# Maths IQ correlations

Maths Coursework Data

Handling Task

Statistics Data

In this piece of coursework I have been set the task to find out about the students in our school.

I need to prove the following hypothesis:

'Pupils in Band A perform better than pupils in Band B'

I must suggest whether the hypothesis is correct or incorrect. I will do this by comparing Band A with Band B in the following areas:

* Mean averages from key stage three- (level tiers 3-5, 4-6, 5-7)

* Range of scores

* Modal and median of the scores.

There are many different techniques and methods which I can use to solve my above problem. I will use some techniques which will enable me to work out means, ranges, modes and medians of scores. To help me with my work I could also use cumulative frequency graphs, box plots and interquartile ranges. These will all help me to compare the differences between the two bands from looking at their scores.

I am going to focus on 96 pieces of data (pupils) which I shall be analysing the levels and scores of both bands A and B.

The first thing that I am going to do is, to compare the overall results from both bands regarding their scores from a maths SATS paper. This would involve me using a process called stratified sampling. This basically involves me reducing the amount of data that I need to compare. This method is seen as time consuming on a very large scale of data, which is handy for me. To compare the results I will need to sample the data. I am aiming to have a stratified sample size of 30 as it is nearly a third of my total data. Stratified sampling ensures that a fair proportion of pupils are chosen from both bands.

I will use the maths levels data first, so now I need to find out how many pupils there are in each band.

Band A = 42

Band B = 54

Then, I need to find out exactly how many pupils from both Bands A and B that were entered into certain math tiers. They range from tier levels 3-5, 4-6 and 5-7.

Maths entry levels

TIERS

3-5

4-6

5-7

Band A

Band B

TIERS

3-5

4-6

5-7

Band A

21

8

3

Band B

21

6

7

As all the pupils are doing different tiers to one another I need to use a certain equation to find out my stratified sample size of 30 pieces of data. This equation will help me work out how many students I need to choose from each tier of each band.

The equation is as follows: number of pupils in tier * 30 (sample size)

Total number of pupils

Band A

Tier 3-5 = 21 * 30 = 6.5625 7 pupils

96

Tier 4-6 = 8 * 30 = 2.5 3 pupils

96

Tier 5-7 = 13 * 30 = 2.5 4 pupils

96

Band B

Tier 3-5 = 21 * 30 = 6.5625 7 pupils

96

Tier 4-6 = 16 * 30 = 5 5 pupils

96

Tier 5-7 = 17 * 30 = 5.3125 5 pupils

96

The sum of my entire stratified sample sizes adds up to make a total of 31 for my sample size. My sample size has gone slightly above of what it should be due to the fact that some working outs required rounding off, which may give me results that are not totally accurate.

7+ 3+ 4+ 7+ 5+ 5= 31

I will use a method called systematic sampling when choosing my sample data. One way in which I will do it is by picking the first student then every third one.

Band A

Tier 3-5 - Pupils chosen= 100, 118, 128, 136, 147, 172, 185

Tier 4-6 - Pupils chosen= 101, 134, 174

Tier 5-7 - Pupils chosen= 98, 130, 142, 163

Band A

Tier 3-5 - Pupils chosen= 97, 107, 153, 158, 168, 181, 191

Tier 4-6 - Pupils chosen= 106, 122, 133, 150, 189

Tier 5-7 - Pupils chosen= 105, 114, 120, 140, 170

I have now picked my sample data from each of the tiers. I will need to record the levels. After this I will be able to find the mean, range, mode and median of Band A where I can then compare and analyse it with Band B.

I am totally aware that I can compare the marks for these students although there are issues regarding the different levels. I.e. a high mark on an easier paper still can give a lower level than a low score on a higher tier paper.

Band A

Tier 3-5 = 4, 3, 4, 3, 4, 5, 5 mean for band A= 68/14 = 4. 86 (2dp)

Tier 4-6 = 6, 5, 5 range = 7 - 3 = 4

Tier 4-6 = 5, 6, 6, 7 mode = 5

Median = 14+1 = 7.5th = 5 is the median

2

The data in order is: 3, 3, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 7 = the value is in between the two highlighted values, therefore it becomes a 5 = the median.

Lower quartile = 1/4 * 14 = 3.5th = 4

Upper quartile = 3/4 * 14 = 10.5th = 5.5

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