Question

1. : The president of a company that manufactures car seats has been concerned about the number and cost of machine breakdowns. The problem is that the machines are old and becoming quite unreliable. However, the cost of replacing them is quite high, and the president is not certain that the cost can be made up in today’s slow economy. To help make a decision about replacement, he gathered data about last month’s costs for repairs in dollars (y) and the ages in months (x) of the plant’s 20 welding machines as recorded in repairs.mtw.

a) Find the estimated regression equation to predict the repair cost of a machine from its age.

b) Interpret the slope in the estimated regression equation.

c) Find and interpret the correlation between the repair cost of a machine and its age.

d) Find the coefficient of determination, and discuss what this statistic tells you.

e) At 5% significance level, test to determine whether the age of a machine and its monthly cost of repair are linearly related.

f) Predict the average monthly repair cost of welding machines that are 120 months old.

g) Predict with 95% confidence the monthly repair cost of a welding machine that is 120 months old.

Ages X
110
113
114
134
93
141
115
115
115
142
96
139
89
93
91
109
138
83
100
137

Cost Y
655.34
753.36
785.04
886.28
685.24
952.32
649.48
677.96
866.90
1052.74
724.84
897.52
670.54
701.88
583.62
935.60
948.96
708.30
840.22
832.08

Answer 1

X	Y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
110	655.34	11.22	18244.18	452.49
113	753.36	0.12	1372.78	12.97
114	785.04	0.42	28.85	-3.49
134	886.28	426.42	9190.87	1979.69
93	685.24	414.12	11060.94	2140.23
141	952.32	764.52	26214.52	4476.78
115	649.48	2.72	19861.55	-232.54
115	677.96	2.72	12645.23	-185.54
115	866.9	2.72	5850.57	126.21
142	1052.74	820.82	68816.50	7515.73
96	724.84	301.02	4299.56	1137.66
139	897.52	657.92	11472.34	2747.35
89	670.54	592.92	14369.06	2918.86
93	701.88	414.12	7837.74	1801.61
91	583.62	499.52	42762.52	4621.78
109	935.6	18.92	21079.85	-631.57
138	948.96	607.62	25137.79	3908.23
83	708.3	921.12	6742.22	2492.07
100	840.22	178.22	2480.94	-664.95
137	832.08	559.32	1736.31	985.47

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	2267	15808.22	7196.55	311204.3	35599.02
mean	113.35	790.41	SSxx	SSyy	SSxy

a)

sample size ,   n =   20          
here, x̅ = Σx / n=   113.35   ,     ȳ = Σy/n =   790.41  
                  
SSxx =    Σ(x-x̅)² =    7196.5500          
SSxy=   Σ(x-x̅)(y-ȳ) =   35599.0          
                  
estimated slope , ß1 = SSxy/SSxx =   35599.0   /   7196.550   =   4.9467
                  
intercept,   ß0 = y̅-ß1* x̄ =   229.7049          
                  
so, regression line is   Ŷ =   229.705   +   4.947   *x

b)

for unit increase in value of variable x , variable y get increase by 4.947

c)

correlation coefficient ,    r = Sxy/√(Sx.Sy) =   0.7522

there is moderate relationship between x and y

d)

R² =    (Sxy)²/(Sx.Sy) =    0.5659

56.59 % of variation in reparing cost is explained by variable ages.

e)

Ho:   ß1=   0          
H1:   ß1╪   0          
n=   20              
alpha =   0.05              
estimated std error of slope =Se(ß1) = Se/√Sxx =    86.637   /√   7196.55   =   1.021
                  
t stat = estimated slope/std error =ß1 /Se(ß1) =    4.9467   /   1.0213   =   4.844
                  
Degree of freedom ,df = n-2=   18              
p-value =    0.0001              
decison :    p-value<α , reject Ho              
reject Ho and conclude  that linear relations exists between X and y

f)

Predicted Y at X=   120   is                  
Ŷ =   229.705   +   4.947   *   120   =   823.306

g)

X Value=   120                      
Confidence Level=   95%                      
                          
                          
Sample Size , n=   20                      
Degrees of Freedom,df=n-2 =   18                      
critical t Value=tα/2 =   2.101   [excel function: =t.inv.2t(α/2,df) ]                  
                          
X̅ =    113.35                      
Σ(x-x̅)² =Sxx   7196.6                      
Standard Error of the Estimate,Se=   86.637                      
                          
Predicted Y at X=   120   is                  
Ŷ =   229.705   +   4.947   *   120   =   823.306
                          
standard error, S(ŷ)=Se*√(1/n+(X-X̅)²/Sxx) =    20.529                      
margin of error,E=t*Std error=t* S(ŷ) =   2.1009   *   20.5286   =   43.1289      
                          
Confidence Lower Limit=Ŷ +E =    823.306   -   43.1289   =   780.178      
Confidence Upper Limit=Ŷ +E =   823.306   +   43.1289   =   866.435