WEE I 7.1.3: Hypothesis test for a population proportion. 13:29.1 CECE1' f c -3::i_..' Jump to level 1 An airline company is interested in improving customer satisfaction rate from the 64% currently claimed. The company sponsored a survey of 140 customers and found that 98 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 pvalue? Ex: 0.456 Does sufficient evidence exist that the customer satisfaction rate is different than the claim by the company at a significance level off: = 0.1? Select v E -m Example 7.1.2: Customer satisfaction. After a series of scandals, the percentage of satisfied customers of an airline dipped to an all time low of 47%. As a result, aggressive changes were implemented to improve customer experience and a public relations firm was hired to rehabilitate the airline's image. To determine if their efforts are successful, a survey was conducted to determine if customers were satisfied or unsatisfied with the airline. 132 of the 240 respondents answered that their experience was satisfactory. Does sufficient evidence exist that the company's efforts were succeeding at the o = 0.05 significance level? That is, does statistically significant evidence exist that suggests that the proporti of satisfied customers is higher? Solution Since the survey asked if the customers were satisfied or unsatisfied, the dataset contains binary categorical data. Thus, the hypothesis test involves a population proportion instead of a population mean. The null hypothesis is that the customer satisfaction proportion hol steady. Since the board of directors want to know if the changes are successful in increasing customer satisfaction, the hypothesis test right-tailed with an alternative hypothesis that the percentage of satisfied customers is higher. Mathematically, Ho : p = 0.47 Ha : p > 0.47 132 of the 240 respondents viewed their experience as satisfactory, so the sample proportion is p = 132 240 = 0.55. Since po = 0.47 a n = 240, the test statistic is p - Po 0.55 - 0.47 Z= ~2.483 Po 1 - Po 0.47(1-0.47) 240 The p-value is P(z > 2.483) ~ 0.006. p-value = 0.006 z = 2.49 Since the p-value is less than the significance level a = 0.05, sufficient evidence exists to support the claim that the customer satisfaction is higher than the all time low of 47%