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Student's t-test
1. Dr Azmi Mohd Tamil
Normally distributed data
Comparison between 2 groups
Student’s
T-Test
1 drtamil@gmail.com 2015
2. Concept
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Basically we are comparing the mean of the
outcome (as measured in continuous data)
among the treated against the mean of the
outcome among the control.
If the treatment is effective, we would expect
those treated to have a significantly higher
(“causative”) or lower (“protective”) mean of the
outcome compared to those in the control group.
n Mean HDL S.D.
Fish Oil + 30 1.4767 0.45549
Control - 30 1.2593 0.44488
3. Research Question
Will those treated with fish oil supplement have a
higher level of HDL compared to control group?
If it is true that fish oil is good for you, then we
expect a significantly higher level of HDL among
those treated with fish oil group (1.48+0.46)
compared to the control group (1.26+0.44).
Student’s t-test will test whether the HDL mean
among treated group (1.48) is significantly higher
than the HDL mean of control group (1.26).
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n Mean HDL S.D.
Fish Oil + 30 1.4767 0.45549
Control - 30 1.2593 0.44488
4. Null hypothesis
Null hypothesis assumes that there is no
association between fish oil usage and HDL
level. As though taking fish oil has no effect on
the HDL level.
If taking fish oil has no effect on HDL, then
there should be no difference of mean of HDL
level between treatment and control groups.
The statistical test conducted is to decide
whether or not to reject the null hypothesis.
So if the mean HDL level of treatment group is
significantly different than the mean HDL of
control group, null hypothesis is rejected.
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5. Significance We expect better HDL level
among the treated group,
compared to the control
group. The better the
improvement, the larger is the
mean difference between the
two groups.
The larger the difference, the
more significant is the p value.
The bigger the mean
difference, the bigger is the t
value, therefore the smaller is
the p value, therefore more
likely to be significant.
The smaller the variance, the
smaller is the p value.
6. Significance The larger the difference, the
more significant is the p value.
But here the difference is only
0.2174, which is even smaller
than the standard deviation
(0.45549 & 0.44488)
The small mean difference,
and the large standard
deviation/variance will lead to
a smaller t value, therefore
larger p value, therefore more
likely to be insignificant.
n Mean HDL S.D.
Fish Oil + 30 1.4767 0.45549
Control - 30 1.2593 0.44488
7. Example:
t = (1.4767 – 1.2593)_____
(0.455492/30 + 0.444882/30)0.5
t = 0.2174/0.1162=1.870
df = (30-1)+(30-1) = 58
Critical value for df = 60 for p=0.05 is 2.00,
Critical value for df = 40 for p=0.05 is 2.02,
The calculated t is smaller than the critical value,
therefore the null hypothesis is NOT
REJECTED.
Conclusion: Although the treated group HDL
level is slightly higher (1.48+0.46) compared to
the control group (1.26+0.44), the difference is
not statistically significant (p>0.05)
Hint: Memorise critical value of 1.96 for
df=infinity.7 drtamil@gmail.com 2015
8. Refer to Table A3.
We don’t have df=58,
so we use df=40 instead.
Calculated t = 1.870 < 2.02
If t=2.02, p=0.05
Therefore if t=1.870, p>0.05.
10. Summary
drtamil@gmail.com 201510
Hypothesis – those treated with fish oil
supplement have a higher level of HDL compared
to control group.
Null hypothesis
No difference of mean of HDL level between
treatment and control groups, or;
There is no association between fish oil treatment
and HDL level.
Suitable statistical test – Student’s t-test
(compare mean between 2 groups, n=60-assume
normally distributed data).
11. Summary
drtamil@gmail.com 201511
Calculation;
t = (1.4767 – 1.2593)_____
(0.455492/30 + 0.444882/30)0.5
= 0.2174/0.1162=1.870
df = (30-1)+(30-1) = 58
p > 0.05
Null hypothesis NOT REJECTED because p >
0.05
Conclusion:
Although the treated group HDL level is slightly
higher (1.48+0.46) compared to the control
group (1.26+0.44), the difference is not
statistically significant (p>0.05); or
There is no association between fish oil usage
and HDL level.