reserch analysis between Wall's & Omore ice cream by:Mian Shahnnawaz
Post on 27-Apr-2015
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A Research Analysis between
walls and Omore
PRESENTED BY:
Research Analysis between walls and Omore
Questionnaire
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 as the different companies clamed? Chi-Square Goodness-of-Fit Test
Test , “whether male selection preferences differed significantly from female selection preference.“ Chi-Square Test for Homogeneity
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
Frequency Male/Female
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
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
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%
High and Low user
Distributions of High and Low user
MONTHLY_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
Distributions of High and Low user
Current position of different companies ?
FAV_ICE_CREEM * MY_FAVOURATE_BCOZ Crosstabulation
Count
MY_FAVOURATE_BCOZ
QUALITY TASTE PRICE Total
WALLS 29 15 15 59 FAV_ICE_CREEM
OMORE 7 24 10 41
Total 36 39 25 100
Current position of different companies ?
How Monthly Wall’s Ice-cream Expense Depends on Age?
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 .000a
Residual 7344.219 98 74.941
Total 10194.110 99
a. Predictors: (Constant), MONTHLY_EXPENCE_ICECREEM
b. Dependent Variable: AGE
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
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.
Problem#1• Different Ice Cream companies
claim that 65% of the market share is covered by Walls, and 35% is cover by Omore. Total 100%
• Suppose a randomly-selected 100 customers of different companies shows that Walls 59% , and Omore got 41%
• Tell us Is this consistent with Telecom companies 's claim? Use a 0.05 level of significance.
• State the hypotheses.
The first step is to state the null hypothesis and an alternative hypothesis.
– Null hypothesis: The proportion of customers are as fellow, 65% of the customers are of walls, and 35% Omore respectively.
–Alternative hypothesis: At least one of the proportions in the null hypothesis is false.
`
• Analyze Sample Data.
Applying the chi-square goodness of fit test 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.
where DF is the degrees of freedom, k is the number of levels of the categorical variable, n is the number of observations in the sample, Ei is the expected frequency count for level i, Oi is the observed frequency count for level i, and Χ2 is the chi-square test statistic.
• DF= K= 2-1=1
(Ei) = n * pi
(E1) = 100 * 0.41 = 41(E2) = 100 * 0.59 = 59
• Χ2 = Σ [ (Oi - Ei)2 / Ei ]
Χ2 = [ (59 - 65)2 /65 ] + [ (41 - 35)2 / 35 ]
Χ2 = [ 0.554 ] + [ 1.029]
Χ2 = 1.58
We use the Chi-Square Distribution Calculator to find P(Χ2 > 1.58) = 0.2088
• Interpret results.
Since the P-value (0.2088) is more than the significance level (0.05),
we can accept the null hypothesis.
• Means Accept:Null hypothesis: The proportions in the null hypothesis is true.
Chi-Square Test for Homogeneity
• This lesson explains how to conduct a chi-square test of homogeneity.
The test is applied to a single categorical variable from two different populations. It is used to determine whether frequency counts are distributed identically across different populations.
• In a survey of Ice Cream companies, How people chose preferences? we might ask respondents to identify their favorite company .
We might ask the same question of two different populations, such as males and females.
We could use a chi-square test for homogeneity to determine whether male selection preferences differed significantly from female selection preference.
• State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
• Null hypothesis: The null hypothesis states that 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 Walls H0: Pboys who prefer Omore = Pgirls who prefer Omore
Alternative hypothesis:At least one of the null hypothesis statements is false.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square test for homogeneity.
Analyze sample data. Applying the chi-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.
FAV_ICE_CREEM * SEX Crosstabulation
Count
SEX
MALE FEMALE Total
WALLS 33 26 59 FAV_ICE_CREEM
OMORE 28 13 41
Total 61 39 100
FAV_ICE_CREEM * SEX Crosstabulation
SEX
MALE FEMALE Total
Count 33 26 59 WALLS
Expected Count 36.0 23.0 59.0
Count 28 13 41
FAV_ICE_CREEM
OMORE
Expected Count 25.0 16.0 41.0
Count 61 39 100 Total
Expected Count 61.0 39.0 100.0
Interpret results. Since the P-value (0.213) is more than the significance level (0.05), we can accept the null hypothesis.
Alternative hypothesis: The null hypothesis statements is true
Chi-Square Tests
Value df
Asymp. 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 .150
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
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