Instructor: Hello students.
Welcome to biostat ER example nine,
chapter seven continuation.
In this example, we have already learned how
to perform hypothesis testing
for multi sample independent continuous outcome data.
And for that purpose we used ANOVA,
our analysis of variance.
For this example, I used problem 14 from our textbook.
And for your convenience I will read the problem again
as we are continuing with post hoc analysis.
Suppose a hypertension trial is mounted
and 18 participants are randomly assigned
to one of the comparison treatments.
Each participant takes the assigned medication
and his or her systolic blood pressure
is recorded after six months on the assigned treatment.
The data is shown in table 7-58.
Is there a difference in mean systolic blood pressure
among treatments?
Run the appropriate test at alpha = 0.05.
Now we have performed the initial analysis of variance
and we have found that there is a significant difference
in means systolic blood pressure among treatments,
so now we are going to proceed
with performing post hoc analysis.
And for our post hoc analysis, first we will compare,
the means between the group
that received the standard therapy
and the placebo group.
So in step one we are going to set up the hypothesis
and determine level of significance.
The null here states that there is no difference
between the two means and the alternative states
that there is a difference.
The alpha here is 0.05 as provided.
In step two we will select the appropriate test statistics,
which is again provided here.
In step three we are going
to determine the decision rule.
So the degrees of freedom here
is going to be still 18 - 3 = 15.
And we will reject null if the T calculated
is less than or equal to 2.131,
or if T calculated is greater than or equal to 2.131.
Now in step four we are going to perform
the calculation for the test statistics.
So for the denominator, I'm going to go through
that calculation first step by step.
So here we have to take the square root of MSW,
which is something we are going to get
from the ANOVA table and the MSW stands
for means square within.
And then we are going to multiply that
after we divide the sample size by one,
and add them for those two groups.
So the sample size for the groups
that we are comparing here,
which is the group that received the standard therapy
and the group that received the placebo,
and that is equal to six.
So that is exactly what I've done here.
The MSW is 56.4.
I have inserted that here.
The sample size, the small n here is six
and it's also six here, I've inserted that.
Then in the next step, again,
I'm doing it step by step algebraically.
So we are going to divide one by six,
and we get 0.17.
We divide one by six again we get 0.17.
Then what we do is we add them and we get 0.34.
We multiply that by 56.4, and then we get 19.2.
We take the square root of 19.2 and we get 4.4.
So now this is the denominator
which we have to insert into our actual formula.
So the numerator here is the difference
between the two means, and that is 146.2 - 121,
and then we are going to divide that by 4.4.
So once we do the subtraction, we get 25.2,
we divide that by 4.4, we get 5.7.
So what is the conclusion here?
The conclusion here is
that our T calculated is greater than 2.131.
So we have statistically significant evidence
to reject null hypothesis,
and conclude that there is a significant difference
in the mean systolic blood pressure
between the standard and the placebo groups.
So we have compared two groups right now.
Now we are going to proceed with the other comparisons.
Next we are going to compare the placebo
and the new treatment group.
So again, for step one we will set up the hypothesis
and determine level of significance.
The null hypothesis here again is going to state
that the means are equal.
The alternative is going to state that they are not equal,
And the alpha is still 0.05 as provided.
The test statistics here will remain the same.
It is going to be the T calculated,
the formula is exactly the same
and I have provided you all the information here.
The step three will be to set up the decision rule.
Again, that will remain the same.
The degrees of freedom will be 15,
and we will reject if our T calculated again
is less than or equal to 2.131,
or if our T calculated is greater than or equal to 2.131.
Now in step four we have to calculate our test statistics.
Now given that our sample size was equal
for each three groups, we do not have
to recalculate the denominator,
because the MSW is still the same, which is 56.4
as given here, and the sample size is six
again for each group.
So we don't have to do that calculation again,
we can just basically use 4.4 here.
And what we have to do right now
is insert that into our T statistics calculation.
However, the numerator here will be different,
it's going to be 146.2,
and from that we are going to deduct 122.7
and then we are going to divide it by 4.4.
So once we do the algebra, we get 5.3.
So in step five we are going to draw the conclusion.
So we have found that our T calculated
is actually greater than 2.131,
so we have statistically significant evidence
to reject the null hypothesis
and conclude that there is a significant difference
in the mean systolic blood pressure between the placebo
and the new treatment groups.
Now given we have three groups,
we will do the last comparison,
and that is going to be between the group
that received the standard treatment
and the group that received the new treatment.
So again, in step one we are going to set up the hypothesis
and determine the level of significance.
So here again the null will be stating
that the means are equal,
and the alternative will state
that the means are not equal,
the alpha is still going to be 0.05.
So in step two we are going
to select the appropriate test statistic.
And again, it is the same formula that we used previously,
I have copied and pasted it here.
I have also provided you with the information
regarding the degrees of freedom.
Now step three, again, it will be identical,
the degrees of freedom again is going to be 15
and we will reject the null if our T calculated
is less than or equal to 2.131,
or if our T calculated is greater than or equal to 2.131.
Now, in step four we have to compute the test statistic.
Now, as I mentioned previously,
again the denominator here will remain the same
because our sample size is equal, six per group,
and the MSW will not change,
which we already obtained from the ANOVA table.
So the denominator will still be 4.4,
however, the numerator will change.
So now it will be 122.7,
and from that we are going to deduct 121,
and then we are going to divide it by 4.4.
So once we do the subtraction
and we do the division, we get 0.39.
And from this,
we are going to draw the conclusion
and in step five, when we draw the conclusion,
we realize that now we are failing to reject the null,
because our T calculated, which is 0.39,
is actually less than 2.131.
So we do not have statistically significant evidence
to reject null hypothesis and conclude
that there is a significant difference
in the mean systolic blood pressure
between the standard and new treatment groups.
So thank you for your attention.
I know this was a long problem,
and as such I divided into two videos
and I hope that was helpful,
and please feel free to reach out
if you have any questions.
Thank you, again.