Question

In a study of office lighting, an HR specialist wanted to determine if the accountants in the building are more productive in fluorescent lighting versus natural lighting. She gives a sample of n = 4 accountants in the office a task to complete in both the natural light setting and the fluorescent light setting, and measures how long it takes (in minutes) for the accountants to complete the tasks. Each accountant was tested in both lighting settings. Shorter times indicate higher productivity. The mean difference scores was MD = 7 with SSD­ = 94, reflecting that when in natural lighting, the accountants completed their tasks faster than in fluorescent lighting. Assume we are doing a two-tailed test at α = .05

5-Compute the test statistic t

7.923

2.500

1.305

3.025

6-What decision do you make regarding the null hypothesis?

Reject

Fail to reject

7-Which of the following statements accurately describes the results of this study?

Although the accountants were more productive in natural lighting, this was not significant.

The accountants were significantly more productive in the fluorescent lighting.

Although the accountants were more productive in fluorescent lighting, this was not significant.

The accountants were significantly more productive in the natural lighting.

8-What should the HR specialist use to report effect size?

Estimated Cohen’s d

Confidence intervals

Nothing, because the effect of lighting on productivity was not significant

R-squared

Answer 1

mean of difference ,    D̅ =    7.000                  
                          
std dev of difference , Sd = √(SSD/(n-1)) = 5.598                  
                          
std error , SE = Sd / √n =    5.5976   / √   4   =   2.7988      
                          
t-statistic = (D̅ - µd)/SE = (   7   -   0   ) /    2.7988   =   2.50

6)

p-value =        0.0438
Decision:   p-value <α , Reject null hypothesis  

7)

The accountants were significantly more productive in the fluorescent lighting.

8)

Estimated Cohen’s d