Question

An HR manager of a company finds that teenagers frequently change jobs. The dissatisfaction with their present jobs is a major factor in the decision they make. Thus, she selects a sample of interviews of 15 teenagers from the past six months. She records the number of months the teenagers spent on their previous jobs：12 5 1 6 20 24 16 7 11 8 23 19 25 14 4

a. Calculate the range of months that the teenagers spent on their jobs with some description

b. Calculate the median months that each spent at their previous job with some explanation

c. Calculate the interquartile range for the months each teenager spent at his or her previous job with some description.

d. Construct a grouped data frequency distribution for the months the teenagers spent at their previous job. Briefly explain.

e. Suppose the HR manager decides to employ the teenagers who worked longer than 90th percentiles of months from her sample. Determine the minimum number of months each teenager should have worked to gain employment in this company

Answer 1

(a)

Range is defined as the difference between maximum and minimum values.

Here,

Minimum observed months = 1

Maximum observed months = 25

So, range = Maximum - Minimum = 25-1 = 24

(b)

Median is the value which divides whole ordered data set into two equal halves.

Rearranging the given data in increasing order we have,

1,    4,    5,    6,    7,    8,    11,    12,    14,    16,    19,    20,    23,    24,    25

There are 15 sample data in total.

As 15 is an odd number, the (15+1)/2 th i.e. 8 th observation divides the whole ordered data set into two equal halves.

Hence, median = 12

(c)

Interquartile range = Third quartile - First quartile

We found that, second quartile = median = 12 is the 8 th observation

This 8 th observation divides the data set into two equal halves. First quartile is the median of the first half while third quartile is the median of the second half.

As 8 is an even number, average of 8/2 th i.e. 4 th and 8/2+1 th i.e. 5 th observation is the value of first quartile.

So, first quartile = (6+7)/2 = 6.5

As 8 is an even number, average of (8-1)+8/2 th i.e. 11 th and (8-1)+8/2+1 th i.e. 12 th observation is the value of third quartile.

So, third quartile = (19+20)/2 = 19.5

Hence, interquartile range = 19.5-6.5 = 13

(d)

We have range 24. So, we can create 5 groups with width 5 as follows.

Class limit	Class boundary	Frequency
1-5	0.5-5.5	3
6-10	5.5-10.5	3
11-15	10.5-15.5	3
16-20	15.5-20.5	3
21-25	20.5-25.5	3
Total	---	15

(e)

We have sample values. But we do not know population standard deviation (or variance). So, we have to use t-distribution.

Suppose, random variable X denotes number of months the teenagers spent on their previous jobs.

Corresponding statistic is given by

Here,

Number of observation

Sample mean is given by

Sample standard deviation is given by

Degrees of freedom

We know,

[Using R-code 'qt(0.90,14)']

Hence, 10.27178 is the minimum number of months each teenager should have worked to gain employment in this company.