Download Just My Cup of t (test)

IB Biology / IHS
Just My Cup of “t”! (test, that is!)
Or: How to conduct an unpaired two-tailed t-test
Adapted from Jennifer Lockwood, Newbury Park High School, NP, CA
Introduction
What is statistics? Statistics is the science of collecting,
organizing, analyzing, and interpreting data in order to
make decisions. There are many different types of statistical tests. A
t-test examines the difference between the means of two sets of
data.* See fig. 1. This is done in order to determine if any observed
difference is due to chance alone or if the difference between the
means is the result of some other factor; in other words, if that
difference is significant.
Fig. 1
In science, researchers are often looking to see if experimental group
data are significantly different from control group data. (If the data are significantly different,
the researchers may propose that their treatment, the independent variable, within the
experimental group caused the observed difference in the data.) How does a researcher
determine if results are significant? By convention, scientists agree that if there is less than a
5% chance of getting the observed difference by chance, they conclude that they have found a
statistically significant difference between the two groups being tested.
Two-tailed (or, two-sample) t-tests are used when the researcher wants to find out if the results
would be interesting in either direction (i.e. if the treatment had a positive or negative effect).
And in an unpaired two-tailed t-test, there is no requirement that the two groups be of equal
sizes. The two-sample t-test is one of the most commonly used hypothesis tests.
How to conduct an unpaired two-tailed (two-sample) t-test
1) Define the null hypothesis.
2) Calculate the t-test statistic (tcalc). It's time-consuming to calculate so in this handout we
often calculate it for you.
3) Specify the degrees of freedom (df) using this formula: df = (n1 + n2) - 2, where n1 and n2
are the number of observations in each data set. For example, if an experimental group has 12
participants and a control group has 10 participants, then df = (12+10) – 2 = 20.
4) Look up the t critical value (tcrit) in a t-distribution table for the particular df at P = 0.05.
5) If tcalc < tcrit, we accept the null hypothesis. If tcalc > tcrit, we reject our null hypothesis.
Let’s look at the example in Table 1.
Two different classes take the same Biology test.
The calculated means, sample standard deviations
(SD), and sample sizes are shown at the right.
_______________________________________
*The data should be normally distributed with a sample size of at least 10.
Table 1. Per. 2 & 4 Bio test results
out of 62 possible points
Period 2
Period 4
17 students
15 students
Mean: 53.2 points
Mean: 53.9 points
Sample SD: 4.4
Sample SD: 5.5
Working through the example:
1) The null hypothesis: there is no difference between period 2's and 4’s mean test scores except
that which is due to random chance.
2) The t-test statistic (tcalc) in this example is 0.3997. We calculated it for you to save you time.
In "real life" one consults a software program or online calculator.**
3) We determine the degrees of freedom (df) by applying the above formula:
df
=
(n1 + n2) – 2
=
(17 + 15) – 2
=
32 – 2
=
30
Table 2. tcrit values
for a t
4) Now we consult a table of t-critical
values. In the row for 30 df, tcrit is
2.04 at P = 0.05.
Note, P ranges from 0 (not likely) to 1
(certain). Table 2 only shows a few of the
values for P: 0.05, 0.025, 0.01 & 0.005.
5) Recall, tcalc was 0.3997, which is < tcrit
(2.04) at P = 0.05. When tcalc < tcrit, we
must accept our null hypothesis:
there is no difference between period
2's and 4’s test scores except that
which is due to random chance.
**If you must see the t-test formula, it is:
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Let's practice some more!
1. An experiment is performed to discover the effect of using different amounts of fertilizer on the growth
of bean plants. After germination, the beans are planted in sterilized soil to which fertilizer is added. As
you can see in Table 3, there are ten trials of each of 3 levels of fertilizer treatment and a control. The
height of each of the bean plants is measured in centimeters ( + 0.5 cm) 25 days after planting.
Table 3. The effect of the amount of fertilizer on bean plant height
Group A: no
Group B: 0.001%
Group C: 0.01%
Group D: 0.1%
Trial Number
fertilizer
fertilizer
fertilizer
fertilizer
1
2
3
4
5
6
7
8
9
10
10
7
8
7
8
10
9
8
7
9
8
8
7
10
10
7
8
8
9
9
7
10
8
10
8
9
9
8
6
7
12
13
15
10
10
15
10
10
14
10
a) Find the Mean and Standard Deviation of each set of data.
(**In Excel: to calculate mean =AVERAGE(data range)
to calculate sample SD =STDEV(data range).
Treatment
Group
Group
Group
Group
A:
B:
C:
D:
Table 4. Statistics generated from Table 3 data
Sample
t-Test score when treatment is
Mean (cm)
Standard
compared to control
(to save time, we'll just use 2 of 4 combo's)
Deviation (cm)
no fertilizer
0.001% fertilizer
0.01% fertilizer
0.1% fertilizer
0.20
4.61
b) Discuss the variability of the bean plant data using the sample standard deviation values.
c) Compare the means of each group of data.
d) Is there a significant difference in bean plant height between the Group B fertilizer treatment and the
control?
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e) Explain how you figured out the answer to (d) using the 5 steps on page 1 of this handout.
f) Is there a significant difference in bean plant height between the Group D fertilizer treatment and the
control?
g) Explain how you figured out the answer to (f) using the 5 steps on page 1 of this handout.
Table 5. Resting heart rates
(bpm). Use for problem 2.
2. Resting heart rates (beats/min) are measured in 20 non-runners and 20
marathon-running athletes. The data are shown in Table 5 at the right.
On average, do male marathon runners have a significantly different resting
heart rate than untrained men? Show ALL steps 1-5 from page 1 of this handout.
Non-runner
males
72
75
68
61
82
78
79
77
71
75
74
79
81
83
77
74
70
69
78
77
Male Marathon
runners
62
48
47
51
55
54
60
48
57
56
55
50
59
50
55
58
61
60
49
53
3. Scientists determined the average size of salmon that spawned in two different streams. 38 salmon were
sampled for one stream and 22 for the other stream. The value of “tcalc” was found to be 1.29. Is the
average size of the salmon in each stream significantly different? Explain. (Note: to use Table 2, use the
degrees of freedom row closest to the d.f. for this problem.)
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