Question

Improvement in the quality (expressed in % of units that fail and require free service under warranty) was regressed on the number hours spent on the task. In the table below, you see the results of a regression analysis.

Regression: Quality
	Intercept	Hours
coefficient	4.55770072	-1.02936616
std error of coef	0.80943691	0.13291009
t-ratio	5.6307	7.7448
p-value	0.0002%	0.0000%
Standard error of regression		2.42986326
R-squared		60.60%
Adjusted R-squared		59.59%
number of observations		41
residual degrees of freedom		39

1. What is the mean influence of each extra hour (spent on a task) on the quality improvement ?
2. Develop a 95% confidence interval for the β-coefficient of Hours.
3. Is the number of extra hours spent on the task is significantly related to the quality improvement? Use α=0.01. Look at the output and answer the question. You need not do a formal hypothesis test showing all steps.
4. Develop a 99% confidence interval for the β-coefficient of Hours.
5. Test the claim that an additional hour spent on task results in an average improvement of better than 1% in quality as defined (% units that require service under warranty). Use α=0.10.

Answer 1

a)

Improvement in quality will decrease by -1.02936616 unit with increase of each extra hour

b)

n =   41                  
alpha,α =    0.05                  
estimated slope=   -1.02936616                  
std error =    0.13291009                  
                      
Df = n-2 =   39                  
t critical value =    2.0227   [excel function: =t.inv.2t(α,df) ]              
                      
margin of error ,E = t*std error =    2.0227   *   0.13291009   =   0.2688  
                      
95%   confidence interval is ß1 ± E                   
lower bound = estimated slope - margin of error =    -1.02936616   -   0.2688   =   -1.2982  
upper bound = estimated slope + margin of error =    -1.02936616   +   0.2688   =   -0.7605  

c)

p-value =    0.0000   
decision:   p value < α , so, reject the null hypothesis      

Slope is significant

d)

n =   41                  
alpha,α =    0.01                  
estimated slope=   -1.02936616                  
std error =    0.13291009                  
                      
Df = n-2 =   39                  
t critical value =    2.7079   [excel function: =t.inv.2t(α,df) ]              
                      
margin of error ,E = t*std error =    2.7079   *   0.13291009   =   0.3599  
                      
99%   confidence interval is ß1 ± E                   
lower bound = estimated slope - margin of error =    -1.02936616   -   0.3599   =   -1.3893  
upper bound = estimated slope + margin of error =    -1.02936616   +   0.3599   =   -0.6695  

THANKS

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