Example of constructing a stem-and-leaf chart, a histogram, and an ogive

Suppose that we wish to collect information on the effect of acid rain upon the yield of barley crops. We take a sample of farms and measure the yield of the crop in 1980 to serve as a baseline. We subsequently took samples after a power plant started upwind of the plots in 1982 and 1986 (years with similar precipitation and degree days) for comparative purposes.

Here is the raw data. It is also available in JMP file in the Sample Program Library at http://www.stat.sfu.ca/~cschwarz/Stat-650/Notes/MyPrograms.

Units are g/400 $m^{2}$ . We will begin with trying to describe the 1980 data.

How would you describe this data set? Try drawing a dot plot.

To draw a stem and leaf plot, we break the data into stems and leafs. For example, for the above data, stems might be groups of 10, but this would give too many leaves. How about, groups of 20, e.g., 120-139, 140-159, etc. The leafs will then be the next digit following the stem:

We get the following stem and leaf plot.

12	788	
14	686789
16	0168911589
18	0369
20	01458123378
22	1551125679
24	1881
26
28
30	5
32
34
36
38	2

What do you notice from the stem-and-leaf plot?

Stem and leaf plots are quick and dirty methods of showing structure. A more traditional way is the histogram. This has the advantage of allowing comparison among different groups more readily as will be seen later.

For the barley yield data above:

• n=50. This suggests 1+3.3log(50)=6.6 classes. Use 6 or 7 classes.

• min values are 127, 138, 138, 146, 148; no evidence of outliers here. [Note that there is some ambiguity in the stem-and-leaf plot because each stem is a group of 20. The raw data needs to be consulted to find the exact minimum value.]

• max values are 392, 315, 251, 248, 248; possible outliers on top

• if we use the outliers, class width is about (392-127)=265/6=44(or 50) - too large

• try (251-127)=124/6 gives around 20.

• class bounds are established starting at 120.

• we will use the left endpoint rule, i.e. include the left endpoint, but not the right

This gives us the following frequency table: Let X represent the actual yield of barley:

				Relative
		Relative	Cumulative	Cumulative
Class bounds	Frequency	Frequency	Frequency	Frequency
120<=X<140	3	.06	3	.06
140<=X<160	6	.12	9	.18
160<=X<180	10	.20	19	.38
180<=X<200	4	.08	23	.46
200<=X<220	11	.22	34	.68
220<=X<240	10	.20	44	.88
240<=X<260	4	.08	48	.96
260<=X<280
280<=X<300
300<=X<320	1	.02	49	.98
320<=X<340
340<=X<360
360<=X<380
380<=X<400	1	.02	50	1.00

Check that the total frequencies give n and that relative frequencies total 100%.

We then draw the histogram:

• What is the overall shape of the data? Symmetric, skewed?

• Where is the middle of the data?

• What fraction of observations were between 200 and 240 g/400 $m^{2}$ ?

• Why did a `hole' appear in the middle of the data? Is this evidence of two different sets of data combined together?

We then draw the ogive:

• What fraction of observations are less than 190 g/400 $m^{2}$ ?

• Where is the bottom cutoff that contains 10% of the yields?

• What is the median?