According to the U.S. Department of Education, the following are the numbers, in millions, of college degrees awarded in various years since 1970.

Question

I have worked this problem over and over and I can't seem to get where I need to go. It may be operator error in Excel or something but I seriously could use some help here:

Year-College graduates: 1970-1.271, 1980-1.731, 1985-1.828, 1990-1.940, 1995-2.218, 1998-2.298, 2000-2.385, 2001-2.416, 2002-2.494, 2003-2.621.

1. Let t be the number of years since 1970. Determine the best linear function and an exponential function to model the number of graduates. Label them, L(t) for linear function and E(t) for exponential function. Show data table and graph of function with equation and coefficient of determination displayed. (Remember to display scientific notation correctly.)
2. Use each function to predict the number of college graduates in 2010. Research the actual number of college graduates in 2010 and calculate the percent difference for each function.
3. Use each function to predict the year when there will be 3 million college graduates.
4. Which function seems to provide a more accurate prediction of graduates in 2010 as well as the year when there will be 3 million college graduates? e. What is the doubling time for the exponential model?

Answer 1

^1.330843062t = years since 1970

I used 1970=0, 1980=10, etc.

 

The linear part is straightforward.  Just do linear regression on the data. You may have a different procedure to do the regression.  The trick required to get the exponential model is to do linear regression on the natural log of the the y-values (number of graduates), then convert that result into exponential form. (see below)

 

LINEAR

I used the SLOPE() and INTERCEPT() functions to get m and b:

 

 

m= 0.037735

b = 1.274937

 

y = 0.03775x + 1.274937

 

2010 prediction: 2.784

 

EXPONENTIAL

Replace the y-values with ln(y).

Get m and b as above.

 

I get this:

m = 0.019883

b = 1.274937

 

ln(y) = 0.019883x + 1.274937

 

You have to convert this to an exponential expression.

e^ln(y) = e ^{0.019883x + 1.274937} = e ^1.274937 * e ^0.019883x

y = 1.330843e ^0.019883x

 

2010 prediction: 2.948

 

I'll let you do the research for the actual 2010 values and calculate the % differences.

 

When will it reach 3 million?

Solve each equation for x when f(x) = 3

 

LINEAR

0.03775x + 1.274937 = 3

X = 45.7

round up to 46, 1970+46 = 2016

 

EXPONENTIAL

1.330843e^0.019883x = 3

ln(1.330843)+0.019883x = ln(3)  -- This is the form before we converted it to exponential!

x = 40.88

round up to 41, 1970+41 = 2011

 

Time to double.

Solve e^0.019883x = 2 for x

0.019883x = ln(2)

x= ln(2) / 0.019883x

x = 34.86 ∼35 years

 

Hope this helps!