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

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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