Aplikasi Sebaran Normal Pertemuan 12

Matakuliah : L0104 / Statistika PsikologiTahun : 2008

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Learning Outcomes

3

Pada akhir pertemuan ini, diharapkan mahasiswa akan mampu :

Mahasiswa akan dapat menghitung peluang Binomial dengan pendekatan pada sebaran normal baku.

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Outline Materi

4

• Metode deskriptif untuk sebaran normal

• pendekatan sebaran Binomial pada sebaran normal baku

• Koreksi kekontinuan

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The Normal Approximation to the Binomial

• We can calculate binomial probabilities using– The binomial formula– The cumulative binomial tables– Do It Yourself! applets

• When n is large, and p is not too close to zero or one, areas under the normal curve with mean np and variance npq can be used to approximate binomial probabilities.

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Approximating the Binomial

Make sure to include the entire rectangle for the values of x in the interval of interest. This is called the continuity correction. continuity correction. Standardize the values of x using

npq

npxz

npq

npxz

Make sure that np and nq are both greater than 5 to avoid inaccurate approximations!

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Example

Suppose x is a binomial random variable with n = 30 and p = .4. Using the normal approximation to find P(x 10).

n = 30 p = .4 q = .6

np = 12 nq = 18

683.2)6)(.4(.30

12)4(.30

Calculate

npq

np

683.2)6)(.4(.30

12)4(.30

Calculate

npq

np

The normal approximation

is ok!

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Example

)683.2

125.10()10(

zPxP

2877.)56.( zP

AppletApplet

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Example

A production line produces AA batteries with a reliability rate of 95%. A sample of n = 200 batteries is selected. Find the probability that at least 195 of the batteries work.

Success = working battery n = 200

p = .95 np = 190 nq = 10

The normal approximation

is ok!

))05)(.95(.200

1905.194()195(

zPxP

0722.9278.1)46.1( zP

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Central Limit Theorem: If random samples of n observations are drawn from a nonnormal population with finite and standard deviation, then, when n is large, the sampling distribution of the sample mean is approximately normally distributed, with mean and standard deviation . The approximation becomes more accurate as n becomes large.

Central Limit Theorem: If random samples of n observations are drawn from a nonnormal population with finite and standard deviation, then, when n is large, the sampling distribution of the sample mean is approximately normally distributed, with mean and standard deviation . The approximation becomes more accurate as n becomes large.

x

n/

Sampling Distributions

Sampling distributions for statistics can be Approximated with simulation techniquesDerived using mathematical theoremsThe Central Limit Theorem is one such theorem.

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Example

Toss a fair coin n = 1 time. The distribution of x the number on the upper face is flat or uniform.uniform.

71.1)()(

5.3)6

1(6...)

6

1(2)

6

1(1

)(

2

xpx

xxp

71.1)()(

5.3)6

1(6...)

6

1(2)

6

1(1

)(

2

xpx

xxp

AppletApplet

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Example

Toss a fair coin n = 2 time. The distribution of x the average number on the two upper faces is mound-mound-shaped.shaped.

21.12/71.12/

:Dev Std

5.3:Mean

21.12/71.12/

:Dev Std

5.3:Mean

AppletApplet

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Example

Toss a fair coin n = 3 time. The distribution of x the average number on the two upper faces is approximately normal.approximately normal.

987.3/71.13/

:Dev Std

5.3:Mean

987.3/71.13/

:Dev Std

5.3:Mean

AppletApplet

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Why is this Important?

The Central Limit TheoremCentral Limit Theorem also implies that the sum of n measurements is approximately normal with mean n and standard deviation .

Many statistics that are used for statistical inference are sums or averages of sample measurements.

When n is large, these statistics will have approximately normalnormal distributions.

This will allow us to describe their behavior and evaluate the reliabilityreliability of our inferences.

n

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How Large is Large?

If the sample is normalnormal, then the sampling distribution of will also be normal, no matter what the sample size.

When the sample population is approximately symmetricsymmetric, the distribution becomes approximately normal for relatively small values of n. (ex. n=3 in dice example)

When the sample population is skewedskewed, the sample size must be at least 30at least 30 before the sampling distribution of becomes approximately normal.

x

x

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The Sampling Distribution of the Sample Proportion

The Central Limit TheoremCentral Limit Theorem can be used to conclude that the binomial random variable x is approximately normal when n is large, with mean np and standard deviation .

The sample proportion, is simply a rescaling of the binomial random variable x, dividing it by n.

From the Central Limit Theorem, the sampling distribution of will also be approximately approximately normal, normal, with a rescaled mean and standard deviation.

n

xp ˆ

p̂

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The Sampling Distribution of the Sample Proportion

The standard deviation of p-hat is sometimes called the STANDARD ERROR (SE) of p-hat.

A random sample of size n is selected from a binomial population with parameter p

he sampling distribution of the sample proportion,

will have mean p and standard deviation

If n is large, and p is not too close to zero or one, the sampling distribution of will be approximately approximately normal.normal.

n

xp ˆ

n

pq

p̂

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Finding Probabilities for the Sample Proportion

0207.9793.1)04.2(

)

100)6(.4.

4.5.()5.ˆ(

zP

zPpP

0207.9793.1)04.2(

)

100)6(.4.

4.5.()5.ˆ(

zP

zPpP

If the sampling distribution of is normal or approximately normal standardize or rescale the interval of interest in terms of

Find the appropriate area using Table 3.

If the sampling distribution of is normal or approximately normal standardize or rescale the interval of interest in terms of

Find the appropriate area using Table 3.

p̂

npq

ppz

ˆ

Example: Example: A random sample of size n = 100 from a binomial population with p = .4.

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Example

The soda bottler in the previous example claims that only 5% of the soda cans are underfilled. A quality control technician randomly samples 200 cans of soda. What is the probability that more than 10% of the cans are underfilled?

)10.ˆ( pP

)24.3()

200)95(.05.

05.10.(

zPzP

0006.9994.1 This would be very unusual,

if indeed p = .05!

n = 200

S: underfilled can

p = P(S) = .05

q = .95

np = 10 nq = 190

n = 200

S: underfilled can

p = P(S) = .05

q = .95

np = 10 nq = 190OK to use the normal

approximation

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