Essentials of Modern Business Statistics with Microsoft Office Excel (Book Only)
Essentials of Modern Business Statistics with Microsoft Office Excel (Book Only)
7th Edition
ISBN: 9781337298353
Author: David R. Anderson, Dennis J. Sweeney, Thomas A. Williams
Publisher: South-Western College Pub
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Chapter 3, Problem 62SE

The average number of times Americans dine out in a week fell from 4.0 in 2008 to 3.8 in 2012 (Zagat.com, April 1, 2012). The number of times a sample of 20 families dined out last week provides the following data. Chapter 3, Problem 62SE, The average number of times Americans dine out in a week fell from 4.0 in 2008 to 3.8 in 2012
a.Compute the mean and median.
b.Compute the first and third quartiles.
c.Compute the range and interquartile range.
d.Compute the variance and standard deviation.
e.The skewness measure for these data is 0.34. Comment on the shape of this distribution. Is it the shape you would expect? Why or why not?
f.Do the data contain outliers?

Expert Solution
Check Mark
To determine

(a)

To find the mean and median of the given data.

Answer to Problem 62SE

Mean = 3

Median = 3

Explanation of Solution

Given:

The data:

6153730313
4124105631

Formula used:

  Mean(x¯)=xinMedian =  ( n 2 ) th term+ ( n 2 +1 ) thterm2

Calculation:

  Mean = 59202.95

Now, let's calculate median,

In order to calculate we need to first arrange the data in ascending order

0011111233
3334455667

  Median =  ( 20 2 ) thterm+ ( 20 2 +1 ) thterm2 10 thterm+ 11 th term23+323

Conclusion:

The mean and median for the given data is 3.

Expert Solution
Check Mark
To determine

(b)

To find the first and third quartile of the given data.

Answer to Problem 62SE

  Q1=1Q3=4.75

Explanation of Solution

Given:

The data:

6153730313
4124105631

Formula used:

  Q1=[14( n+1)]th termQ3=[34( n+1)]th term

Calculation:

We need to first arrange the data in ascending order;

0011111233
3334455667

  Q1=[14( n+1)]th term[14( 20+1)]th term5.25thterm5th term+(6 th term5 th term)×141+(11)×141

  Q1=[34( n+1)]th term[34( 20+1)]th term15.75thterm15th term+( 16 th term 15 th term)×344+(54)×344.75

Conclusion:

The value of first quartile is 1 and third quartile is 4.75.

Expert Solution
Check Mark
To determine

(c)

To find the value of Range and Interquartile Range.

Answer to Problem 62SE

  Range = 7Interquartile Range = 3.75

Explanation of Solution

Given:

The data:

6153730313
4124105631

Formula used:

  Q1=[14( n+1)]th termQ3=[34( n+1)]th term

Calculation:

We need to first arrange the data in ascending order;

0011111233
3334455667

  Range = Highest Value - Lowest Value7 - 07

  Interquartile Range = Q3Q14.7513.75

Conclusion:

The value of range is 7 and interquartile range = 3.75.

Expert Solution
Check Mark
To determine

(d)

To find the value of variance and Standard deviation.

Answer to Problem 62SE

  StandardDeviation = 2.09Variance = 4.366

Explanation of Solution

Given:

The data:

6153730313
4124105631

Formula used:

  StandardDeviation= ( x i x ¯ ) 2 n1Variance =  ( x i x ¯ )2n1

Calculation:

x(xix¯)(xix¯)2
0-2.958.7025
0-2.958.7025
1-1.953.8025
1-1.953.8025
1-1.953.8025
1-1.953.8025
1-1.953.8025
2-0.950.9025
30.050.0025
30.050.0025
30.050.0025
30.050.0025
30.050.0025
41.051.1025
41.051.1025
52.054.2025
52.054.2025
63.059.3025
63.059.3025
74.0516.4025
Sum590.0082.95
Mean2.95

  Standard Deviation =  82.95 192.09

  Variance = (StandardDeviation)22.0924.366

Conclusion:

The value of Standard Deviation is 2.09 and Variance is 4.366.

Expert Solution
Check Mark
To determine

(e)

To identify the shape of the distribution when skewness is equal to 0.34.

Answer to Problem 62SE

Left Skewed or Negatively Skewed

Explanation of Solution

Given:

The data:

6153730313
4124105631

It has been given that the value of skewness is 0.34 and as we have calculated the value of mean and median where the value of median is 3 and mean is 2.95, When the median is greater than mean then the shape of the distribution is negatively skewed or left skewed.

Conclusion:

The shape of the given data is negatively skewed or left skewed.

Expert Solution
Check Mark
To determine

(f)

To identify whether there is any outlier in the given data or not.

Answer to Problem 62SE

No outliers exist

Explanation of Solution

Given:

The data:

6153730313
4124105631

Formula used:

  Outlier<Q11.5(IQR)Outlier>Q3+1.5(IQR)

  Q11.5(IQR)11.5(3.75)4.625

  Q3+1.5(IQR)4.75+1.5(3.75)10.375

Since in the given data there is no value which is less than -4.625 and greater than 10.375. Therefore, there is no outliers exists in the given data set.

Conclusion:

The data contains no outliers.

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Chapter 3 Solutions

Essentials of Modern Business Statistics with Microsoft Office Excel (Book Only)

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