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. Complete the F-table. (Round your answers to two decimal places.) Compute a Bonferroni procedure and interpret the results. (Assume experimentwise alpha equal to o.05. Select all that apply.) There are no significant differences between any of the groups. There is a significant difference in decision making for the 12-hour and 48-hour sleep deprivation conditions. There is a significant difference in decision making for the 12-hour and 24-hour sleep deprivation conditions. There is a significant difference in decision making for the 24-hour and 48-hour sleep deprivation conditions.

Explanation / Answer

We shall do this in the open source stats package R

The complete R snippet is as follows

# read the data into R dataframe
data.df<- read.csv("C:\Users\586645\Downloads\Chegg\sleep.csv",header=TRUE)
str(data.df)

library(reshape2)
mdata <- melt(data.df)

# perform anova analysis
a<- aov(lm(value~ variable,data=mdata))

#summarise the results
summary(a)

The results are

> summary(a)
Df Sum Sq Mean Sq F value Pr(>F)
variable 2 165.3 82.67 4.977 0.022 *
Residuals 15 249.2 16.61   
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Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

As the p value is less than 0.05 , hence the result is significant

The table is thus

ANOVA is an omnibus test , it only tells whether the group means are different or not , it doesnt tell which groups are different

Lets perform a test to check the difference between the actual groups

TukeyHSD(a)
Tukey multiple comparisons of means
95% family-wise confidence level

Fit: aov(formula = lm(value ~ variable, data = mdata))

$variable
diff lwr upr p adj
X24hours-X12hours -4.666667 -10.77875 1.445417 0.1506186
X48hours-X12hours -7.333333 -13.44542 -1.221250 0.0182755
X48hours-X24hours -2.666667 -8.77875 3.445417 0.5092494

as the p value is for 12-48 hours is less than 0.05 , hence this difference in signifcant , hence B

the bonferroni correction

> pairwise.t.test(mdata$value,mdata$variable,p.adj = "bonferroni")

   Pairwise comparisons using t tests with pooled SD

data: mdata$value and mdata$variable

X12hours X24hours
X24hours 0.198 -   
X48hours 0.021 0.825   

P value adjustment method: bonferroni

this also tells us that the correct pair is 12-48 , as the p value is 0.021 which is less than 0.05

Anova: Single Factor SUMMARY Groups Count Sum Average Variance 12hours 6 145 24.16667 30.16667 24hours 6 117 19.5 7.5 48hours 6 101 16.83333 12.16667 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 165.3333 2 82.66667 4.976589 0.021991 3.68232 Within Groups 249.1667 15 16.61111 Total 414.5 17

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