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Lecture Notes

All lecture notes, handwritten from Statistics for Business. Including...
Module

Statistics for business I (MA20228)

3 Documents
Students shared 3 documents in this course
Academic year: 2022/2023
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University of Bath

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10/02/

Pre lecture notes

lecture 1

random

exp

=

outcome

can't

be

predicted

with

certainty

can be

repeated

indefinitely

,

under

same

conditions

random variable

=

a

number

given

to each outcome

often

annotated

as

✗ or

Y

>

depends

how

many

RV's there are

make

quantitative

treatment

of

RE's

possible

RV is

often

capital

letter

,

the values of

RV

=

lower

case

>

symbol

convention

discrete RV

finite

number of

values

continuous RV

=

infinite

number of

values

(

probability

)

distribution

=

the

pattern

the RV

's

follow

normal

,

uniform

,

binomial

, Poisson

,

hypergeometric

distribution

for

DRV

=

✗

→

Pi

=

f-

Gci)

lgas

signs

probability

probability functions

valyespea.e

,

¥÷¥¥¥Ñ¥¥

(

specific

)

discrete

uniform distribution

=

finite

number

of

values are

equally

likely

to

be

observed

flock

In

e.

g. throwing

a

dice

> PCX -41=

,

PC✗=2)=Y

,

PCX-31= .. .

>

fG)=f

(2)

if

(3)

=fGD=fG)=fG)

Independent

random variables

= 2

variables

that aren't

correlated

correlation

between

RV

's ✗ andY

=

Pxx

> =

measures

linear

relationship

between 2 Variables

>

between

  • and

=

XP

,

YI * 2

IRV

'S

=

P✗Y=O

=

✗

F

,

YA but

1=

=

perfect

correlation PXY-

=/

2

IRV

's

0

no correlation

mean

aka

expected

value

of t/---ECXJ

Mean

of

prob

dist

=

μ✗=¥xifGc

;)

←

for

DRV

variance

= ok

-_

spread

of

Rv

around

μ

####### %¥¥£Éμ×B=§cxi

μ×Y

fix;)

""

Rules

i

mean

of

sum of

2 RVs

= sum of

means of

the individual RV

's

μ×+y= Mx

My

mean

of

the

product

of

RV and

a constant (a)

=

product

of

constant

and

mean of

RV

μ

ax

= a

Mx

:

μ✗ 2 ✗

=

2 ×μ✗

variance of

product

of

RV

by

constant

=

product

of

var of RV

and

square

of

constant

Otay

=

of0k

23/02/

Pre

lecture notes

lecture

2

Binomial Distribution

binomial

experiment

=

n

repeated

trials

,

with each

trial

having

2 outcomes

(

success or failure

)

PCs

)

=p

and

PCF )

= 1-

p

trials are

independent

binomial

RV

=

number of successful

outcomes

bbc.n

=

p41-p)n-x

x=number

of

successful

outcomes

a-

number

of

trials

p

probability

of

success

1%

number of

combinations

'

n

choose X

'

n

!

(E)

=

so :(n

x):

e.

g.

if

A-

4 and x=

:

4

!

2

! (4-2)

!

→

4 × 3 × 2 × 1

→

2 ¥

=

f

2 × 1 × 2 × 1

anything

choose 0=

☒

I choose 1

=

x

coin

example

=

coin

is

flipped

5

times

,

calculate

probability

distribution where

head

is success.

bloc :-b

,

)

(E)

pxc1-pln-xpb

)

=

(F)

0--1× 1 ×0=0.

Pfx= 1)

=

(7)

0.540=5×0× 0 -0625=0.

Pbc

:

=

(E)0.540=10×0×0--0.

PGc=

3)=

(3)0-512=10×0×025--

PCx= 47=(110-5)

'

=

5 ×0×0=0-

Pfx=5)=(

E.)

0-59=1×0×1--0.

mean

=

μ×=np

e.

g.

coin

example

:

μ×=

5 ×0-5--

variance

ok

=

npctp

)

e.

g.

coin

example

:

02--2-5×0=1.

binomial

proportion

=

number

of

successes

÷

number

of

trials

:

p= In

μp=p

o§=pa

pl

n

need

to calculate 2

values

z=

X

M

2cg

ample

)=fn

'

0

example

: time

taken

by

bus =

✗

,

mean

= 32 mins

,

var

=

100min

,

bus is late

if

it

takes

over

37 Mins .

What is

prob

bus

is

late

?

2--37-

=

51-0--0-

Foo

PIX >

37 )

=

Plz

>

0=

Plz e-

0%1-0-6915=

7

7

9

7

Pllate

)

Pllate

plnot late)

2 score at

05 on

2 table

= bus is

late 31%

of

times

B)

how

long

does slowest 20%

journeys

last

?

→

all

times

greater

than

xo

=

20%

slowest

PCX

>

xo)

= 0. so ☒

e-

Xo

)

=

→

on z

table

,

0=0.

so

,

DC -

=

0-

-1> 10 (0-84)-132=40.

10

=

20%

of

slowest

journeys

are

at least

404

mins

213122

pre

lecture notes lecture 3

Population

and sample

a

sample

is a small

part

of

population

population

full

set

of

individuals we want

to draw conclusions

from

sample

mean will

be

different

to

population

mean

different samples

have different means

I

=L

?§xi

=

'ñfX

,

-1×2+14....

  • '

XD

Central

Limit Theorem

central

limit theorem

=

when

sample

size

is

large

cnzz

,

the

sampling

distribution

of

the

sample

mean

is

very

close to normal distribution

if

pop

.

mean

=

μ

,

and

pop

. var

= o?

the

sampling

distribution

of

sample

mean

=

I~NCM

example

:

Time taken

for

bus

journey

is norm

.

dist

. with

μ

-32mins

and

0=10 mins

.

One week

,

a sample

of

35

buses

gave

mean

of

335

.

A- 35

,

I

=

335

sampling

dist

of

E= 33 NC

,

¥1s)

→ INN

(32,1-69)

E)

z=✗¥✗n

=

335,2=13*

,

= 0.

=

0-

1- 0-

=

018943

Point Estimation lecture

4

recap

=

central

limit theorem

states

if

a

sample

size is above

30

,

Jc

is close

to

μ

,

and

§

is

close to

p

sample

variance

=s

5=?§(

Xi

  • x-P

n

l

the S2 is

= 02 .

so

S2 is

a

point

estimator

for

02

>

5C is

point

estimator

for

μ

>

§

is

point

estimator

for p

parameter

point

estimator

M

ñ=¥

point

estimators are

helpful

when

parameter

is

02

5=?§CXi

  • IT

Unknown

n

I

p

§

In

Interval

Estimation

used because

point

estimation

doesn't

provide

sufficient statistical

information

.

interval estimation uses

sample

data

to estimate

an

interval

of

possible

values

of

a

parameter

of

interest

point

est

only

gives

a

single

value

focusing

on

5C

Where n

>

30

I can have

any

value

,

but

it's

likely

dose to

μ

.

We want

to

quantify

this closeness

by

finding

an

interval Jc

and

μ

are

both in

¥¥'¥÷¥g

95%

probability

that distance

between Jc and

μ

is not

greater

i than distance between Xu or xrand

μ

"

÷

"

"

"

"

→

si.

probability

that

I is outside

shaded area

to

find

interval

of

951

.

confidence

.

Must

find

either

μ

Xe

or

Xr

M

4

>

these

lengths

are

identical to

I

Ice or Er

I

>

where Ir and

Ecu are

the

extremes of

the

interval we are

looking

for

5C

za

,

%sμ<

I

-12%

Ern

example

:

insurance

company

estimates

mean

age

of

policy

holders .

A

survey

of

36

samples

has mean

age

of

39s and variance of

  1. What is

estimation

for

average age

at 90%

confidence

?

g-

fμz%nEμEÑ+t%%

two

, two tailed

.

00s one

tailed

df=

36-1=

895-4%

e-

ME

395+4%

3732M£

41-

Interval Estimation for

Proportion

use when

trying

to

find

a

proportion

within a

sample

⑧ -2%

a-

PED

-12%

example

:

⑧

~

Nlp ,

JET

☒

=

In

=

7*0=

o

=

if

=

=

034-1-96×0.

EPE

0341-1-96×0.

EPE

example

:

E- 2180

,

0=

,

n=3O

Ho

:p

-2000 vs Hi

:μ

✗

St.

=

Two

Sample Hypothesis Te s t i n g

: lecture

7

(6)

  • 2

independent populations

example

:

30 CEO

's

.

12 women

and 18 men

showed

average salary

for

women was 76 million

and

average

salary

for

men was

  1. Million .

Means are

unknown

,

but

of

=

12

million

and on

.-1

.

Test at a-8.

critical

if

average

Womens

salary

differs

from

mens.

awe

I. ~N(

μ

,,

%-) Xin

Nlpk

,

%)

difference of

difference

-_

I

,

Ez~N(

μ

,

μ

,

o÷Y

sample

means

test

hypothesis

:

Ho

=

Mm

μf=O

Hi

=

Mm

Mf

=/

0

←

%n÷¥¥ =

E#--2-Hceo=MwdeepaidCV--

1-

214>1-

,

so

reject

tf

"

"

"

/

testtaistiu

Two

sample

testing

with unknown

and

equal

variances

when

the

population

variances are unknown we use

the

sample

standard deviations

G) and calculate

the mean

weight

.

The

mean

weight

is

called the

pooled

variance

estimate Gip

)

5p=

In

,

-11s

:

(

Nz

  • Ps

N ,

  • Nz -

We must use

the t

score .

E-

Ii-Ez_

szp%+ntÑ

the

CV

can

be

found

on table

where

DF

= At Nz

example

:

independent

random

samples of checking

account

balances

a

2 branches showed

:

Branch aNE-oud-ntssmaemakespsa ,

A

,

Yo

'

a

¥ 8

test at £-

.

It= 1000 ÑB

= 920

Sa

= 150 515-120 AA=

12 NB=lO

§p= (12-1)

is021-(10-1) 1202 = 18855

12+10-

t.io/z-7to--10oo-92O---

1 D.

F- 20

.

I sided test so E- oros

=

1-

FEE :o)

as

taboos ,

we can't

reject

null

hypothesis

.

When we

are

using

proportions

instead

of

means

,

we use

the

pooled probability

of success

④

⑤

=p ,§

, the$ 2

N

,

1- Nz

and the

pooled

variance

becomes

:

5-a

-0-(1-8)

ftp)

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Lecture Notes

Module: Statistics for business I (MA20228)

3 Documents
Students shared 3 documents in this course

University: University of Bath

Was this document helpful?
10/02/22
Pre
lecture
notes
lecture
1
random
exp
=
outcome
can't
be
predicted
with
certainty
↳
can
be
repeated
indefinitely
,
under
same
conditions
random
variable
=
a
number
given
to
each
outcome
↳
often
annotated
as
✗
or
Y
>
depends
how
many
RV
's
there
are
↳
make
quantitative
treatment
of
RE
's
possible
↳
RV
is
often
capital
letter
,
the
values
of
RV
=
lower
case
>
symbol
convention
discrete
RV
-
-
finite
number
of
values
continuous
RV
=
infinite
number
of
values
(
probability
)
distribution
=
the
pattern
the
RV
's
follow
↳
normal
,
uniform
,
binomial
,
Poisson
,
hypergeometric
distribution
for
DRV
=
✗
→
Pi
=
f-
Gci
)
lgas
signs
probability
*
probability
↳
functions
valyespea.e.ie
,
¥÷¥¥¥Ñ¥¥
(
specific
)
discrete
uniform
distribution
↳
=
finite
number
of
values
are
equally
likely
to
be
observed
↳
flock
In
↳
e.
g.
throwing
a
dice
>
PCX
-41=8
,
PC✗=2)=Y6
,
PCX
-31=46
.
.
.
>
fG)=f
(2)
if
(3)
=fGD=fG)=fG)

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