Question

I am trying to find an appropriate statistical test to run for a research study using someone else's gathered data (so that no IRB process is needed). In their data they present:

Likelihood of Falling Asleep:

Never 17

Seldom 22

Moderate 15

High 12

Use of napping during duty:

Never 27

Rarely 19

Sometimes 16

Often 4

Both of these seem to be independent variables, but is there a way to show a relationship (or lack thereof) without a dependent variable. In this case the dependent variable could be "pilot" of which 66 were surveyed for the study that I am taking the data from. Trying accurately to show whether or not the likelihood of falling asleep in the cockpit is related to whether or not the pilot naps on duty outside of the cockpit.

Thanks!

Answer 1

To find the whether or not the likelihood of falling asleep in the cockpit is related to whether or not the pilot naps on duty outside of the cockpit. we use Chi-Square test for association using contingency table.

Column and Row Totals
	Likelihood of Falling Asleep	Use of napping during duty:	Row Totals
Never	17	27	44
Seldom	22	19	41
Moderate	15	16	31
High	12	4	16
Column Totals	66	66	132  (Grand Total

Hypothesis:

H₀ : likelihood of falling asleep in the cockpit is not related tothe pilot naps on duty outside of the cockpit.

H₁ : likelihood of falling asleep in the cockpit is related to the pilot naps on duty outside of the cockpit.

Level of significance = 0.05

Eij = Expected value of two nominal variables

Oij = Observed value of two nominal variables
Degree of freedom is calculated by using the following formula:
DF = (r-1)(c-1) = 1*3 = 3
Where
DF = Degree of freedom
r = number of rows
c = number of columns

Results
	Likelihood of Falling Asleep	Use of napping during duty:	Row Totals
Never	17  (22.00)	27  (22.00)	44
Seldom	22  (20.50)	19  (20.50)	41
Moderate	15  (15.50)	16  (15.50)	31
High	12  (8.00)	4  (8.00)	16
Column Totals	66	66	132  (Grand Total)

values in the () represents Expected counts

The chi-square statistic is 6.5245. The p-value is .088701. The result is not significant at p < .05.