Teacher Guide and Answer Key, Algebra 1 Activity

Based on AP* Statistics Problem 6, 1997

Algebra I TEKS addressed: (a)(3); (a)(4); (a)(5); (a)(6); (A.1)(B); (A.1)(C); (A.1)(D); (A.1)(E); (A.2)(B); (A.2)(C); (A.2)(D); (A.3)(A); (A.5)(A); (A.5)(C); (A.6)(B); (A.6)(D); (A.7)(B); (A.7)(C)

This problem is an example of a good Pre-AP* problem because it requires students to incorporate many concepts into one problem. Students are asked to interpret equations (linear) and graphs and explain the reasons for their interpretations. The problem begins as a typical Algebra 1 scatter plot problem (in questions 1-4). Questions 5-10 extend the problem to concepts the students will encounter in an AP Statistics class. Students are asked to evaluate the linear model they found for the given data.

1. a. & b. See the table of values below for problem 2

2.

3.

4.
a. We will use y = .5x + 1. (Answers will vary because the students are not using a graphing calculator to find the equations. The students will use their knowledge of linear functions and the graphs of linear functions to write a linear function that models this data. They will use their model to help them answer the remaining questions.)

What does the slope represent?

• the predicted change in the asking price (in thousands of $) divided by the change in the years (after 1979); or
• a predicted change of $500 per year; or
• for every year after 1979, an expected increase in asking price of 0.5 thousands of dollars.

b. The asking price-intercept is 1. This represents the asking price ($1000) for the year 0 (1979).

5. Using the model written in #4, the asking price might be $5500. When I plotted the price (9, 5.5) on my scatter plot, the price appears too high because year 8 is $3400 and year 11 is $6000. I think the price should be closer to year 8 than year 11.

6. For this example we will let the year 2002 represent the present year so that using the model written in #4 the asking price might be $12,500. When I extended my graph to include this data, this price appears to be too low. If I use the rate of change that I found in #4a, then from year 14 to year 23 the price should have increased about $4500. Therefore, the predicted price in 2002 would be $8900 + $4500 = $13,400.

7. Answers to this question will vary. Teachers should discuss that question #5 is an example of interpolation (predicting within the range of the given data set) and question #6 is an example of extrapolation (predicting outside of the range of the given data set). Interpolation is more reliable than extrapolation.

8.

9.

10. The points (year, difference) would lie on the line y = 0 (difference = 0). This would indicate that this model was a good model since the difference between the actual and predicted was 0, meaning that the predicted price was the same as the actual price. On my graph the early year prices are negative, indicating that the predicted price is more than the actual price and the later year prices are positive, indicating that the predicted price is less than the actual price. This graph would help support my answers to problems #5 and #6.