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.)

?????−???? lower value=

?????−???? upper value=

Answer 1

I am using excel data analysis function for the Anova analysis

Steps:

1. Input data
2. Click on 'Data' on toolbar and then click on data analysis
3. Select 'Anova- Single Factor'
4. Input your range and other info accordingly and you will get your analysis.

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Column 1	20	438.2	21.91	3.751
Column 2	20	363.6	18.18	4.402
Column 3	20	374.5	18.725	5.704
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	162.36	2.00	81.18	17.58	0.00	3.16
Within Groups	263.29	57.00	4.62
Total	425.65	59.00

	Good	Bad	None
Mean	22	18.2	21.3
n	20	20	20

$1-\alpha$ Confidence interval for difference is

(11 - 12+tn-k,a/2V 02 (1/n1 +1/n2))

Where = 4.62

MSST Error variance = n -k

When $\alpha$ = 0.01 (1 - 0.99)

= 2.92 Using excel funct 'tinv' with df =57, p= 0.005

In-k,a/2=t57,0.005

Arranging Means in ascending order

Bad < None < Good

We will subtract smaller mean from bigger mean

(1)

For Good and bad

99% Confidence interval for diff is

(( 22 - 18.2)+2.924.62(1/20 +1/20)

(1.81517, 5.78483)

(2)

For Good and None

99% Confidence interval for diff is

((22 - 21.3)2.92.4.62(1/20 +1/20)

(-1.28483, 2.68483)

(3)

For Bad and None

99% Confidence interval for diff is

((21.3-18.2)-2.92.4.62(1/20 +1/20))

(1.11517, 5.08483)