**The following
data is based on the first class letter postage for the US Mail
from 1933 to 1966.**

YEAR |
RATE (IN CENTS) |

1919 | 2 |

1932 | 3 |

1958 | 4 |

1963 | 5 |

1968 | 6 |

1971 | 8 |

1974 | 10 |

1975 | 13 |

1978 | 15 |

1981 | 20 |

1985 | 22 |

1988 | 25 |

1991 | 29 |

1994 | 32 |

1997 | 33 |

1999 | 34 |

2002 | 37 |

**Lets look at the scatter plot
of this data:**

**Does this function represent
a linear model, log model, power model or an exponential model?
Even though when we look at the scatter plot it appears to be
exponential, but we can't just use our eyes to determine which
model the data fits best.**

**Since a power function and an
exponential function are pretty similar lets examine both models.**

**A power regression is in the
form y=ax^b, by transforming the equation into a linear form we
can look at the correlation coefficient. By looking at the correclation
coefficient we are able to decide which model best fits the data,
power or exponential.**

**Lets begin with our form for
power function**

**y=ax^b**

**Take the log of both sides**

**log y = log (ax^b)**

**Use log properties to perform
the following:**

**log y = log a + (b log x)**

**Let log y = Y and Let log a =
A and log x = X**

**therefore, Y = A + bX, so now
we have a linear form of the power function and we can use (log
x, log y) to graph our data. By using (log x, log y) we can look
at the correlation coefficient and determine if it is a good fit
for the graph.**

log x | log y |

3.28307497473547 | 0.301029995663981 |

3.28600712207947 | 0.477121254719662 |

3.29181268746712 | 0.602059991327962 |

3.29292029960001 | 0.698970004336019 |

3.29402509409532 | 0.778151250383644 |

3.29468662427944 | 0.903089986991943 |

3.29534714833362 | 1 |

3.29556709996248 | 1.11394335230684 |

3.29622628726116 | 1.17609125905568 |

3.29688447553855 | 1.30102999566398 |

3.29776051109913 | 1.34242268082221 |

3.29841638006129 | 1.39794000867204 |

3.29907126002741 | 1.46239799789896 |

3.29972515397564 | 1.50514997831991 |

3.3003780648707 | 1.51851393987789 |

3.30081279411812 | 1.53147891704226 |

3.3014640731433 | 1.56820172406699 |

**The graph below shows the power
function of the stamp data using (log x, log y). We have drawn
the trend line for the data and found the correlation coefficient
which is 0.96. The closer the correlation coeffiicient is to one
the better fit the data is.**

**Now lets look at the exponential
model and see if its correlation coefficient better fits the model.**

**Exponential Function**

**Lets begin with the form y =
ab^x**

**Take the log of both sides:**

**log y = log (ab ^x)**

**Use the properties of log to
obtain the following:**

**log y = log a + x log b**

**Let log y= Y and log a = A and
log b= B**

**then Y = A + Bx**

**therefore, Y = A + BX, so now
we have a linear form of the exponential function and we can use
( x, log y) to graph our data. By using ( x, log y) and we can
look at the correlation coefficient to determine if it is a good
fit for the graph.**

year | log y |

1919 | 0.301029995663981 |

1932 | 0.477121254719662 |

1958 | 0.602059991327962 |

1963 | 0.698970004336019 |

1968 | 0.778151250383644 |

1971 | 0.903089986991943 |

1974 | 1 |

1975 | 1.11394335230684 |

1978 | 1.17609125905568 |

1981 | 1.30102999566398 |

1985 | 1.34242268082221 |

1988 | 1.39794000867204 |

1991 | 1.46239799789896 |

1994 | 1.50514997831991 |

1997 | 1.51851393987789 |

1999 | 1.53147891704226 |

2002 | 1.56820172406699 |

**The graph below shows the exponential
function of the stamp data using (x, log y). We have drawn the
trend line for the data and found the correlation coefficient
which is 0.9598. The closer the correlation coeffiicient is to
one the better fit the data is.**

**Since the correlation coefficient
of the exponential model and the power model are very close either
model will work.**

**Now we need to find the equation
that best fits the power function and the exponential function
so we can use the equations to predict years and rates.**

**In order to find the equation
we need to find the slope and y-intercept for our data.**

**To find the slope: Correlation
Coefficient * (Standard Deviation of X divided by the Standard
Deviation of y).**

**To find the y-intercept: Mean
of y-Slope*Mean of x**

**The following spreadsheet shows
how the standard deviation of x,y, slope and the y-intercept were
calculated.**

year | rate(in cents) | X1-Mean | SQRED | Y1-Mean | SQRED |

1919 | 2 | -56 | 3136 | -15.52 | 240.8704 |

1932 | 3 | -43 | 1849 | -14.52 | 210.8304 |

1958 | 4 | -17 | 289 | -13.52 | 182.7904 |

1963 | 5 | -12 | 144 | -12.52 | 156.7504 |

1968 | 6 | -7 | 49 | -11.52 | 132.7104 |

1971 | 8 | -4 | 16 | -9.52 | 90.6304 |

1974 | 10 | -1 | 1 | -7.52 | 56.5504 |

1975 | 13 | 0 | 0 | -4.52 | 20.4304 |

1978 | 15 | 3 | 9 | -2.52 | 6.3504 |

1981 | 20 | 6 | 36 | 2.48 | 6.1504 |

1985 | 22 | 10 | 100 | 4.48 | 20.0704 |

1988 | 25 | 13 | 169 | 7.48 | 55.9504 |

1991 | 29 | 16 | 256 | 11.48 | 131.7904 |

1994 | 32 | 19 | 361 | 14.48 | 209.6704 |

1997 | 33 | 22 | 484 | 15.48 | 239.6304 |

1999 | 34 | 24 | 576 | 16.48 | 271.5904 |

2002 | 37 | 27 | 729 | 19.48 | 379.4704 |

Total Sum Squared | 8204 | 2412.2368 | |||

Divide by Sample | 512.75 | 150.7648 | |||

Sx | 22.6439837484485 | Sy | 12.2786318456089 | ||

Sx/Sy | 1.84417808377784 | ||||

SLOPE | |||||

POWER | 1.77041096042673 | ||||

EXPON. | 1.77004212480997 | ||||

Y-INTERCEPT | |||||

POWER | 1943.98239997332 | ||||

EXPON. | 1943.98886197333 |

**Now let's take a look at the
equation of the power function and the exponential function:**

__Power Function__

**Y = 1943.9823 + 1.77041096 X**

__Exponential Function:__

**Y = 1943.98886 + 1.77004212 X**

**Now we can use these equations
to predict which year the stamp will reach $1.00.**

**Since the functions are so close
it doesn't matter which one you choose.**

**When the cost of the stamp will
be 64 cents?**

**How soon should we expect the
next 3 cent increase?**

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Weaver's Home Page