We are asked to determine if the data
contains any outliers.
Notice how we have the number of text messages per day,
and we have a total of 13 data values
which means n is equal to 13.
An outlier is a data value
that is significantly greater than
or less than the other data values.
Before we determine the outliers though,
we do need to discuss the methods used
to determine quartile 1, quartile 3,
and as well as the outliers.
In this video, we'll be using the Moore and McCabe method
to determine the quartiles
where quartile 1 has the median
of the lower half of the data
not including the median,
and quartile 3 is the median of the upper half of the data,
again not including the median.
And then we'll use Tukey's method to determine
the outliers where outliers, according to this method,
are values that are less than quartile 1 minus 1.5
times interquartile range or greater than
quartile 3 plus 1.5 times the interquartile range,
where the interquartile range is Q3 minus Q1.
If we're going back to our data,
again the data has to be ordered from least to greatest
which it already is.
Before we can determine quartile 1 and quartile 3,
we need to determine the median of the data set
which is the data value in the middle,
and because we have an odd number of data values,
the median is a data value in the list.
Notice how 34 is in the middle because
there are six data values to the right,
and six data values to the left of 34
when the data is ordered from least to greatest.
So this is the median.
And now we need to find quartile 1 and quartile 3.
Quartile 1 is the median of the lower half of the data,
and again we don't include the median in the lower half,
and therefore we have six data values, which is even
and therefore we have two middle values.
We'll have to find the mean of the two middle values
to determine the median of the lower half or quartile 1.
Notice if we draw a vertical line here,
there are three data values to the left and right,
and therefore 25,
and in this case 25 are both in the middle.
So the mean of 25 and 25 of course is still 25,
and therefore 25 is quartile 1.
And now we need to find the median
of the upper half of the data.
And again we have six data values,
there are two data values in the middle, 36 and 37.
We need to find the mean of these two data values
which would be 36.5.
The quantity 36 plus 37 divided by two is 36.5,
which is quartile 3.
And now to determine the outliers,
we need to find quartile 1 minus 1.5 times
the interquartile range,
and as well as quartile 3 plus 1.5 times
the interquartile range,
where the interquartile range is equal to
Q3 minus Q1 and in our case,
36.5 minus 25, what is equal to 11.5.
So Q1 minus 1.5 times interquartile range
is 25 minus 1.5 times 11.5,
and Q3 plus 1.5 times interquartile range
is 36.5 plus 1.5 times 11.5.
And now we need to compute these values
which are 7.75 and 53.75.
If there are any data values less than 7.75
or greater than 53.75, they are considered outliers.
We'll notice how we do have a value that is less than 7.75.
We have the data value of 7,
and we also have a data value
greater than 53.75 which is 55.
The given data set has two outliers, 7 and 55.
7 is considered significantly less than
the rest of the data values,
and 55 is considered significantly greater than
the rest of the data values.
I hope you found this helpful.