Data Analysis Exercises

Exhibit 1.1: Frequency Distribution and Graphing with Excel

Figure 1: TechTicket ticket agency sales in the last 240 days

Q1 (b)

1. The shape of the histogram columns is normal distribution with a bell-shaped symmetrical density curve. Thus, the mean of the distribution equals its median (Berenson, 2014). The chart also has one peak at the center.
2. The shape of the graph can be altered by changing the class width. For instance, increasing the class width to 1,500 gives a new graph as shown below. 

Figure 2: New graph after increasing the class width to1,000

Increasing the class width from 500 to 1,000 reduces the number of histogram columns. However, the shape of the distribution does not change; the shape of the histogram columns is normal distribution with a bell-shaped symmetrical density curve, mean of the distribution equals its median and the chart also has one peak at the center.

Q1 (c)

The assumption made here is that the data is normally distributed.  With normal distribution, the data lean towards to a central value with no bias left or right (Gupta, 2016).

To obtain the mean and standard deviation of the data in the Exhibit 1;

Step 1: We first calculate the midpoints of the classes using the formula

Midpoint of interval = 0.5 (Lower class limit + Upper class limit)

Step 2: We multiply the frequency of each interval by its mid-point

Step 3: We obtain the sum of all the frequencies (f) and the sum of all the fx. 

Table 1: Frequency table for Calculation of mean (See excel calculations)

To obtain the mean, we divide ‘sum of fx’ by ‘sum of f’ which in this case is 240 days. Thus the mea of the data in Exhibit 1 is given by

Mean=692,500/240= 2885.416667 ~ $2,885/day

The Standard Deviation of the same data is calculated using the formula  

Table 1: Frequency table for calculation of standard deviation (See excel calculations)

In our case, from Excel (see the attached worksheet), S if given by the

3.66865E+12/240 =123,636.7573~$123,636/day

Q2

From the report, two measures of central tendency could have been conducted; mode and median. This is because using mean would have skewed the results to the middle point of the 7 point scale and in this case the score would have been 4 and not 3.1 as given. For mode, the doctor could have looked at the most repeated score. With median, the doctor could easily picked the value that is at the middle which would have been 4 and not 3.1 as given.

Q3 (a)

Table 3: Tabulations of statistics (Please see the attached Excel doc.)

Q3 (b) (i)

Table 3: A percentage frequency table (Please see the attached Excel doc.)

Q3 (b) (ii)

The average completion time for Sydney factory is 62.4 minutes whereas for Melbourne factory is 67.5 minutes.

Q4 (a)

Minimum-10

Maximum-200

Range-190

Median-45

1^st Quartile (Q1)-30

3^rd Quartile (Q1)-60

IQR-30

Q4 (b)

-The shape of the graph is skewed to the right.

 -The highest sales was realized in 1997 i.e. 3500 which is the highest point of the graph.

-Sales declined in year 2001.

References

Berenson, M. L. (2014). Basic Business Statistics (5th ed.). Pearson Education Limited.

Gupta, B. (2016). Introduction to Basic Statistics. Interview Questions in Business Analytics, 6(2), 23-35. doi:10.1007/978-1-4842-0599-0_3

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