Question

FILE TOOLS VIEW Document3 - Word (Product -5. a. Postgraduate student from UTP has interested to identify the customer satisfaction from different airline services operating from KLIA in Malaysia. The randomly selected and surveyed passengers from the three airlines are given below. Is there a difference in mean satisfaction level among the three airlines? Use the 0.05 significance level and use the 5-step procedure for testing a hypothesis. Airline 1: Airline 2: Airline 3: 8 3 3 11 2 4 10 1 5 2 3 4 [10 marks]

Answer 1

Step 1: Null hypothesis: H:1 = 4= Hz There is no difference in mean satisfaction level among the three airlines. Alternative hypothesis: H, :Not all the means are equal. There is a difference in mean satisfaction level among the three airlines.

Step 2: Rejection rule: If p-values a(=0.05). then reject the null hypothesis H. Step 3: EXCEL (Data Analysis) output: Anova: Single Factor Count SUMMARY Groups Airline 1 Airline 2 Airline 3 Sum Average Variance 29 9.666667 2.333333 11 2.2 164 0.666667 0.7 ANOVA Source of Variation Between Groups Within Groups S 107.2 9.466667 116.6667 df MS F P-value Fcrit 2 53.6 50.95775 1.23E-05 4.256495 91.051852 11 Total

Step 4: Test Statistic: F = 50.96. Critical value: Fuit = 4.26 Step 5: Conclusion: Since F(= 50.96) > Ferit (=4.26). then reject the null hypothesis. Therefore, there is a difference in mean satisfaction level among the three airlines.