Question

Researchers investigated whether providing a fancy, foil-wrapped piece of chocolate with the dinner bill would lead to higher tips than not providing such a treat (Strohmetz et al., 2002). Ninety-two dinner parties at a restaurant in Ithaca, New York, were randomly assigned to receive such a piece of chocolate or not with their dinner bill. Of the 46 parties who received the chocolate, the average tip (as a percentage of the bill) was 17.84%, with a standard deviation of 3.06%. Of the 46 parties who did not receive the chocolate, the average tip (as a percentage of the bill) was 15.06%, with a standard deviation of 1.89%.

a. Identify the explanatory and response variables.

b. Explain why, in this case, looking at the percentage of the bill is a more useful measure than looking at the exact tip amount (as the waitress in Activity 22-4 did).

c. Is this an observational study or an experiment? Explain.
d.   Do you have enough information to check whether the technical conditions for the two-sample t-test and confidence interval are satisfied? Explain.
e.   Conduct the appropriate test of the researchers’ conjecture. Report all aspects of the test, including your test decision at the α ? 0.05 level, and summarize your conclusion.
f.   Produce and interpret a 95% confidence interval for estimating how much the chocolate adds to the tip percentage on average.
g.   Summarize your conclusions. Be sure to address issues of generalizability and causation.
h.   Describe what Type I error and Type II error mean in the context of this study. Also describe a potential consequence of each type of error for the restaurant manager and wait staff.

Answer 1

a.
Explanatory Variables vs. Response Variables. The response variable is the focus of a question in a study or experiment.
An explanatory variable is one that explains changes in that variable. It can be anything that might affect the response variable.
b.
the percentage of the bill is a more useful measure than looking at the exact tip amount because Researchers investigated whether providing a fancy,
foil-wrapped piece of chocolate with the dinner bill would lead to higher tips than not providing such a treat
c.
a observational study is where nothing changes and just record what you see,
but an experimental study is where you have a control group and a testable group.
d.
yes,
enough information to check whether the technical conditions for the two-sample t-test and confidence interval are satisfied
e.
Given that,
mean(x)=17.84
standard deviation , s.d1=3.06
number(n1)=46
y(mean)=15.06
standard deviation, s.d2 =1.89
number(n2)=46
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.679
since our test is right-tailed
reject Ho, if to > 1.679
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =17.84-15.06/sqrt((9.3636/46)+(3.5721/46))
to =5.242
| to | =5.242
critical value
the value of |t α| with min (n1-1, n2-1) i.e 45 d.f is 1.679
we got |to| = 5.24238 & | t α | = 1.679
make decision
hence value of | to | > | t α| and here we reject Ho
p-value:right tail - Ha : ( p > 5.2424 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 5.242
critical value: 1.679
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that the percentage of the bill is a more useful measure than looking at the exact tip amount.
f.
TRADITIONAL METHOD
given that,
mean(x)=17.84
standard deviation , s.d1=3.06
number(n1)=46
y(mean)=15.06
standard deviation, s.d2 =1.89
number(n2)=46
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((9.364/46)+(3.572/46))
= 0.53
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.05
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 45 d.f is 2.014
margin of error = 2.014 * 0.53
= 1.068
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (17.84-15.06) ± 1.068 ]
= [1.712 , 3.848]
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DIRECT METHOD
given that,
mean(x)=17.84
standard deviation , s.d1=3.06
sample size, n1=46
y(mean)=15.06
standard deviation, s.d2 =1.89
sample size,n2 =46
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 17.84-15.06) ± t a/2 * sqrt((9.364/46)+(3.572/46)]
= [ (2.78) ± t a/2 * 0.53]
= [1.712 , 3.848]
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g.
interpretations:
1. we are 95% sure that the interval [1.712 , 3.848] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion
h.
Type 1 error is possible only when its reject the null hypothesis
Type 2 error is possible only when its fails to reject the null hypothesis.
In this context,
Type 1 error is possible because it reject the null hypothesis.