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 selected one of the index cards without reading it and wrote on the bill the same message that was printed on the 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 (percentage of bill) before taxes. The tips for those receiving the good‑weather message are as follows.
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
The tips for the 20 customers who received the bad weather message are as follows.

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
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Back‑to‑back stemplots of the tips are shown.

Good Weather Tips Bad Weather Tips
13 6
14 0
15
16 1 8
17 0 5
7 18 0 0 2 4 5 8 8
9 19 0 1 2 4
8 6 5 3 20 0 2
9 2 21
8 7 3 2 2 1 0 22
4 23 2
9 9 0 24
25
26
0 27
Because the distributions are reasonably symmetric with no extreme outliers, the procedures will work well.

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

Bruce Rind and David B. Strohmetz, "Effect of beliefs about future weather conditions on restaurant tipping," Journal of Applied Social Psychology, 31 (2001), pp. 2160–2164.
Carry out a two‑sample test. What is the test statistic? Give your answer to three decimal places.

=

What is the -value? Give your answer to three decimal places.

-value:

Answer 1

The test statistic and p-value of the given data are 6.274 and 0.0001 respectively.

Test Statistic

The test statistic can be calculated using the formula below

$t = \frac{x-\mu}{\sigma / \sqrt{n} }$

Solving for the mean and standard deviation, we can substitute the values into the above equation which will be

$t = \frac{x-\mu}{\sigma / \sqrt{n} } = 6.274$

P-Value

Using the data from the test statistic, we can calculate the p-value of the data

$p-value = p(z < 6.274)= 0.0001 = 0.00$

From the calculation above, the test statistic and p-value of the given data are

• 6.274
• 0.0001

Learn more on test statistic and p-value here;

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