Research and Analysis of Customer Satisfaction Rate in an Airline Company

An airline company is interested in improving customer satisfaction rate from the 72% currently claimed. The company sponsored a survey of 300 customers and found that 227 customers were satisfied. Determine whether sufficient evidence exists that the customer satisfaction rate is higher than the claim by the company. What is the test statistic z? What is the p-value? Does sufficient evidence exist that the customer satisfaction rate is different than the claim by the company at a significance level ax=0.05?

  Research and Analysis of Customer Satisfaction Rate in an Airline Company In the competitive landscape of the airline industry, customer satisfaction is a key metric that companies strive to improve to attract and retain loyal customers. When an airline company claims a certain satisfaction rate, it is essential to have data-backed evidence to support this claim. In this case, the airline company aims to determine if there is sufficient evidence that the customer satisfaction rate is higher than the claimed 72%. Thesis Statement: Based on the survey data collected from 300 customers, where 227 customers reported being satisfied, there is statistical evidence to suggest that the customer satisfaction rate is higher than the claimed 72% by the airline company. Methodology: To analyze the data and test the hypothesis, we will perform a one-sample z-test. The formula for the z-test is: [ z = \frac{(p - P)}{\sqrt{\frac{P(1-P)}{n}}} ] where: - ( p ) is the sample proportion of satisfied customers - ( P ) is the claimed proportion of satisfied customers by the company (0.72 in this case) - ( n ) is the sample size (300 in this case) Calculation of Test Statistic (z): Given: - ( p = \frac{227}{300} = 0.757 ) - ( P = 0.72 ) - ( n = 300 ) Plugging these values into the formula, we get: [ z = \frac{(0.757 - 0.72)}{\sqrt{\frac{0.72(1-0.72)}{300}}} ] Calculating further, we find: [ z \approx \frac{0.037}{\sqrt{\frac{0.72*0.28}{300}}} = \frac{0.037}{\sqrt{0.2016/300}} ] [ z \approx \frac{0.037}{\sqrt{0.000672}} \approx \frac{0.037}{0.0259} \approx 1.43 ] Therefore, the test statistic ( z = 1.43 ). Calculation of p-value: Using the z-table or statistical software, we can find that the p-value corresponding to a z-score of 1.43 is approximately 0.0775. Conclusion: At a significance level of ( \alpha = 0.05 ), since the p-value of 0.0775 is greater than ( \alpha = 0.05 ), we fail to reject the null hypothesis. This means that there is not enough statistical evidence to conclude that the customer satisfaction rate is different than the claimed rate by the company. In summary, while there is a higher proportion of satisfied customers in the sample data compared to the claimed rate, this difference is not statistically significant at the 95% confidence level. To further validate this result, conducting additional surveys or analyzing more extensive data may be necessary for a more conclusive inference on the customer satisfaction rate in the airline company.      

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