MGMT2262
Worksheet #1
Solutions
1) If we use one row for each stem unit:
2 | 8 9
3 | 2 8 9
4 | 1 3 5 5 6 6 7 8 8 9 9 9
5 | 1 2 2 3 3 7 8 9
6 | 2 4
7 | 1 3 8
2) Here is the table:
|
Class |
Frequency |
Percent |
Cum. % |
|
$20,000
to less than $30,000 |
2 |
6.67% |
6.67% |
|
$30,000
to less than $40,000 |
3 |
10.00% |
16.67% |
|
$40,000
to less than $50,000 |
12 |
40.00% |
56.67% |
|
$50,000
to less than $60,000 |
8 |
26.67% |
83.33% |
|
$60,000
to less than $70,000 |
2 |
6.67% |
90.00% |
|
$70,000
and over |
3 |
10.00% |
100.00% |
3) Using Excel, here is the histogram (using the frequency column):
4) Once again, using Excel (with the cumulative % column):
5) If we use the descriptive statistics tool in Excel, here is the output:
|
Column1 |
|
|
|
|
|
Mean |
5.0167 |
|
Standard
Error |
0.2160 |
|
Median |
4.9000 |
|
Mode |
4.9000 |
|
Standard
Deviation |
1.1832 |
|
Sample
Variance |
1.4001 |
|
Kurtosis |
0.4048 |
|
Skewness |
0.3832 |
|
Range |
5.0000 |
|
Minimum |
2.8000 |
|
Maximum |
7.8000 |
|
Sum |
150.5000 |
|
Count |
30.0000 |
The output gives the mean as 5.0167 (which translates into $50,167), the standard deviation as 1.1832 ($11,832) and the median as 4.9 ($49,000). You should be able to use your calculator to compute the mean and standard deviation. To compute the median by hand, there are 30 values. To find the position of the median, we multiply 30 by 0.5 to get 15. Since this is an integer, we take the average of the 15th and 16th values which are both 4.9 to derive the same answer as that given by Excel.
6) The coefficient of skewness = 3(5.0617 4.9)/1.1832 = 0.2958. Thus the data is slightly skewed right.
7) For this data, the coefficient of variation is 1.1832/5.0617 = 23.59%. Since this is greater than 20.2%, this companys annual salaries have more variation relative to its mean.
8) To construct the box plot, we need Q1 and Q3. To find Q1, its position is 30(0.25) = 7.5. Since this is not an integer, we round up and take the 8th value which is 4.5. To find Q3, its position is 30(0.75) = 22.5. Since this is not an integer, we round up and take the 23rd value which is 5.7. The IQR = 5.7 4.5 = 1.2. The lower inner fence is 4.5 1.5(1.2) = 4.5 1.8 = 2.7. The upper inner fence is 5.7 + 1.8 = 7.5. There are no outliers on the lower end, but 7.8 is an outlier on the upper end. To determine if 7.8 is a normal outlier or an extreme outlier, we compute the upper outer fence which is 5.7 + 3(1.2) = 5.7 + 3.6 = 9.3. Since 7.8 is outside the inner fence but not the outer fence, it is an outlier, but not an extreme outlier. We would draw the left whisker to the minimum value of 2.8 and the right whisker to the maximum value inside the fence which is 7.3. On the right side of the upper inner fence. we would indicate the outlier of 7.8 by an asterisk.
9) Since there are so many units in the 10s row, we could arbitrarily decide to split the data from 10-11 and then the remainder on the next row:
0 | 6 7 9
1 | 0 0 0 0 1 1 1 1
1 | 4 5 5
2 |
3 | 2
10) Here is the table:
|
Class |
Frequency |
Percent |
Cum. % |
|
6 to 9 |
3 |
20.00% |
20.00% |
|
10 to 13 |
8 |
53.33% |
73.33% |
|
14 to 17 |
3 |
20.00% |
93.33% |
|
30 to 33 |
1 |
6.67% |
100.00% |
11) We use the percent column in creating the polygon in Excel:
Note that we include the classes that are excluded from the frequency table.
12) The mean is 12.1333 and the standard deviation is 6.0459. To find the median, its position is 15(0.5) = 7.5. The median is the 8th value of 11. The coefficient of skewness = 3(12.1333 11)/6.0459 = 0.5624. Note that the mean and standard deviation are pulled straight from my calculator. The data is slightly skewed right.
13) First, we need Q1 and Q3. To find Q1, its position is 15(0.25) = 3.75. Since this is not an integer, we round up and take the 4th value which is 10. To find Q3, its position is 15(0.75) = 11.25. Since this is not an integer, we round up and take the 12th value which is 14. The IQR = 14 - 10 = 4. The lower inner fence is 10 1.5(4) = 10 6 = 4. The upper inner fence is 14 + 6 = 20. There are no outliers on the lower end, but 32 is an outlier on the upper end. To determine if 32 is a normal outlier or an extreme outlier, we compute the upper outer fence which is 14 + 3(4) = 14 + 12 = 26. Since 32 is outside the outer fence, it is an extreme outlier.
14) Since both 10 and 11 occur 4 times, both of these values are modes.
15) The range is 32 6 = 26
16) This table summarizes the results:
|
|
With 32 |
Without 32 |
Difference |
|
Mean |
12.1333 |
10.7143 |
1.419 |
|
Median |
11 |
10.5 |
0.5 |
Note that for the data without 32, the sample size is 14. Thus, the median is the average of the 7th and 8th values of 10 and 11. By computing the difference between the old and new values, we see that the mean is affected more than the median.