MGMT2262 Tutorial sheet #1 Solutions
1) We look in the cumulative relative frequency column for 30 to under 40 to find the percentage who are under 40 which is 70.24%
2) P(at least 20) = 100% - P(Under 20) = 100% - 3.4% = 96.6%
3) There are 744 clients between the ages of 20 and 30 and 1087 clients between the ages of 30 and 40. So, there are 1831 clients between the ages of 20 and 40. Since there are 2739 clients in total, the percentage between 20 and 40 is 1831/2739 = 66.85%.
4) Since
the median is the 50th percentile, we look to see in which the class
the cumulative relative frequency is more than 50%. We see that 30.56% of the
clients are under the age of 30, but 70.24% are under the age of 40. That means
the median is between the ages of 30 and 40. So we would find the median in the
30-40 age class.
5) To
do this using Excel, I would recreate this frequency table in Excel with the
class in column A and the frequency in column B, with the titles in row 1. To
create the histogram, click the graph icon and choose the Column type and click
Next.
·
The data range is B2:B8.
·
I click the Series tab. The Category (X) axis label is
A2:A8. I click Next.
·
Under the Titles tab, I enter “Ages of Outreach
Clients” as the chart title.
·
I click the Legends tab and uncheck the Show legend
box. I click Finish.
After the graph is created, I can adjust the size of the graph. I can also adjust the point size of the labels by right clicking on them and choosing the Format option. Here is what my graph looks like after adjusting it:
6) For each class, we need the midpoints. This Excel spreadsheet summarizes the analysis:
|
Class |
M = Midpoint |
F = Frequency |
M*F |
M2*F |
|
0-13 |
6.5 |
2 |
13 |
84.5 |
|
13-20 |
16.5 |
91 |
1501.5 |
24774.75 |
|
20-30 |
25 |
744 |
18600 |
465000 |
|
30-40 |
35 |
1087 |
38045 |
1331575 |
|
40-50 |
45 |
577 |
25965 |
1168425 |
|
50-60 |
55 |
164 |
9020 |
496100 |
|
60-80 |
70 |
74 |
5180 |
362600 |
|
|
|
2,739 |
98,325 |
3,848,559.25 |
|
|
|
Mean |
35.898 |
|
|
|
|
Std. Dev. |
10.7924 |
|
The second column shows the class midpoints. In the third column, we see the total number of observations is 2,739. To compute the group mean, we multiply the midpoint by the frequency for that class and compute the total which is 98,325. The group mean is then 98,325/2739 = 35.898 after rounding to 4 decimals. To compute the group standard deviation, we first need the sum of M2*F = 3,848,559.25. Then the group standard deviation = sqrt((3,848,559.25 – (2739)(35.898)2)/2738) = 10.7924 after rounding to 4 decimals.
7) Since the number of times a person goes to the movies is a counting number, the data is discrete.
8) Here is the table:
|
Class |
Frequency |
Percent |
Cumulative Percent |
|
1 to 2 |
5 |
25.00% |
25.00% |
|
3 to 4 |
9 |
45.00% |
70.00% |
|
5 to 6 |
4 |
20.00% |
90.00% |
|
7 to 8 |
2 |
10.00% |
100.00% |
9) The percentage who go less than 5
times per month is the same as the percentage who go no more than 4 times per
month which is 70% from the cumulative percent column.
10) From the cumulative percent column, we
see that 25% go no more than twice per month. Then, the percentage who go more
than twice per month is 100% - 25% = 75%.
11) There are 2 approaches. In the first
approach, 9/20 go 3 to 4 times per month and 4/20 go 5 to 6 times per month for
a total of 13/20 = 65% who go at least 3 but no more than 6 times per month. In
the second approach, P(at least 3 but no more than 6) = P(no more than 6) –
P(no more than 2) = 90% - 25% = 65%.
12) Be
sure to be able to get the mean and standard deviation on your calculator. To
get these values from Excel, choose Data Analysis from the Tools menu. (If it
is not there, click Add-Ins and check the Analysis ToolPak box.) From Data
Analysis choose Descriptive Statistics. Suppose the data is in column A from A1
through A20.
1. The
input range is A1:A20.
2. I
choose B1 for the output range.
3. I
check the Summary statistics box and click OK.
Here is the output:
|
Column1 |
|
|
|
|
|
Mean |
3.85 |
|
Standard
Error |
0.424729 |
|
Median |
4 |
|
Mode |
4 |
|
Standard
Deviation |
1.8994459 |
|
Sample
Variance |
3.6078947 |
|
Kurtosis |
-0.151232 |
|
Skewness |
0.494213 |
|
Range |
7 |
|
Minimum |
1 |
|
Maximum |
8 |
|
Sum |
77 |
|
Count |
20 |
From this we see that the mean is 3.85 and the standard deviation is 1.9 if we round to 2 decimals.
13) First we need Q1 and Q3 to determine if there are any outliers. We have n=20. The position of Q1 is 20(0.25) = 5. Since this is an integer, we take the average of the 5th and 6th values which are 2 and 3 respectively. So Q1=2.5. The position of Q3 is 20(0.75) = 15. Since this is an integer, we take the average of the 15th and 16th values which are both 5. So Q3=5. The IQR = 5 – 2.5 = 2.5. So, the lower inner fence = 2.5 – 1.5(2.5) = 2.5 – 3.75 = -1.25. Since the lowest value is 1, there are no outliers on the lower end. The upper inner fence = 5 + 1.5(2.5) = 5 + 3.75 = 8.75. Since the highest value is 8, there are no outliers on the upper end. Finally, we need the median (AKA Q2). The position of Q2 is 20(0.5) = 10. Since this is an integer, we take the average of the 10th and 11th values which are both 4. So Q2=4. When we draw our box, the left corner is at Q1=2.5, the right corner is at Q3=5, and the middle line at Q2=4. The left whisker is drawn to the smallest value inside the inner fence which is 1. The right whisker is drawn to the largest value inside the inner fence which is 8.
14) The coefficient of skewness = 3(3.85 – 4)/1.9 = -0.2368. Since this is a negative value between –1 and 0, the data is slightly skewed left.