Introduction

For most of the work you do in this book, you will use a histogram to display the data. One advantage of a histogram is that it can readily display large data sets.

A histogram consists of contiguous (adjoining) boxes. It has both a horizontal axis and a vertical axis. The horizontal axis is more or less a number line, labeled with what the data represents, for example, distance from your home to school. The vertical axis is labeled either frequency or relative frequency (or percent frequency or probability). The graph will have the same shape with either label. The histogram (like the stemplot) can give you the shape of the data, the center, and the spread of the data. The shape of the data refers to the shape of the distribution, whether normal, approximately normal, or skewed in some direction, whereas the center is thought of as the middle of a data set, and the spread indicates how far the values are dispersed about the center. In a skewed distribution, the mean is pulled toward the tail of the distribution.

The relative frequency is equal to the frequency for an observed value of the data divided by the total number of data values in the sample. Remember, frequency is defined as the number of times an answer occurs. If

• f = frequency,
• n = total number of data values (or the sum of the individual frequencies), and
• RF = relative frequency,

then

For example, if three students in Mr. Ahab’s English class of 40 students received from 90 to 100 percent, then f = 3, n = 40, and RF = $$ = $\frac{3}{40}$ = 0.075. Thus, 7.5 percent of the students received 90 to 100 percent. Ninety to 100 percent is a quantitative measures.

To construct a histogram, first decide how many bars or intervals, also called classes, represent the data. Many histograms consist of five to 15 bars or classes for clarity. The width of each bar is also referred to as the bin size, which may be calculated by dividing the range of the data values by the desired number of bins (or bars). There is not a set procedure for determining the number of bars or bar width/bin size; however, consistency is key when determining which data values to place inside each interval.

Example 2.9

The following data are the heights (in inches to the nearest half inch) of 100 male semiprofessional soccer players; the heights are continuous data since height is measured:

60, 60.5, 61, 61, 61.5,

63.5, 63.5, 63.5,

64, 64, 64, 64, 64, 64, 64, 64.5, 64.5, 64.5, 64.5, 64.5, 64.5, 64.5, 64.5,

66, 66, 66, 66, 66, 66, 66, 66, 66, 66, 66.5, 66.5, 66.5, 66.5, 66.5, 66.5, 66.5, 66.5, 66.5, 66.5, 66.5, 67, 67, 67, 67, 67, 67, 67, 67, 67, 67, 67, 67, 67.5, 67.5, 67.5, 67.5, 67.5, 67.5, 67.5,

68, 68, 69, 69, 69, 69, 69, 69, 69, 69, 69, 69, 69.5, 69.5, 69.5, 69.5, 69.5,

70, 70, 70, 70, 70, 70, 70.5, 70.5, 70.5, 71, 71, 71,

72, 72, 72, 72.5, 72.5, 73, 73.5,

74

 

The smallest data value is 60, and the largest data value is 74. To make sure each is included in an interval, we can use 59.95 as the smallest value and 74.05 as the largest value, subtracting and adding .05 to these values, respectively. We have a small range here of 14.1 (74.05 – 59.95), so we will want a fewer number of bins; let’s say eight. So, 14.1 divided by eight bins gives a bin size (or interval size) of approximately 1.76.

NOTE

We will round up to two and make each bar or class interval two units wide. Rounding up to two is a way to prevent a value from falling on a boundary. Rounding to the next number is often necessary even if it goes against the standard rules of rounding. For this example, using 1.76 as the width would also work. A guideline that is followed by some for the width of a bar or class interval is to take the square root of the number of data values and then round to the nearest whole number, if necessary. For example, if there are 150 values of data, take the square root of 150 and round to 12 bars or intervals.

The boundaries are as follows:

• 59.95
• 59.95 + 2 = 61.95
• 61.95 + 2 = 63.95
• 63.95 + 2 = 65.95
• 65.95 + 2 = 67.95
• 67.95 + 2 = 69.95
• 69.95 + 2 = 71.95
• 71.95 + 2 = 73.95
• 73.95 + 2 = 75.95

The heights 60 through 61.5 inches are in the interval 59.95–61.95. The heights that are 63.5 are in the interval 61.95–63.95. The heights that are 64 through 64.5 are in the interval 63.95–65.95. The heights 66 through 67.5 are in the interval 65.95–67.95. The heights 68 through 69.5 are in the interval 67.95–69.95. The heights 70 through 71 are in the interval 69.95–71.95. The heights 72 through 73.5 are in the interval 71.95–73.95. The height 74 is in the interval 73.95–75.95.

The following histogram displays the heights on the x-axis and relative frequency on the y-axis:

Figure 2.5

Example 2.10

The following data are the number of books bought by 50 part-time college students at ABC College; the number of books is discrete data since books are counted:

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,

2, 2, 2, 2, 2, 2, 2, 2, 2, 2,

3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,

4, 4, 4, 4, 4, 4,

5, 5, 5, 5, 5,

6, 6

Eleven students buy one book. Ten students buy two books. Sixteen students buy three books. Six students buy four books. Five students buy five books. Two students buy six books.

Calculate the width of each bar/bin size/interval size.

Solution 2.10

The smallest data value is 1, and the largest data value is 6. To make sure each is included in an interval, we can use 0.5 as the smallest value and 6.5 as the largest value by subtracting and adding 0.5 to these values. We have a small range here of 6 (6.5 – 0.5), so we will want a fewer number of bins; let’s say six this time. So, six divided by six bins gives a bin size (or interval size) of one.

Notice that we may choose different rational numbers to add to, or subtract from, our maximum and minimum values when calculating bin size. In the previous example, we added and subtracted .05, while this time, we added and subtracted .5. Given a data set, you will be able to determine what is appropriate and reasonable.

The following histogram displays the number of books on the x-axis and the frequency on the y-axis:

Figure 2.6

Example 2.11

Using this data set, construct a histogram.

Number of Hours My Classmates Spent Playing Video Games on Weekends
9.95	10	2.25	16.75	0
19.5	22.5	7.5	15	12.75
5.5	11	10	20.75	17.5
23	21.9	24	23.75	18
20	15	22.9	18.8	20.5

Table 2.16

Solution 2.11

Figure 2.7

Some values in this data set fall on boundaries for the class intervals. A value is counted in a class interval if it falls on the left boundary but not if it falls on the right boundary. Different researchers may set up histograms for the same data in different ways. There is more than one correct way to set up a histogram.

Frequency Polygons

Frequency polygons are analogous to line graphs, and just as line graphs make continuous data visually easy to interpret, so too do frequency polygons.

To construct a frequency polygon, first examine the data and decide on the number of intervals and resulting interval size, for both the x-axis and y-axis. The x-axis will show the lower and upper bound for each interval, containing the data values, whereas the y-axis will represent the frequencies of the values. Each data point represents the frequency for each interval. For example, if an interval has three data values in it, the frequency polygon will show a 3 at the upper endpoint of that interval. After choosing the appropriate intervals, begin plotting the data points. After all the points are plotted, draw line segments to connect them.

Example 2.12

A frequency polygon was constructed from the frequency table below.

Frequency Distribution for Calculus Final Test Scores
Lower Bound	Upper Bound	Frequency	Cumulative Frequency
49.5	59.5	5	5
59.5	69.5	10	15
69.5	79.5	30	45
79.5	89.5	40	85
89.5	99.5	15	100

Table 2.17

Figure 2.8

Notice that each point represents frequency for a particular interval. These points are located halfway between the lower bound and upper bound. In fact, the horizontal axis, or x-axis, shows only these midpoint values. For the interval 49.5−59.5 the value 54.5 is represented by a point, showing the correct frequency of 5. For the interval occurring before 49.5–59.5, (or 39.5–49.5), the value of the midpoint, or 44.5, is represented by a point, showing a frequency of 0, since we do not have any values in that range. The same idea applies to the last interval of 99.5–109.5, which has a midpoint of 104.5 and correctly shows a point representing a frequency of 0. Looking at the graph, we say that this distribution is skewed because one side of the graph does not mirror the other side.

Frequency polygons are useful for comparing distributions. This comparison is achieved by overlaying the frequency polygons drawn for different data sets.

Example 2.13

We will construct an overlay frequency polygon comparing the scores from Example 2.12 with the students’ final numeric grades.

Frequency Distribution for Calculus Final Test Scores
Lower Bound	Upper Bound	Frequency	Cumulative Frequency
49.5	59.5	5	5
59.5	69.5	10	15
69.5	79.5	30	45
79.5	89.5	40	85
89.5	99.5	15	100

Table 2.19

Frequency Distribution for Calculus Final Grades
Lower Bound	Upper Bound	Frequency	Cumulative Frequency
49.5	59.5	10	10
59.5	69.5	10	20
69.5	79.5	30	50
79.5	89.5	45	95
89.5	99.5	5	100

Table 2.20

Figure 2.9

Suppose that we want to study the temperature range of a region for an entire month. Every day at noon, we note the temperature and write this down in a log. A variety of statistical studies could be done with the data. We could find the mean or the median temperature for the month. We could construct a histogram displaying the number of days that temperatures reach a certain range of values. However, all of these methods ignore a portion of the data that we have collected.

One feature of the data that we may want to consider is that of time. Since each date is paired with the temperature reading for the day, we don't have to think of the data as being random. We can instead use the times given to impose a chronological order on the data. A graph that recognizes this ordering and displays the changing temperature as the month progresses is called a time series graph.

Constructing a Time Series Graph

To construct a time series graph, we must look at both pieces of our paired data set. We start with a standard Cartesian coordinate system. The horizontal axis is used to plot the date or time increments, and the vertical axis is used to plot the values of the variable that we are measuring. By using the axes in that way, we make each point on the graph correspond to a date and a measured quantity. The points on the graph are typically connected by straight lines in the order in which they occur.

Example 2.14

The following data show the Annual Consumer Price Index each month for 10 years. Construct a time series graph for the Annual Consumer Price Index data only:

Year	Jan	Feb	Mar	Apr	May	Jun	Jul
2003	181.7	183.1	184.2	183.8	183.5	183.7	183.9
2004	185.2	186.2	187.4	188.0	189.1	189.7	189.4
2005	190.7	191.8	193.3	194.6	194.4	194.5	195.4
2006	198.3	198.7	199.8	201.5	202.5	202.9	203.5
2007	202.416	203.499	205.352	206.686	207.949	208.352	208.299
2008	211.080	211.693	213.528	214.823	216.632	218.815	219.964
2009	211.143	212.193	212.709	213.240	213.856	215.693	215.351
2010	216.687	216.741	217.631	218.009	218.178	217.965	218.011
2011	220.223	221.309	223.467	224.906	225.964	225.722	225.922
2012	226.665	227.663	229.392	230.085	229.815	229.478	229.104

Table 2.21

Year	Aug	Sep	Oct	Nov	Dec	Annual
2003	184.6	185.2	185.0	184.5	184.3	184.0
2004	189.5	189.9	190.9	191.0	190.3	188.9
2005	196.4	198.8	199.2	197.6	196.8	195.3
2006	203.9	202.9	201.8	201.5	201.8	201.6
2007	207.917	208.490	208.936	210.177	210.036	207.342
2008	219.086	218.783	216.573	212.425	210.228	215.303
2009	215.834	215.969	216.177	216.330	215.949	214.537
2010	218.312	218.439	218.711	218.803	219.179	218.056
2011	226.545	226.889	226.421	226.230	225.672	224.939
2012	230.379	231.407	231.317	230.221	229.601	229.594

Table 2.22

Solution 2.14

Figure 2.10 The annual amounts are plotted for each year. Then, consecutive points are connected with a line.