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. Here are the tips for those receiving the good-weather message:

21.0   19.0   19.9   20.9   22.4   23.6   23.0   25.4   22.5   20.6

24.9   22.4   27.5   20.9   22.6   24.4   21.4   22.6   22.0   22.8

The tips for the 20 customers who received the bad-weather message are listed here:

18.5   19.1   19.2   18.9   18.4   19.0   19.0   16.4   17.1   14.0

17.4   13.9   17.7   20.0   20.5   18.9   18.1   23.5   18.6   19.9

(a) What degrees of freedom would you use in the conservative two-sample t procedures to compare the lack-of-control and in-control groups?
df =

What is the two-sample t test statistic? (Round your answer to three decimal places.)
t =

Answer 1

Ans Given data are

Here are the tips for those receiving the good-weather message:

21.0   19.0   19.9   20.9   22.4   23.6   23.0   25.4   22.5   20.6

24.9   22.4   27.5   20.9   22.6   24.4   21.4   22.6   22.0   22.8

and

The tips for the 20 customers who received the bad-weather message are listed here:

18.5   19.1   19.2   18.9   18.4   19.0   19.0   16.4   17.1   14.0

17.4   13.9   17.7   20.0   20.5   18.9   18.1   23.5   18.6   19.9

now first we calculates sample mean and sample standard deviation

for good-weather message data

ΣΧί 60 η

$\bar{Xg}$ = 22.49

s1 = (E(Xi – Xg)^2 1 n

s1=1.9976

for bad-weather message

ΣΧι Xb η.

$\bar{Xb}$ = 18.405

s2 = (E(Xi – Xb)^2 n - 1

s2 = 2.1089

Here we use two sample t test

Test of Hypothesis :

To Test :

H0= µ1 = µ 2 versus   H1 : µ1 ≠ µ2

a) For use in the conservative two-sample t procedures to compare the lack-of-control and in-control groups

degrees of freedom = n1+n2-2 = 20+20-2 = 38

degrees of freedom = 38

Test Statistic :

t = Xg - āb-A (n1 - 1)s12 + (n2-)s22 nl + n2 - 2 1 n1 + 1 n2

t= 22.49 – 18.405 (20- 1)1.99762 + (20– 1)2.10892 20 + 20 - 2 1 1 + 2020

t = 6.289

the two-sample t test statistic

t = 6.289