5. An organizational psychologist is hired as a consultant to a company planning

Question

5. An organizational psychologist is hired as a consultant to a company planning to open a coffee house for college students. The company wants to know if their customers will drink morecoffee if the coffee house is decorated in a Paris theme or in a San Francisco theme. The psychologist sets up two'similar rooms with the two themes. Eight students spend an afternoon in each room drinking all the coffee they like. The order in which they sit in the rooms is rotated so that halt spend their first afternoon in the Paris room and half in the San Francisco room. The amount of coffee each participant drinks in each room is shown below. Use the 05 significance levet to determine if there a significant difference between the numbers of cups of coffee consumed in the two rooms. Participant-Paris . SanFrancisco Dnee score 8.5 4.3 20 8.4r 4.6 7.3 7.0 3.3 3.0 3.5 3.5 45.5 A) What is the appropriate test for this data? B) Conduct the appropriate hypothesis test. Only hand calculate if you choose a z-test and then follow the five steps of hypothesis testing and provide a drawing. Otherwise, use SPSS. C) Give an interpretation of this data and, when appropriate, report the statistics. D) Explain why the test you chose is the most appropriate. MParis = 5.69 MSan Francisco-5.39

Explanation / Answer

Given that,
null, H0: Ud = 0
alternate, H1: Ud != 0
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.365
since our test is two-tailed
reject Ho, if to < -2.365 OR if to > 2.365
we use Test Statistic  
to= d/ (S/n)
where
value of S^2 = [ di^2 – ( di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = 0.3
We have d = 0.3
pooled variance = calculate value of Sd= S^2 = sqrt [ 3.46-(2.4^2/8 ] / 7 = 0.63
to = d/ (S/n) = 1.36
critical Value
the value of |t | with n-1 = 7 d.f is 2.365
we got |t o| = 1.36 & |t | =2.365
make Decision
hence Value of |to | < | t | and here we do not reject Ho
ANSWERS
---------------
null, H0: Ud = 0
alternate, H1: Ud != 0
test statistic: 1.36
critical value: reject Ho, if to < -2.365 OR if to > 2.365
decision: Do not Reject Ho
p-value: 0.2171 > 0.05
Do not reject HO

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