Questions: Freight car loadings over an 18-week period at a busy port are as follows: Week Number 1 300 2 305 3 310 4 305 5 310 6 315 7 365 8 395 9 425 10 420 11 450 12 480 13 485 14 490 15 540 16 545 17 550 18 555 a. Determine a linear trend line for expected freight car loadings. Note: Round your intermediate calculations and final answers to 2 decimal places. b. Use the above trend equation to predict expected loadings for Weeks 20 21. Note: Round your final answers to 2 decimal places. The forecasted demand for Week 20 is and for Week 21 is c. The manager intends to install new equipment when the volume exceeds 880 loadings per week. Assuming the current trend continues, in which week (at the earliest) should the loading volume reach that level? Note: Use the rounded answers, as required, from any previous part of this problem. Do not round any other intermediate calculations. Round your final answer to 2 decimal places. It should reach 880 loadings in Week

Solution Steps

The means of the independent variable \( x \) (weeks) and the dependent variable \( y \) (freight car loadings) are calculated as follows:

\[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i = 9.5 \]

\[ \bar{y} = \frac{1}{n} \sum_{i=1}^{n} y_i = 419.17 \]

The correlation coefficient \( r \) is determined to be:

\[ r = 0.98 \]

The slope \( \beta \) is calculated using the following formulas:

Numerator for \( \beta \):

\[ \sum_{i=1}^{n} x_i y_i - n \bar{x} \bar{y} = 80275 - 18 \cdot 9.5 \cdot 419.17 = 8597.5 \]

Denominator for \( \beta \):

\[ \sum_{i=1}^{n} x_i^2 - n \bar{x}^2 = 2109 - 18 \cdot 9.5^2 = 484.5 \]

Thus, the slope \( \beta \) is:

\[ \beta = \frac{8597.5}{484.5} = 17.75 \]

The intercept \( \alpha \) is calculated as follows:

\[ \alpha = \bar{y} - \beta \bar{x} = 419.17 - 17.75 \cdot 9.5 = 250.59 \]

The linear trend line equation is given by:

\[ \hat{y} = 250.59 + 17.75t \]

Using the trend line equation, the forecasted demand for Week 20 and Week 21 is calculated as follows:

For Week 20:

\[ \hat{y}_{20} = 250.59 + 17.75 \cdot 20 = 605.59 \]

For Week 21:

\[ \hat{y}_{21} = 250.59 + 17.75 \cdot 21 = 623.34 \]

To find the earliest week when loadings exceed 880, we solve the equation:

\[ 880 = 250.59 + 17.75t \]

Rearranging gives:

\[ t = \frac{880 - 250.59}{17.75} \approx 35.46 \]

Final Answer