Instructor: Hello students.
Now we are going to work on problem 1 E.
Again, we are using the same dataset and the same problem
but because it's been a while
I'm just gonna read the problem again.
A study is run to estimate the mean total cholesterol level
in children two to six years of age.
A sample of nine participants is selected
and their total cholesterol levels are measured as follows
185, 225, 240, 196, 175, 180, 194, 147 and 223.
So one is asking us which measures the mean
or median is a better measure of atypical value
and we need to justify our answer.
Now to answer this question what we have to do
is we have to check for outliers.
So how do we do that?
To find out if we have any outliers or not
we have to calculate the tukey fences
for upper and lower limits.
And here we are going to use some of the values
that we calculated before.
So basically for the lower limit of the tukey fences
we will be using the formula.
And again I always encourage students
to write the formula first.
So it's gonna be Q1 minus 1.5, and Q3 minus Q1.
Now please keep in mind that another name
for Q3 minus Q1 is the interquartile range.
So here what is our Q1?
We have already calculated it, so it is going to
be 177.5, and this is the difference.
So it's going to be 224 minus 177.5.
So 177.5 minus 69.75.
And this will give us the lower limit,
which is going to be 107.75.
And if we do round it up, it's going to be 107.8.
Okay, simple.
So now that is our lower limit.
We need to calculate the upper limit.
So basically the upper limit, it's very similar
except the formula is a little bit different.
For the upper limit we are going to use this formula.
It's Q3 plus 1.5, multiplied by Q3 minus Q1.
Now as I said before, Q3 minus Q1 has a name.
And what is that name?
That name is called IQR.
So this is called IQR, our interquartile range.
Now as you can see, looking at this formula, this part
we have already calculated that from here, right?
The only thing that's gonna be different here
now we are going to add this part instead of deducting it.
So let's do it.
We don't need to recalculate that part
as we have already done it.
So here the Q3 is again
we calculated it before is 224.
And now instead of deducting we are going to add 69.75.
So when we add 224 plus 69.75, what do we get?
We get 293.75 and we round it up.
It ends up being 293.8.
Now there are no outliers. Why?
Because if you look at our ordered set
and we have done that previously, none of our values
are below 107.8 or above 93.8.
So the best measure here will be mean
because we have no outliers.
The best measure is mean.
Okay?
So the best measure of a typical value
in absence of any outliers will be mean.
Okay, now we are going to move to the next one.