31076449 reserch analysis between wall s omore ice cream by mian shahnnawaz

7/30/2019 31076449 Reserch Analysis Between Wall s Omore Ice Cream by Mian Shahnnawaz

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A Research Analysis between

walls and Omore

PRESENTED BY:


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Research Analysis between

walls and OmoreQuestionnaire

Descriptive statistics

Distributions of High and Low user

Current position of different companies

Analysis of Quality, Taste and Price

How Monthly Ice-cream Expense Depends on Age?

Test ,whether the proportion of market share is same asthe different companies clamed?Chi-Square Goodness-of-Fit Test

Test , whether male selection preferences differedsignificantly from female selection preference.

Chi-Square Test for Homogeneity


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This Data is Belongs to Students,Businessmen, and Job Holder

PROFESSION

Frequency Percent Valid Percent

Cumulative

Percent

STUDENT 40 40.0 40.0 40.0

BUSINESS MAN 27 27.0 27.0 67.0

JOB HOLDER 31 31.0 31.0 98.0

OTHER 2 2.0 2.0 100.0

Valid

Total 100 100.0 100.0


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Frequency Male/FemaleSEX


Cumulative

Percent

MALE 61 61.0 61.0 61.0

FEMALE 39 39.0 39.0 100.0

Valid

Total 100 100.0 100.0


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SEX


Cumulative

Percent

MALE 61 61.0 61.0 61.0

FEMALE 39 39.0 39.0 100.0

Valid

Total 100 100.0 100.0


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SEX * FAV_ICE_CREEM Crosstabulation

FAV_ICE_CREEM

WALLS OMORE Total

Count 33 28 61

Expected Count 36.0 25.0 61.0

MALE

% within SEX 54.1% 45.9% 100.0%

Count 26 13 39


SEX

FEMALE

% within SEX 66.7% 33.3% 100.0%

Count 59 41 100


Total

% within SEX 59.0% 41.0% 100.0%


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High and Low user


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Distributions of High and

Low userMONTHLY_EXPENCE_ICECREEM * FAV_ICE_CREEM Crosstabulation

Count

FAV_ICE_CREEM

WALLS OMORE Total

MONTHLY_EXPENCE_ICEC

REEM

100.00 8 2 10

200.00 6 4 10

500.00 10 11 21

1000.00 12 6 18

1500.00 8 8 16

2000.00 10 8 18

2500.00 5 2 7

Total 59 41 100


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Distributions of High and

Low user


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Current position of different

companies ?

FAV_ICE_CREEM * MY_FAVOURATE_BCOZ Crosstabulation

Count

MY_FAVOURATE_BCOZ

QUALITY TASTE PRICE Total

WALLS 29 15 15 59FAV_ICE_CREEM

OMORE 7 24 10 41

Total 36 39 25 100


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Current position of differentcompanies ?


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How Monthly Walls

Ice-cream ExpenseDepends on Age?


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Model Summary

Model R R Square

Adjusted R

Square

Std. Error of the

Estimate

1 .529a

.280 .272 8.657

a. Predictors: (Constant), MONTHLY_EXPENCE_ICECREEM

ANOVAb

Model Sum of Squares df Mean Square F Sig.

1 Regression 2849.891 1 2849.891 38.028 .000

a

Residual 7344.219 98 74.941

Total 10194.110 99

a. Predictors: (Constant), MONTHLY_EXPENCE_ICECREEM

b. Dependent Variable: AGE


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Y= a+ bX

Y = 14.320 +3.072(X)Monthly expence of

wall ice cream =14.320+ 3.072(Age)

Rs.107 = 14.320+ 3.072(30 years)

NOTE : Age zero , Makes no sense

Coefficientsa

Model

Unstandardized Coefficients

Standardized

Coefficients

B Std. Error Beta t Sig.

1 (Constant) 14.320 2.182 6.563 .000

MONTHLY_EXPENCE_ICEC

REEM

3.072 .498 .529 6.167 .000

a. Dependent Variable: AGE


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Chi-Square Goodness-of-

Fit Test

The test procedure described in this

lesson is appropriate when the

following conditions are met:

The sampling method is simple

random sampling.

The population is at least 10 times

as large as the sample. The variable under study is

categorical.

The expected value for each level

of the variable is at least 5.
http://stattrek.com/Help/Glossary.aspx?Target=Simple%20random%20samplinghttp://stattrek.com/Help/Glossary.aspx?Target=Simple%20random%20samplinghttp://stattrek.com/Help/Glossary.aspx?Target=Categorical%20variablehttp://stattrek.com/Help/Glossary.aspx?Target=Levelhttp://stattrek.com/Help/Glossary.aspx?Target=Levelhttp://stattrek.com/Help/Glossary.aspx?Target=Categorical%20variablehttp://stattrek.com/Help/Glossary.aspx?Target=Simple%20random%20samplinghttp://stattrek.com/Help/Glossary.aspx?Target=Simple%20random%20sampling


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Problem#1

Different Ice Cream companies

claim that 65% of the market

share is covered by Walls, and

35% is cover by Omore. Total100%

Suppose a randomly-selected 100

customers of different companiesshows that Walls 59% , and

Omore got 41%

Tell us Is this consistent with

Telecom companies 's claim? Use

a 0.05 level of significance.


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State the hypotheses.The first step is to state the nullhypothesis and an alternative

hypothesis.

Null hypothesis: The proportionof customers are as fellow, 65%

of the customers are of walls, and35% Omore respectively.

Alternative hypothesis: Atleast one of the proportions inthe null hypothesis is false.


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`

Analyze Sample Data.

Applying the chi-square goodness

of fit test to sample data, wecompute the degrees of freedom,

the expected frequency counts, and

the chi-square test statistic.

Based on the chi-square statistic

and the degrees of freedom, we

determine the P-value.
http://stattrek.com/Help/Glossary.aspx?Target=Degrees%20of%20freedomhttp://stattrek.com/Help/Glossary.aspx?Target=P-valuehttp://stattrek.com/Help/Glossary.aspx?Target=P-valuehttp://stattrek.com/Help/Glossary.aspx?Target=P-valuehttp://stattrek.com/Help/Glossary.aspx?Target=P-valuehttp://stattrek.com/Help/Glossary.aspx?Target=Degrees%20of%20freedom


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where DF is the degrees of freedom, k is the numb

of levels of the categorical variable, n is the numbe

of observations in the sample, Ei is the expected

frequency count for level i, Oi is the observed

frequency count for level i, and 2 is the chi-square

test statistic.

DF= K= 2-1=1

(Ei) = n * pi

(E1) = 100 * 0.41 = 41

(E2) = 100 * 0.59 = 59


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2= [ (Oi - Ei)2 / Ei ]

2 = [ (59 - 65)2 /65 ] + [ (41 - 35)2 / 35 ]

2 = [ 0.554 ] + [ 1.029]

2 = 1.58

We use the Chi-Square Distribution Calculator

o find P(2 > 1.58) = 0.2088
http://stattrek.com/Tables/ChiSquare.aspxhttp://stattrek.com/Tables/ChiSquare.aspxhttp://stattrek.com/Tables/ChiSquare.aspxhttp://stattrek.com/Tables/ChiSquare.aspx


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Interpret results.

Since the P-value (0.2088) is more tha

the significance level (0.05),

we can accept the null hypothesis.

Means Accept:

Null hypothesis: The proportions inthe null hypothesis is true.


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Chi-Square Test for

Homogeneity

This lesson explains how toconduct a chi-square test ofhomogeneity.

The test is applied to a singlecategorical variable from two

different populations. It is used to

determine whether frequency

counts are distributed identicallyacross different populations.
http://stattrek.com/Help/Glossary.aspx?Target=Categorical%20variablehttp://stattrek.com/Help/Glossary.aspx?Target=Categorical%20variable


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State the hypotheses. The first step is tostate the null hypothesis and an alternative

hypothesis.

Null hypothesis: The null hypothesis statesthat the proportion of boys who prefer the

Walls is identical to the proportion of girls.

Similarly, for the Omore. Thus

H0: Pboys who prefer Walls = Pgirls who prefer WallsH0: Pboys who prefer Omore = Pgirls who prefer Omore

Alternative hypothesis:At least one of the null hypothesis statemen

is false.


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Formulate an analysis plan. For thisanalysis, the significance level is 0.05.

Using sample data, we will conduct a chi-

square test for homogeneity.

Analyze sample data.Applying thechi-square test for homogeneity to sample

data, we compute the degrees of

freedom, the expected frequency counts,and the chi-square test statistic. Based on

the chi-square statistic and the degrees of

freedom, we determine the P-value.
http://stattrek.com/Help/Glossary.aspx?Target=Chi-square%20test%20for%20homogeneityhttp://stattrek.com/Help/Glossary.aspx?Target=Chi-square%20test%20for%20homogeneityhttp://stattrek.com/Help/Glossary.aspx?Target=Degrees%20of%20freedomhttp://stattrek.com/Help/Glossary.aspx?Target=Degrees%20of%20freedomhttp://stattrek.com/Help/Glossary.aspx?Target=P-valuehttp://stattrek.com/Help/Glossary.aspx?Target=P-valuehttp://stattrek.com/Help/Glossary.aspx?Target=P-valuehttp://stattrek.com/Help/Glossary.aspx?Target=P-valuehttp://stattrek.com/Help/Glossary.aspx?Target=Degrees%20of%20freedomhttp://stattrek.com/Help/Glossary.aspx?Target=Degrees%20of%20freedomhttp://stattrek.com/Help/Glossary.aspx?Target=Chi-square%20test%20for%20homogeneityhttp://stattrek.com/Help/Glossary.aspx?Target=Chi-square%20test%20for%20homogeneityhttp://stattrek.com/Help/Glossary.aspx?Target=Chi-square%20test%20for%20homogeneity


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FAV_ICE_CREEM * SEX Crosstabulation

Count

SEX

MALE FEMALE Total

WALLS 33 26 59FAV_ICE_CREEM

OMORE 28 13 41

Total 61 39 100

FAV_ICE_CREEM * SEX Crosstabulation

SEX

MALE FEMALE Total

Count 33 26 59WALLS


Count 28 13 4

FAV_ICE_CREEM

OMORE


Count 61 39 100Total



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nterpret results. Since the P-value (0.213) ismore than the significance level (0.05), we can

ccept the null hypothesis.

Alternative hypothesis:he null hypothesis statements is true

Chi-Square Tests

Value dfAsymp. Sig. (2-

sided)Exact Sig. (2-

sided)Exact Sig. (1

sided)

Pearson Chi-Square 1.554a

1 .213

Continuity Correctionb

1.077 1 .299

Likelihood Ratio 1.570 1 .210

Fisher's Exact Test .297 .1

N of Valid Cases 100

a. 0 cells (.0%) have expected count less than 5. The minimum expected count is 15.99.

b. Computed only for a 2x2 table

31076449 reserch analysis between wall s omore ice cream by mian shahnnawaz

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