Question

3. A study was conducted concerning the effects of varying lengths of unemployment on ability to concentrate. Data represent participants' proofreading scores (number of errors found). Finding fewer errors on the proofreading task is a behavioral indicator of stress. Length of unemployment was regarded by the researchers as an indicator of exposure to chronic stress. Fifteen participants were included in the study and followed across three periods: employment (control), short-term unemployment and long-term unemployment. (20%) Data were collected and are shown below: Length of Unemployment Short 0 Control 8 10 9 1 Long 6 7 7 6 9 8 10 2 7 5 Please analyze the data fully (including appropriate follow-up tests) and provide a brief written summary of the results. Use paper and pencil to calculate the statistics. Show all work./

Answer 1

		Observed data				squared data
	Control	Short	Long		Control	Short	Long
	8	0	6		64	0	36
	10	1	7		100	1	49
	9	2	7		81	4	49
	8	7	6		64	49	36
	10	5	9		100	25	81
Total	45	15	35	GT=95	409	79	251	∑x2=739
Mean	9	3	7

We have to test

H₀: the proofreading scores is same for all the lengths of unemployment i.e. µ_control = µ_short = µ_long

H₁: the proofreading scores for at least one length of unemployment are significantly different from others i.e. at least one µ_i is different.

To test the above hypotheses, we need to perform one-way ANOVA as below:

We have n= 15

t=3

r=5

Correction factor=

(GT) 72

Correction factor=

(ss] 15

Correction factor=601.667

Total sum of squares= sum of squares of every observation - correction factor

Total sum of squares= 739 – 601.667

Total sum of squares= 137.33

Treatment sum of squares=

* + + 3) - CF

Treatment sum of squares=

-452 163 352. + +4 601.667

Treatment sum of squares=

695 - 601.667

Treatment sum of squares=

93.33

Error sum of squares= Total SS – Treatment SS

Error sum of squares=137.33–93.33

Error sum of squares= 44

ANOVA Table

Source of Variation	df	SS	MS	F	p value
Groups	3-1=2	93.33	46.667	12.727	0.001
Error	12	44	3.667
Total	15-1=14	137.33

The p-value of above ANOVA table is 0.001 which is less than 0.05 and it indicates that we have strong evidence against null hypothesis to reject it, so we fail to reject the null hypothesis and we can conclude that the proofreading scores significantly differs with the length of unemployment.

Pairwise comparison:

Now, we have established that the proofreading scores differs significantly with the length of unemployment, we perform the pairwise test to test which pair of means are significantly different.

To perform the pairwise comparison we need to calculate the least significant difference as below:

2 MSE LSD = (a/2 error df)* r V

We are comparing the pairwise mean at 0.05 level of significance, so we need the table value of t at 0.025 level i.e. t_{(0.025, 12)} = 2.178813

Also the MSE = 3.667 and r=5

So

LSD = 2.178813* 2 + 3.667 5 V

LSD = 2.64

Now if any pair of mean difference is greater than this LSD, then that pair of mean is said to be significantly different. The results are shown in below table:

Pair of treatments	Pairwise difference	Result	Decision
control-short	6	more than LSD	Significantly different
control-long	2	less than LSD	Significantly not different
long-short	4	more than LSD	Significantly different

In short, we can report that average proofreading of short length of unemployment is significantly different from other two averages.

The lowest average is of the short length of unemployment which indicates that people with short length of unemployment has high behavioural stress.