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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
Frequency Percent Valid Percent
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
Frequency Percent Valid Percent
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
Expected Count 23.0 16.0 39.0
SEX
FEMALE
% within SEX 66.7% 33.3% 100.0%
Count 59 41 100
Expected Count 59.0 41.0 100.0
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%20sampling7/30/2019 31076449 Reserch Analysis Between Wall s Omore Ice Cream by Mian Shahnnawaz
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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%20freedom7/30/2019 31076449 Reserch Analysis Between Wall s Omore Ice Cream by Mian Shahnnawaz
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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.aspx7/30/2019 31076449 Reserch Analysis Between Wall s Omore Ice Cream by Mian Shahnnawaz
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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%20variable7/30/2019 31076449 Reserch Analysis Between Wall s Omore Ice Cream by Mian Shahnnawaz
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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%20homogeneity7/30/2019 31076449 Reserch Analysis Between Wall s Omore Ice Cream by Mian Shahnnawaz
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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
Expected Count 36.0 23.0 59.0
Count 28 13 4
FAV_ICE_CREEM
OMORE
Expected Count 25.0 16.0 41.0
Count 61 39 100Total
Expected Count 61.0 39.0 100.0
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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