Computer R Us Company

Computer R Us Company Introduction The Computer R Us Company received numerous complaints about the services offered in their CompleteCare division. After thorough investigations into the complaints, the management established that the division was experiencing problems as a result of inadequacy of trained operators and problems with distribution and availability of parts. In response to these problems, the management came up with four initiatives that aimed at improving customer satisfaction. In this paper, analysis will be carried using various tools to establish the effectiveness of the initiatives that were put in place.Advertising We will write a custom case study sample on Computer R Us Company specifically for you for only $16.05 $11/page Learn More Research design Sampling technique This research was conducted using research survey study approach. Data was collected using a questionnaire that had three sections. The first part required personal information, that is, age a nd gender. In the second section a Likert scale of ten points was used to collect some data. The final section focused on determinants of customer satisfaction. Four questions were asked in this section and each had a Likert scale of ten points. The random sampling technique was used to select a sample of 500 customers (Kothari, 2004). The questionnaires were sent to the 500 customers and only 420 responded. In order to collect the data necessary for this study, several steps will be taken to ensure that appropriate care is taken to protect the participants. There are no universally accepted determinants of customer satisfaction ((Verbeek, 2008). Besides, the results of previous studies do not give conclusive result on the most effective determinant. Therefore, the attributes used by the management of Computer R Us to improve the level of customer satisfaction are a sample of what other companies use (Zikmund, Babin, Carr, Griffin, 2012). Analysis The first test show that the overal l satisfaction is statistically different from 6 out of 10. The calculated mean is 4.4881 and it is less than the goal. The result of the second question shows that the overall satisfaction of female customers is higher than that of male customers. Therefore, there is a need to improve the level of satisfaction of the male customers. The results of the third question indicate that there is no difference in the level of satisfaction across the different age groups. Further, tests on question five shows that there is no difference in gender composition across the five age groups.Advertising Looking for case study on business economics? Let's see if we can help you! Get your first paper with 15% OFF Learn More The fifth test reveals that customers tend to be more satisfied with the loyalty rewards program than response times in the CompleteCare division. Therefore, the management needs to improve the response time in the division. The final test shows that all the four initiatives have a potential of improving customer satisfaction. Further, response time of the CompleteCare division and level of advice CompleteCare staff provides on Computers R Us products have more impact than the other two initiatives (Baltagi, 2011). Recommendations The results of hypothesis testing show that the management did not achieve their goal. For the company to achieve the target of 6 out of 10, the management needs to consider the recommendations listed below. Decrease the response time of the CompleteCare division. This can be achieved by increasing the number of well trained personnel and equipment that can facilitate service delivery at the division. The company should introduce a rating system that can be used by customers continuously. The management should also focus on improving the level of satisfaction of the male customers. The management should train the CompleteCare staff on a continuous basis. This will improve the quality of advice they give c lients. References Baltagi, G. (2011). Econometrics. New York: Springer Publisher Kothari, J. (2004). Research methodology: methods and techniques. New Delhi: New Age International (P) Limited Publishers. Verbeek, M. (2008). A guide to modern econometrics. England: John Wiley Sons. Zikmund, W., Babin, B., Carr, J., Griffin, M. (2012). Business research methods. USA: Cengage Learning.Advertising We will write a custom case study sample on Computer R Us Company specifically for you for only $16.05 $11/page Learn More Appendix: Hypothesis testing Does the current level of customer satisfaction differ from management’s goal of 6 out of 10? Hypothesis H0: The current level of customer satisfaction = 6. H1: The current level of customer satisfaction ≠  6. Statistical technique In this case a one sample t-test will be used to test the hypothesis. Justification One sample t-test is most suitable for evaluating a hypothesis that compares the actual mea n and hypothesized mean. Results of the test Variable 1 Variable 2 Mean 4.488095238 6 Variance 5.505824526 0 Observations 420 420 Pearson Correlation Hypothesized Mean Difference 0 df 419 t Stat -13.20498454 P(T=t) one-tail 7.85063E-34 t Critical one-tail 1.64849841 P(T=t) two-tail 1.57013E-33 t Critical two-tail 1.965641842 t Stat -13.20498454 t Critical two-tail 1.965641842 P(T=t) two-tail 1.57013E-33 Interpretation The results show that t-calculated is greater than t-critical. Also, the p-value (1.57013E-33) is less than alpha (5%). Therefore, the null hypothesis will be rejected at the 95% confidence level. This implies that the current level of customer satisfaction differ from management’s goal of 6 out of 10. Is there any difference between the overall satisfaction of male and female customers at Computers R Us? Hypothesis H0: Overall satisfaction of male customers = overall satisfaction of female customers at compute r R Us. H1: Overall satisfaction of male customers ≠  overall satisfaction of female customers at computer R Us.Advertising Looking for case study on business economics? Let's see if we can help you! Get your first paper with 15% OFF Learn More Statistical technique In this case, a paired sample t-test will be used to test the hypothesis. Justification A paired sample t-test is the most suitable for testing hypothesis that compared the mean of two related variables. Results of the test t-Test: Two-Sample Assuming Equal Variances Female Male Mean 3.589430894 5.75862069 Variance 4.27564294 4.507873231 Observations 246 174 Pooled Variance 4.371757391 Hypothesized Mean Difference 0 Df 418 t Stat -10.47338477 P(T=t) one-tail 2.98994E-23 t Critical one-tail 1.648507149 P(T=t) two-tail 5.97988E-23 t Critical two-tail 1.965655464 Interpretation In the results, the mean and variance of overall satisfaction for the male is greater than that of the female group. Further, t-calculated is greater than t-critical. Also, the p-value is less than alpha (5%). Therefore, the null hypothesis will be rejected at the 95% confidence level. This implies that there is a difference between the overall satisfact ion of male and female customers of the company. Are there any differences in the overall customer satisfaction across the following age groups: under 20, 21-30, 31-40, 41-50, 51 and over? Hypothesis H0: There is no difference in the overall satisfaction across the various age groups. H1: The overall satisfaction of at least one age group is different from the others. Statistical technique In this case, analysis of variance (ANOVA) will be used to test the hypothesis. Justification ANOVA is the most suitable technique for testing hypothesis that entails comparing mean for more than one group. One way ANOVA will be used because there is only one independent variable. Results of the test Anova: Single Factor SUMMARY Groups Count Sum Average Variance Under 20 47 180 3.829787 6.579093432 21-30 109 501 4.59633 6.150356779 31-40 105 466 4.438095 5.786996337 41-50 107 485 4.53271 4.647504849 over 50 52 253 4.865385 4.236425339 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 29.52282 4 7.380705 1.344941124 0.252492 2.393438 Within Groups 2277.418 415 5.487753 Total 2306.94 419 Interpretation In the results above, the value of F-calculated is less than the F-critical. Besides, the p-value is greater than alpha (5%). Therefore, the null hypothesis will not be rejected at the 95% confidence level. This implies that there is no difference in the overall satisfaction across the various age groups. Are there any differences in the gender compositions across the five age groups? Hypothesis H0: There are no differences in gender composition across the five age groups. H1: Gender composition is different in at least one of the age groups. Statistical technique Analysis of variance (ANOVA) will be used to test the hypothesis. Justification ANOVA is the most suitable technique for testing hypothesis that entail comparing mean for more than one group. One way ANOVA will be used because there is only one independent variable (Verbeek, 2008). Results of the test Anova: Single Factor SUMMARY Groups Count Sum Average Variance Under 20 47 20 0.425532 0.249769 21-30 109 47 0.431193 0.247537 31-40 105 43 0.409524 0.244139 41-50 107 41 0.383178 0.238582 over 50 52 23 0.442308 0.251508 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 0.18386 4 0.045965 0.18751 0.944872 2.393438 Within Groups 101.7304 415 0.245134 Total 101.9143 419 Interpretation In the results above, the value of F-calculated is less than the F-critical. Besides, the p-value is greater than alpha (5%). Therefore, the null hypothesis will not be rejected at the 95% confidence level. This implies that there are no differences in gender composition across the five age groups. Is there any difference in customer satisfaction based upon ‘response times in the CompleteCare division’ and the ‘loyalty rewards p rogram’? Hypothesis H0: Customer satisfaction based upon response times in the CompleteCare division = the customer satisfaction based upon loyalty reward program. H1: Customer satisfaction based upon response times in the CompleteCare division ≠  the customer satisfaction based upon loyalty reward program. Statistical technique In this case, a paired sample t-test will be used to test the hypothesis. Justification A paired sample t-test is the most suitable for testing hypothesis that compared the mean of two related variables (Verbeek, 2008). Results of the test t-Test: Paired Two Sample for Means Response time Loyalty reward program Mean 3.242857143 5.645238095 Variance 4.222502557 7.842817366 Observations 420 420 Pearson Correlation -0.011950135 Hypothesized Mean Difference 0 Df 419 t Stat -14.09404771 P(T=t) one-tail 1.69112E-37 t Critical one-tail 1.64849841 P(T=t) two-tail 3.38224E-37 t Critical two-tail 1.965641842 Interpr etation The results show that t-calculated is greater than t-critical. Also, the p-value is less than alpha (5%). Therefore, the null hypothesis will be rejected at the 95% confidence level. This implies that there are differences in customer satisfaction based upon response times in the CompleteCare division and the loyalty rewards program. Are any of the initiatives proposed by management related to the overall satisfaction of Computers R Us customers? Hypothesis H0: The initiatives proposed by the management are determinants of the overall satisfaction of Computer R Us customers. H1: The initiatives proposed by the management are not determinants of the overall satisfaction of Computer R Us customers Statistical technique In this case, a multiple regression analysis will be used. Justification Multiple regression analysis is used to model the relationship between one dependent variable and other explanatory variables. Results of the test SUMMARY OUTPUT Regression Sta tistics Multiple R 0.965602191 R Square 0.932387592 Adjusted R Square 0.931735906 Standard Error 0.613066164 Observations 420 ANOVA Df SS MS F Significance F Regression 4 2150.962676 537.7407 1430.732 3.3062E-241 Residual 415 155.9778006 0.37585 Total 419 2306.940476 Coefficients Standard Error t-Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0% Intercept 0.896561298 0.103883791 8.630425 1.32E-16 0.69235727 1.10076533 0.69235727 1.100765326 Response time 0.86471784 0.03836737 22.53785 4.92E-74 0.789299227 0.94013645 0.789299227 0.940136454 Level of advice 0.271037316 0.041145932 6.58722 1.36E-10 0.190156892 0.35191774 0.190156892 0.351917739 Level of communication -0.01775345 0.021480889 -0.82648 0.409009 -0.05997837 0.02447146 -0.05997836 0.024471457 Loyalty reward program 0.007973903 0.010696359 0.745478 0.456405 -0.01305189 0.0289997 -0.01305189 0.0289997 Interpretation The F-test will be used to test the overall significance of the regression model. The p-value for the F test is less than alpha (0.05). Therefore, reject the null hypothesis and conclude that the four determinants are significant determinants of the overall customer satisfaction. Further, the p-value for response time and level of advice are greater than alpha (0.05). This implies that they are significant determinants of overall customer satisfaction of the company.