The Relationship Between Price, Date of Release/ Re-Release of a Sample of 52 Randomly Selected Films

To determine a population for my course-work, I am going to use the HMV superstore Internet site to search for my sample of films. This film site is split up into categories corresponding to letters of the alphabet; therefore there are a total of 26 categories as there is 26 letters of the alphabet. Each category contains roughly 500 films giving a total population of 130,00 films. Obviously this population is far too vast, so I what I propose to do is to number each film in a category and then the random number generator function on my calculator to give two random numbers. I then match up the numbers with the corresponding films, and get two films for each category, complete with total running time and year of release. I will therefore end up with a sample of 52 films, fulfilling the criteria of a sample of at least 50 items of single variable data

Hypotheses

* The later the year cheaper the film

* The Price Has a relation between the videos

Using the help of Microsoft Excel I have calculated the prices with those shown below.

Mean

0.5861538

Standard Deviation

4.91910426

st Quartile

5.99

Median

0.99

3rd Quartile

2.99

I have now decided to construct a frequency table so that I can draw a cumulative frequency graph, which will enable me to draw a box and whisker plot, and therefore visually see any or all outliers in my data. Here follows my table:

Class

Frequency

Cumulative Frequency

0=<x<1.99

0

0

2=<x<3.99

4=<x<5.99

6

7

6=<x<7.99

2

9

8=<x<9.99

3

22

0=<x<11.99

8

30

2=<x<13.99

1

41

4=<x<15.99

7

48

6=<x<17.99

2

50

8=<x<19.99

51

20=<x<21.99

0

51

22=<x<23.99

0

51

24=<x<25.99

0

51

26=<x<27.99

0

51

28=<x<29.99

52

Total

52

From my sorted stem and leaf diagram it is possible to calculate median, first and third quartiles of my data, and so will now do so, so that I can compare these results to those obtained from my graph.

First Quartile = 1/4 x 52 + 1/2 = position 13.5

Value = £5.99

Median = 1/2 x 52 + 1/2 = position 26.5

Value = 10.99

Third Quartile = 3/4 x 52 + 1/2 = position 39.5

Value = 12.99

Standard Deviation

In most data sets

· About 2/3 of the values lie within 1 standard deviation of the mean

· About 95% of the values lie within 2 standard deviations of the mean

· About 99.5% lies within 3 standard deviations.

To see if this is so with my data, I will now perform the relevant calculations:

Mean = 10.59

0.59 + 4.92 = 15.51

0.59 - 4.92 = 5.67

Therefore about 2/3 of my data should lie between £5.67 and £15.51 that is, about 38 values should be within this range. About 40 values are in this range, so this piece of information is pretty relevant for my information.

Mean = 10.59

0.59 + 2(4.92) = 15.51

0.59 - 2(4.92) = 5.67

Therefore about 95% of my data should lie in the range of. About 49 values should be in this range. About 49 values are in this range

Conclusion

The quality of the investigation could be improved by using a more varied population; in an ideal world this would be an Internet site of every film ever made. As this does not exist, the site that contains the most comprehensive list of films will have to do. Which in this case is the site that I used, the HMV superstore site. My method of collection would not vary even if I used another site. If I used three randomly picked numbers for each category, then I would end up with a sample of 78 films, which I think is too much to produce an accurate report on.

I think that this investigation was a success, as it helped me to gain some information that will help me in another topic, and because it was completed as accurately as I could.