Business Statistics Assignment

Sheet1

HYPOTHESIS TEST FOR A PROPORTION

THE PROCEDURE IS BASICALLY THE SAME AS FOR THE MEAN. THE ONLY

DIFFERENCE IS IN THE FORMULA FOR Z, SINCE WE ARE USING THE

NORMAL APPROXIMATION TO THE BINOMIAL DISTRIBUTION

and

SINCE WE ARE TESTING AN HYPOTHESIS CONCERNING THE

POPULATION PROPORTION, p, WE USE THAT VALUE TO

CALCULATE SIGMA.

[REMEMBER THE REQUIREMENT FOR USING THE NORMAL APPROXIMATION:

np and nq must be greater than 5.]

1.

po

0.500

x

16.6666666667

p1

n

25

p

0.5

p-hat

0.667

q

0.5

2.

n

25

Z

1.667

-1.667

sigma

0.1000

p-value

0.0478

0.096

2-tail

3.

a

0.05

Z alpha

1.645

4.

0.04

5.

COMPARE THE ACTUAL AND CRITICAL Z VALUES:

SINCE THE ACTUAL Z VALUE OF 2.361 IS GREATER THAN

THE CRITICAL VALUE OF 1.96, IT IS IN THE REJECTION REGION.

6.

THEREFORE, WE DO REJECT THE NULL HYPOTHESIS.

AND WE CONCLUDE THAT THE PROPORTION IS NOT EQUAL

TO 0.36.

II. P-VALUE APPROACH

0.9908871347

0.0091128653

ACCORDING TO THE NORMAL CURVE THE AREA ABOVE A Z-VALUE

OF 2.361 = .009

COMPARE THE P-VALUE WITH THE ALPHA VALUE, AND

THE P-VALUE OF .009 IS LESS THAN THE ALPHA VALUE OF .025

OR, DOUBLING THE P-VALUE OF .009 = .018. COMPARE THIS TO ALPHA OF .05.

THEREFORE, WE DO REJECT THE NULL HYPOTHESIS.

AND WE CONCLUDE THAT THE PROPORTION IS NOT EQUAL

TO 0.36.

Sheet2

Sheet3

ˆ

pp

z

s

-

=

ˆppz

pq

n

s

=

pqn

96

ˆ

0.4364

220

p

==

96ˆ0.4364220p

ˆ

pp

z

s

-

=

ˆppz

.4364.36

2.361

0.03236

z

-

==

.4364.362.3610.03236z

.36*.64

0.03236

220

pq

n

s

===

.36*.640.03236220pqn

0

1

:0.36

:0.36

Hp

Hp

=

¹

01:0.36:0.36HpHp

.05

a

=

.05

1.96

Z

=±

1.96Z

MBD01489F17.unknown

MBD0148DBE6.unknown

MBD0148F3C6.unknown

MBD0148B9FF.unknown

MBD014886A9.unknown

MBD01488BE2.unknown

MBD01484159.unknown

MBD014853CF.unknown

MBD01484158.unknown