Question

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 :

Ages X	Cost Y
110	655.34
113	753.36
114	785.04
134	886.28
93	685.24
141	952.32
115	649.48
115	677.96
115	866.9
142	1052.74
96	724.84
139	897.52
89	670.54
93	701.88
91	583.62
109	935.6
138	948.96
83	708.3
100	840.22
137	832.08

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.

Answer 1

ΣX = 2267
ΣY = 15808.22
ΣX * Y = 1827460.76
ΣX2 = 264161
Sxx =Σ (Xi - X̅ )2 = 7196.55
Syy = Σ( Yi - Y̅ )2 = 311204.2692
Sxy = Σ (Xi - X̅ ) * (Yi - Y̅) = 35599.023

X̅ = Σ (Xi / n ) = 2267/20 = 113.35
Y̅ = Σ (Yi / n ) = 15808.22/20 = 790.411

part a)

Equation of regression line is Ŷ = a + bX
b = ( n Σ(XY) - (ΣX* ΣY) ) / ( n Σ X2 - (ΣX)2 )

b = 4.9467

a =( ΣY - ( b * ΣX ) ) / n
a =( 15808.22 - ( 4.9467 * 2267 ) ) / 20
a = 229.7049
Equation of regression line becomes Ŷ = 229.7049 + 4.9467 X

Part b)

Slope = 4.9467

For every unit increase in ages, there would be increase of 4.9467 of cost.

part c)

r = 0.7522

There is moderate and positive correlation between age and cost.

Part d)

= 0.5659

There is 56.59% of variation in cost is exaplined by the variable age.

Part e)

To Test :-

H0 :- ß = 0

H1 :- ß 0

Test Statistic :-
t = ( b - β) / ( S / √(S(xx)))
t = ( 4.9467 - 0 ) / ( 86.6367 / √(7196.55))
t = 4.8437

Test Criteria :-
Reject null hypothesis if |t| > t(α, n-2)
t(α/2,n-2) = t(0.05/2 , 20 - 2 ) = 2.1009
| t | > t(α/2, n-2) = 4.8437 > 2.1009
Result :- Reject null hypothesis

Decision based on P value
P - value = P ( t > 4.8437 ) = 0.0001
Reject null hypothesis if P value < α level of significance
P - value = 0.0001 < 0.05 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis

There is statistically lienar relation between variables.

Part f)

Ŷ = 229.7049 + 4.9467 X

Ŷ = 229.7049 + 4.9467 ( 120 )
Ŷ = 823.3089

part g)

Predictive Confidence Interval of
Ŷ = 229.7049 + 4.9467X
Ŷ = 823.3089

t(0/2) = t(0.05/2) = 2.101
X̅ = (Xi / n ) = 2267/20 = 113.35


95% Predictive confidence interval is ( 636.2490 < < 1010.3638 )