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

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The Relationship Between Price, Date of Release/ Re-Release of a Sample of 52 Randomly Selected Films The title of my investigation is 'The Relationship Between Price, Date of Release/ Re-Release of a Sample of 52 Randomly Selected Films. As you can interpret from this title, I am going to investigate to see if the Price and Release/ Re-Release a set of selected feature length films is effected by I will also be seeing if there is a relationship between the length of the title and the film and also to see if modern films are longer than earlier ones. I have chosen to investigate into this topic because I am interested in Films and I am currently studying Media, And I wish to find out if an audience after they have finished watching a film how popular the film is on video, In short, my aim for this investigation is to find if here is a relationship between the price of a video and the time it was released / re-released. ...read more.


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 10.5861538 Standard Deviation 4.91910426 1st Quartile 5.99 Median 10.99 3rd Quartile 12.99 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 1 1 4=<x<5.99 16 17 6=<x<7.99 2 19 8=<x<9.99 3 22 10=<x<11.99 8 30 12=<x<13.99 11 41 14=<x<15.99 7 48 16=<x<17.99 2 50 18=<x<19.99 1 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 1 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. ...read more.


Mean = 10.59 10.59 + 2(4.92) = 15.51 10.59 - 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. ...read more.

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