Question

The Anxiety Correlation Coefficient is a number that determines the level of anxiety that a person has toward stressful situations. Scores in the range: (5,7.5) are considered normal and do not impede performance on the job among Federation members. Scores below this range indicate possible pathological tendencies, and scores above this range indicate excessive anxiety which may impede job performance and over-all mental health.

The random sample of 10 crew members were given the Anxiety Correlation Coefficient test before their journey to the nebulae and again 2 weeks after they returned from their journey. The following are the results of the two tests:

before:

5.2

5.6

6.6

7.0

6.7

6.1

7.0

6.5

7.1

6.9

after

6.7

8.3

7.2

7.6

6.6

5.9

8.0

7.9

7.1

7.4

Use the Classical Method, with α= .002, to determine if there is a difference between the pre-and-post test scores.

1. List all computer/calculator steps used in calculations
2. State the conclusion of the experiment. Is there sufficient statistical evidence to support the claim that the difference between pre and post-tests were different?

iii. Find the correlation coefficient, the Least Squares Regression Line, and sketch a scatterplot for the pre and post test scores for the sample of 10 crew members. Describe whether the correlation is weak, moderate, strong, or zero, and if it is negative or positive. Interpret what this correlation tells us about differences in individual reactions to stressful situations before and after the trip to the nebulae?

Answer 1

	sample 1	sample 2	Di	(Di - Dbar)²
sum =	64.7	72.70	-8.000	7.120

Ho :   µd=   0                      
Ha :   µd ╪   0                      
                              
Level of Significance ,    α =    0.002                      
                              
sample size ,    n =    10                      
                              
mean of sample 1,    x̅1=   6.470                      
                              
mean of sample 2,    x̅2=   7.270                      
                              
mean of difference ,    D̅ =ΣDi / n =   -0.8000                      
                              
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    0.8894                      
                              
std error , SE = Sd / √n =    0.8894   / √   10   =   0.2813          
                              
t-statistic = (D̅ - µd)/SE = (   -0.8   -   0   ) /    0.2813   =   -2.844  
                              
Degree of freedom, DF=   n - 1 =    9                      
t-critical value , t* =    ±   4.2968   [excel function: =t.inv.2t(α,df) ]                   

|test stat| < |critical value | , do not reject Ho

there is not sufficient statistical evidence to support the claim that the difference between pre and post-tests were different?

...........

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	64.70	72.70	3.72	4.80	0.70
mean	6.47	7.27	SSxx	SSyy	SSxy

correlation coefficient ,    r = SSxy/√(SSx.SSy) =   0.166

weak positive

since correlation is weak, so no or less difference  in individual reactions to stressful situations before and after the trip to the nebulae

...........

Sample size,   n =   10      
here, x̅ = Σx / n=   6.470          
ȳ = Σy/n =   7.270          
SSxx =    Σ(x-x̅)² =    3.7210      
SSxy=   Σ(x-x̅)(y-ȳ) =   0.7      
              
estimated slope , ß1 = SSxy/SSxx =   0.701/3.721=   0.1884      
intercept,ß0 = y̅-ß1* x̄ =   7.27- (0.1884 )*6.47=   6.0511      
              
Regression line is, Ŷ=   6.051   + (   0.188   )*x

Please let me know in case of any doubt.

Thanks in advance!

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