Question

Favorable weather has been shown to be associated with increased tipping. Will just the belief that future weather will be favorable lead to higher tips? The researchers gave 60 index cards to a waitress at an Italian restaurant in New Jersey. Before delivering the bill to each customer, the waitress randomly selected a card and wrote on the bill the same message that was printed on the index card. Twenty of the cards had the message "The weather is supposed to be really good tomorrow. I hope you enjoy the day!" Another 20 cards contained the message "The weather is supposed to be not so good tomorrow. I hope you enjoy the day anyway!" The remaining 20 cards were blank, indicating that the waitress was not supposed to write any message. Choosing a card at random ensured that there was a random assignment of the diners to three experimental conditions. The table shows the percentage tips for the three messages.

Good weather report	20.8	18.7	19.9	20.6	22.0	23.4	22.8
	24.9	22.2	20.3	24.9	22.3	27.0	20.4
	22.2	24.0	21.2	22.1	22.0	22.7
Bad weather report	18.0	19.0	19.2	18.8	18.4	19.0	18.5
	16.1	16.8	14.0	17.0	13.6	17.5	19.9
	20.2	18.8	18.0	23.2	18.2	19.4
No weather report	19.9	16.0	15.0	20.1	19.3	19.2	18.0
	19.2	21.2	18.8	18.5	19.3	19.3	19.4
	10.8	19.1	19.7	19.8	21.3	20.6

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(a) Find the Tukey simultaneous 99% confidence intervals for all pairwise differences among the three population means. These populations are “Good and Bad,” “Good and None,” and “None and Bad.” (Enter your answers rounded to five decimal places.)

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Answer 1

R code:

x1=c(20.8, 18.7, 19.9, 20.6, 22, 23.4, 22.8,24.9,22.2,20.3, 24.9, 22.3, 27, 20.4,22.2, 24, 21.2, 22.1, 22, 22.7)
x2=c(18, 19, 19.2, 18.8, 18.4, 19, 18.5, 16.1, 16.8,14,17,13.6,17.5, 19.9, 20.2, 18.8, 18, 23.2,18.2, 19.4)
x3=c(19.9, 16,15, 20.1, 19.3, 19.2, 18, 19.2, 21.2, 18.8, 18.5,19.3, 19.3, 19.4, 10.8, 19.1, 19.7, 19.8, 21.3, 20.6)
y=c(x1,x2,x3)
g1=replicate(20,"Good")
g2=replicate(20,"Bad")
g3=replicate(20,"None")
group=c(g1,g2,g3)
TukeyHSD(aov(y~group), conf.level=.99)   

Output:

Tukey multiple comparisons of means
    99% family-wise confidence level

Fit: aov(formula = y ~ group)

$group
                    diff      lwr             upr     p adj
Good-Bad   4.040 1.97145 6.10855 0.0000006
None-Bad   0.545 -1.52355 2.61355 0.7048975
None-Good -3.495 -5.56355 -1.42645 0.0000109

$\mu Good-\mu Bad~lower~value=1.97145\\ \mu Good-\mu Bad~upper~value=6.10855.\\ \mu Good-\mu None~lower~value=1.42645\\ \mu Good-\mu None~upper~value=5.56355\\ \mu None-\mu Bad~lower~value=-2.61355\\ \mu None-\mu Bad~upper~value=1.52355$