Answer to Question #252571 in Statistics and Probability for rasha

Question

The following data represent the time (in minutes) recorded for a sample of 20 employees in order to get to work using public transportation. 28 29 32 37 33 25 29 32 41 34 29 31 33 32 34 30 31 32 35 33 1) Does the sample contain any extreme values? Justify your answer with a suitable test. Comment on the results 2) According to your conclusion in part (1), calculate the best central and the best absolute dispersion measure. 3) Comment on the results obtained in part (2). 4) Without any calculation, if all values are multiplied by 10 minutes, will the location measures and outlier existence be the same? Justify your answer.

Answer 1

Least to Greatest Values:

25, 28, 29, 29, 29, 30, 31, 31, 32, 32, 32, 32, 33, 33, 33, 34, 34, 35, 37, 41

1)

extreme values bounds:

high = $Q_3+1.5IQR$

low = $Q_1-1.5IQR$

Q₁=((n+1)/4)th value of the observation

=(21/4)th value of the observation

=(5.25)th value of the observation

=5th observation +0.25[6th-5th]

=29+0.25[30-29]

 =29.25

Q₃=(3(n+1)/4)th value of the observation

=(3⋅21/4)th value of the observation

=(15.75)th value of the observation

=15th observation +0.75[16th-15th]

=33+0.75[34-33]

=33.75

$IQR=Q_3-Q_1=33.75-29.25=4.5$

$Q_3+1.5IQR=33.75+1.5\cdot4.5=40.5$

$Q_1-1.5IQR=29.25-1.5\cdot4.5=22.5$

so, extreme value is 41

2)

best central tendency measure is mean:

$\mu=\sum x_i/n=32$

best absolute dispersion measure is sample standard deviation:

$s=\sqrt{\frac{\sum (x_i-\mu)^2}{n-1}}=3.43$

3)

8 values are less than the mean, 4 values are equal to the mean,

8 values more than the mean

16 values are in the interval of one standard deviation from the mean: $\mu \pm s$

4)

if all values are multiplied by 10, then:

$\mu_1=\sum 10x_i/n=10\sum x_i/n=10\mu =320$

mean increases by 10 times

$s_1=\sqrt{\frac{\sum (10x_i-10\mu)^2}{n-1}}=10\sqrt{\frac{\sum (x_i-\mu)^2}{n-1}}=10s=34.3$

standard deviation increases by 10 times