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Dataset Travel
Question 1. Run a crosstabs using the variables BRAND and loyalty. What do the results tell you?
Null hypothesis: Ho: Customers are not loyal to the brandAlternative hypothesis: H1: Customers are loyal to the brand
brand * loyalbin Crosstabulation
loyalbinTotal
01
brand348Count23479313
Expected Count264.049.0313.0
% within loyalbin18.5%33.6%20.9%
349Count22081301
Expected Count253.847.2301.0
% within loyalbin17.4%34.5%20.1%
350Count26238300
Expected Count253.047.0300.0
% within loyalbin20.7%16.2%20.0%
351Count28621307
Expected Count258.948.1307.0
% within loyalbin22.6%8.9%20.5%
352Count26316279
Expected Count235.343.7279.0
% within loyalbin20.8%6.8%18.6%
TotalCount12652351500
Expected Count1265.0235.01500.0
% within loyalbin100.0%100.0%100.0%
Chi-Square Tests
ValuedfAsymp. Sig. (2-sided)
Pearson Chi-Square91.486a4.000
Likelihood Ratio94.3314.000
Linear-by-Linear Association79.0521.000
N of Valid Cases1500
a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 43.71.
Symmetric Measures
ValueApprox. Sig.
Nominal by NominalPhi.247.000
Cramer's V.247.000
N of Valid Cases1500
Expected Count is different from observed count for all the brands. Also, the Chi square value is less than alpha value. Hence, there is a relationship between brand and loyalty. P value is significant at the confidence interal of 95%(=.05) . Hence we accept the alternative hypothesis.
Question 2. Delete the brands associated with UK and AirUSA. Rerun the crosstabs. What do the results tell you?Assuming brand 349 to be UK and brand 350 to be AirUSA- Null hypothesis-Ho: Customers are not loyal towards the brandAlternative hypothesis- H1: Customers are loyal towards the brands.
brand * loyalbin Crosstabulation
loyalbinTotal
01
brand348Count23479313
Expected Count272.640.4313.0
% within loyalbin29.9%68.1%34.8%
351Count28621307
Expected Count267.439.6307.0
% within loyalbin36.5%18.1%34.1%
352Count26316279
Expected Count243.036.0279.0
% within loyalbin33.6%13.8%31.0%
TotalCount783116899
Expected Count783.0116.0899.0
% within loyalbin100.0%100.0%100.0%
Chi-Square Tests
ValuedfAsymp. Sig. (2-sided)
Pearson Chi-Square65.184a2.000
Likelihood Ratio62.0202.000
Linear-by-Linear Association62.8031.000
N of Valid Cases899
a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 36.00.
Symmetric Measures
ValueApprox. Sig.
Nominal by NominalPhi.269.000
Cramer's V.269.000
N of Valid Cases899
Analysis:Expected Count is different from observed count. P value is significant at the confidence interval of 95%(=.05) . Hence we accept the alternative hypothesis.
DataSet: FAST
Question 1: What statistical analysis is suitable to measure brand equity with the collected data? Why?
Question 2
Question 3 Analyse a Fast Food Brand to determine relationship between Loyalty and Respondents profile (eg. Age, Region, Income).Brand 263Independent variable- Region, Age , IncomeDependent variable- Loyalty Null Hypothesis: There is no relation between independent and dependent variableAlternative Hypothesis: There is a relation between independent and dependent variableFor Region and Income We use CrossTabs. For Age we will use Discriminant AnalysisRegion:
If there is NO relationship between the variables, the observed and expected frequencies will be the same. For our example, this is clearly not the case.And hence, there is a relationship between loyalty and relationship.
Chi-Square Tests
ValuedfAsymp. Sig. (2-sided)
Pearson Chi-Square75.067a27.000
Likelihood Ratio80.09227.000
Linear-by-Linear Association7.8901.005
N of Valid Cases283
a. 15 cells (37.5%) have expected count less than 5. The minimum expected count is .82.
Symmetric Measures
ValueApprox. Sig.
Nominal by NominalPhi.515.000
Cramer's V.297.000
N of Valid Cases283
Pearsons Chi square value is less than the alpha value . there is a relationship between the variables based on the level of confidence. We accept alternative hyphothesis.
Income:
Chi-Square Tests
ValuedfAsymp. Sig. (2-sided)
Pearson Chi-Square16.586a27.941
Likelihood Ratio17.62627.915
Linear-by-Linear Association.4681.494
N of Valid Cases283
a. 11 cells (27.5%) have expected count less than 5. The minimum expected count is 2.37.
Symmetric Measures
ValueApprox. Sig.
Nominal by NominalPhi.242.941
Cramer's V.140.941
N of Valid Cases283
Here, though the observed and expected counts are different but the Chi square value is greater than our alpha value. Thus the difference is due to sampling error.Null hypothesis is accepted. Income and loyalty are not related.
Age:
Question 4 Ariel created binary variables for familiarity, uniqueness, relevance, loyalty and popularity by splitting responses into high and low. Why would they would choose to do (or not do) this? In other words, what information is gained and what information is lost? Ariel research created binary variables by splitting responses into high and low. They have considered various factors while calculating brand equity .if they would have chosen exact numbers(like 1 to 7) or (8,9,10),then it was even more difficult to analyse the data. In fact the data consists of 125000 records. By creating binary variables, they became somewhat comfortable in analysing data. By creating binary variables for responses, exact information for a brand got lost. If we are measuring a data on a scale of 1 to 7(i.e. for low) there can be a huge difference between 1 and 7 but in this data sheet they are clubbed into same category as low. Now this creates confusion while analysing data.1 and 7 can be extreme values but they have been grouped into one category.
Question 5 Do You agree Ariel's Measure of BRAND EQUITY?Brand Equity is the The value premium that a company realizes from a product with a recognizable name as compared to its generic equivalent. The variables that Ariel has used can fairly capture the associations that he customers have with the brands and whether they think that the brand is reliable, unique and worthy of a premium price and hence can measure brand equity .