Question

Researchers gave 40 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!"

After the customers left, the waitress recorded the amount of the tip, percent of bill, before taxes. Given are the tips for those receiving the good‑weather message.

20.8	18.7	19.9	20.6	21.9	23.4	22.8	24.9	22.2	20.3
24.9	22.3	27.0	20.5	22.2	24.0	21.2	22.1	22.0	22.7

Given are the tips for the 20 customers who received the bad‑weather message.

18.0	19.1	19.2	18.8	18.4	19.0	18.5	16.1	16.8	14.0
17.0	13.6	17.5	20.0	20.2	18.8	18.0	23.2	18.2	19.4

Stemplots for both data sets are shown.

Good weather tips

18	7
19	9
20	3 5 6 8
21	2 9
22	0 1 2 2 3 7 8
23	4
24	0 9 9
25
26
27	0

Bad weather tips

13	6
14	0
15
16	1 8
17	0 5
18	0 0 2 4 5 8 8
19	0 1 2 4
20	0 2
21
22
23	2

Neither stemplot suggests a strong skew or the presence of strong outliers. Because of this, t procedures are reasonable here.

Is there good evidence that the two different messages produce different percent tips?

Let μ1 be the mean tip percent when the forecast is good, and let μ2 be the mean tip percent when the forecast is bad. Select the correct hypotheses statements that we want to test.

H0:μ1=μ2 versus Ha:μ1>μ2

H0:μ1=μ2 versus Ha:μ1≠μ2

H0:μ1=μ2 versus Ha:μ1<μ2

H0:μ1≠μ2 versus Ha:μ1<μ2

What degrees of freedom (df) would you use in the conservative two‑sample t procedures to compare the percentage of tips when the forecast is good and bad? (Enter your answer as a whole number.)

df=

What is the two‑sample t test statistic (rounded to three decimal places)?

t=

Test whether there is good evidence that the two different messages produce different percent tips at α=0.1 . The null hypothesis of no difference in tips due to the weather "forecast" is

not rejected.

rejected.

Answer 1

(a)

Correct option:

H0:μ1=μ2 versus Ha:μ1≠μ2

(b)

df= n1 + n2 - 2 = 20 + 20 - 2 = 38

(c)

From the given data, the following statistics are calculated:

n1 = 20

1 = 22.22

s1 = 1.9549

n2 = 20

2 = 18.19

s2 = 2.1046

Pooled standard deviation is given by:

Test Statistic is given by:

t = (22.22-18.19)/0.6423

= 6.274

The two‑sample t test statistic = t = 6.274

(d)

= 0.10

ndf = 38

From Table, critical values of t = 1.6860

Since calculated value of t is greater than critical value of t, the difference is significant. Reject null hypothesis.

Correct option:

The null hypothesis of no difference in tips due to the weather "forecast" is rejected.