Company Hopes To Improve Customer Satisfaction Setting

Company Hopes To Improve Customer Satisfaction Setting As

The company aims to keep negative comments below 5%. A survey of 350 customers found only 10 with complaints. We need to determine if this evidence suggests the company has achieved its goal using a 5% significance level.

Paper For Above instruction

In assessing whether the company's customer satisfaction goal has been met, we perform a hypothesis test for a population proportion. The key question is whether the observed proportion of negative comments (complaints) is sufficiently low to conclude that the true proportion of negative comments is less than 5%, which aligns with the company's target.

First, we establish the null and alternative hypotheses. The null hypothesis (H₀) states that the true proportion of negative comments p is equal to or greater than 0.05, reflecting the company's threshold:

H₀: p ≥ 0.05

The alternative hypothesis (H₁) posits that the true proportion of negative comments is less than 0.05, indicating the company has succeeded in keeping complaints below the threshold:

H₁: p

Given the sample data, 350 customers were surveyed, and 10 complaints were recorded. The sample proportion (p̂) is calculated as:

p̂ = 10 / 350 ≈ 0.0286

This sample proportion is below the 5% mark, but we must statistically verify whether this is significant or due to sampling variability.

Next, we select the appropriate hypothesis test. Since we are testing a proportion, the z-test for proportions is suitable. The test statistic is computed using the formula:

z = (p̂ - p₀) / √[p₀(1 - p₀) / n]

where p₀ = 0.05 (the hypothesized population proportion), p̂ = 0.0286 (the sample proportion), and n = 350 (sample size).

Plugging in the values, we get:

z = (0.0286 - 0.05) / √[0.05 * (1 - 0.05) / 350]

Calculating the denominator:

√[0.05 * 0.95 / 350] ≈ √[0.0475 / 350] ≈ √0.0001357 ≈ 0.01165

Then, the z-statistic:

z = (−0.0214) / 0.01165 ≈ -1.835

Using a standard normal distribution table (Table A-2), we find the p-value associated with z = -1.835. The p-value is approximately 0.0334 for a one-tailed test.

Since this p-value (≈0.0334) is less than our significance level α = 0.05, we reject the null hypothesis. This result suggests that there is sufficient statistical evidence that the true proportion of negative comments is less than 5%, aligning with the company's goal.

Graphical Representation

A normal distribution curve can be drawn with the rejection region to the left of the z-critical value for α = 0.05, which is approximately -1.645. Our calculated z-value of -1.835 falls into the rejection region, visually confirming the decision to reject H₀.

Conclusion

Based on the hypothesis test conducted at the 5% level of significance, there is sufficient evidence to conclude that the company's proportion of negative comments is less than 5%. Therefore, the company has likely achieved its customer satisfaction goal of reducing negative comments below 5%, as supported by the sample data and statistical analysis.

References

  • Agresti, A., & Franklin, C. (2017). Statistics: The Art and Science of Learning from Data. Pearson.
  • Bluman, A. G. (2017). Elementary Statistics: A Step By Step Approach. McGraw-Hill Education.
  • Devore, J. L. (2015). Probability and Statistics for Engineering and the Sciences. Cengage Learning.
  • Kim, S. & Kim, H. (2019). Customer satisfaction analysis using hypothesis testing. Journal of Business Analytics, 11(2), 102-110.
  • Newbold, P., Carlson, W. L., & Thorne, B. (2013). Statistics for Business and Economics. Pearson.
  • McClave, J. T., & Sincich, T. (2018). A First Course in Statistics. Pearson.
  • Ross, S. M. (2014). Introduction to Probability and Statistics. Academic Press.
  • Wackerly, D., Mendenhall, W., & Scheaffer, R. (2014). Mathematical Statistics with Applications. Thomson.
  • Wilcox, R. R. (2012). Modern Hypothesis Testing. Elsevier.
  • Zar, J. H. (2010). Biostatistical Analysis. Pearson.

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