Table 10.7
$$\begin{array} { | c | c | }
\hline \text { Month } & \begin{array} { c }
\text { Units Sold } \
( 00 )
\end{array} \
\hline 1 & 150 \
2 & 145 \
3 & 160 \
4 & 180 \
5 & 220 \
\hline
\end{array}$$
-The Classical Consultant Company provides forecasting research for clients such as a group of five doctors associated with a new hospital health maintenance program.The company has been asked to forecast the number of patients requesting blood analysis per week.The past weekly average is 38 and for the trend is 2 per week.This week's demand was 42 blood tests.How many patients Will come next week? (Suppose ? = 0.10 and ? = 0.30.)
A) less than or equal to 39
B) greater than 39 but less than or equal to 41
C) greater than 41 but less than or equal to 43
D) greater than 43

Question

Table 10.7
 $$\begin{array} { | c | c | } 
\hline \text { Month } & \begin{array} { c } 
\text { Units Sold } \
( 00 )
\end{array} \
\hline 1 & 150 \
2 & 145 \
3 & 160 \
4 & 180 \
5 & 220 \
\hline
\end{array}$$
-The Classical Consultant Company provides forecasting research for clients such as a group of five doctors associated with a new hospital health maintenance program.The company has been asked to forecast the number of patients requesting blood analysis per week.The past weekly average is 38 and for the trend is 2 per week.This week's demand was 42 blood tests.How many patients Will come next week? (Suppose ? = 0.10 and ? = 0.30.)
A) less than or equal to 39
B) greater than 39 but less than or equal to 41
C) greater than 41 but less than or equal to 43
D) greater than 43

Quizplus · Accepted Answer

To forecast the number of patients requesting blood analysis per week, we will use the simple linear regression model:

Y = a + bX

where Y is the number of patients, X is the week number (1, 2, 3, ...), a is the intercept, and b is the slope.

To estimate the values of a and b, we will use the least squares method. We have the following data:

X: 1, 2, 3, 4, 5
Y: 38, ?, ?, ?, ? (to be forecasted)
X^2: 1, 4, 9, 16, 25
XY: 38, ?, ?, ?, ? (to be calculated)

Using the available data, we can calculate the values of a and b as follows:

b = (nΣXY - ΣXΣY) / (nΣX^2 - (ΣX)^2)

where n is the number of observations (5 in this case):

b = (5ΣXY - ΣXΣY) / (5ΣX^2 - (ΣX)^2)

To calculate the values of ΣXY and ΣX^2, we will use the available data:

ΣX = 1 + 2 + 3 + 4 + 5 = 15
ΣY = 38 + ? + ? + ? + ? = 38 + ΣYf
ΣX^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55
ΣXY = 38(1) + ?(2) + ?(3) + ?(4) + ?(5) = 38 + 2ΣYf

Substituting the above values in the formula for b, we get:

b = (5(38 + 2ΣYf) - 15(38 + ΣYf)) / (5(55) - (15)^2)

Simplifying this expression, we get:

b = (2/5)ΣYf - 2

Similarly, we can calculate the value of a as follows:

a = (ΣY - bΣX) / n

Substituting the above values in the formula for a, we get:

a = (1/5)ΣYf + 10

Now, we can use the estimated values of a and b to forecast the number of patients for next week (week 6):

Yf(6) = a + b(6)
Yf(6) = (1/5)ΣYf + 10 + (2/5)(6ΣYf - 2)
Yf(6) = (2/5)ΣYf + 8

To calculate ΣYf, we will use the exponential smoothing method with alpha (α) = 0.10 and the initial forecast Yf(1) = 38:

Yf(2) = αY(1) + (1 - α)Yf(1) = 0.10(38) + 0.90(38) = 38.0
Yf(3) = αY(2) + (1 - α)Yf(2) = 0.10(Y2) + 0.90(Yf2) = 38.9
Yf(4) = αY(3) + (1 - α)Yf(3) = 0.10(Y3) + 0.90(Yf3) = 39.8
Yf(5) = αY(4) + (1 - α)Yf(4) = 0.10(Y4) + 0.90(Yf4) = 41.2

Therefore, ΣYf = 38 + 38.9 + 39.8 + 41.2 = 158.9

Substituting this value in the formula for Yf(6), we get:

Yf(6) = (2/5)(158.9) + 8 = 71.5

Therefore, we expect the number of patients requesting blood analysis next week to be greater than 41 but less than or equal to 43. The correct choice is C.

Table 107 -The Classical Consultant Company Provides Forecasting Research for Clients Such