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Box plots & t-tests
Box Plots
Box plots are a graphical representation of your sample (easy to visualize descriptive statistics); they are also
known as box-and-whisker diagrams. Any data that you can present using a bar graph can, in most cases, also
be presented using box plots. A box plot provides more information about the data than does a bar graph.
Things to know about box plots
Your sample is presented as a box.
The spacings between the different parts of the box help indicate the degree of dispersion (spread) and
skewness in the data, and identify outliers.
A box plot shows a 5-number data summary: minimum, first (lower) quartile, median, third (upper)
quartile, maximum.
The box is divided at the median.
The length of the box is the interquartile range (IQR).
The 1st quartile is the bottom line.
The 3rd quartile is the top line.
Example
Quartiles divide frequency distributions
• Q
1
:1
st
or lower quartile: cuts off lowest 25% of the data
• Q
2
:2
nd
quartile or median: 50% point, cuts data set in half
• Q
3
:3
rd
quartile or upper quartile: cuts off lowest 75% of the data (or highest 25%)
Q
1
is the median of the first half of your data set.
Q
3
is the median of the second half of your data set.
The difference between the upper and lower quartiles is called the interquartile range. The interquartile range
spans 50% of a data set, and eliminates the influence of outliers because the highest and lowest quarters are
removed.
Example:
A biologist samples 12 red oak trees in a forest plot and counts the number of caterpillars on each tree.
The following is a list of the number of caterpillars on each tree: 34, 47, 1, 15, 57, 24, 20, 11, 19, 50, 28, 37
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Calculate the median, 1
st
and 3
rd
quartile.
Step 1: Arrange the values in ascending order
1, 11, 15, 19, 20, 24, 28, 34, 37, 47, 50, 57
Step 2: Calculate the median
Median = (24 + 28)/ 2 = 26
Step 3: Determine Q1
Lower quartile = value of middle of first half of data
Q
1
= the median of 1, 11, 15, 19, 20, 24
= (3
rd
+ 4
th
observations) ÷ 2
= (15 + 19) ÷ 2 = 17
Step 4: Determine Q3
Upper quartile = value of middle of second half of data Q
3
= the median of 28, 34, 37, 47, 50, 57
= (3
rd
+ 4
th
observations) ÷ 2
= (37 + 47) ÷ 2 = 42
Outliers
Observations that are 1.5 x IQR greater than Q3 or less than Q1 are called outliers and are distinguished by a
different mark, e.g., an asterisk. The actual symbols used don’t matter as long as you are consistent and you
explain your symbols in the figure legend. In the figure below the arrows are pointing to the outliers.
Do not remove outliers from the dataset unless there is good reason to do so. A good reason could be that the
outlier is a typo, for example a student records a caterpillar mass of 10 grams instead of 0.10 grams. Or if the
equipment used to measure the observation failed, for example if the balance measures a caterpillar as 10
grams. Don’t remove outliers just because you want to make your dataset look “prettier”. Outliers can point us
to interesting patterns or let us know that we may need to increase are sample size.
How do you determine if there are any outliers in your sample?
1. Calculate IQR x 1.5
2. Add this value to Q3. Are there any values greater than Q3 + (IQR x 1.5)? If so, then these values are
outliers.
3. Subtract this value from Q1. Are there any values smaller than Q1 – (IQR x 1.5)? If so, then these
values are outliers.
Whiskers
The two vertical lines (called whiskers) outside the box extend to the smallest and largest observations within
1.5 x IQR (interquartile range) of the quartiles. If there are no outliers, then the whiskers extend to the min and
max values.
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Comparing populations using t-tests
The t-test is a parametric test for comparing two sets of continuous data. It is used for comparing two sample
means. For example, we want to know if the average mass of the caterpillars fed leaves from plants receiving
the fertilizer treatment is larger than the average mass of caterpillars fed leaves from plants receiving the control
treatment. A t-test allows you to determine if there is a statistically significance difference between the two
treatments. When you are comparing two samples, then you use a t-test. A t-test doesn’t work if you are
comparing more than two samples. For example, if you were comparing caterpillars fed leaves from a high,
low, and control fertilizer treatments, then you would not be able to use a t-test. The two-sample t-test is a
hypothesis test for answering questions about the mean when the data are collected from two random samples
of independent observations, each from an underlying normal distribution.
H
1
: The means are the two samples (or treatments) are different.
H
0
: (null hypothesis): The samples are from the same populations = the means are equal.
A t-test uses 3 pieces of information
1. Sample size = the number of replicates in each sample.
2. The difference between the two means
The figures above show frequency histograms of hypothetical caterpillar mass data. The difference
between the two means is greater in figure A). These data will produce a more significant t-test.
3. The variance of each sample
The figures above show frequency histograms of hypothetical caterpillar mass data. The difference between the
two means is equal; however, the two samples in A) are more clearly distinguishable because of their smaller
variances. The data in sample A) are more likely to produce a significant t-test. Variance is represented by s
2
.
Variance is a measure of data scattering around the mean. Variance is a way to measure how variable or
dispersed the data are. Variance is equal to the standard deviation (standard deviation = s) squared.
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Types of t-tests
The type of t-test that you decide to use depends on your hypothesis and methods used to collect your data. This
handout provides background information about the t-test and explains how you could calculate it by hand.
However, I strongly recommend that you use EXCEL or another statistical software package to do t-tests. The
point of doing one by hand is so that you understand what goes into a t-test. If you have to do multiple t-tests,
then doing it by hand is tedious.
A paired t-test compares two different measures taken from the same individual. For example, I could compare
your scores on the ecology pretest with your score on the final to test the hypothesis that you learned something
about ecology this semester.
A t-test for independent samples compares the averages of two samples. For example, I could compare your
scores on the ecology final to the scores from last year’s class to test the hypothesis that changes made in the
class this year improved student learning. We call this an independent test because the individuals in first
sample are not connected to the individuals in the second sample, i.e. the samples are independent of one
another.
Additionally, a t-test can also be one-tailed or two-tailed. This distinction depends on your hypothesis. If your
hypothesis is directional, then you would do a one-tailed test. For example “I hypothesize that students in this
year’s class will score higher on the final than students in last year’s class.” A one-tailed test completely
disregards the possibility of testing for significance in the other direction.
If the difference between the two samples could go in either direction (sample A could be greater than B or
sample B could be greater than A) then you would do a two-tailed test. For example “I hypothesize that there
will be a difference in exam scores between this year’s class and last year’s class.”
Only use a one‐tailed test when you have a very good reason to expect that the difference between the two
samples will be in a particular direction. If the difference between the two samples could go in either direction,
then do a two-tailed test.
t-test assumptions
The data in each sample are normally distributed.
The data in each sample have approximately equal variances. The t-test is fairly robust with regard to
this assumption, but if there is a large difference between the variances in each population then you can
also do a t-test that assumes unequal variance.
A t-test is fairly robust to deviations from normality; this means that is it usually still OK to do a t-test even
when your data do not have a perfect normal distribution. In this class, because of time constraints, we are going
to assume that the data are normally distributed without first testing this assumption. However, if you are
collecting data for “real” research purposes then you need to test this assumption first.
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How to conduct an independent t-test
There are two parts to a t-test: the t-statistic and the p-value. First, you calculate the t statistic and then you
determine the p-value that goes along with your t-statistic.
t – statistic:
In English: The mean of sample 1 minus the mean of sample 2 divided by the square root of the variance of sample 1
divided by the sample size of sample 1 plus the variance of sample 2 divided by the sample size of sample 2.
You also need to calculate the degree of freedom (df) for the t-test:
Degree of freedom = n
1
+ n
2
- 2
Example
You are researching the impact of urbanization on robin populations. You conduct point-counts in urban and
rural settings. From these point-counts you are able to estimate robin density (# of robins per hectare). You
conduct 15 point-count surveys in each setting.
The first step is to establish the specific hypotheses we wish to examine.
H
0
: Null hypothesis is that the difference between the two groups is 0. There is no difference in robin density
between the urban and rural settings.
H
1
: Alternative or research hypothesis - the difference between the two groups is not 0. There is a difference
in robin density between the urban and rural settings. This hypothesis would be tested using a 2-tailed t-test.
Or you could make a directional prediction – I predict that robin density will be higher in the rural environment.
This hypothesis would be tested using a 1-tailed t-test. For this example, we are going to use the 2-tailed
hypothesis.
The data
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Information you need for a t-test (calculated from the data):
Sample size of a = n
a
= 15
Sample size of b = n
b
= 15
Mean
a
= 21.17
Mean
b
= 19.49
Var
a
= s
a
2
= 35.64
Var
b
= s
b
2
= 7.79
Degree of freedom: 15 + 15 – 2 = 28
If you plug these numbers into the above t-stat formula, then you will calculate t = 0.988. Is this a high or low
value?
Now you need to consult a statistical table to determine the significance of the t-statistic. There is a copy of part
of a statistical table included at the end of these notes – additionally, if you have taken a statistics class
previously and still have your textbook, then there is probably a copy of the table at the end of the book.
1. First, look down the left hand column to find the degrees of freedom that matches the one you
calculated. Then, reading across that row, compare your t-value with the number in the second column.
The number in the chart is the critical value of t for p = 0.05
2. If your calculated t-statistic is smaller than the p = 0.05 t-critical value, then your two samples are NOT
significantly different. You must accept the null hypothesis that the two samples came from the same
population. There is no difference between the means.
3. If your calculated t-statistic is larger than p = 0.05 t-critical value, then the difference between the means
is statistically significant. You can reject the null hypothesis with no more than a 5% error rate and
accept the alternative hypothesis.
4. The next column in the table presents the t-critical values for p = 0.025. If your value is larger than this
number, then you can reject your null with a 2.5% error rate.
5. Then compare the computed p-value to alpha
Step 5: Comparing the computed p-value to alpha
All hypothesis tests are based on the same basic principles and are setup to minimize the probability of drawing
an incorrect conclusion.
Two possibilities (reality):
Null hypothesis is true.
Null hypothesis is false.
Two outcomes of a hypothesis test:
We reject the null hypothesis.
We fail to reject the null hypothesis.
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Two possible types of mistakes:
Type 1 error: The null hypothesis is true but is rejected. The probability of a type 1 error is denoted by
alpha (α).
Type 2 error: The null hypothesis is false, but is it not rejected. The probability of a type 2 error is
denoted by beta (β).
A p-value represents the probability, if H
0
is actually true, that random chance could produce your observed
results. If you calculate p = 0.01, then this means that there is a 1% chance of the null hypothesis being true (no
difference between the means) given your observed results. If you calculate p = 0.57, then this means that there
is a 57% chance of the null hypothesis being true given your observed results.
To determine if you should accept your hypothesis (i.e. reject your null hypothesis), you calculate a t-statistic
and p-value.
You accept your hypothesis if p < α (greek letter alpha)
For most scientific studies, the accepted value of α is 0.05
The data support the research hypothesis if p < 0.05.
There is a statistically significant difference between the two populations.
The data fail to support the null hypothesis.
The data fail to support the research hypothesis if p > 0.05.
There is no statistically significant difference between the two populations.
The data support your null hypothesis.
For a t-test, as the difference between your calculated t-statistic and t-critical increases, the p-value decreases.
Back to the example:
t = 0.988 Is this a high or low value?
By looking at the table, you can see that this value is less than the t-critical value (2.05) and therefore it is not
significant at p = 0.05. Therefore, we reject our research hypothesis and support our null hypothesis. There is
no significant difference in robin density between urban and rural settings.
In this example, p > 0.05
But what if I’m using a computer program to analyze my data?
If you are using a computer program to calculate your statistics (which is what we normally do), then you don’t
need to figure out the t-critical value using a table. Most computer statistical programs, including the statistics
you will do in EXCEL, compute the p-value for you. In this example, p = 0.331. This value is much larger than
the accepted p-value of 0.05. This large p-value indicates that we cannot reject our null hypothesis (another
way of saying this is that we fail to support our research hypothesis).
Reporting statistics in your results section
Report the t-statistic, degrees of freedom and p-value using the following format: Sentence stating your results
(t
df
= t-stat; p-value).
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Continuing with the robin example:
If you used a table to compare your t-statistics and t-critical:
Robin density did not differ between the urban and rural settings (t
28
= 0.988; p>0.05).
If you used a computer program to analyze your statistics:
Robin density did not differ between the urban and rural settings (t
28
= 0.988; p= 0.331).
Report your statistics, don’t discuss p-values and t-statistics.
This is wrong: My p-value was 0.046, this is less than the accepted value of 0.05, so I can reject my null
hypothesis and accept my alternative hypothesis.
In the results section you just state the results (including the statistics). In the discussion section, you interpret
and explain your results.
Don’t ignore your statistical results.
In this example, average robin density was higher in the urban setting (21.19 robins per count) than in the rural
setting (19.49 robins per count). However, these means are NOT significantly different. So you cannot make
any statements about differences in robin density between the two environments. It is incorrect to say that
robin density was higher in the urban setting.
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Practice problems with answers
Problem 1 – Brown trout box plots
Figure 1: Brown trout eat a variety of freshwater invertebrates and the size of the food that they eat can vary
with trout age. This figures shows box plots of the age-related variation in prey size of Salmo trutta, brown
trout, in the Furelos River (NW Spain) during summer. From: Ontogenetic Dietary Shifts in a Predatory Freshwater Fish Species: The
Brown Trout as an Example of a Dynamic Fish Species By Javier Sánchez-Hernández, María J. Servia, Rufino Vieira-Lanero and Fernando Cobo
a) What is the independent variable? What is the dependent variable?
b) Describe how prey size changes as trout age class increases.
c) Visually estimate the minimum, maximum, median, Q1, Q3 and the IQR for teach treatment.
Age class
Min.
Max.
Median
Q1
Q3
IQR
0+
1+
2+
3+
d) Which of the box plots would most likely have normal? Explain your answer.
2. Creating and interpreting box plots
a) Use the data below to construct box plots of the two samples. Round to whole numbers.
Sample A
Sample B
11
14
38
16
15
17
35
20
11
11
8
13
32
17
10
16
16
22
16
15
27
13
18
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b) Do you think that samples A and B were collected from the same population? Why or why not?
c) Which sample would you predict would have a greater standard deviation? Explain your answer.
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3. A researcher hypothesized that phosphorus-fertilization of host plants will impact caterpillar growth. 8
caterpillars were reared on plants fertilized with phosphorus and 8 caterpillars are reared on plants that
receive a water only (no phosphorus) treatment. The data are presented below.
Pupal Mass
-
Phosphorus
Pupal Mass -
No
phosphorus
12
8
7
7
3
4
11
14
8
6
5
7
14
12
7
5
Sample Mean
Sample Size
Variance
13.70
11.84
a) What would be the null hypothesis in this study?
b) What would be the alternate (research) hypothesis?
c) What type of t-test would you use?
d) How many degrees of freedom for your t-test?
e) What is your t-critical?
f) What is your computed t-statistic?
g) Is there a significant difference between the two groups?
h) Write these results up as you would in the results section of a lab report (include a statement of the
results and the statistics).
i) Interpret these results as you would in the discussion section of your lab report (i.e. what do these results
mean).
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Answers
1. Brown trout
a) What is the independent variable? What is the dependent variable?
Independent variable: Brown trout age class
Dependent variable: Prey size
b) Describe how prey size changes as trout age class increases.
Older fish eat larger prey.
c) Visually estimate the minimum, maximum, median, Q1, Q3 and the IQR for teach treatment. (Since
these are estimates, it’s OK if your values are slightly different).
Age
class
Min.
Max.
Median
Q1
Q3
IQR
0+
6
15
11
9
12
3
1+
4
17
9
6
11
5
2+
5
24
12
10
20
10
3+
12
26
24
18
25
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d) Which of the box plots would most likely have normal distributions (you may want to review the pine
needle handout)? Explain your answer.
The box plots for age classes 0+ and 1+ most likely were produced from data that are normally
distributed. When data are normally distributed (bell curve) then the mean = median and there are
approximately an equal number of replicates above and below the median (the median looks like it is in
the middle of the distribution of data).
2. Use the data below to construct box plots of the two samples. Round to whole numbers.
Sample A
Sample B
11
14
38
16
15
17
35
20
11
11
8
13
32
17
10
16
16
22
16
15
27
13
18
13
a) Do you think that samples A and B were collected from the same population? Why or why not?
Answers may vary – a correct answer depends on a correct explanation. The two samples have the same
median but the range of values is greater in sample A. There is much more variation within the sample
A population and the mean is greater.
b) Which sample would you predict would have a greater standard deviation?
Sample A (greater range of data). If you actually calculated the values:
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Problem 2: A researcher hypothesized that phosphorus-fertilization of host plants will impact caterpillar growth.
8 caterpillars were reared on plants fertilized with phosphorus and 8 caterpillars are reared on plants that receive
a water only (no phosphorus) treatment. The data are presented below.
Pupal Mass
-
Phosphorus
Pupal Mass -
No
phosphorus
12
8
7
7
3
4
11
14
8
6
5
7
14
12
7
5
Sample Mean
8.375
7.875
Sample Size
8
8
Variance
13.70
11.84
a) What would be the null hypothesis in this study?
There will not be a difference in mass between the caterpillars fed phosphorus-treated leaves and the
caterpillars fed control leaves.
b) What would be the alternate (aka research) hypothesis?
There will be a difference in mass between the caterpillars fed phosphorus-treated leaves and the
caterpillars fed control leaves.
c) What type of t-test would you use?
A 2-tailed t-test. In the description above, the researcher hypothesizes that phosphorus will impact
caterpillar mass, so there could be a positive or negative effect of phosphorus on caterpillar mass.
d) How many degrees of freedom for your t-test?14
e) What is your t-critical? t
critical
= 2.145
f) What is your computed t-statistic? t
stat
= 0.279
g) Is there a significant difference between the two groups? No
h) Write these results up as you would in the results section of a lab report (include a statement of the
results and the statistics).
Pupal mass of caterpillars fed phosphorus treated leaves was not significantly different than pupal mass
of caterpillars fed control leaves (t
14
= 0.279 , p > 0.05).
i) Interpret these results as you would in the discussion section of your lab report (i.e. what does these
results mean).
Phosphorus had no effect on caterpillar growth. Perhaps phosphorus is not a limiting nutrient for
caterpillars, so adding more phosphorus to the plants did not increase caterpillar mass. Other
explanations are possible. However, it is important that you don’t discuss you t-stat or p-value in the
discussion.
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Critical values for the t-distribution
