APPLIED STAT.IN BUS.+ECONOMICS
APPLIED STAT.IN BUS.+ECONOMICS
6th Edition
ISBN: 9781259957598
Author: DOANE
Publisher: RENT MCG
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Textbook Question
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Chapter 4, Problem 91CE

The self-reported number of hours worked per week by 204 top executives is given below. (a) Estimate the mean, standard deviation, and coefficient of variation using an Excel worksheet to organize your calculations. (b) Do the unequal class sizes hamper your calculations? Why do you suppose that was done?

Chapter 4, Problem 91CE, The self-reported number of hours worked per week by 204 top executives is given below. (a) Estimate

a.

Expert Solution
Check Mark
To determine

Estimate the mean, standard deviation and coefficient of variation from the grouped data and frequencies.

Answer to Problem 91CE

The mean, standard deviation and coefficient of variation from the grouped data and frequencies are:

MeasuresValues
Mean60.196
Standard deviation8.536
Coefficient of variation14.2%

Explanation of Solution

Calculation:

The given information is about the distribution of number of hours worked per week by 204 top executives.

The mean for the grouped data is:

x¯=j=1kfjmjn

The standard deviation for the grouped data is:

s=j=1kfj(mjx¯)n1

Where,

fj is the frequency for jth interval,

mj is the mid-point for jth interval,

n is the total sample size (j=1kfj), when j lies between 1 and k.

If xl and xu are the lower and upper limits of the jth interval, then mj=xl+xu2.

Coefficient of variation:

The coefficient of variation (CV) is a measure of relative variability which is the ratio of the standard deviation to the mean. If the coefficient of variation is high, then level of dispersion around the mean is high. The CV is often expressed as a percentage.

CV=100×sx¯,

Where, s be the sample standard deviation and x¯ be the sample mean.

The table below gives the mean for the grouped data:

From

xl

To

xu

Frequency

fj

Mid-point

mj=xl+xu2

fj×mj(mjx¯)2fj(mjx¯)2
40501240+502=45540(4560.196)2=230.922771.04
506011650+602=556,380(5560.196)2=26.9983131.768
60807460+802=705,180(7060.196)2=96.127112.88
80100280+1002=90180(9060.196)2=888.281776.56
  fj=204 j=14fjmj=12,280 j=14fj(mjx¯)2=14,792.248

The mean for the grouped data is:

Substitute the values fj=204 and j=14fjmj=12,280 to get the mean,

x¯=12,280204=60.196

Thus, the mean from the grouped data and frequencies is 60.196.

The standard deviation for the grouped data is:

Substitute the values fj=204 and j=14fj(mjx¯)2=14,792.248 to get the standard deviation,

s=14,792.2482041=14,792.248203=72.868=8.536

Thus, the standard deviation from the grouped data and frequencies is 8.536.

The coefficient of variation is:

Substitute x¯=60.196,s=8.536,

CV=100×sx¯=100×8.53660.196=100×0.141814.18%

Thus, the CV is 14.18%.

b.

Expert Solution
Check Mark
To determine

State whether the unequal class size hamper the calculations.

Answer to Problem 91CE

No, the unequal class sizes hamper the calculations.

Explanation of Solution

The unequal sample sizes in the observations disturb the calculations. The unequal class sizes means that none of the class size will hold the value zero.

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Chapter 4 Solutions

APPLIED STAT.IN BUS.+ECONOMICS

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