Question

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

Control: 8 10 9 8 10. Short: 0 1 2 7 5 Long: 6 7 7 6 9

Please analyze the data fully (including appropriate follow-up tests) and provide a brief written summary of the results.

Answer 1

As there are more than 2 lengths of unemployement, we shall use an ANOVA to test the mean number of errors across the lengths of unemployment.

						Total
Control	8	10	9	8	10	45
Short	0	1	2	7	5	15
Long	6	7	7	6	9	35
					Total	95

Null hypothesis: The avaerage number of errors made in proof reading is the same across three employment levels.

Alternate Hypothesis: The average number of errors vary for at least two periods.

Level of significance:

g= 0,05

Test Statistic: We shall use the F-test by constructing ANOVA table:

1. Calculate the correction factor:

2 5 3 vi=1 -j=1 Yij CF n

$CF=\frac{95^2}{15}=\frac{9025}{15}=601.6667$

2. Calculate the total sum of squares:

3 5 TSS = 4; -CF i=1 j=1

We need the squares of individual values and total them:

  

Control	64	100	81	64	100	409
Short	0	1	4	49	25	79
Long	36	49	49	36	81	251
					Total	739

$TSS=739-601.6667$

$TSS=137.3333$

3. The sum of squares due to periods: SSP

$SSP=\sum_{i=1}^{3}\frac{y_{i\cdot }^{2}}{5}-CF$

  $SSP=\frac{45^2+15^2+35^2}{5}-601.6667$

$SSP=\frac{2025+225+1225}{5}-601.6667$

$SSP=\frac{3475}{5}-601.6667$

$SSP=695-601.6667$

$SSP=93.3333$

4. Error sum of squares ESS=TSS-SSP

ESS=137.3333-93.3333

$SSP=44$

We shall concstruct the ANOVA table:

Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	93.3333	2	46.6667	12.7273	0.0011	3.8853
Within Groups	44.0000	12	3.6667
Total	137.3333	14

From the ANOVA table, since the p-value for teh periods is p=0.0011<0.05, we reject the null hypothesis. Hence, we conclude that the average number of errros differes between at least two periods of employment.  

We shall do a post-hoc analysis of the means to find out which pairs are differing significantly.

We shall calculate the Least significant difference using the formula

$LSD=t_{1-\alpha /2,12}*\sqrt{MS_{errror}*\left ( \frac{1}{5}+\frac{1}{5} \right )}$

$LSD=2.1788*\sqrt{3.6667*0.4}$

$LSD=2.1788*\sqrt{1.4667}$

$LSD=2.1788*1.2111$

$LSD=2.6387$

We shall form the differenc table as below:

Groups	Average	Comparison	Absolute difference	LSD	Significant
Control	9	Control Vs. Short	6	2.6387	Yes
Short	3	Control Vs. Long	2	2.6387	No
Long	7	Short Vs. Long	4	2.6387	Yes

We can see from the above that the difference between Control and Short and Short Vs. Control are statistically different.