Question

The following data was recorded for the "Age" of the various participants in a sports activity at a community centre. Age: 20 32 36 37 29 20 27 30 25 37 22 20 20 36 38 32 35 25 24 32 20 27 23 26 28 Use MiniTab to obtain the answers to following questions and draw the required diagrams. a. Find values of Mean, Variance and Standard Deviation using MiniTab. Simply write down the actual formulas for these three quantities. b. Calculate the values (by showing your calculations) of First, Second and Third Quartile and the Inter Quartile Range (IQR). c. Draw the boxplot, histogram, stem and leaf diagrams. Based on the boxplot, what can you say about the data in an approximate way? d. Now calculate the value of the median by simply using the stem and leaf diagram drawn above. It was found that the 14th observation was recorded incorrectly. It should have been '56' instead of 36. In light of this, answer the following by using MiniTab. e. Find values of Mean, Median, Standard Deviation and draw the boxplot for this slightly altered data. Can you see something unusual in the boxplot and comment on the values of Mean, Standard Deviation and Median when you compare them with what you found in part 'a' above?

Answer 1

I am using EXCEL to perform the analysis as I don't have MINITAB.

Functions of Excel used for calculation

The data of Age of 25 participants is stored from B5:B29

mean	=AVERAGE(B5:B29)
variance	=VAR(B5:B29)
Standard deviation	=STDEV.S(B5:B29)
Q1	=QUARTILE(B5:B29,1)
Q2	=MEDIAN(B5:B29)
Q3	=QUARTILE(B5:B29,3)
IQR	=B36-B34

The interquartile range is a measure of where the “middle fifty” is in a data set. The interquartile range formula is the first quartile subtracted from the third quartile: IQR = Q3 - Q1

Count	Age
1	20
2	32
3	36
4	37
5	29
6	20
7	27
8	30
9	25
10	37
11	22
12	20
13	20
14	36
15	38
16	32
17	35
18	25
19	24
20	32
21	20
22	27
23	23
24	26
25	28
mean	28.04
variance	38.04
Standard deviation	6.167658
Q1	23
Q2	27
Q3	32
IQR	9

BOXPLOT: EXCEL> INSERT> boxplot

Boxplot shows that there is no outlier in the data. Data is roughly equally distributed about the mean/ median value. So we can say data is roughly normal and follows the bell curve.

Stem and Leaf Plots

A Stem and Leaf Plot is a special table where each data value is split into a "stem" (the first digit or digits) and a "leaf" (usually the last digit).

STEM	LEAF
2	000002345567789
3	0222566778

Histogram

14th observation in changed from 36 to 56

Count	Age
1	20
2	32
3	36
4	37
5	29
6	20
7	27
8	30
9	25
10	37
11	22
12	20
13	20
14	56
15	38
16	32
17	35
18	25
19	24
20	32
21	20
22	27
23	23
24	26
25	28
mean	28.84
variance	67.30667
Standard deviation	8.204064
Q1	23
Q2(median)	27
Q3	32
IQR	9

Yes due to the introduction of 56 data point, we have an outlier(dot) in the boxplot indicating that this is outside the permissible range of values.

The outlier is calculated by following formula

Lower limit = Q1 - 1.5*IQR = 23 - 1.5*9 = 9.5

Upper Limit = Q3 + 1.5*IQR = 32 + 1.5*9 = 45.5

Since 56 is beyond 45.5, it is an outlier.

Also, standard deviation and variance have increased as due to change from 36 to 56. Mean is almost same as before as mean is not much affected by one data point because it is a measure of central tendency but standard deviation and variance indicate spread so they change considerably due to the inclusion of a larger or smaller value.  

Median remain unaffected as it is also a measure of central tendency.