1. Making and Using Graphs

2. Types of graphs
Scatter diagram: A graph of the value of one variable against the value of another variable
Time-series graph: A graph that measures time on the x-axis and the variable or variables of interest on the y-axis.
Cross-section graph: A graph that shows the values of a variable for different groups in the population at a point in time.

3.
1991
1999
Scatter diagram
1959

4. Source: U.S. News and World Report
115 U.S. universities
Source: U.S. News and World Report

5. Recessions are shaded

8. Source: Authors’ calculations using the March Demographic Files of the Current Population Survey

9. Cross-section graph Source: The Economist
Unemployment rates in industrialized countries, May 2000
Cross-section graph
Source: The Economist

12. Relationships among variables
Variables X and Y have a positive or direct relationship if, ceteris paribus, they move in the same direction.
Variables X and Y have a negative or inverse relationship if, ceteris paribus, they move in opposite directions.
The relationship between X and Y is linear if it can be described by a straight line.

13. Direct, linear Distance traveled in 5 hours (miles) 300 200 40 6040 60
Speed (MPH)

14. Direct, nonlinear Recovery time (MINUTES) 30 20 100 200 400
100
200
400
Distance sprinted (Yards)

16. Inverse, linear
Time partying (Hours) 5 5
Time studying (Hours)

21. Advertising ($1,000’s per Mo.)
Linear graphs
Advertising ($1,000’s per Mo.)
Sales ($1,000’s per Mo.) 40 2 46 3 49 6 58 7 61 11 73 12 76
The Jewelry Mart

22. We want to describe the (causal) relationship between sales and advertising using a simple linear equation—because, as it turns out, the relationship is linear
Let Y denote sales per month (in $1,000’s)
Let X denote advertising per month (in $1,000’s)
Thus we have:
Y = f(X)
Where Y is the dependent variable and X is the independent (explanatory variable)

23. Relationship is linear76 73 F E
Sales ($1,000’s per mo.) 49 46 B 40 A 2 3 11 12
Advertising ($1,000’s per mo.)

24. Where a is the intercept and b is the slope coefficient.
Now we want to describe the relationship by a linear equation like this:
Y = a + bX
Where a is the intercept and b is the slope coefficient.

25. The intercept (a) Y
a > 0
a = 0 X
a < 0

26. Computing the slope of a line

27. The slope (b)
b > 0 Y a
b = 0
b < 0 X

28. Now let’s compute the slope of this line76 73 F E
Sales ($1,000’s per mo.) 49 46 B 40 A
Notice that a = $40 2 3 11 12
Advertising ($1,000’s per mo.)

29. Moving from point A to B (or from point E to F), the vertical change is $3 and the horizontal change is $1. Thus our slope is equal to 3
Interpretation: A $1,000 increase in monthly advertising expenditures will result in a $3,000 increase in monthly jewelry sales(and vice-versa), other things being equal.

30. Thus we have:
Y = X
Suppose that the management of the Jewelry Mart have set a monthly sales target of $64,000. How much advertising is necessary to meet the target?
Y = X Hence: 64 = X  24 = 3X  X = 8
The Jewelry Mart needs to spend $8,000 per month on advertising to achieve the sales target.