Name: _____________________________________
Date: ________________________
The Chi-Square Statistical Test: An Introduction
A friend asks you to help him make a decision by tossing a coin. If the coin lands heads, your
friend will go to the movie tonight. If it lands tails, he will go to the ballgame. You flip the coin,
and it lands heads. Your friend asks you to flip the coin again. It lands heads. Your friend asks
you to flip the coin again. It lands heads. Your friend becomes angry, convinced that there is
something wrong with the coin. To determine whether this is true, you decide to flip the coin
500 times. You find that the coin lands heads 325 times and tails 175 times. Your friend
believes that this is proof that the coin is rigged; you aren’t so sure. To find out, you decide to
use a well-known statistical test called the chi-square analysis.
Scientists often employ statistics to interpret experimental results and determine whether the
results have met a minimal level of acceptance, or should be rejected. This allows us to come to
a conclusion about a hypothesis. Such an analysis is always provisional; we can not provide a
hypothesis that is definitely true because we can never be certain that we have examined all the
evidence or considered all possible alternatives. However, statistical tests allow us to generalize,
allowing us to determine whether the results we see can be attributed to chance, or to some other
outside factor.
In the above example, we can use the chi-square test as follows:
You observed 500 coin flips. Of the 500 flips, if the coin really is fair, you expected to find 250
heads and 250 tails. You actually observed 325 heads and 175 tails.
Your friend’s hypothesis is that the two groups of results are different. Your hypothesis is that
there is no difference between the heads group and the tails group (the “null hypothesis”). The
chi-square formula is as follows:
2
Chi-square(x ) =
(observed-expected)2
∑ -------------------------(expected)
Note: ∑ is the symbol for sum.
To calculate the chi-square value for this experiment, analyze the two sets of results.
Heads: (325 – 250)2 / (250)
Tails: (175-250) 2 / (250)
= 22.5
= 22.5
The sum of the two “runs” of the formula is 45.0. This is called the chi-square number for the
experiment.
A second number that is needed in a chi-square analysis is the number of degrees of freedom. A
degree of freedom is either: the number of different treatment groups minus one, or the number
of possible outcomes to the experiment, minus one. In this experiment, the coin can land either
heads or tails, so there are two possible outcomes. That means that there is one degree of
freedom in the experiment.
Now that we have the chi-square number and the degree of freedom, we can use a chart to
determine whether the null hypothesis should be accepted or rejected. Look in the far left
column of the chart to find the number of degrees of freedom. Then, look to the right of this
number until you find a place for the number 45. On the chart, the number 45 would be to the
right of the last number on the table, 10.83. That means that (looking up to the column headings)
there would be less than a 1 in 1000 chance that randomness caused the results you saw. (For
example, if you found that the chi-square number was 2, this is between the number 1.64 and
3.84. Looking up to the column headings, that means that there would be between a 1 in 2 and a
1 in 5 chance that the results can be attributed to randomness).
Randomness can be rejected if there is less than a 5% (1 in 20) chance that it occurred. When
you reject randomness, you reject the null hypothesis. In your coin flip experiment, your friend
is correct. It is safe to say that the null hypothesis can be rejected. There is a statistically
significant difference between the “heads” group and the “tails” group. Perhaps your coin was
loaded or altered!
This experiment:
ENVIRONMENTAL TEMPERATURE AND METABOLIC RATE IN Carassius auratus
INTRODUCTION
Any change in the metabolic rate of Carassius auratus
can be easily observed by counting the number of times
the gill cover (or operculum) beats per minute. The
operculum opens and closes as water (containing
oxygen) passes over the gills. When the metabolic rate
increases, the animal uses more oxygen and the
operculum can be seen to open and close more
frequently.
gill operculum
Carassius auratus
http://www.dnr.cornell.edu/sarep/fish/Cyprinidae/goldfishpic.html
QUESTION
What is the effect of temperature changes on the swimming behavior and metabolic rate of a
goldfish (C. auratus)?
HYPOTHESIS
As the environmental temperature increases, the metabolic rate of a goldfish will (circle one):
increase / decrease / stay the same
As the environmental temperature decreases, the metabolic rate of a goldfish with (circle one):
increase / decrease / stay the same
Explain your reasoning:
MATERIALS
live Carassius auratus specimens
ice cubes
500 mL beaker, jar or bowl
water
thermometer
stopwatch or online clock
One of the basic features of life is that living things respond to the environment. In this
experiment, we will be exploring the respiration rate of goldfish as a function of temperature.
Specifically, we will see if temperature change has a statistically significant effect on the number
of breaths per minute a goldfish takes.
PROCEDURE
1) Obtain a beaker from the back of the room. The beaker will already contain water and a
goldfish. (It is important that the goldfish be calm and “acclimated” before you begin).
2) Use the thermometer to measure the temperature of the water. Record this information.
3) Count and record the number of breaths your fish takes in a single minute. Record in the
data table. Perform the count a second and then a third time. Calculate the average.
4) Place two ice cubes into the beaker.
5) Wait a few minutes for the water temperature to cool down by 5 degrees Celsius.
6) Count and record the number of breaths your fish takes in a single minute. Record in the
data table. Perform the count a second and then a third time. Calculate the average.
7) Pour warm water into the beaker so that the temperature rises to 5 degrees Celsius above
where you started.
8) Count and record the number of breaths your fish takes in a single minute. Record in the
data table. Perform the count a second and then a third time. Calculate the average.
9) On the board, write the number of breaths your fish took at each temperature (room temp,
warm temp, and cool). When all groups are finished, record the information.
10) Determine the average respiration rate for each temperature.
Treat your goldfish well! Dead goldfish = lower lab grade. Mistreated goldfish = lower lab
grade and report on proper handling of laboratory animals.
GROUP DATA:
Temperature
Start (X) = _______
X - 5ºC = _______
X + 5ºC = _______
Trial One
Trial Two
Trial Three
Average
beats/min
CLASS DATA:
Temperature
Average
beats/min
Temperature
Average
beats/min
ANALYSIS
1. Use a sheet of graph paper or Microsoft Excel to plot “Temperature ºC” on the x-axis, and
“Average Operculum Beats/Min” on the y-axis. Use your data and the data that you obtained
from your classmates. Use a linear scale with equal intervals. Draw a “line of best fit”
through all of your data points.
2. Summarize the general trend of your graph in a few sentences.
3. What happens to the metabolic rate of Carassius auratus: (a) when environmental
temperature decreases, and (b) when environmental temperature increases?
4. Are the results of your experiment statistically significant? (That is, do the results from each
experimental group differ enough from the control group that they can be considered
different?). Perform a chi-square calculation and show all work to document your reasoning.
5. Why does temperature have this effect on metabolic rate? Do some research to come up with
a physiological reason for this question.
6. Use specific information from your data chart and graph to predict the goldfish metabolic
rate at 5ºC? At 35ºC?
7. Why did you take three measurements at each temperature instead of one?
8. Restate your hypothesis. Did your results support or refute your hypothesis? Were you
surprised by the results?
9. List sources of error in the experimental setup and suggest improvements for these errors.
10. What other factors aside from temperature may have affected the metabolic rate of your
goldfish?
Critical Values Table
Degr. Probability
Of
= 0.9
Freed. (9 in 10)
Probability
= 0.5
(1 in 2)
Probability
= 0.2
(1 in 5)
Probability
= 0.05
(1 in 20)
Probability
= 0.01
(1 in 100)
Probability
= 0.001
(1 in 1000)
1
2
3
4
5
6
0.46
1.39
2.37
3.37
4.35
5.35
1.64
3.22
4.64
5.99
7.29
8.56
3.84
5.99
7.82
9.49
11.07
12.59
6.64
9.21
11.35
13.28
15.09
16.81
10.83
13.82
16.27
18.47
20.52
22.46
0.016
0.21
0.58
1.06
1.61
2.20
Chi-Square Analysis Example
You are performing a breeding experiment with fruit flies, Drosophila melanogaster. According
to your hypothesis, a cross between a wingless female and an eyeless male should produce
offspring in the following proportions:
¼ normal, ¼ wingless, ¼ eyeless, ¼ wingless and eyeless
In counting the offspring, you observe the following: 32 normal, 28 wingless, 42 eyeless, 38
wingless and eyeless.
Chi-square analysis can be used to determine the probability that deviation from expectation is
due to chance.
Chi-square(x2) =
(observed-expected)2
∑ -------------------------(expected)
Note: ∑ is the symbol for sum.
To calculate the chi-square value for this experiment, analyze the four sets of results. Since there
are 140 offspring, we would expect 35 of each type.
Normal:
(32 – 35)2 / (35)
= 0.257
2
Wingless:
(28 – 35) / (35)
= 1.400
Eyeless:
(42 – 35) 2 / (35)
= 1.400
Wingless and Eyeless: (38 – 35) 2 / (35)
= 0.257
Adding the four values, the chi-square number is 3.314.
Since there are four possible outcomes, there are three degrees of freedom. Looking across this
row in the table for 3.314, we find that it would fall between 2.37 and 4.64. This represents a 1
in 2 (50%) to 1 in 5 (20%) possibility that the results we saw were due to random chance.
A 5% or less possibility would be required to reject your proposed hypothesis. Since this criteria
is not met, we can accept the hypothesis, and assume that deviation is due to chance.