In a recent study by Lu et al (2017) it was suggested that the number countries you have visited, the less precise your moral compass becomes, e.g. the more you are likely to cheat on a test. They suggest that exposure to a variety of moral codes means people start to view morality as relative rather than absolute leading to more lax following of rules for example. Imagine the table contains some of their data:

Covariable 1: Number of countries visited	Covariable 2: No. of times the Pps cheated on a test
2	4
4	7
6	12
7	7
8	15
13	12
21	16
24	15

a. Explain which statistical test Lu et al should use to analyse this data and why (3)

Lu et al should use a Spearmans rho test. This is because they are doing a test of association as it’s a correlational study; they are looking for a relationship between the pairs of scores relating to number of countries visited and number of times Pps cheat on a test. Secondly, they have ratio data as they have the number of countries and the number of cheats for each Pps and these can be put into rank order.

b. Calculate the observed value (7)

Covariable 1: Number of countries visited	Covariable 2: No. of times the Pps cheated on a test	Covariable 1: Ranked (R1)	Covariable 2: Ranked (R2)	D	D2
2	4	1	1	0	0
4	7	2	2.5	-0.5	0.25
6	12	3	4.5	-1.5	2.25
7	7	4	2.5	1.5	2.25
8	15	5	6.5	-1.5	2.25
13	12	6	4.5	1.5	2.25
21	16	7	8	-1	1
24	15	8	6.5	1.5	2.25

Step 1: Rank the data for co-variable 1, in this example the number of countries visited were already sorted into ascending order so that was rather easy 😉

Step 2: Rank the co-variable 2 data; here you had to be careful with your ranking, there were several tied ranks which need to be handled with care and you needed t write the ranks against the correct numbers.

Step 3: Do the co-variable 1 ranks minus the co-variable 2 ranks for each Pp (R1 -R2) to find the difference.

Step 4: Do DXD to give the D2 (D squared) column)

Step 5: Look up the Spearman’s formula in front of booklet and remember first stage is to find the sum of the D2 column.

∑D²= 12.5  (here the ∑ means sum total, you just had to add up the D² column on your calculator).

6X12.5=75

N=8    8²=64    64-1= 63   63×8=504

75/504=0.14881

1-0.14881=0.85119

r=0.851

c. State whether you should use a one or a two tailed test when checking for significance and why (2)

Lu et al would use a one tailed test because in the scenario it says that they believed that the more countries a person had visited the more they would cheat on the test and this is a directional hypothesis.

d. State the critical value (1)

The critical value for a one tailed test at the 0.05 level of significance when N=8 is 0.548.

e. State whether Lu would reject the null hypothesis based on this analysis and why (3)

Yes Lu et al would reject the null hypothesis. This is because their observed value of R=0.851 is greater than the critical value of 0.548, (for a one tailed test at the 0.05 level when N=8), and this means that the directional hypothesis that “the more countries you have visited, the more you will cheat on a test” can be retained and the null rejected, therefore the probability that the null hypothesis is true or that the results arose due to chance alone is less than 1 in 20, (p<0.05). The results are actually still significant at the 0.5% level (critical value 0.81) meaning the probability that the null is actually correct or the results are due to chance alone is less than 1 in 200, (p<0.005).