A supermarket chain (FRESHM) funds the purchase of merchandise monthly by bank loans. Monthly loan data were collected over the last 25 months. Answer the
A supermarket chain (FRESHM) funds the purchase of merchandise monthly by bank loans. Monthly loan data were collected over the last 25 months. Answer the following questions:
a. Use the linear regression approach
1.A test for the presence of linear trend was run. From the computer output answer. Which statement is correct?
a.R sq = 0.78. This is clearly more than 0.05. There is a linear trend
b.P value for the intercept = 7.81E-18. This is clearly small, so the intercept is not zero. There is linear trend.
c.P value for 'Loans' = 4.42E-09 < 0.05. Beta1 is not zero, so there is a linear trend.
d.None of the above
2.For the regression equation you have created in question 1, what does the estimated slope of this equation tell you in this particular problem? Answer in the Answers sheet.
3.Use the regression equation of part '2' (not the template results) to make a prediction of expected loans for each of the next three months (months 26, 27, and 28). Answer in the Answers sheet. Show your calculations.
4.FRESHM wants to be 95% confident that it does not borrow less than the amount of funds actually needed, in any of the next three months (26, 27, 28). What loan is the minimum amount FRESHM have to borrow monthly, to meet the confidence requirement in each of the next three months? (Use the template!).
a.Already calculated above in a-3.
b.1424.83 in each month
c.1401.82, 1424.83, and 1447.70, in the three months respectively
d.None of the above
5.
b.Use the Holt's approach.
1.Determine the best value of alpha and gamma, for the MAE criterion. Show your work on a spreadsheet.
2.For the model you obtained in question '5' using the optimal alpha and gamma, you should get L25 =1766.89, and T25 =66.88.Which statement is correct?
a.The local slope is L25/T25 = 1766.89/66.88 = 26.42
b.The intercept (namely, the time series at t=0) is estimated to be 1766.89 as of t=25
c.The average slope at t=25 is 66.88/25.
d.At t=25 the monthly loan is estimated to rise by 66880 dollars per month.
3.Use L25 =1766.89 andT25 = 66.88 from the template of the model you obtained in question '5' to calculate the forecast of loans for months 26, 27, and 28.
Show your calculations.
As a matter of policy, management will approve borrowing up to a total of $1,900,000 for next month (month 26). How confident can management be that this loan amount will be sufficient to purchase the merchandise required that month? You can use trial and error, or goal seek with the template.
Month Loan in $1000
1 1156
2 1101
3 1195
4 1072
5 1237
6 1106
7 1148
8 1005
9 1350
10 1115
11 1196
12 1298
13 1260
14 1264
15 1235
16 1357
17 1373
18 1358
19 1489
20 1428
21 1516
22 1681
23 1598
24 1699
25 1781
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