I'll do your homework if you do mine
In an experiment in which telephone callers to an airline were put on hold with an advertisement, Muzak, or classical music in the background. The table shows the data classified by factor type and partial results of a one-way ANOVA table. The variable of interest is how long customers will stay on hold before hanging up. The customers were randomly assigned to each group so that there or may not be the same number in each group.
> fitaov = aov(y~type)
> summary(fitaov)
Df Sum Sq Mean Sq F value Pr(>F)
type 2 189.23 xxxxxxx xxxxx xxxxxx **
Residuals 15 89.54 xxxxxx
--- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
> TukeyHSD(fitaov)
Tukey multiple comparisons of means
95% family-wise confidence level
Fit: aov(formula = weight_gain ~ source)
$source
diff lwr upr p adj
musak-classical music -9.22 -17.422553 -1.022553 0.0064091
musak-advertisment -0.5 -13.222553 12.222553 0.9950816
classical music-advertisment 4.2 -8.522553 16.922553 0.7079603
1. When we reject a null hypothesis that means that the alternative is true
2. We should reject the null hypothesis that advertisements and classical music have the same mean
3. We should reject the null hypothesis that advertisements and musak have the same mean
4. We should reject the overall ANOVA null hypothesis that all three means are equal at the 0.05 level.
5. We should reject the null hypothesis that musak and classical music have the same mean at the 0.05 level
Answer with which of the 5 statements above are true in this case (there is more than one right).
Thanks.