In: Statistics and Probability
Use this study to answer the following six questions: A large company had several members of its board of directors arrested for embezzlement, widely reported in the national news in a negative manner. Their human resources department was concerned that the company's resulting reputational damage had a negative effect on their employees. The HR department administers a standard stress questionnaire to their employees annually, which has a mean score of 129.4 (higher numbers indicate more stress). They obtained a sample of 61 current employees and gave them the stress questionnaire after the negative news was publicized; results showed a mean of =131.2 and standard deviation of = 12.30. Did the news about the arrests increase employee stress?
Identify the null and alternative hypotheses:
What is the correct CV and decision rule for rejecting the null hypothesis?
What is the standard error?
What is the observed value of t?
What is eta2?
Write an APA style conclusion about the results, including properly formatted statistics, in the context of the study.
Ho : µ = 129.4
Ha : µ > 129.4
degree of freedom= DF=n-1=
60
CV , critical t value, t* =
1.671 [Excel formula =t.inv.rt(α,df) ]
decision rule, reject Ho if t >1.671
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Standard Error , SE = s/√n = 12.3000 / √
61 = 1.5749
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t-test statistic= (x̅ - µ )/SE = (
131.200 - 129.4 ) /
1.5749 = 1.143
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decision: fail to reject Ho beause t <1.671
there is not enough evidence to conlcude that news about the arrests increase employee stress
t(60) = 1.143
p value=0.1288