Calculating
Chi – Square manually:
1. Draw a contingency table.
2. Enter the Observed
frequencies or counts (O)
3. Calculate totals (in the
margins).
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4.
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Calculate the Expected frequencies
(E)
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a.
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For each cell: Column total/N times
Row total
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b.
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Write the Expected frequency into
the appropriate box in the table.
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CHECK: Expected frequencies (E)
marginal totals are the same as for Observed frequencies (O)
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Eyeball the contingency table, noting
where the differences between O (observed) and E (Expected) values occur. If
they are close to each other, the levels of the independent (predictor)
variable are not having an effect.
5. Calculate Chi-square statistic
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O = Observed
frequency
E = Expected frequency  = Sum of above across all
cells
6. Find the
probability value (p) associated with the obtained Chi-square statistic
a. Calculate degrees of freedom (df)
df = (# rows - 1)(#
columns - 1)
b. Use the abbreviated table of Critical
Values for Chi-square test to find the p value.
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We use this test for comparing
the means of two samples (or treatments), even if they have different numbers
of replicates. In simple terms, the t-test compares the actual difference
between two means in relation to the variation in the data (expressed as the
standard deviation of the difference between the means).
T- test
There are two types of
t-test:
1.
Paired
sample t-test
2.
Independent
sample t-test
3.
Single
sample t - test
Procedure to conduct a t – test:
1. We need to construct a null
hypothesis - an expectation - which the experiment was designed to test. For
example:
If we are analysing the heights
of pine trees growing in two different locations, a suitable null hypothesis
would be that there is no difference in height between the two locations. The
student's t-test will tell us if the data are consistent with this or depart
significantly from this expectation. [NB: the null hypothesis is simply
something to test against. We might well expect a difference between trees
growing in a cold, windy location and those in a warm, protected location, but
it would be difficult to predict the scale of that difference - twice as high? Three
times as high? So it is sensible to have a null hypothesis of "no
difference" and then to see if the data depart from this.
2. List the data for sample 1.
3. List the data for sample 2.
4. Record the number (n) of
replicates for each sample (the number of replicates for sample 1 being termed
n1 and the number for sample 2 being termed n2)
5. Calculate mean of each sample
(1 and 2).
6. Calculate s 2 for each sample;
call these s 12 and s 22 [Note that actually we are using S2 as an estimate of
s 2 in each case]
5. Calculate the variance of the
difference between the two means (sd2)
6. Calculate sd (the square root
of sd2)
7. Calculate the t value.
(When doing this, transpose 1 and
2 if 2 > 1 so that you always get a positive value)
8. Enter the t-table at (n1 + n2
-2) degrees of freedom; choose the level of significance required (normally p =
0.05) and read the tabulated t value.
9. If the calculated t value
exceeds the tabulated value we say that the means are significantly different
at that level of probability.
10. A significant difference at p
= 0.05 means that if the null hypothesis were correct (i.e. the samples or
treatments do not differ) then we would expect to get a t value as great as
this on less than 5% of occasions. So we can be reasonably confident that the
samples/treatments do differ from one another, but we still have nearly a 5%
chance of being wrong in reaching this conclusion.
Now compare your calculated t
value with tabulated values for higher levels of significance (e.g. p = 0.01).
These levels tell us the probability of our conclusion being correct. For
example, if our calculated t value exceeds the tabulated value for p = 0.01,
then there is a 99% chance of the means being significantly different (and a
99.9% chance if the calculated t value exceeds the tabulated value for p =
0.001). By convention, we say that a difference between means at the 95% level
is "significant", a difference at 99% level is "highly
significant" and a difference at 99.9% level is "very highly
significant".
What does this mean in
"real" terms? Statistical tests allow us to make statements with a
degree of precision, but cannot actually prove or disprove anything. A
significant result at the 95% probability level tells us that our data are good
enough to support a conclusion with 95% confidence (but there is a 1 in 20
chance of being wrong). In biological work we accept this level of significance
as being reasonable.
Written By: Priyesh Bhadauriya
Priyesh Bhadauriya 2013214
Group members:
Nikita Agarwal 2013171
Nimisha Agarwal 2013173
Parth Mehta 2013193
Nihal Moidu
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