estimating and comparing

Analisis Statistik Bisnis - MM Prasetiya Mulya

1

Sampling Distributions and the Central Limit Theorem

Distribution of a population

0

10

20

30

40

50

0 20 40 60 80 100

y

Distribution of the Sample

means for sample size n=50

0

20

40

60

80

45 50 55 60 65

Sample mean

= 52.24 = 15.02 20.52)y(E 21.2y

)y(En

y

For large sample sizes:

Distribution of the sample means is approximately

normal, regardless of its population distribution.

Std dev of sample means

= std error of estimate

For large population:

11N

nN


2

Estimating a Population Mean () using large sample size (n30)

y

95%

1.96 1.96 yy %95)96.1y96.1(P yy

y96.1y

The 95% confidence interval for the population mean () is:

ny

s

Using large sample size (n 30),

the distribution of the sample means

can be assumed normal.

Where


3

0

/2 /2

z /2-z /2

z

y2/zy

The 100(1- )% confidence interval for the population mean () is:

ny

Large Sample

n 30

Commonly Used Values of z /2

Confidence Coefficient: 90% 95% 99%

: 10% 5% 1%

/2 : 0.05 0.025 0.005

z /2 : 1.645 1.96 2.576

Std normal (z) distribution

Where z/2 is a z value with an area /2 to its right

s


4

0

/2 /2

t /2-t /2

t

Small Sample

Normal Population

y2/ sty

The 100(1- )% confidence interval for the population mean () is:

n

ssy

Confidence Coefficient: 90% 95% 99%

: 10% 5% 1%

/2 : 0.05 0.025 0.005

df=10 t /2 : 1.812 2.228 3.169

df=15 t /2 : 1.753 2.131 2.947

Student’s t distribution

df = degree of freedom = n-1

Estimating a Population Mean () using small sample size (n<30)

Where t /2 is a t value based on (n-1) degree of freedom, such that the

probability that t > t /2 is /2.


5

0 t.025

t

z.025

0.0250.025

1.96 2.228

Std normal distribution

t distribution for df=10

The smaller the sample size, the higher are the the tails of the t distribution.

or

The smaller the sample size, the greater will be the spread of the t distribution.

For n 30 the two distributions will be nearly identical.

About Student’s t distribution


6

Example 4: Confidence Intervals for the Population Mean

Sebuah pabrik TV melakukan pengujian sample secara acak sebanyak 35 unit untuk

mempekirakan rata-rata umur tabung TV merk X. Dari uji sample dapat dihitung rata-rata

umurnya 8900 jam dengan deviasi standardnya 500 jam.

Dengan tingkat keyakinan 95% perkirakan rata-rata umur tabung TV merk X.

Jawab:

Diketahui n=35, y = 8900, S=500 dan (1-)=95%

y2/zy

46.8435

500

n

sy

96.1zz 025.02/ = 1-95%=0.05 (dari tabel normal)

= 8900 (1.96)(84.46) = 8900 165.54 =8734 s/d 9066 jam

dengan tingkat keyakinan 95%, interval rata-ratanya = 8734 s/d 9066 jam

n 30 large sample


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umurnya 8900 jam dengan deviasi standardnya 500 jam. Diasumsikan distribusi umur tabung

TV merk X mendekati normal.

Dengan tingkat keyakinan 95% , perkirakan rata-rata umur tabung TV merk X.

Jawab:


y2/ sty

099.12915

500

n

ssy

145.2tt 025.02/

= 1-95%=0.05 dan df=15-1=14

(dari tabel student’s t)

= 8900 (2.145)(129.099) = 8900 276.92 =8623 s/d 9177 jam

dengan tingkat keyakinan 95%, interval rata-ratanya = 8623 s/d 9177 jam

n < 30, distribusi populasi normal


8




umurnya 8900 jam dengan deviasi standardnya 500 jam. Diasumsikan distribusi umur tabung

TV merk X mendekati normal.

Dengan tingkat keyakinan 95% , perkirakan minimum rata-rata umur tabung TV merk X.

Jawab:


ysty

099.12915

500

n

ssy

761.1tt 5.0

= 1-95%=0.05 dan df=15-1=14


= 8900 - (1.761)(129.099) = 8900 - 227.34 =8672 jam

dengan tingkat keyakinan 95%, minimum rata-ratanya = 8672 jam

n < 30, distribusi populasi normal dan tidak diketahui min


9

Estimating About (1- 2)

POPULATION

1 2

Sample size

Population mean

Population variance

Sample mean

Sample variance

n1 n2

1 2

12 2

2

ÿ1 ÿ2

s12 s2

2

Large sample (n130 and n2 30) Distribution of ÿ1 and ÿ2 normal

Distribution of (ÿ1-ÿ2) normal

)2y1y(2/2121 z)yy()(

The two samples are randomly and

independently selected from two

populations

2121)2y1y( n

sn

snn

22

21

22

21

Where:


10

Small sample (n1<30 or n2<30)

Both populations are approximately normal.

The population variances are equal Distribution of ÿ1 and ÿ2 approximately normal

Distribution of (ÿ1-ÿ2) approximately normal

21

2p2/2121

n

1

n

1st)yy()(

Where:

2nn

s)1n(s)1n(s

21

222

2112

p

t/2 based on (n1+n2-2) df


11

Example 7: Confidence Interval for (1 - 2)

Sebuah lembaga konsumen menguji kualitas 2 merk lampu (diukur dari umur lampu).

Diambil secara acak dari pasar 10 lampu merk A dan 8 lampu merk B.

Dari hasil pengujian didapat:

- rata-rata umur lampu dari sampel A dan B adalah 4000 dan 4600 jam.

- deviasi standar umur lampu dari sampel A dan B adalah 200 dan 250 jam.

Dengan tingkat keyakinan 90% perkirakan perbedaan umur lampu merk B terhadap

umur lampu merk A.

Ay = 4000, sA=200

= 1-90%=0.10 dan df=10+8-2=16


Jawab:

Diketahui nA=10,

nB=8, By

(1-)=90%= 4600, sB=250

746.1tt 05.02/

75.498432810

250)18(200)110(s

222

p

Dengan tingkat keyakinan 90%, (B- A) =

9.18460010

1

8

175.49843746.1)40004600(

= 415 s/d 785 jam


12

The 100(1- )% confidence interval for the population proportion is:

Using large sample size (np and n(1-p) 5) the distribution of the

sample proportion can be assumed normal.

n = sample size p = sample proportion

^ ^

^

p = p z/2 p (1-p)

n

^ ^^

Example:

A electronic manufacturer wants to know about customer preference

about color of a product. 48 customers prefer red products and 32

customers prefer yellow products. 48/80(1-48/80)

Proportion of red preference = p = 48/80 1.96 80

= 0.6 0.107

Estimating a Population Proportion (p)


13

The 100(1- )% confidence interval estimator p1 – p2

Using large sample sizes (n1p1 and n1(1-p1) 5) (n2p2 and n2(1-p2) 5)

the distribution of the p1-p2 can be assumed normal.

^ ^

^

p1 – p2 = (p1 – p2) z/2 p1 (1-p1) p2 (1-p2)

n1

^ ^^

Estimating the difference between two population proportion

^ ^

^

^

+ n2

^ ^


14

Sample Sizes for Estimating and p

z/2

B

p (1-p)^ ^

Parameter Sample Size

z/2

Bn =

2

p n =

2

Notes:

B = bound on the error estimation

Because we have not taken the sample we use

Range/4 p = 0.5^


15

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