Assignment #12

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.

Power Regression.

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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