Air traffic controllers perform the vital function of regulating the traffic of

Question

Air traffic controllers perform the vital function of regulating the traffic of passenger planes. Frequently, air traffic controllers work long hours with little sleep. Researchers wanted to test their ability to make basic decisions as they become increasingly sleep deprived. To test their abilities, a sample of 6 air traffic controllers is selected and given a decision-making skills test following 12-hour, 24-hour, and 48-hour sleep deprivation. Higher scores indicate better decision-making skills. The table lists the hypothetical results of this study.

(a) Complete the F-table. (Round your answers to two decimal places.)

Sleep Deprivation 12 Hours 24 Hours 48 Hours 24 19 18 19 22 21 35 24 22 27 22 15 23 15 16 21 22 15

Explanation / Answer

Step 1                          
   Null Hypothesis Ho :           µ1 =µ2 =µ3           
   Alternative Hypothesis :           µ1 µ2 µ3           
Step 2                          
   Degrees of freedom between = k - 1 = 3 - 1 = 2                      
   Degrees of freedom Within = n - k = 18 - 3 = 15                      
                          
   Degrees of freedom Total F( k-1,n - k,) at 0.05 is = F Crit = 3.682                      
                          
Step 3                          
   Grand Mean = G / N = 24.833+20.667+17.833 / 3 = 21.111                      
    SST = ( Xi - GrandMean)^2 = (24-21.111)^2 + (19-21.111)^2 + (35-21.111)^2 + ……..& so on = 407.778                      
   SS Within = (Xi - Mean of Xi ) ^2 =,(24-24.833)^2 + (19-24.833)^2 + (35-24.833)^2 + ……..& so on = 259                      
                          
   SS Between = SST - SS Within = 407.778 - 259 = 148.778                      
Step 4                          
   Mean Square Between = SS Between / df Between = 148.778/2 = 74.389                      
   Mean Square Within = SS Within / df Within = 259/15 = 17.267                      
                          
Step 5                          
   F Cal = MS Between / Ms Within = 74.389/17.267 = 4.308                      
   We got |F cal| = 4.308 & |F Crit| =3.682                      
                          
MAKE DECISION                          
   Hence Value of |F cal| > |F Crit|and Here We Reject Ho                      

ONE WAY ANOVA Treatments Mean = X /n 12 24 19 35 27 23 21 24.833 24 19 22 24 22 15 22 20.667 48 18 21 22 15 16 15 17.833

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