疵品,第二台機器生產的 400 樣本種,有 17 個瑕疵品,請用 5% 的顯著水準測驗這兩台機器的品質是否相當。

10002375.

0)0425.0575(.

z Failed to reject H0

21

21

/ˆˆ/ˆˆ

0)ˆˆ(

nqpnqp

ppz

Page 55: Population proportion and sample proportion

© 蘇國賢 2007社會統計(上) Page 55

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© 蘇國賢 2007社會統計(上) Page 56

Page 57: Population proportion and sample proportion

© 蘇國賢 2007社會統計(上) Page 57

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© 蘇國賢 2007社會統計(上) Page 58

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© 蘇國賢 2007社會統計(上) Page 59

Page 60: Population proportion and sample proportion

© 蘇國賢 2007社會統計(上) Page 60

例題例題

• 兩家銀行,信用卡部門上個月申請件數與核准件數如下表:

• 當顯著水準等於 5% 時,檢定兩家銀行信用卡核准率是否相同?若不同,則求核准率差的 95% 信賴區間。

AB

申請件數

核准件數

銀行 350 273

銀行 450 378

Page 61: Population proportion and sample proportion

• 因檢定統計量小於左尾臨界值,故拒絕虛無假設,兩家銀行信用卡核准率不同。

AB

申請件數

核准件數

銀行 350 273

銀行 450 378

0:,0: 10 BABA ppHppH欲檢定

81375.0450350

378273ˆ

p混和樣本比例

16.2

4501

3501

18625.081375.0

450378

350273

Z檢定統計量

96.1025.0 Z左尾臨界直

450

1

350

118625.081375.006.0

%95

025.0 Z

信賴區間核准率差的

( . ~ . )0 114 0 006