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cusomer satsfaction survey analysis

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  • #28773

    Sridhar
    Member

    Hi,
            We did a customer satisfaction survey in our company. Before doing the survey we categorized our population into different stratas i.e different plot, different grade. We selected the sample also in the similar manner i.e. stratified sampling plan. Now i have with me the no. of responses on a scale of 1 to 7 for all the combinations of plot and Grade ( i.e we have 4 plots and 3 grades, so i have all 12 combinations responses).I did all the simple analysis like bar charts etc.
          Can i apply ANOVA technique here, if so what kind of transformations i have to make. How to test the hypothesis that Is there any difference between plots or grades.
           Please provide me some infromation. If anybody has done similar analysis please give me their personal id.
    thanks in advance
    A.Sridhar
     

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    #72166

    Erik L
    Participant

    Sridhar,
    There’s a couple different approaches that you can take here.  First you can take all 12 responses as your rows and the satisfaction scores as columns and perform a 12*7 contingency table analysis.  The test would be for independence utilizing the Chi-square distribution.  A ‘yes’ implies that what is in the rows is not related to the column variable.  A ‘no’ response tells us that there is some association, but not its strength.  Since r+c>7, I’d go for Chi-sqaure over the Fisher-Irwin calculation.  FI calculates a p-value via the hypergeometric distribution.  It calculates a probability that the observed deviation from independence would occur by chance alone.  A small p-value indicates that factors other than chance are occurring.  However, it should be used in low count situations.  In large count situations the Chi-square will For the Chi-square approximation to hold you should look for the following attributes:  all cell entries at least 5 and the expected values for all cells should be at least 1.  Or, you can take the major classifications of ‘plot’ and ‘grade’ and perform a 2*7 analysis.  This would be a similar test for independence and probably where I’d start the analysis.
     
    Regards,
    Erik

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    #72193

    Sridhar
    Member

    Hi Erik,
         Tahnks for the reply. I did try this method, but  the problem is most of the cells i got or less thank 5 and i dont know what to do. Can you suggest what to od if i have celss less than 5.
          Another question is because i am having all the combinations can i apply ANOVA(DOE) to fine out which plot and which grade gives me optimal satisfaciton i.e what i mean to say that if i get plot and grade for which i have optimum response then i will provide the similar kind of facilities to other plots and other grades also.
            Can i think of such analysis if so what changes i have to make in the response column, because my response is of attribute type.
    thanks in advance
    sridhar
     
     

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    #72213

    Erik L
    Participant

    Sridhar,
    If you have cell entries less than 5 you could always go with the Fisher-Irwin analysis. There certainly could be ways to model the data for DOE application. When you have data in an attribute type response there are three transformations that could be applied. The easiest transformation is a rank transformation. Even if the data is continuous in nature, I usually push my BBs to make a parallel analysis with a rank transform. It’s non-parametric and is a good gauge to see if the response from the standard factorial approach is satisfying the prerequisites for analysis. If your response is a proportion or a count, we start right off with a violation of one of the core DOE principles. Our variance is going to be tied to the mean response. So, if we achieve the desired effect to our process when we ‘kick it’ the variance is going to fluctuate. This is a violation of the standard analysis. The following transformations can be used to steady the variance. For proportions, with an underlying distribution of binomial, you can start with the transform
    With this transform, you’ll typically lose linearity of the variance function below phat=.2 and above phat=.8 To extend the linearity you could use Freeman-Tukey’s Modification:
    Stability should now extend from phat =0.05 to phat=.95
    For counts use:
    Freeman-Tukey’s modification is:
    This transform is recommended when you have a lot of 0 in the data set. You could also add a small value 0.00001 to the zeroes and use the square root of the observed count. Hope this helps.
    Regards,
    Erik
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    #72214

    Erik L
    Participant

    Ok,
    Cut and paste didn’t work well.  I’ll send an e-mail to the forum via another format so that you can view the transforms.
    Regards,
    Erik

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    #72236

    Sridhar
    Member

    Hi Erick,
           Thanks for the reply. Can you please give me your e – mail id , i will be in touch with you if i need any clarifications.
    thanks
    sridhar
     

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    #72243

    Erik L
    Participant

    Sridhar,
    Business:  [email protected]
    Home:  [email protected]
    Look forward to hearing from you.
     
    Regards,
    Erik
     

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