Session:2 Descriptive Statistics
2.1 Display Data
Introductory Business Statistics | Leadership Development – Micro-Learning Session
Rice University 2020 | Michael Laverty, Colorado State University Global Chris Littel, North Carolina State University| https://openstax.org/details/books/introductory-business-statistics
Stem-and-Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs
One simple graph, the stem-and-leaf graph or stemplot, comes from the field of exploratory data analysis. It is a good choice when the data sets are small. To create the plot, divide each observation of data into a stem and a leaf. The leaf consists of a final significant digit. For example, 23 has stem two and leaf three. The number 432 has stem 43 and leaf two. Likewise, the number 5,432 has stem 543 and leaf two. The decimal 9.3 has stem nine and leaf three. Write the stems in a vertical line from smallest to largest. Draw a vertical line to the right of the stems. Then write the leaves in increasing order next to their corresponding stem.
EXAMPLE 2.1
For Susan Dean’s spring pre-calculus class, scores for the first exam were as follows (smallest to largest):
33; 42; 49; 49; 53; 55; 55; 61; 63; 67; 68; 68; 69; 69; 72; 73; 74; 78; 80; 83; 88; 88; 88; 90; 92; 94; 94; 94; 94; 96; 100
Stem | Leaf |
---|---|
3 | 3 |
4 | 2 9 9 |
5 | 3 5 5 |
6 | 1 3 7 8 8 9 9 |
7 | 2 3 4 8 |
8 | 0 3 8 8 8 |
9 | 0 2 4 4 4 4 6 |
10 | 0 |
The stemplot shows that most scores fell in the 60s, 70s, 80s, and 90s. Eight out of the 31 scores or approximately 26% (831)(831) were in the 90s or 100, a fairly high number of As.
TRY IT 2.1
For the Park City basketball team, scores for the last 30 games were as follows (smallest to largest):
32; 32; 33; 34; 38; 40; 42; 42; 43; 44; 46; 47; 47; 48; 48; 48; 49; 50; 50; 51; 52; 52; 52; 53; 54; 56; 57; 57; 60; 61
Construct a stem plot for the data.
The stemplot is a quick way to graph data and gives an exact picture of the data. You want to look for an overall pattern and any outliers. An outlier is an observation of data that does not fit the rest of the data. It is sometimes called an extreme value. When you graph an outlier, it will appear not to fit the pattern of the graph. Some outliers are due to mistakes (for example, writing down 50 instead of 500) while others may indicate that something unusual is happening. It takes some background information to explain outliers, so we will cover them in more detail later.
EXAMPLE 2.2
The data are the distances (in kilometers) from a home to local supermarkets. Create a stemplot using the data:
1.1; 1.5; 2.3; 2.5; 2.7; 3.2; 3.3; 3.3; 3.5; 3.8; 4.0; 4.2; 4.5; 4.5; 4.7; 4.8; 5.5; 5.6; 6.5; 6.7; 12.3
Problem
Do the data seem to have any concentration of values?
NOTE
The leaves are to the right of the decimal.
Solution
The value 12.3 may be an outlier. Values appear to concentrate at three and four kilometers.
Stem | Leaf |
---|---|
1 | 1 5 |
2 | 3 5 7 |
3 | 2 3 3 5 8 |
4 | 0 2 5 5 7 8 |
5 | 5 6 |
6 | 5 7 |
7 | |
8 | |
9 | |
10 | |
11 | |
12 | 3 |
TRY IT 2.2
The following data show the distances (in miles) from the homes of off-campus statistics students to the college. Create a stem plot using the data and identify any outliers:
0.5; 0.7; 1.1; 1.2; 1.2; 1.3; 1.3; 1.5; 1.5; 1.7; 1.7; 1.8; 1.9; 2.0; 2.2; 2.5; 2.6; 2.8; 2.8; 2.8; 3.5; 3.8; 4.4; 4.8; 4.9; 5.2; 5.5; 5.7; 5.8; 8.0
EXAMPLE 2.3
Problem
A side-by-side stem-and-leaf plot allows a comparison of the two data sets in two columns. In a side-by-side stem-and-leaf plot, two sets of leaves share the same stem. The leaves are to the left and the right of the stems. Table 2.4 and Table 2.5 show the ages of presidents at their inauguration and at their death. Construct a side-by-side stem-and-leaf plot using this data.
Solution
Ages at Inauguration | Ages at Death | |
---|---|---|
9 9 8 7 7 7 6 3 2 | 4 | 6 9 |
8 7 7 7 7 6 6 6 5 5 5 5 4 4 4 4 4 2 2 1 1 1 1 1 0 | 5 | 3 6 6 7 7 8 |
9 8 5 4 4 2 1 1 1 0 | 6 | 0 0 3 3 4 4 5 6 7 7 7 8 |
7 | 0 1 1 1 2 3 4 7 8 8 9 | |
8 | 0 1 3 5 8 | |
9 | 0 0 3 3 |
President | Age | President | Age | President | Age |
---|---|---|---|---|---|
Washington | 57 | Lincoln | 52 | Hoover | 54 |
J. Adams | 61 | A. Johnson | 56 | F. Roosevelt | 51 |
Jefferson | 57 | Grant | 46 | Truman | 60 |
Madison | 57 | Hayes | 54 | Eisenhower | 62 |
Monroe | 58 | Garfield | 49 | Kennedy | 43 |
J. Q. Adams | 57 | Arthur | 51 | L. Johnson | 55 |
Jackson | 61 | Cleveland | 47 | Nixon | 56 |
Van Buren | 54 | B. Harrison | 55 | Ford | 61 |
W. H. Harrison | 68 | Cleveland | 55 | Carter | 52 |
Tyler | 51 | McKinley | 54 | Reagan | 69 |
Polk | 49 | T. Roosevelt | 42 | G.H.W. Bush | 64 |
Taylor | 64 | Taft | 51 | Clinton | 47 |
Fillmore | 50 | Wilson | 56 | G. W. Bush | 54 |
Pierce | 48 | Harding | 55 | Obama | 47 |
Buchanan | 65 | Coolidge | 51 |
President | Age | President | Age | President | Age |
---|---|---|---|---|---|
Washington | 67 | Lincoln | 56 | Hoover | 90 |
J. Adams | 90 | A. Johnson | 66 | F. Roosevelt | 63 |
Jefferson | 83 | Grant | 63 | Truman | 88 |
Madison | 85 | Hayes | 70 | Eisenhower | 78 |
Monroe | 73 | Garfield | 49 | Kennedy | 46 |
J. Q. Adams | 80 | Arthur | 56 | L. Johnson | 64 |
Jackson | 78 | Cleveland | 71 | Nixon | 81 |
Van Buren | 79 | B. Harrison | 67 | Ford | 93 |
W. H. Harrison | 68 | Cleveland | 71 | Reagan | 93 |
Tyler | 71 | McKinley | 58 | ||
Polk | 53 | T. Roosevelt | 60 | ||
Taylor | 65 | Taft | 72 | ||
Fillmore | 74 | Wilson | 67 | ||
Pierce | 64 | Harding | 57 | ||
Buchanan | 77 | Coolidge | 60 |
Another type of graph that is useful for specific data values is a line graph. In the particular line graph shown in Example 2.4, the x-axis (horizontal axis) consists of data values and the y-axis (vertical axis) consists of frequency points. The frequency points are connected using line segments.
EXAMPLE 2.4
In a survey, 40 mothers were asked how many times per week a teenager must be reminded to do his or her chores. The results are shown in Table 2.6 and in Figure 2.2.
Number of times teenager is reminded | Frequency |
---|---|
0 | 2 |
1 | 5 |
2 | 8 |
3 | 14 |
4 | 7 |
5 | 4 |
TRY IT 2.4
In a survey, 40 people were asked how many times per year they had their car in the shop for repairs. The results are shown in Table 2.7. Construct a line graph.
Number of times in shop | Frequency |
---|---|
0 | 7 |
1 | 10 |
2 | 14 |
3 | 9 |
Bar graphs consist of bars that are separated from each other. The bars can be rectangles or they can be rectangular boxes (used in three-dimensional plots), and they can be vertical or horizontal. The bar graph shown in Example 2.5 has age groups represented on the x-axis and proportions on the y-axis.
EXAMPLE 2.5
Problem
By the end of 2011, Facebook had over 146 million users in the United States. Table 2.8 shows three age groups, the number of users in each age group, and the proportion (%) of users in each age group. Construct a bar graph using this data.
Age groups | Number of Facebook users | Proportion (%) of Facebook users |
---|---|---|
13–25 | 65,082,280 | 45% |
26–44 | 53,300,200 | 36% |
45–64 | 27,885,100 | 19% |
Solution
TRY IT 2.5
The population in Park City is made up of children, working-age adults, and retirees. Table 2.9 shows the three age groups, the number of people in the town from each age group, and the proportion (%) of people in each age group. Construct a bar graph showing the proportions.
Age groups | Number of people | Proportion of population |
---|---|---|
Children | 67,059 | 19% |
Working-age adults | 152,198 | 43% |
Retirees | 131,662 | 38% |
EXAMPLE 2.6
Problem
The columns in Table 2.10 contain: the race or ethnicity of students in U.S. Public Schools for the class of 2011, percentages for the Advanced Placement examine population for that class, and percentages for the overall student population. Create a bar graph with the student race or ethnicity (qualitative data) on the x-axis, and the Advanced Placement examinee population percentages on the y-axis.
Race/ethnicity | AP examinee population | Overall student population |
---|---|---|
1 = Asian, Asian American or Pacific Islander | 10.3% | 5.7% |
2 = Black or African American | 9.0% | 14.7% |
3 = Hispanic or Latino | 17.0% | 17.6% |
4 = American Indian or Alaska Native | 0.6% | 1.1% |
5 = White | 57.1% | 59.2% |
6 = Not reported/other | 6.0% | 1.7% |
Solution
TRY IT 2.6
Park city is broken down into six voting districts. The table shows the percent of the total registered voter population that lives in each district as well as the percent total of the entire population that lives in each district. Construct a bar graph that shows the registered voter population by district.
District | Registered voter population | Overall city population |
---|---|---|
1 | 15.5% | 19.4% |
2 | 12.2% | 15.6% |
3 | 9.8% | 9.0% |
4 | 17.4% | 18.5% |
5 | 22.8% | 20.7% |
6 | 22.3% | 16.8% |
EXAMPLE 2.7
Problem
Below is a two-way table showing the types of pets owned by men and women:
Dogs | Cats | Fish | Total | |
Men | 4 | 2 | 2 | 8 |
Women | 4 | 6 | 2 | 12 |
Total | 8 | 8 | 4 | 20 |
Given these data, calculate the conditional distributions for the subpopulation of men who own each pet type.
Solution
Men who own dogs = 4/8 = 0.5
Men who own cats = 2/8 = 0.25
Men who own fish = 2/8 = 0.25
Note: The sum of all of the conditional distributions must equal one. In this case, 0.5 + 0.25 + 0.25 = 1; therefore, the solution “checks”.
Histograms, Frequency Polygons, and Time Series Graphs
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 rule of thumb is to use a histogram when the data set consists of 100 values or more.
A histogram consists of contiguous (adjoining) boxes. It has both a horizontal axis and a vertical axis. The horizontal axis is labeled with what the data represents (for instance, 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 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%, then, f = 3, n = 40, and RF = fn�� = 340340 = 0.075. 7.5% of the students received 90–100%. 90–100% are 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 number of bars needs to be chosen. Choose a starting point for the first interval to be less than the smallest data value. A convenient starting point is a lower value carried out to one more decimal place than the value with the most decimal places. For example, if the value with the most decimal places is 6.1 and this is the smallest value, a convenient starting point is 6.05 (6.1 – 0.05 = 6.05). We say that 6.05 has more precision. If the value with the most decimal places is 2.23 and the lowest value is 1.5, a convenient starting point is 1.495 (1.5 – 0.005 = 1.495). If the value with the most decimal places is 3.234 and the lowest value is 1.0, a convenient starting point is 0.9995 (1.0 – 0.0005 = 0.9995). If all the data happen to be integers and the smallest value is two, then a convenient starting point is 1.5 (2 – 0.5 = 1.5). Also, when the starting point and other boundaries are carried to one additional decimal place, no data value will fall on a boundary. The next two examples go into detail about how to construct a histogram using continuous data and how to create a histogram using discrete data.
EXAMPLE 2.8
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. Since the data with the most decimal places has one decimal (for instance, 61.5), we want our starting point to have two decimal places. Since the numbers 0.5, 0.05, 0.005, etc. are convenient numbers, use 0.05 and subtract it from 60, the smallest value, for the convenient starting point.
60 – 0.05 = 59.95 which is more precise than, say, 61.5 by one decimal place. The starting point is, then, 59.95.
The largest value is 74, so 74 + 0.05 = 74.05 is the ending value.
Next, calculate the width of each bar or class interval. To calculate this width, subtract the starting point from the ending value and divide by the number of bars (you must choose the number of bars you desire). Suppose you choose eight bars.
NOTE
We will round up to two and make each bar or class interval two units wide. Rounding up to two is one 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:
- 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.
TRY IT 2.8
The following data are the shoe sizes of 50 male students. The sizes are continuous data since shoe size is measured. Construct a histogram and calculate the width of each bar or class interval. Suppose you choose six bars.
9; 9; 9.5; 9.5; 10; 10; 10; 10; 10; 10; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5
11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11.5; 11.5; 11.5; 11.5; 11.5; 11.5; 11.5
12; 12; 12; 12; 12; 12; 12; 12.5; 12.5; 12.5; 12.5; 14
EXAMPLE 2.9
Create a histogram for the following data: 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.
Because the data are integers, subtract 0.5 from 1, the smallest data value and add 0.5 to 6, the largest data value. Then the starting point is 0.5 and the ending value is 6.5.
Problem
Next, calculate the width of each bar or class interval. If the data are discrete and there are not too many different values, a width that places the data values in the middle of the bar or class interval is the most convenient. Since the data consist of the numbers 1, 2, 3, 4, 5, 6, and the starting point is 0.5, a width of one places the 1 in the middle of the interval from 0.5 to 1.5, the 2 in the middle of the interval from 1.5 to 2.5, the 3 in the middle of the interval from 2.5 to 3.5, the 4 in the middle of the interval from _______ to _______, the 5 in the middle of the interval from _______ to _______, and the _______ in the middle of the interval from _______ to _______ .
Solution
- 3.5 to 4.5
- 4.5 to 5.5
- 6
- 5.5 to 6.5
Calculate the number of bars as follows:
where 1 is the width of a bar. Therefore, bars = 6.
The following histogram displays the number of books on the x-axis and the frequency on the y-axis.