GCSE Maths Statistics Coursework

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Maths Statistics Coursework

1st of February 2006

GCSE Mathematics Coursework- Data Handling

Hypothesis:

Children between Year 7-11 change physically and mentally as they mature.

Aims:

To investigate my hypothesis I have set some aims:

  • To investigate the height change between year groups
  • To observe correlation between IQ and age
  • To investigate changes in favourite TV genre choice

Introduction:

I have chosen to investigate this hypothesis for my coursework because I think it will be interesting and lead to many fascinating questions and hopefully answers.

Method:

To investigate my aims I intend to take a sample of the data.

I will take a stratified sample of the data taking a proportion of boys and girls from every year group. To take a stratified sample I will divide the whole population of children into year group and gender e.g. Year 7 Boys. I will then take 10% of each stratum to gain fair representation of each stratum.

 

Having taken this sample I will attempt to answer aim 1. I will group together the height data within the stratum creating a frequency table. From this, I will plot a cumulative frequency graph. After this, I will find the median, the upper and lower quartiles and create a “box and whisker” plot. I will do this to each strata, place all 5 “box and whisker” diagrams on the same page and compare them. I will use “box and whisker” diagrams as they should be able to see the jump in heights between the year groups clearly. I will also be able to asses the skew of the data. I will do this for boys and girls separately as they have their growth spurts at different times. I will take that standard deviation of the children’s heights as this will allow me to see how far spread the children’s heights are in relation to the mean average.

From this information, I should be able to infer a judgment on some of the physical changes of children between these ages.

From my stratified sample, I will answer aim 2 by plotting a scatter diagram showing IQ against age this will allow a direct comparison of the two variables. I will plot the whole population sample on one graph to show any correlation between the two variables. If the variables show correlation, I will plot a line of best fit using the mean average of the data. From this information, I should be able to infer a judgment on some of the mental changes that take place during this time of children’s development.


To answer aim 3 I will use the stratified sample. I will rank the students favourite television genre with the most popular being one. I will then use a technique to see how closely the children (of different ages) choices correlate. To do this I will use Spearman’s rank:

rs= 1-( 6 ∑ d2 / n (n2-1)

d = difference between ranks

n = number of pieces of data

This equation will give a number between -1 and 1. The number line below shows the degrees of correlation.

-1                                        0                                        1

From this information, I should be able to infer a judgement on the change in maturity of the children between these ages.


Results for Aim 1:

I have placed the children’s heights into these categories and calculated the cumulative frequency.


I have plotted the cumulative frequency of all the boys and all the girls on to two graphs.

Now I am going to plot the cumulative frequency of each stratum on to graph paper separately, and then plot a “box and whisker” diagram from each graph before imposing these “box and whisker” diagrams on to one page for comparison.

Join now!

I am going to calculate the standard deviation of each stratums height. The formula for calculating the standard deviation is:

Standard deviation = √ {(∑fx2/∑f) – Mean2}

First, I will calculate the boys’ height deviations.

Mean= ∑fx/∑f = 23.625÷15 = 1.575

Variance = (∑fx2/∑f) – Mean2 = 37.299÷15 – 2.481=0.006

Standard deviation = √ Variance = 0.080

Mean= ∑fx/∑f = 23.375÷15 =1.558

Variance = (∑fx2/∑f) – Mean2 = 36.639÷15 - 2.428 = 0.0146

Standard deviation = √ Variance = 0.121


Mean= ∑fx/∑f = 19.8 ÷ 12 = 1.65

Variance ...

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