Question

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

Answer 1

(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.