Problem Nikolai works for his dad\'s automotive supply store. Nikolai is a smart

Question

Problem Nikolai works for his dad's automotive supply store. Nikolai is a smart kid, but quite rebellious and persnickety. When he's managing the store alone, he sets the store music to punk rock. His father gets furious when Nikolai changes the station from oldies (dad's favorite) or from classical (mom's favorite). Nikolai is acing his marketing research class with Dr. Bechkoff and makes a deal with his dad. "Dad, if I play punk rock every day for one week, oldies every day the next week, and easy-listening every day the week after that, and there is a significant difference in customer satisfaction levels between the 3 weeks, and if satisfaction is lowest for my favorite music, then I won't change the store music ever again. HOWEVER, if there is no significant difference among satisfaction levels, or if there is a difference and our customers are more satisfied while listening to my music, then I get to permanently have the store music set to punk rock!" "Deal!" exclaimed his father (who really just wanted the free customer satisfaction research for his store.) Method Nikolai created a questionnaire that had these 3 items on it. On a scale of 1 to 7 (1 = not at all satisfied and 7 = extremely satisfied), please answer the following questions: 1) "Overall, how satisfied are you shopping at this automotive supply store," 2) "How satisfied are you with your shopping experience today? and 3) "How satisfied are you with the current atmosphere at this automotive supply store?" The answers for all three questions were averaged, to form one overall satisfaction score per customer Data Nikolai gave out 10 surveys each week to randomly selected customers, which yielded the following sample data: 15 16 17 18 19 3.7 6.8 Punk rock Punk rock Punk rock Punk rock 4.3 5.8 6.6 6.0 4.9 3.6 5.4 4.8 4.7 4.6 3.2 3.6 21 Punk rock Punk rock Punk rock Punk rock Punk rock 5.0 4.6 5.2 3.8 4.9 Classical 25 Classical 27 5.2 6.4 5.5 13 30 5.9 When you are finished, complete the MRA7 Assessment or at assessment 1. Understand what the Levene's test results mean. 2. Interpret & report your ANOVA results 3 Know whether or not to Reject or Fail to Reject the null hypothesis (Ho). 4. Interpret your multiple comparisons table 5. Understand the results of Scheffe's post hoc test. 6. Explain the entire ANOVA process and what it means with regard to the data provided.

Explanation / Answer

I can use R-software to analysis this data set

1.

x=read.csv("Book11.csv")     # File name in csv format
library(car)                          # package
leveneTest(x[,1]~x[,2])         # command

Levene's Test for Homogeneity of Variance (center = median)
      Df F value Pr(>F)
group 2 0.5336 0.5925
      27             

From the output above we can see that the p-value is not less than the significance level of 0.05. This means that there is no evidence to suggest that the variance across groups is statistically significantly different. Therefore, we can assume the homogeneity of variances in the different treatment groups.

2.

res.aov <- aov(x[,1]~x[,2], data=x) # Compute the analysis of variance
summary(res.aov) # Summary of the analysis

          Df Sum Sq Mean Sq F value Pr(>F)
x[, 2]       2 15.54   7.769   8.901 0.00107 **
Residuals   27 23.57   0.873                 
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Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

As see that p-value is less than the significance level 0.05, we can conclude that there are significant differences between the groups in the model.

3. We reject the null hypothesis

4.Tukey HSD are performing multiple pairwise-comparison between the means of groups.

TukeyHSD(res.aov)

Tukey multiple comparisons of means
    95% family-wise confidence level

Fit: aov(formula = x[, 1] ~ x[, 2], data = x)

$`x[, 2]`
         diff        lwr          upr    p adj
Ol-Cl 1.38 0.3440372 2.4159628 0.0073704
PR-Cl -0.26 -1.2959628 0.7759628 0.8092606
PR-Ol -1.64 -2.6759628 -0.6040372 0.0015162

It can be seen that only the difference between PR and Ci is not significant with an adjusted p-value of 0.8092.

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