Answer:
The answers to the question are
(a) The time series plot is given as attached
(b) The parameters for the line that minimizes MSE for the time series are;
y-intercept, b₀ = 19.993
Slope, b₁ = 1.77
MSE = 19.44
T[tex]_t[/tex] = 19.993 + 1.774·t
(c) The average cost increase that the firm is realizing per year is $ 1.77
(d) The estimate of the cost/unit for next year is $35.96.
Step-by-step explanation:
(a) Using the provided data, the time series plot is given as attached
(b) Given hat the y-intercept, = b₀
Slope = b₁
Therefore the linear trend forecast equation is given s
T[tex]_t[/tex] = b₀ + b₁·t
The linear trend line slope is given as
b₁ = [tex]\frac{\Sigma^n_{t=1}(t-\overline{\rm t)}(Y_t-\overline{\rm Y)}}{\Sigma^n_{t=1}(t-\overline{\rm t} )^2}[/tex]
b₀ = [tex]\overline{\rm Y}[/tex] - b₁·[tex]\overline{\rm t}[/tex]
Where:
Y[tex]_t[/tex] = Time series plot value at t
n = Time period number
[tex]\overline{\rm Y}[/tex] = Time series data average value and
[tex]\overline{\rm t}[/tex] = Average time, t
Therefore, [tex]\overline{\rm t}[/tex] = [tex]\frac{\Sigma^n _{t=1} t}{n} = \frac{36}{8} =4.5[/tex]
[tex]\overline{\rm t}[/tex] = 4.5
[tex]\overline{\rm Y}[/tex] = [tex]\frac{\Sigma^n _{t=1} Y_t}{n} = \frac{223.8}{8} =27.975[/tex]
[tex]\overline{\rm Y}[/tex] = 27.975
Therefore the linear trend line equation T[tex]_t[/tex], is
b₁ = [tex]\frac{\Sigma^n_{t=1}(t-\overline{\rm t)}(Y_t-\overline{\rm Y)}}{\Sigma^n_{t=1}(t-\overline{\rm t} )^2}[/tex] = [tex]= \frac{74.5}{42}[/tex] = 1.774
b₀ = [tex]\overline{\rm Y}[/tex] - b₁·[tex]\overline{\rm t}[/tex] = 27.975 - 1.774×4.5 = 19.993
Therefore the trend equation for the linear trend is
T[tex]_t[/tex] = 19.993 + 1.774·t
MSE = [tex]\frac{1}{2} \Sigma(Y-\overline{\rm Y)^ }^2[/tex] = [tex]\frac{155.495}{8}[/tex] = 19.44
(c) From the linear trend equation, the average is given as the slope of the curve or b₁ which is equal to 1.774
Therefore the average cost increase that the firm has been realizing per year is $ 1.77
(b) From the equation of the future trend, we have when y = 9
T[tex]_t[/tex] is given as
T[tex]_t[/tex] = 19.993 + 1.774×9 = 35.96
The cost/unit for 9th year is $35.96
