Question

A student hourly employee does small secretarial projects. Her manager wants to learn about the relationship between the number of projects (y) the student completes in a day and the number of hours (x) she works in a day. A random sample of nine days provided the following information:

Working Hours (X)	1	2	3	4	4	6	7	7	8
Number of Projects (Y)	2	3	5	6	5	8	9	8	10

a) define variable types (circle all that apply)

Working Hours (X): response      predictor      dependent     independent

Number of Projects (Y): response      predictor      dependent     independent

b) Sketch scatter plot for the number working hours and the number of projects.

c) Does the scatter plot in part b. reveal a linear the number of projects and the number of working hours? Explain.

d) Using provided output, write out the estimated regression (least squares) equation of the number of projects as a linear function of the working hours.

	Coeffecients	Standard Error
Intercept	1.1018	0.3777
Working Hours	1.0972	0.0725

e) Using the regression equation, predict the number of projects completed in 5 hours.

f) Formally, test if there is a linear relationship between the number of projects and the number of working hours. Use alpha = 0.05

i) State H₀ and H_a

ii) Calculate the test statistic

iii) Specify rejection region for H₀. Use appropriate distribution critical values.

iv) Make decision and conclusion

g) Compute the 95% confidence interval for the regression slope.

h) Do the values in the 95% CI support the existence of non-trivial positive linear relationship between the number of projects and the number of working hours? Explain.

Answer 1

a) Working Hours (X): predictor, independent

Number of Projects (Y): response, dependent

b) Scatterplot:

c) Yes, there is a linear relationship between the number of projects and the number of working hours.

d) Regression equation using output:

e) Predicted value at X =   5          
ŷ = 1.1018 + 1.0972   *   5
= 6.5878   

f) Null and alternative hypothesis:

Ho:   β₁ =   0          
H1:   β₁ ╪   0          
n=   9          

df = 9-1=8
alpha=   0.05              
Standard error of slope = Se(b1) = 0.0725

t stat = b/se(b1) = 1.0972/0.0725 =  15.134   

t-critical value= T.INV.2T(0.05) =+/-  2.3646

Reject Ho if t >2.3646 or if t < -2.3646

decision : t > critical t , reject Ho              

g) 95% confidence interval:

t-critical value= T.INV.2T(0.05) =+/-  2.3646

Lower limit = b - tc*se(b1) =  0.9257
Upper limit = b + tc*se(b1) = 1.2688