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Service without a smile. 鈥淪ervice with a smile鈥 is a slogan that many businesses adhere to. However, some jobs (e.g., judges, law enforcement officers, and pollsters) require neutrality when dealing with the public. An organization will typically provide 鈥渄isplay rules鈥 to guide employees on what emotions they should use when interacting with the public. A Journal of Applied Psychology (Vol. 96, 2011) study compared the results of surveys conducted using two different types of display rules: positive (requiring a strong display of positive emotions) and neutral (maintaining neutral emotions at all times). In this designed experiment, 145undergraduate students were randomly assigned to either a positive display rule condition(n1=78)or a neutral display rule condition(n2=67). Each participant was trained to conduct the survey using the display rules. As a manipulation check, the researchers asked each participant to rate, on a scale of 1= 鈥渟trongly agree鈥 to5= 鈥渟trongly disagree,鈥 the statement, 鈥淭his task requires me to be neutral in my expressions.鈥

a. If the manipulation of the participants was successful, which group should have the larger mean response? Explain.

b. The data for the study (simulated based on information provided in the journal article) are listed in the table above. Access the data and run an analysis to determine if the manipulation was successful. Conduct a test of hypothesis using伪=0.05 .

c. What assumptions, if any, are required for the inference from the test to be valid?

The data is given below

Positive Display Rule:

243333444444444444454444444444444445555555555555555555555555555555555555555555


Neutral Display Rule:

3321211122122232212222212222221222222232122212122322222222222122222


Short Answer

Expert verified

The definition of service is to restore, preserve, and supply something to someone. Clients who are satisfied with the service they get are more inclined to trust and remain committed to the company.

Step by step solution

01

Step-by-Step SolutionStep 1: Calculate the means and standard deviation of both groups

The mean of students displaying Positive Rule =4.4872and the mean of students displaying Neutral Rule = role="math" localid="1652701137112" 1.8955

The standard deviation of students displaying Positive Rule = 0.6595and the standard deviation of students displaying Neutral Rule = 0.4965

02

(a) Find the group having a larger mean response 

The mean response of the positive and neutral groups is4.4872 and1.8955 respectively.

Thus, if the manipulation of the participants is successful, then the positive emotions group has the larger mean response than the neutral emotions group.

Hence, the corresponding members of the groups of positive emotions must disagree with the given statement, which results in a higher response.

03

(b) Conduct z-test

Null Hypothesis: The manipulation was not successful. H0:12=0

Alternate Hypothesis: The manipulation was successful Ha:12>0

Let the confidence level be 90%.

Level of significance ()=0.10

So,role="math" localid="1652701587818" 2=0.05and z0.05=1.645

z=x1x212n1+22n2=4.48721.8955(0.6595)278+(0.4965)267=2.59170.0962=26.940

The critical value is1.645

As the value role="math" localid="1652701582886" zis more than the critical value, the null hypothesis should be rejected.

Therefore, the data provide sufficient evidence to indicate that12>0 .

04

(c) State the assumption

The assumption is that independent random samples are selected for each population.

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Most popular questions from this chapter

Shared leadership in airplane crews. Refer to the Human Factors (March 2014) study of shared leadership by the cockpit and cabin crews of a commercial airplane, Exercise 8.14 (p. 466). Recall that each crew was rated as working either successfully or unsuccessfully as a team. Then, during a simulated flight, the number of leadership functions exhibited per minute was determined for each individual crew member. One objective was to compare the mean leadership scores for successful and unsuccessful teams. How many crew members would need to be sampled from successful and unsuccessful teams to estimate the difference in means to within .05 with 99% confidence? Assume you will sample twice as many members from successful teams as from unsuccessful teams. Also, assume that the variance of the leadership scores for both groups is approximately .04.

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4.134 Refer to Exercise 4.133. Find the following probabilities:

a.P(20x30)b.P(20<x30)c.P(x30)d.P(x45)e.(x40)f.(x<40)g.P(15x35)h.P(21.5x31.5)

Let t0 be a particular value of t. Use Table III in Appendix D to find t0 values such that the following statements are true.

a.=P(-t0<t<t0).95wheredf=10b.P(t-t0ortt0)wheredf=10c.P(tt0)=.05wheredf=10d.P(t-t0ortt0)=.10wheredf=20e.P(t-t0ortt0)=.01wheredf=5

Question: Two independent random samples have been selected鈥100 observations from population 1 and 100 from population 2. Sample means x1=26.6,x2= 15.5 were obtained. From previous experience with these populations, it is known that the variances are12=9and22=16 .

a. Find (x1-x2).

b. Sketch the approximate sampling distribution for (x1-x2), assuming (1-2)=10.

c. Locate the observed value of (x1-x2)the graph you drew in part

b. Does it appear that this value contradicts the null hypothesis H0:(1-2)=10?

d. Use the z-table to determine the rejection region for the test againstH0:(1-2)10. Use=0.5.

e. Conduct the hypothesis test of part d and interpret your result.

f. Construct a confidence interval for 1-2. Interpret the interval.

g. Which inference provides more information about the value of 1-2鈥 the test of hypothesis in part e or the confidence interval in part f?

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