Help calculating outliers

The data displayed below shows the shaft diameter for various parts coming off an
assembly line.

a) Using the stem-and-leaf plot below, calculate:
Stem-and-Leaf Display: Diameter
Stem-and-leaf of Diameter N = 15
Leaf Unit = 1.0
5 234
5 5689
6 24
6 567
7 12
7
8
8 9
MEAN = 943/15 = 62.87
_______
MEDIAN = 62
_______
RANGE = 52 to 89
_______
STANDARD DEVIATION = 9.33714

b) Describe the distribution of values within the stem-leaf plot.
The stem and leaf display is unimodel and skewed to right. there is 1 outlier of 89.
b) What is the best measure of centre and spread for the shaft diameters. Explain your you choice.
The best measure of centre and spread is the median. The mean is pulled to right by outlier of 89. The median is more realistic as most numbers are closer to it.

c)
Identify any outlier(s) by performing the appropriate calculations. This is part of question I don not know how to calculate.

d) [FONT=Arial,Bold][FONT=Arial,Bold]Suppose that there is an error in one of our data points. The “89” was actually a “79”. [/FONT][FONT=Arial,Bold][FONT=Arial,Bold][/FONT][/FONT]Indicate whetherthis mistake would make the values in part (a) greater, lesser, or almost unchanged once we correct the data:
Mean would be 62.2 and Standard Deviation would be 7.92104 so the the values would be lessor.

I believe my calculations for the above questions are correct, could you please explain how to calculate the outlier of 89.
[/FONT]

Sue0113 said:

The data displayed below shows the shaft diameter for various parts coming off an
assembly line.

a) Using the stem-and-leaf plot below, calculate:
Stem-and-Leaf Display: Diameter
Stem-and-leaf of Diameter N = 15
Leaf Unit = 1.0
5 234
5 5689
6 24
6 567
7 12
7
8
8 9
MEAN = 943/15 = 62.87
_______
MEDIAN = 62
_______
RANGE = 52 to 89
_______
STANDARD DEVIATION = 9.33714

b) Describe the distribution of values within the stem-leaf plot.
The stem and leaf display is unimodel and skewed to right. there is 1 outlier of 89.

If you claimed this, then how? If it is an outlier, it cannot be stated by seeing
its position in the stem-and-leaf plot alone.



b) What is the best measure of centre and spread for the shaft diameters. Explain your you choice.
The best measure of centre and spread is the median. The mean is pulled to right by outlier of 89. The median is more realistic as most numbers are closer to it.

c)
Identify any outlier(s) by performing the appropriate calculations. This is part of question I don not know how to calculate.

d) Suppose that there is an error in one of our data points. The “89” was actually a “79”. Indicate whetherthis mistake would make the values in part (a) greater, lesser, or almost unchanged once we correct the data:
Mean would be 62.2 and Standard Deviation would be 7.92104 so the the values would be lessor.

I believe my calculations for the above questions are correct, could you please explain how to calculate the outlier of 89.

Click to expand...

For c)

1) Determine Q1.

2) Determine Q3.

3) Determine Q3 - Q1.

That is the IQR (interquartile range).

Look at the median (which is also Q2).

4) Determine Q2 - 1.5*IQR (This is the lower fence.)

5) Determine Q2 + 1.5*IQR. (This is the upper fence.)

If any of the data values are equal to or less than the quantity in step 4,
then they are outliers.

If any of the data values are equal to or greater than the quantity in step 5,
then they are outliers.

work for part c

Q1 = 55
Q2 = 62
Q3 = 67
IQR= Q3-Q1 = 67-55 = 12
62-1.5 * 12 = 44
62+1.5*12 = 80

So now I'm confused are 44 and 80 the outliers?

Sue0113 said:

Q1 = 55
Q2 = 62
Q3 = 67
IQR= Q3-Q1 = 67-55 = 12
62-1.5 * 12 = 44
62+1.5*12 = 80

So now I'm confused are 44 and 80 the outliers?

Click to expand...

Sue0113,

no, they are not.

First, you did well in your calculations.


- - - - - - - - - -


Second, 44 is a "lower fence." This indicates that if you have (had had) any data values
that are (were) equal to 44 or less, then those would be outliers.

But, your smallest data value is 52, so you do not have any outliers on the lower end.


- - - - - - - - -


Third, 80 is an "upper fence." This indicates that if you have (had had) any data values
that are (were) equal to 80 or greater, then those would be outliers.

It turns out that you have just one data value that is equal to or greater than 80.

That value is 89.


- - - - - - - - -


Therefore, there is one total outlier, and that outlier is 89.

Thanks

Thanks for explaining that. This adult learning is something else. Coming back to school after
30 years is difficult.