Question #1
a)
The first step in making histogram is to determine the frequency values for each bin. We calculate the number of score values that is equal or below the bin value. For example, for bin with value of 10, we calculate the number of countries with scores less or equal to 10. The following table represents frequency distribution of Donating Money scores, Helping a Stranger scores and Volunteering Time scores:
In order to provide descriptive statistics, use Data Analysis plug-in in Excel:
Frequency histograms are given below on one common graph:
b)
Calculate 1.5*IQR. The calculations of lower and upper limits are given in the table below:
For Donating, a value is considered as an outlier the score is 74 or above. For Helping, outliers are below 14 and above 83. Similarly, for Volunteering, outliers are above 53:
c)
Donating Money scores distribution is positively skewed. The boxplots show the outliers that we have calculated above: values 77, 78 and 91 are extremely high.
Helping score distribution is approximately symmetric and bell-shaped. There are no outliers in this data. This is also confirmed by the boxplots.
Volunteering scores distribution is also positively skewed. There are no outliers in this data. This is also confirmed by the boxplots.
We know that for skewed distributions (Donating and Volunteering) the most appropriate measure of central tendency is median value and the best measure of dispersion is IQR. For symmetric distributions (Helping), the best measure of central tendency is mean value and the best measure of variability is standard deviation.
d)
In the given 135 countries, on average, people are more likely to help a stranger (mean percentage value is 28.77037 with the standard deviation of 19.05097). On average, there are 23% (median value) of people who donate money to a charity (with IQR of 24). The median value of volunteering percent is 19% with IQR of 16%. In Australia, 66% of people donate money, 65% help a stranger and 37% are volunteers. Generally, Australian data is not in a general trend – its indicators are higher than averages.
Question #2
a)
A 100% stacked column chart is prepared according to the given instructions:
b)
The probability that a respondent chooses to participate in a fundraiser is:
P=2761570=0.175796
Approximately 17.58% of all respondents participate in a fundraiser.
c)
This is a conditional probability. It is represented in percentages in the table below (it is marked with yellow color):
In other words, given that the respondent chooses to participate in a fundraiser, the probability that he or she is Generation Y is 0.3977, Generation X – 0.2375, Baby boomers – 0.1292, matures – 0.1082.
d)
Similarly, for Mobile category:
P=1441570=0.09172
Given that the respondent chooses to participate in a mobile, the probability that he or she is Generation Y is 0.3068, Generation X – 0.1313, Baby boomers – 0.0653, matures – 0.012.
e)
Matures are most likely to donate via mail and least likely to give by mobile/text and social networks or through the workplace. Baby boomers also prefer mail most of all, however, they also widely use workplace and website ways. Generation X people usually make donation via workplace, less likely by website, fundraiser and mobile ways and very rarely – by mail. Generation Y people usually make their donations via mobile way, less often by workplace, website or fundraiser ways and almost never by mail.
Generally, baby boomers are most likely to make a donation and Generation Y people are least likely to donate money for a charity.
Question #3
a)
Let the number of radio ads is x1, the number of flyers bundles ads is x2 and the number of daily newspaper ads is x3. These are our decision variables. The objective function is the volume of audience reached through the various media. Hence, we have to maximize the following function:
5000x1+2400x2+2800x3→to max
Constraints are the following:
that the number of radio spots would be capped at 12
x1≤12
the maximum number of bundles of flyers to be printed will be 25
x2≤25
the local newspaper limits the number of advertisements from a single company to 20.
x3≤20
at least five combined print-based advertisements will be placed:
x2+x3≥5
no more than $2,400 be spent on print advertising:
240x2+320x3≤2400
Joey Wombat has budgeted up to $8000 for promotion:
800x1+240x2+320x3≤8000
Also, non-negativity constraints:
x1,x2,x3≥0,
b)
The solution is obtained in .xls file by using Solver.
c)
Shadow price shows that we can increase total audience by 6.25 by increasing total cost available by $1. We cannot increase total audience if change constraint on the number of print advertising. We can increase total audience by 3.75 for each additional dollar, spent on print advertising.
d)
Yes, it will be changed. We can see it in Allowed Increase section. We can see that this constraint may be increased up to 4000 units (exactly to 12000) and it will affect solution. The new objective function value will be initial value (59000) plus 6.25*4000 = 84 000 (new value).
e)
Yes, it will affect the objective function value. We can see that the allowable increase of print ads cost is by 3600 (up to 6000). The new solution will be 73200.
f)
In the initial conditions, Joey should make 7 Radio ads, 10 Flyer ads and no Newspaper ads. This will give him the maximum possible volume of covered audience of 59000. I recommend Joey to increase total cost by $4000 (up to $12,000) and spend $6400 on print advertising. It will increase the volume of audience covered.
Question #4
a)
b)
Interpretation of the slope: a change in number of unique visits by 1000 causes the corresponding change in the number of online orders by 75.2.
Interpretation of the intercept: when there are no unique visits, the number of online-orders is usually 79.5
Interpretation of R-square: approximately 83.88% of variance in on-line orders is explained by this model.
c)
Substitute 80 in the equation:
y=0.0752*80+0.795=6.811
The number of online orders is estimated at the level of 681.1 orders.
d)
The number of unique visits is significantly linearly associated with the number of on-line orders placed. It is appeared that an additional 1000 unique visits causes an increase in the number of orders by 75.2. If all visits during the considered time frame are not unique, the regular level of orders is approximately 79.5.
For example, when 80000 unique visitors were attracted to the website, the estimated number of orders is 681.1
