Question

In a study, you measure how many cups of coffee a random sample of 10 college students drink per day and their stress levels and obtain the following data:

Participant	Cups of Coffee	Stress Level
1	3	40
2	4	30
3	1	50
4	5	90
5	6	50
6	4	30
7	7	100
8	1	10
9	2	20
10	0	70

1. Construct a scatterplot of the data (using a graph from Excel, another computer program that you are familiar with, or by hand).

2. Compute the statistic that describes the nature of the relationship formed by the data.

3. What is the proportion of variance in stress level that is accounted for by coffee consumption? (Be sure to interpret this value)

4. What is the proportion of variance in stress level that is not accounted for by coffee consumption? (Be sure to interpret this value)

5. Compute the linear regression equation for predicting stress level from coffee consumption.

6. If a student drinks 3 cups of coffee per day, what would you predict that student’s stress level to be?

7. Compute the linear regression equation for predicting coffee consumption from stress level.

8. If a student has a stress level of 60, how many cups of coffee per day do you predict that student drinks?

Answer 1

A)

..............

B)

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	33.00	490.00	48.10	7890.00	293.00
mean	3.30	49.00	SSxx	SSyy	SSxy

correlation coefficient ,    r = SSxy/√(SSx.SSy) =   0.47562

........

C)

R² =    (SSxy)²/(SSx.SSy) =    0.2262  
Approximately    22.62%   of variation in observations of variable Y, is explained by variable x  
..........

D)

unexplained variation =    1-R² =    0.7738

............

E)

sample size ,   n =   10      
here, x̅ = Σx / n=   3.300          
ȳ = Σy/n =   49.000          
SSxx =    Σ(x-x̅)² =    48.1000      
SSxy=   Σ(x-x̅)(y-ȳ) =   293.0      
              
estimated slope , ß1 = SSxy/SSxx =   293/48.1=   6.091476      
intercept,ß0 = y̅-ß1* x̄ =   49- (6.0915 )*3.3=   28.898129      
              
Regression line is, Ŷ=   28.898   + (   6.091   )*x
...........

F0

Predicted Y at X=   3   is          
Ŷ=   28.8981   +   6.0915   *3=   47.17
.......

G)

COOFEE FROM STRESS LEVEL;

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	490.00	33.00	7890.00	48.10	293.00
mean	49.00	3.30	SSxx	SSyy	SSxy

sample size ,   n =   10      
here, x̅ = Σx / n=   49.000          
ȳ = Σy/n =   3.300          
SSxx =    Σ(x-x̅)² =    7890.0000      
SSxy=   Σ(x-x̅)(y-ȳ) =   293.0      
              
estimated slope , ß1 = SSxy/SSxx =   293/7890=   0.037136      
intercept,ß0 = y̅-ß1* x̄ =   3.3- (0.0371 )*49=   1.480355      
              
Regression line is, Ŷ=   1.480   + (   0.037   )*x

...........

H)

Predicted Y at X=   60   is          
Ŷ=   1.4804   +   0.0371   *60=   4

..................

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