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Studies have established that rudeness in the workplace can lead to retaliatory and counterproductive behaviour. However, there has been little research on how rude behaviours influence a victim鈥檚 task performance. Such a study was conducted, and the results were published in the Academy of Management Journal (Oct. 2007). College students enrolled in a management course were randomly assigned to two experimental conditions: rudeness condition (students) and control group (students). Each student was asked to write down as many uses for a brick as possible in minutes. For those students in the rudeness condition, the facilitator displayed rudeness by generally berating students for being irresponsible and unprofessional (due to a late-arriving confederate). No comments were made about the late-arriving confederate to students in the control group. The number of different uses for brick was recorded for each student and is shown below. Conduct a statistical analysis (at $\mathbf{\text{伪=0.01}}$) to determine if the true mean performance level for students in the rudeness condition is lower than the actual mean performance level for students in the control group.

The data is given below

Control Group: $\begin{array}{l}\mathbf{\text{124516217201920191023160491317130212117311119912}}\\ \mathbf{\text{185213015421211101311361013161228191230}}\end{array}$

Rudeness Condition: $\begin{array}{l}\mathbf{\text{4111811965119127573111911107891071141354783815}}\\ \mathbf{\text{91610071513921310}}\end{array}$

Step by step solution

01

Step-by-Step Solution Step 1: Calculate the mean of both groups

Let ${\mathrm{渭}}_1$be the average level of performance of students in the Control Unit =$12.15$ and

${渭}_2$be the mean level of performance for students in the Rudeness Group is 8.5

02

Calculate the standard deviation of both groups

Let ${蟽}_1$be the standard deviation of the performance level of students in Control Group = $8.05$and

${蟽}_2$be the standard deviation of the performance level of students in the Rudeness Group = $3.95$

03

Calculate a pooled estimate of variance

The pooled estimate of variance is${s_p}^2=\frac{(n_1鈭�1){{蟽}_1}^2+(n_2鈭�1){{蟽}_2}^2}{n_1+n_2鈭�2}$

$\begin{array}{l}=\frac{(53鈭�1)64.8025+(45鈭�1)15.6025}{53+45鈭�2}\\ =\frac{3369.73+686.51}{96}\\ =42.2525\end{array}$

04

Conduct a t-test

Null Hypothesis: The mean level of performance of students in the control group is the same as that of the students in the rudeness group. $H_0:{渭}_1鈭�{渭}_2=0$

Alternate Hypothesis: The mean level of performance of students in the control group is higher than that of the students in the rudeness group. $H_a:{渭}_1鈭�{渭}_2>0$

The level of significance is$0.01$.

Degree of freedom =$n_1+n_2鈭�2=96$

From the t-distribution table, the critical value at 0.01 the level of the significance for $96$degrees of freedom about the right-tailed test is role="math" localid="1652700633536" $2.366$.

$\begin{array}{l}t=\frac{\stackrel{炉}{x_1}鈭�{\stackrel{炉}{x}}_2}{\sqrt{{s_p}^2(\frac{1}{n_1}+\frac{1}{n_2})}}\\ =\frac{12.15鈭�8.5}{\sqrt{42.2525\left(\frac{1}{53}+\frac{1}{45}\right)}}\\ =\frac{3.65}{\sqrt{42.2525\left(0.041\right)}}\\ =\frac{3.65}{1.3176}\\ =2.77\end{array}$

The corresponding p-value is$0.003365$

As the p-value is less than$0.01$ , the null hypothesis should be rejected.

Therefore, the data provide sufficient evidence to indicate that${\mathrm{渭}}_1鈭�{\mathrm{渭}}_2>0$ .

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