Solution 1
(a) Plot the data.
Using the data provided, the data can be plotted as following:
(b) Fit a three-year moving average to the data and plot the results.
Let R represents the three-month Treasury bill rates, and
MA represents the three-year moving average,
year rate(R) three-year moving average(MA)
1991 5.38 -
1992 3.43 -
1993 3 3.9367
1994 4.25 3.56
1995 5.49 4.2467
1996 5.01 4.9167
1997 5.06 5.1867
1998 4.78 4.95
1999 4.64 4.8267
2000 5.82 5.08
2001 3.4 4.62
2002 1.61 3.61
2003 1.01 2.0067
2004 2.17 1.5967
2005 3.89 2.3567
(c) Using a smoothing coefficient of W = 0.40, exponentially smooth the series and plot the results.
Let R represents the three-month Treasury bill rates, and
EM represents the exponentially smooth result,
year rate exponentially mooth (w=0.40)
1991 5.38
1992 3.43 5.38
1993 3 4.6
1994 4.25 3.96
1995 5.49 4.076
1996 5.01 4.6416
1997 5.06 4.78896
1998 4.78 4.897376
1999 4.64 4.8504256
2000 5.82 4.76625536
2001 3.4 5.187753216
2002 1.61 4.47265193
2003 1.01 3.327591158
2004 2.17 2.400554695
2005 3.89 2.308332817
(d) Repeat (c), using W = 0.15.
Similarly as the (c),
Let R represents the three-month Treasury bill rates, and
EM represents the exponentially smooth result,
year rate exponentially smooth(w=0.15)
1991 5.38
1992 3.43 5.38
1993 3 5.0875
1994 4.25 4.774375
1995 5.49 4.695719
1996 5.01 4.814861
1997 5.06 4.844132
1998 4.78 4.876512
1999 4.64 4.862035
2000 5.82 4.82873
2001 3.4 4.97742
2002 1.61 4.740807
2003 1.01 4.271186
2004 2.17 3.782008
2005 3.89 3.540207
(e) Using results in (c) and (d), what is your exponentially smoothed forecast for 2006? Check the actual value for 2006.
According the results in (c)，in the 2006, the three-month Treasury bill rates will be:
rc =0.4*3.89 +(1-0.4)* 2.308332817=2.94100
According the results in (d)，in the 2006, the three-month Treasury bill rates will be:
rd =0.15*3.89 +(1-0.15)* 3.540207=3.59268
The actual value for 2006 is 4.73
Available at: http://www.federalreserve.gov/releases/h15/data/Annual/H15_TB_M3.txt
(f) The smoothing coefficient called W indicates that the current forecast should be adjusted to correct the previous forecast. The greater value of (1-w) is, the faster change it responses. Thus the prediction will become unstable. However, a smaller value of (1-w) will cause a delay in prediction.
In the case (d), the value of W is much less than that in the case (c) so 1-W in the case (d) is more than that in the case (c). In addition, as seen from the given data, the three-month Treasury bill rates are volatile, in which greater adjustment alteration to correct the previous forecast will be needed. As a result, the case (d) is more appropriate than the case (c).
Solution2
(a) Plot the data.
Using the data provided, the data can be plotted as following:
(b) Compute a linear trend forecasting equation and plot the trend line.
According to the method of least squares,
y = 10928x - 2E+07, R2 = 0.5715,
(c) Compute a quadratic trend forecasting equation and plot the results.
Using the same method above, a quadratic trend forecasting equation can be drown as following:
y = -1056.5x2 + 4E+06x - 4E+09
(d) Compute an exponential trend forecasting equation and plot the results.
Using the same method above, a quadratic trend forecasting equation can be drown as following:
y = 3E-16e0.0246x, R2 = 0.5043
(e) Which model is the most appropriate?
Theoretically, linear trend forecast, quadratic trend forecast and exponential trend forecast have their application respectively. As shown in the (b), (c), (d), the three models appropriate for the serial to some extent respectively, and the exponential trend forecasting is the most appropriate in this case.
(f) Using the most appropriate model, forecast the number of barrels, in millions, for 2005.
Solution 3
(a) By using 2000 as the Base period, Let
P0 represents the price per kg in 2000, and
Pt represents the price per kg in 2008, and
Q0 represents the quantity in 2000, and
Qt represents the quantity in 2008, then
Laspeyres price index can be calculated as following:
Paasche price index can be calculated as following:
Fisher price index can be calculated as following:
(b) 1. Laspeyres price index chooses the base period (2000) quantity , to measure the price change degree on the condition of invariant quantity . That means if there is no quantity change, one can buy shellfish at the current wholesale price which is about 1.078702 times of the price in the base period .
2. Paassche price index chooses the current period (2008) quantity , to measure the price changing degree on the occasion of invariant current quantity . If no change is on the current quantity, the shellfish price will be 1.071991 times of the base wholesale price.
3. Fisher price index is known as “ideal” price index. Fisher price index is composed as the geometric mean of PP and PL. when measuring the price change degree, no consideration on the quantity variance, the current period price is about 1.075341 times of the base period price.
Solution 4
(a) According Poisson distribution probability formula,
,
and λ=4
The probability of exactly four people arrive at the information booth can be calculated as,
Note: In the Microsoft Excel 2003, “POISSON (4, 4, false)” can be used to get the result.
(b) Similarly as solution (a),
The probability that more than one person will arrive can be calculated as,
1-[P(x=0) + P(x=1)] =1-0.018316-0.073263=0.908422
Note: In the Microsoft Excel 2003, “POISSON (1, 4, true)” can be used to get the probability P(x>1) by using to P(x=0) and P(x=1) calculated.
(c) Similarly as solution (a) and (b), the probabilities can be calculated as following:
P(x=0) = 0.018315639
P(x=1) = 0.073262556
P(x=2) = 0.146525111
P(x=3) = 0.195366815
P(x=4) = 0.195366815
P(x=5) = 0.156293452
P(x=6) = 0.104195635
Six people arrive over a 10-minute period can be formed as follows:
abbreviation first 5-minute second 5-minute probability result
(0,6) 0 6 p(x=0)* p(x=6) 0.00190841
(1,5) 1 5 p(x=1)* p(x=5) 0.011450458
(2,4) 2 4 p(x=2)* p(x=4) 0.028626144
(3,3) 3 3 p(x=3)* p(x=3) 0.038168192
(4,2) 4 2 p(x=4)* p(x=2) 0.028626144
(5,1) 5 1 p(x=5)* p(x=1) 0.011450458
(6,0) 6 0 p(x=6)* p(x=6) 0.00190841
Then, the probability six people arrive over a 10-minute period can be calculated as,
P=0.00190841+0.011450458+0.028626144+0.038168192+0.028626144+0.011450458+0.00190841=0.122138215
Solution 5
The number of problems per new car is distributed as a Poisson random variable, and the Lexus had 0.78 problems per car, as a result the probability that a 2004 Lexus will be calculated in the expression as the following formula with setting k as the random variable:
(a) A 2004 Lexus will have zero problems,
Note: In the Microsoft Excel 2003, “POISSON (0, 0.78, false)” can be used to get the result.
(b) Because the number of a 2004 Lexus which contained one or more problems is just the opposite of those which have zero problems, so the probability can be calculated:
1- P(x=0) =1-0.458406= 0.541594
(c)
The probability of a 2004 Lexus contained two or less problems can be calculated:
1-P(x=0) -P(x=1) =1-0.458406-0.357557=0.184037
