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• Level: GCSE
• Subject: Maths
• Word count: 7463

Extracts from this document...

Introduction

John Saunders (R)

In this investigation, I will be comparing magazines and newspapers. I will compare things such as the length of words in articles and the number of articles in them, as you will see in my hypotheses.

I will use several different magazines and papers to try and make my final results a little more accurate, and so I can extend on what I can compare, again as you will se in my hypotheses.

Papers/magazines I will be using:

 Newspaper Details Magazine Details Daily Mirror For Thursday June 27th 2002 Shoot Monthly For July 2002 Daily Express For Friday June 21st 2002 What’s on TV For 6-12 July 2002 The TV mag* (Supplement from ‘The Sun’ – for 6th July 2002) F1 Racing For July 2002

*A supplement

Calculations:

Throughout this piece, I will be using some calculations quite a lot, to work out certain things which are vital in working out whether my hypotheses are correct. These include:

1. X = fX ÷f

Mean = The total of (frequency x no. of words) ÷ The total number of words

This equation is for working out the mean average of a frequency distribution. It isn’t possible to work out the mean using the normal method in a frequency distribution, because the norm doesn’t account for all the measures for all the numbers (in other words, no matter what differences there are in the number of letters in each word, the total of all the words taken into account will always equal the number the distributed numbers add up to).

For example, in hypothesis one, no matter how the tally ends up like, whether there are 100 twelve letter words or 100 one letter words, the total number of words used in the normal mean would always have to be taken as 100, because that’s what the distribution would always add up to.

2)        X =

Middle

80

60.5

5

10

14

12

60

72.5

6

5

4

4.5

27

77

7

13

10

11.5

80.5

88.5

8

8

4

6

48

94.5

9

1

3

2

18

96.5

10

3

0

1.5

15

98

11

1

0

0.5

5.5

98.5

12

-

2

1

12

99.5

13

-

1

0.5

6.5

100

∑fX

448.5

Mean = fX ÷f

(Mean = The total of fX ÷ The total number of words)

Mean = 448.5 ÷ 100

Mean = 4.485 letters per word

Standard Deviation of the results:

 No. of letters (X) Average frequency (f) X - X (X – X)² f(X – X)² 1 5 -3.485 12.145225 60.726125 2 15.5 -2.485 6.175225 95.7159875 3 20 -1.485 2.205225 44.1045 4 20 -0.485 0.235225 4.7045 5 12 0.515 0.265225 3.1827 6 4.5 1.515 2.295225 10.3285125 7 11.5 2.515 6.325225 72.7400875 8 6 3.515 12.355225 74.13135 9 2 4.515 20.385225 40.77045 10 1.5 5.515 30.415225 45.6228375 11 0.5 6.515 42.445225 21.2226125 12 1 7.515 56.475225 56.475225 13 0.5 8.515 72.505225 36.2526125 ∑f 100 ∑f(X – X)² 565.9775

S.D. = √ ( ∑f(X – X)² ÷ ∑f )

S.D. = √ 565.9775 ÷ 100

S.D. = √ 5.659775

S.D. = 2.379028163

This means that 67% of all the results are within 2.37 (approx.) letters of the mean, which was 4.485

This shows that the standard deviations of both averages are very similar, as the differences between them is only 0.05 (2.37 – 2.32). This also reflects the small difference between the averages:

4.53 – 4.485 = 0.045 letters per word

As you can see, I proved my hypothesis was correct, although I was expecting a much larger difference between the two mean averages. On the next few pages there are several graphs to represent this data in many different ways, mainly to compare the two.

This also shows that the mean AND standard deviation between magazines and newspapers have a very close relationship with each other. This means that the results were almost identical.

Graph 1.1 (page 10):

This graph is a cumulative frequency graph. As you can see, the graph seems to be correct, because it is in the ‘S’ shape that all cumulative frequency graphs end up being in. The lines are in very similar positions, which reflect on the tiny difference between the averages of the difference in letters per word. The red line that represents magazines, shows there are slightly more words that are between 3-6 letters and 8-10 letters long. After these (7 and 11-12 letters), there are more letters per word, because the lines even out again.

Apart from this, the blue line (representing newspapers), and the red line are virtually identical, meaning there are the same numbers of letters per word up to that point on the graph.

Graph 1.2 (page 11):

This is a line graph representing each individual result, and not the cumulative frequency, and again, the blue line represents newspapers and the red line represents magazines. Unlike the cumulative frequency graph, these results are fairly different in most letter categories, but never the less follow the same trend.

The only two major exceptions to the trend are in the five-letter category - where in the newspapers there are more words than in the four-letter category; whereas in magazines there are fewer words than in the four-letter category - and in the seven-letter category - where in newspapers there are more words than in the six letter category; whereas in magazines there are less words in the six-letter category.

The green line represents the general trend of both graphs. It shows that the general trend is positively skewed, because the mean is greater than the median and the mode – which is telling us that as you go further along the ‘x’ axis, there is a less and less number being represented on the ‘y’ axis. This is why the mean is higher – because the mean is affected by smaller amounts of possible anomalous results, whereas the median and mode aren’t, and because the anomalous results appear to be the higher letter frequencies, then this will make the mean a greater number.

Survey results:

 No. who thought answer was newspapers % who thought answer was newspapers No. who thought answer was magazines % who thought answer was magazines No response Year 7 10 63% 6 37% 0 Year 8 12 86% 2 14% 0 Year 9 10 77% 3 23% 0 OVERALL TOTAL (of 43) 32 74% 11 26% 0

*Percentages to nearest whole number

This shows that the year seven’s had the best idea of what the answer turned out to be, because the answers they gave were the most similar in number (with a range of four), which reflects upon the similarity of the number of letters per word on average.

This also shows that the whole of KS3 would be correct, as the majority (74%) thought that newspapers would have the longer words on average.

Hypothesis two:

(There will be more articles in newspapers than in magazines)

For this, I am going to use a similar method to the one I used in hypothesis one, except I am going to use the first 40 pages from the magazines and newspapers, and I am going to use class intervals of five pages to make my results table easier to understand.

Here are the newspapers and magazines I will be using (details of them can be found on page one):

Newspaper 1:                Daily Express                        Magazine 1:                F1 Magazine

Newspaper 2:                Daily Mirror                        Magazine 2:                Shoot Monthly

From these, I will count the number of articles on each page, and input the data into the relevant box, as a tally at first, but then as overall totals later.

There are only two major problems I see with this. The first one is that if an article begins on one page in one interval and finishing on another page in a different interval. To solve this, I am simply going to say that any article will be classed to be on the page it begins on. For example, if there were a double page spread on pages 20 and 21, then it would go in the 16-20 category because the article starts on page 20.

The second problem is that if the magazines start on page 4 or 5, due to the contents or an introduction etc. at the beginning. To solve this, I am going to simply put a ‘0’ in the 1-5 if this occurs, because if I didn’t, it would make the graphs a lot harder to compare and the tables more confusing.

 Tally of no. of articles in ‘x’ pages Pages Newspaper 1 Total Newspaper 2 Total Magazine 1 Total Magazine 2 Total 1-5 //// /// 8 //// 4 0 / 1 6-10 //// //// / 11 //// //// 10 / 1 /// 3 11-15 //// //// 10 //// / 6 0 /// 3 16-20 /// 3 //// //// 10 /// 3 /// 3 21-25 //// 5 /// 3 //// //// 10 //// 4 26-30 //// //// 9 //// /// 8 // 2 // 2 31-35 //// 5 //// 4 / 1 / 1 36-40 //// 5 // 2 / 1 // 2 Σf 56 47 18 19

As you can see, it doesn’t take too much common sense to realise that hypothesis two is again correct. For the sake of preciseness though, I will work out the mean averages:

Mean average for newspapers:

 Page no’s. Value of X Frequency for newspaper one Frequency for newspaper two Average frequency (f) 1-5 3 8 4 6 6-10 8 11 10 11.5 11-15 13 10 6 8 16-20 18 3 10 6.5 21-25 23 5 3 4 26-30 28 9 8 8.5 31-35 33 5 4 4.5 36-40 38 5 2 3.5 ∑f 52.5

X = X ÷ N

(Mean = The total of the single frequencies ÷ the number of frequencies)

Mean = 52.5 ÷ 8

Mean = 6.56 articles per interval, which means…6.56 ÷ 5 = 1.31 articles per page

Mean average for magazines:

 Page no’s. Value of X Frequency for newspaper one Frequency for newspaper two Average frequency (f) 1-5 3 0 1 0.5 6-10 8 1 3 2 11-15 13 0 3 1.5 16-20 18 3 3 3 21-25 23 10 4 7 26-30 28 2 2 2 31-35 33 1 1 1 36-40 38 1 2 1.5 ∑f 18.5

X = X ÷ N

(Mean = The total of the single frequencies ÷ the number of frequencies

Mean = 18.5 ÷ 8

Mean = 2.31 articles per interval, which means…2.31 ÷ 5 = 0.46 articles per page

The difference between the no. of articles per page: 1.31 – 0.46 = 0.85

Although this looks as if it was pretty fair, I think that part of the problem with the magazines is that some articles were several pages long, and there were a lot more pictures in them. Never the less, the point of the hypothesis was to find the number of articles despite these kind of problems, so this hypothesis must be correct.

Graph 2.1 (page 16):

This is a graph to show the correlation between the two variables. Unlike in the graphs on pages 6 and 7, these points aren’t joined up together. This is because we are looking to see how well the lines would fit together supposing we put them on a line of best fit. You can see a blue line on this as well – this is a regression line, which is simply a line of best fit.

There is a point on the graph at (6.56, 2.31) that this line goes through, this is the co-ordinate that shows the average frequencies for both the ‘x’ and ‘y’ axis (i.e. 6.36 and 2.31 are the average frequencies of articles in newspapers and magazines per page). The working out for this is on page 15. This has given me some idea of where to place my line of best fit to make it correct. I don’t think it shows the line of best fit too well, but only because the circled anomalous result made the average co-ordinates larger than they would have been supposing it wasn’t there.

Coefficient of rank correlation for newspapers and magazine (for basic details on correlation, see page 1):

 Page no’s. Frequency for newspapers (average frequency) Frequency for magazines (average frequency) Rank for newspapers Rank for magazines Difference (d) d² 1-5 6 0.5 5 8 -3 9 6-10 11.5 2 1 3.5 -2.5 6.25 11-15 8 1.5 3 5.5 -2.5 6.25 16-20 6.5 3 4 2 2 4 21-25 4 7 7 1 6 36 26-30 8.5 2 2 3.5 -1.5 2.25 31-35 4.5 1 6 7 -1 1 36-40 3.5 1.5 8 5.5 2.5 6.25 ∑d² 71

R = 1 - __6∑d²__

n(n² - 1)

R = 1 - __6 x 71__        = 1 – (426 ÷ 504)        = 1 – 0.84        = 0.16

8(64 – 1)

This shows that the correlation between the number of magazine and newspaper articles per page is very low. This is what I was expecting to happen on the graph, as the results seemed to be scattered in no particular order. Especially with taking the anomalous result (circled on graph) into account, the results of this formula seem to have come out right.

Regression line equation:

 Page no’s. Frequency for newspapers (average frequency) Frequency for magazines (average frequency) 1-5 6 0.5 6-10 11.5 2 11-15 8 1.5 16-20 6.5 3 21-25 4 7 26-30 8.5 2 31-35 4.5 1 36-40 3.5 1.5 ∑f 52.5 18.5

Conclusion

Despite these problems, I am satisfied that the results I got from the data were accurate, even to say that some results didn’t match what I thought the results should have been prior to doing the tests.

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