reserch analysis between wall's & omore ice cream by:mian shahnnawaz

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