Question

Question 1 Test, at the 5% level of significance, whether location (Parramatta or Sydney CBD) of the business is related to the type of business (privately held, publicly traded or franchise)?

Column 1 Profit What was your profit?

Column 2 Type Is the business privately held, publicly traded or a franchise? (Private = 1, Public = 2, Franchise = 3)

Column 3 Employee How many employees work at your business?

Column 4 Age How old is the company (in years)?

Column 5 Location Where is the company? (1 = Parramatta CBD, 2 = Sydney CBD)

(PLEASE ANSWER IN EXCEL)

3 1 3 1 1 1 1 1 1 2 3 3 2 1 2 1 1 3 1 3 1 1 1 1 1 3 1 Profit Type 173000 98000 130000 111000 22000 116000 109000 155000 77000

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Answer #1

Given information summarised into the contingency table as below,

We will use Chi-square test:

Parramatta Sydney Total
Private 28 11 39
Public 7 9 16
Franchise 5 10 15
Total 40 30 70

1590233366828_image.png

To test the hypothesis is that the location of the business is related to the type of business at a 5% level of significance.

Hypothesis:

H0: The location of the business is not related to the type of business.

HA: The location of the business is related to the type of business.

1590233375850_image.png

Chi square test :

Oij= Observed frequencies

Parramatta Sydney Total
Private O11= 28 O12=11 R1=39
Public O21=7 O22=9 R2=16
Franchise O31=5 O32=10 R3=15
Total C1=40 C2=30 N=70

Now we will calculate Expected frequencies as,

Eij = \frac{(Ri*Cj)}{N}

Therefore expected frequencies are

E11=(39*40)/70=22.286 E12=(39*30)/70=16.714
E21=(16*40)/70=9.143 E22=(16*30)/70=6.857
E31=(15*40)/70=8.571 E32=(15*30)/70=6.429

1590233390553_image.png

The Formula for Chi-square:

ChiSquare = x = [ (Oij – Eij)! Eij

Chi Square = \chi ^2 =[(28-22.286)^2/22.286]+[(11-16.714)^2/16.714]+[(7-9.143)^2/9.143]+[(9-6.857)^2/6.857]+[(5-8.571)^2/8.571)]+[(10-6.429)^2/6.429]

\mathbf{Chi Square = \chi ^2 = 8.0629}

1590233400308_image.png

Degrees of freedom:

Formula for Chi square Degrees of Freedom (df) is,

df = (R-1)(C-1) where, R = number of rows and C = number of columns

= (3 -1) * (2 - 1)

df = 2

1590233402532_image.png

P-Value:

P-value calculate using excel function =CHIDIST(8.0629,2)

Here,

8.0629 is a test statistic value

2 indicates degrees of freedom

therefor,

\mathbf{P-Value=0.0177}

1590233406968_image.png

Decision:

P-value(0.0177)<0.05 hence we reject H0

1590233409713_image.png

Conclusion:

P-value < 0.05 we reject H0, Hence we concluded that there is sufficient evidence to indicate that the location of the business is related to the type of business.

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