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Case Study on the Chevrolet Cars’ Market

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ANOVA in marketing researches – Case study on the Chevrolet cars’ market In order to make an analysis of the variation regarding ANOVA, we considered the commercial society S.C. RUCOM S.A. from Craiova, Dolj, 107 Caracal Street, dealer in the sale of cars mark Opel, even from 2002. Starting with 2004, S.C. RUCOM S.A. becomes dealer for cars mark Chevrolet too, cars that first appear on the market in Oltenia that year, being, as a matter of fact, not that known on the Romanian market, in general. The Chevrolet cars, being them from the variety Kalos, Lacetti or Evanda, but especially the ones from the variety Kalos, present a few attractive characteristics for the potential clients. Thus, in the first place, the Chevrolet Kalos cars are …show more content…

In order to apply this method, the formulation of the work hypotheses is imposed, respectively the null hypothesis, that is about to be verified, and the alternative hypothesis. H0: there are no differences between the car sale averages from the three counties. H1: at least two averages are different from one another. The response variable is concretized in the monthly sales and the experimental unity is represented by the months when we register the sales’ figure in those three counties. The registered sales’ figure represents the responses. As we can observe, there is a single factor, the approach of publicity, that defines the populations and there are three levels of this factor. These levels are: - publicity by emphasizing the advantage, the convenience; - publicity by emphasizing the quality; - publicity by emphasizing the price. By applying the variation analysis method, more explicitly presented in subchapter 3.2. of this chapter, we will proceed to the effectuation of the calculi in order to obtain the value of the signification test F of Snedecor. According to the relation (3.30), F = MST . MSE We will calculate the averages of the samples and the general average and we will obtain: x 1 = 8,08; x 2 = 6,42; x 3 = 7,67 x = 7,39 Therefore, the value of SST, respectively the factorial variation, will be: SST = ∑ n j x j − x = 12(8,08 − 7,39) + 12(6,42 − 7,39) + 12(7,67 − 7,39) = 18,05 k 2 2 2 2 j =1 ( ) In

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