Question

According to the Census Bureau, the number of people below the poverty level has been steadily increasing (CNN, September 16, 2010). This means many families are finding themselves there for the first time. The following table shows a portion of the percent of families in the United States who were below the poverty level from 1986-2009 Year 1986 1987 Poverty Rate 10.9 10.7 2009 11.1 Source: U.S. Census Bureau. Click here for the Excel Data File Plot the given series. Poverty Rate 105 (2009, 10.1) 1986 1988 1980 1929 1998 1999 2000 2002 2004200820092010 Year a. Comment on smoothing. The smoother appearance of the graph suggests the presence of random variations. The smoother appearance of the graph suggests the absence of random variations. The jagged appearance of the graph suggests the presence of random variations. The jagged appearance of the graph suggests the absence of random variations. b-1. Use the 3-period moving average to make in-sample forecasts. (Round intermediate calculations to at least 4 decimal places and final answers to 2 decimal places.) - t Yt 1986 10.9 1987 10.7 1988 10.4 1989 10.3 1990 10.7 2007 9.8 2008 10.3 2009 11.1 2010

Year	Poverty Rate
1986	10.9
1987	10.7
1988	10.4
1989	10.3
1990	10.7
1991	11.5
1992	11.9
1993	12.3
1994	11.6
1995	10.8
1996	11
1997	10.3
1998	10
1999	9.3
2000	8.7
2001	9.2
2002	9.6
2003	10
2004	10.2
2005	9.9
2006	9.8
2007	9.8
2008	10.3
2009	11.1

Answer 1

$\left ( a \right )\, \, The\, \, jagged\, \, appearance\, \, of\, \, the\, \, graph\, \, suggests\, \, the\, \, presence\, \, of$

(6) Forecast using 3 - period moving average

$Forecast\, \, for\, \, 1989=\frac{10.9+10.7+10.4}{3}=10.6667$

$Forecast\, \, for\, \, 1990=\frac{10.7+10.4+10.3}{3}=10.4667$

  

10.9 Absolute Squared Year Poverty Rate Forecast Forecast Error Forecast Error 1986 1987 10.7 1988 10.4 1989 10.3 10.6667 0.3667 0.1345 1990 10.71 10.4667 0.23331 0.0544 1991 11.5 10.4667 1.03331 1.0677 1992 11.9 10.8333 1.06671 1.1378 1993 12.3 11.3667 0.9333 0.871 1994 11.6 11.9 0.3 1995 10.8 11.9333 1.1333 1.2844 1996 11 11.5667 L 0.5667 0.3211 1997 10.3 11.1333 0.8333 0.6944 1998 10 10.7 0.71 0.49 9.3 10.4333 1.1333 1.2844 2000 8.7 9.8667 1.1667 1.3612 2001 9.2 9.3333 0.1333 0.0178 2002 9.6 9.0667 0.5333| 0.2844 2003 10 9.1667 0.8333 0.6944 0.09 1999

  

2004 2005 2006 2007 2008 2009 2010 10.2 9.9 9.8 9.8 10.3 11.1 9.6 9.9333 10.0333 9.9667 9.8333 9.9667 10.4 0.6 0.0333 0.2333 0.1667 0.4667 1.1333 0.36 0.0011 0.0544 0.0278 0.2178 1.2844 13.5998 Sum 1 1.733 Sum

  

9.8 + 10.3 + 11.1 Forecast for 2010 = = 10.4

$Mean\, \, Absolute\, \, Deviation\, \, MAD=\frac{\sum Absolute\, \, forecast\, \, error}{n}$

$=\frac{13.5998}{21}=0.6476$

$=\textbf{0.65}$

$Mean\, \, Squared\, \, Error=\frac{Sum\, \, of\, \, squared\, \, error}{n}$

  $=\frac{11.733}{21}=0.5587$

  $=\textbf{0.56}$

$\left ( C1 \right )\, \, Forecast\, \, using\, \, exponential\, \, smoothing\, \, \, method\, \, \alpha =0.6$

$F_{t+1}=\alpha Y_{t}+\left ( 1-\alpha \right )F_{t}$

$Forecast\, \, for\, \, 1987=F_{2}=\alpha Y_{1}+\left ( 1-\alpha \right )F_{1}$

  $=\left ( 0.6 \right )\left ( 10.9 \right )+\left ( 1-0.6 \right )\left ( 10.9 \right )=10.9$

$Forecast\, \, for\, \, 1988=F_{3}=\alpha Y_{2}+\left ( 1-\alpha \right )F_{2}=\left ( 0.6 \right )\left ( 10.7 \right )+\left ( 0.4 \right )\left ( 10.9 \right )$

  $=10.78$

0.2934 Absolute Squared Year Poverty Rate Forecast Forecast Error Forecast Error 1986 10.9 1987 10.7 10.9 0.21 0.04 1988 10.4 10.78 0.381 0.1444 1989 10.3 10.552 0.252 0.0635 1990 10.7 10.4008 0.2992 0.08951 1991 11.5 10.5803 0.9197 0.8458 1992 11.9 11.1321 0.7679 0.5897 1993 12.3 11.5928 0.7072 0.5001 1994 11.6 12.0171 0.4171 0.174 1995 10.8 11.7668 0.9668 0.9347 1996 11 11.1867 0.1867 0.0349 1997 10.3 11.0747 0 .7747 0.60021 1998 10 10.6099 0.6099 0.3721 9.3 10.244 0.944 0.8911 2000 8.7 9.6776 0.9776 0.9557 2001 9.2 9.091 0.109 0.0119 2002 9.6 9.1564 0.4436 0.1968 2003 109.4226 0.5774 0.3334 1999

10.21 2004 2005 2006 2007 2008 2009 2010 9.8 9.8 10.3 11.1 9.769 10.0276 9.951 9.8604 9.8242 10.1097 10.7039 0.431 0.1276 0.151 0.0604 0.4758 0.9903 0.1858 0.0163 0.0228 0.0036 0.2264 0.9807 11.76891 8.2133 Sum Sum

$Forecast\, \, for\, \, 2010=\left ( 0.6 \right )\left ( 11.1 \right )+\left ( 0.4 \right )\left ( 10.1097 \right )=10.7039$

$Mean\, \, Absolute\, \, Deviation\, \, MAD=\frac{\sum Absolute\, \, forecast\, \, error}{n}$

$=\frac{11.7689}{23}=0.5117$

$=0.51$

$Mean\, \, Squared\, \, Error=\frac{Sum\, \, of\, \, squared\, \, error}{n}$

$=\frac{8.2133}{23}=0.3571$

$=0.36$