assignment 1 by ricky

8/13/2019 Assignment 1 by Ricky

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Home Assignment 1Bus 173; Sec 9

Prepared for

M. Siddique Hossain

Prepared By

Ehsan Karim

ID# 101 0347 030


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Answer to the question no. 1

The mean of the population is

= 5.5

SAMPLE AND SAMPLE MEANS

If n = 2 then

SAMPLE SAMPLE MEAN SAMPLE SAMPLE MEAN

2,4 3 4,8 6

2,6 4 6,6 6

2,6 4 6,7 6.5

2,7 4.5 6,8 7

2,8 5 6,7 6.5

4,6 5 6,8 7

4,6 5 7,8 7.5

4,7 5.5


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SAMPLING DISTRIBUTIONS

SAMPLE MEAN PROBABLITY

3 1/15

4 2/15

4.5 1/15

5 3/15

5.5 1/1

6 2/15

6.5 2/15

7 2/15

7.5 1/15


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If n = 3 then

SAMPLE SAMPLE MEAN SAMPLE SAMPLE MEAN

2,4,6 4 4,6,6 4.33

2,4,6 4 4,6,7 5.67

2,4,7 4.33 4,6,8 6

2,4,8 4.67 4,7,8 6.33

2,6,6 4.67 6,6,7 6.33

2,6,7 5 6,6,8 6.67

2,6,8 5.33 6,7,8 7

2,7,8 5.67

0

0.05

0.1

0.15

0.2

0.25

3 3.5 4 4.5 5 5.5 6 6.5 7 7.5

Column12

Column1

Column2

Series 1


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SAMPLING DISTRIBUTIONS

SAMPLE MEAN PROBABLITY

4 2/15

4.33 2/15

4.67 2/15

5 1/15

5.33 1/15

5.67 2/15

6 1/15

6.33 2/15

6.67 1/15

7 1/15


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If n = 4 then

SAMPLE SAMPLE MEAN

2,4,6,6 4.5

2,4,6,7 4.75

2,4,6,8 5

2,4,7,8 5.25

4,6,6,7 5.75

4,6,6,8 6

6,6,7,8 6.75

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

4 4.33 4.67 5 5.33 5.67 6 6.33 6.67 7

Column1

Column2

Series 1


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MEAN PROBABLITY

4.5 1/7

4.75 1/7

5 1/7

5.25 1/7

5.75 1/7

6 1/7

6.75 1/7

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

4.5 4.75 5 5.25 5.75 6 6.75

Column1

Column2

Series 1


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Answer to the question 2

SOME IMPORTANT TOPICS ABOUT CHAPTER 6 AND 7

Distribution of sample statistics:

Any aggregate characteristics about sample observations is called statistic.

Purpose of sampling:

1) TIME and 2) COST

we deal with sample in order to draw conclusion about population in a less

expensive matter.

The basic objective of its study is to draw influence about the population.

In other words, sampling is only a tool which helps to know the characterisitics of

the population by examining a small part of it.

The values obtained from the study of a sample such as average and variance

is known as STATISTICS

On the other words, such values from the population are called parameters.

PRINCIPLES OF SAMPLING:

There are two important principles on which the theory of sampling is based.

They are

1) PRINCIPLES OF STATISTICAL REGULARITY

2) PRINCIPLES OF INERTIA OF LARGE NUMBERS

A moderately large number of items are chosen at random from a large group

are almost sure on the average to possess the characteristics of the large

group.


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This principle points out that if a sample is taken at random from a population,

it is likely to possess almost the same characteristics of that population.

PRINCIPLES OF INERTIA OF LARGE NUMBERS:

Other things being constant larger the size of the sample, more accurate the

results are likely to be.

This difference in the aggregate result is likely to be insignificant when the

number is as the sample size is large.

SAMPLE

N = 5

X = 200

S^2 = 25

POPULATION

N = 5

= 250

= 20

There are three types of distributions

1) POPULATION DISTRIBUTION

2) SAMPLE DISTRIBUTION

3) SAMPLING DISTRIBUTION

* THE POPULATION DISTRIBUTION:


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When we talk of the population distribution we assumed that we have

investigated the population and have full knowledge of its mean and standard

deviation.

* THE SAMPLE DISTRIBUTION:

When we talk a sample distribution we take a sample from a population. A

sample distribution may take any shape. The mean and the standard deviation

of the sample distribution are symbolized by X bar and S respectively.

It may be noted that several sample distributions are possible from a given

population.

* THE SAMPLING DISTRIBUTION:

Sampling distribution constitute the basis of statistical influence and are of

considerable importance in business decision making.

If we take numerous different samples of equal size from the same popualtion,

the probability distribution of the possible values of a given statistic from all

distinct possible samples of equal size is called a sampling distribution.

It is interesting to note that sampling distribution closely approximate a normal

distribution.

It can be proved that the mean of sampling distribution of sample means is the

same as the mean of the population from which sampling was taken.

Let . represent independent random variablescorrespondent to n observation in a sample from a population having thr same

mean

X = sample mean =

= E (X)

= E(

= 1/n E ( ++ )


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= 1/n { E () + E () + .+ E() }= 1/n n =

The mean of the sampling distribution is denoted by the same symbol as the

mean of the population.

E (X) = { / n }

However , the standard deviation of the sampling distribution of mean is given a

special name, standard error of mean as denoted by X bar as a suffix, indicates that in this case we are dealing with a samplingdistributions of means.

The standard of the mean denoted by and is calculated by = This formula holds only when population is infinite on sample are from a finite

population with replacement.

The proportion of the sampling distribution of the mean

1) It has been equal to the mean

E(X) =

2) It has a standard deviation equal to the population standard deciation divided

by square of the sample size. That is

=

= / When is a measure of the spread of X bar unless around or a measureof average sampling error or simply standard error of the mean.

3) It is normally distributed.


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The distribution of the sample means for large sample is distributed normally

whatever the shape of the parent population.

Sample of n = 3 or more items are considered large in statistical analysis.

In practice, the standard deviation of the population is rarely known, and,

therefore the standard deviation of samples which closely approximate the

standard deviation of the population, is used in place of .

Hence the formula for the standard deviation of error will be

= /Where is the standard deviation of the sample.

FITNITE POPULATION CONNECTIM FACTOR

=

=

= n/METHODS OF SAMPLING

1) RANDOM SAMPLING and

2) NON RANDOM SAMPLING

RANDOM SAMPLING

SIMPLE RANDOM SAMPLING

STRATIFIED SAMPLING


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SYSTEMATIC SAMPLING

MULTI STAGE SAMPLING

Under non-random sampling,

1) Judgment sampling

2) Quota sampling

3) Convenience sampling

SIZE OF SAMPLE

1) The size of the sample should increase as the variation in the individual

items increase.

2) The greater the accuracy expected the larger shall the sample size will be.

WHY SAMPLING??

1) Less time

2) Lower cost

3) More reliable results

4) More detailed information.

5) The destructive nature of certain test.


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# The sample mean x bar for a random sample of size n drawn from a normal

population with a normal distribution with mean and variance is also distributed

with and variance

CENTRAL LIMIT THEORM tells us that the mean of a random sample drawn

from a population with any probability distribution will be approximately normally

distributed with mean at variance /nThis CLT enables us to use the n ormal distribution to compute probablities for

sample means obtained from many different population.

CLT STATEMENT

Let ,,.., be a set of n independent randomobservations having identical distributions with mean and variance ^2 and X

as the sum of Xbar as the mean of these random observations.

As n becomes large the CLT states that the distribution of

Z =

Sampling distribution of sample proportion

A population proportion is defined as

P = X/N

When, X is the number of elements which possess a certain characteristics at N

is the total number of items in the population.

A sample proportion is defined as

P bar = x/n

Where x is the number of items in the sample which possess a certain

characteristics and n is the sample size.

A proportion may be considered as a proportion of success and is obtained by

dividing the number of success by sample size n.


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If a random sample of size n is obtained with replacement then the sampling

distribution obeys the binomial probability law.

Suppose that a population is finite and that the probability of occurrence of an

event is P while the probability of non occurrence of the event is ( 1P )

Now if we consider all possible samples of size n drawn from the population

and for each sample determine the proportion P bar of success. Then we obtain

a sampling distribution of proportions where mean n at standard deviation in are given by,

V ( p bar ) = P ( 1P )/n

This is exactly the mean and variance of sampling distribution from bionomical

population. We know that if np (1p) > q ; it can be approximated to the normal

population.

SAMPLING DISTRIBUTION OF THE SAMPLE PROPORTIONS

Let P bar be the sample proportion of successes in a random sample from a

population with proportion of success P then

1) The sampling distribution of p bar has mean P

E ( pbar) = p

2) The sampling distribution of p bar has standard error

= )/n

3) If the sample size is large then the random variable Z =

is approximately distributed as a standard normal. The approximation is good if

np( 1p ) q


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CONFIDENCE INTERVAL

Generally we assume that a random sample is taken from a population that is

normally distributed with unknown mean and variance.

We try to find a range of values rather than a single point to estimate a

population mean. It is hard to believe that population mean will be known and

population variance be known. However, you may be surprised in many

companies, some records can provide a good estimation of population variance.

In order to draw confidence interval we need to know the population parameters

or we have to estimate parameters from the sample observations.

INTERVAL ESTIMATOR

In interval estimate of a population parameter given by two numbers which

parameter may be considered to lie in is called as interval estimate of the

parameter.

There are three steps

1) The particular statistics say that the mean of the sample x bar or S.D of thesample is determined.

2) The confidence level is decided if 90%, 95% , 94%

3) The standard error of the particular statistic is calculated.

Answer to the question no. 3

6.6) a

Probablity function: f(x)


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6.6) b

Cumulative density function F(x)

0

0.05

0.1

0.15

0.2

0.25

0.3

1 2 3 4


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6.8)

a) P(380 < X < 460) = P(X < 460)-P(X


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6.12)

W = a + bX if available funds = 60003X where X = number of units produced,

then mean and variance for the number of units are 1000 and 900

respectevely.

= a + b = 6000 -2(1000) = 4000;

w = x = (-3)^2 (900) = 8100

6.14)

= 20 + = 20 + 4 = 24 millionBid = 1.1 = 1.1(24) = 26.4 million = 1 million

6.18)

P(Z < ) = .7; closest value of = .52P(Z < ) = .25; closest value of = -.67P(Z >) = .2; closest value of = .84P(Z > ) = .6; closest value of = -.25

6.28)


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P(Z > 1.5) = 1 - (1.5) = .0668

6.30)

P (Z > .67) = .25; .67 = 17.8 -

P(Z > 1.03) = .15; 1.04 = 19.2 -

= 15.265, = = 14.317

6.34)

a. P(Z > - 1.28) = .9; -1.28 =

X = 98.8Z

b. P(Z < .84) = .8; .84 =

X = 183.6

c. P (X 1) = 1P(X = 0) = 1- [ ]^2 = 1{ P(Z < .75) }^2 = 1(.2266)^2 = .9487

6.36)

a. P (

< Z <

) = P 9-.25 < Z < .75) = (.75){ 1 - .25} = .372

b. P( Z > 1.28) = .1, 1.28 =

; = 522.4c. 400439 = -39

d. 520559 = -59

e. P (X = 1P( X = 0) = 1{ P ( Z < -120/80 ) }^2 = .2922

7.18)


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= 1.6/= .16 P( Z > 1.645) = .05, Difference = b) P(Z < -1.28) = .1, -1.28 =

, Difference = -2.048

c) P(Z > 1.44) = .075, 1.44 =

DIfference = .2304

7.20)

a) P( Z> 1.96) = .025, 1.96 =

, n = 67.266 that n = 68

b) Ans will be smaller

c) Ans will be larger

7.26)

E() = 0.4 = = .04899

a) Probability that the sample proportion is greater than .45Z =

= P(Z >1.02) = .1539

b) Probability that the sample proportion is less than .29

Z == P(Z < -2.25) = .0122

c) Probability that the sample proportion is between .35 and .51

P(< Z 1.22) = .1112

b) Probability that the sample proportion is less than .48

Z = = P(Z < -2.45) = .0071c) Probability that the sample proportion is between .52 and .66

P( < Z < z = ) = P(-1.63 < Z


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assignment 1 by ricky

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