Question

# A study was done to measure customer satisfaction for 4 different smart phones. The data is...

A study was done to measure customer satisfaction for 4 different smart phones. The data is given below (the data is a satisfaction score, out of 100; the higher the score, the more satisfied the customer was with their phone)

 1. iPhone 2. Nokia 3. Blackberry 4. Android Sample Size 5 4 3 4 Sample Mean 83.0 66.5 73.0 77.0

Overall Mean = 75.5

MSE = 20.54

a) Run a one-factor ANOVA and test H0: μ1 = μ2 = μ3 = μ4 at the 5% level of significance. Set up an ANOVA table (show your work calculating SSC, sum of squares between groups), state H0 and Ha in words, calculate the value of your test statistic, and make your decision/draw your conclusion. Make sure your conclusion is given in words that are in the context of the question.

b) Use the Tukey-Kramer procedure to test H0: μ1 – μ3 = 0 vs Ha: μ3 – μ4 ≠ 0 at the 5% level of significance. What is your decision? You must also clearly state your conclusion in words

(a) The hypothesis being tested is:

H0: µ1 = µ2 = µ3 = µ4

Ha: At least one means is not equal

 xi n n*(xi - xgrand)² iPhone 83 5 281.25 Nokia 66.5 4 324 Blackberry 73 3 18.75 Android 77 4 9 xgrand 75.5 SSB 633 Source SS df MS F p-value Between 633 3 211 10.27264 0.001238 Error 246.48 12 20.54 Total 879.48 15

The p-value is 0.001238.

Since the p-value (0.001238) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that at least one means is not equal.

(b)

 Group Mean Difference HSD Significant Difference μ1 – μ3 10 5.099229 Yes

Therefore, we can conclude that μ3 – μ4 ≠ 0.

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