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# My aim is that within the limits of a small-scale survey I will collect sample data of a population, and by using estimation techniques I will determine the population's parameters (such as the mean and the variance).

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Introduction

Mathematics Coursework – Statistics.

Aim.

My aim is that within the limits of a small-scale survey I will collect sample data of a population, and by using estimation techniques I will determine the population’s parameters (such as the mean and the variance). My population is smarties, and in this investigation I am looking at the individual weight of random smarties, which will be my sample. I decided to stick with weight, as it is a property that will vary a lot, I think, and so I hope will prove an interesting investigation. An important factor to help me decide on how large my sample should be is that the size of the sample must be quite small, because it is stated so in my aim. However, to make accurate estimates of population parameters the sample must be large enough.

Therefore to help me decide on the size of my sample, I have accordingly looked at the Central Limit Theorem, which states that:

• If the sample size is large enough, the distribution of the sample mean is approximately Normal.
• The variance of the distribution of the sample mean is equal to the variance of the sample mean divided by the sample size.

The

Middle

1.00

49

0.99

448

0.98

2

0.97

2

0.96

4

0.95

02344555779

0.94

19

0.93

24499

0.92

12

0.91

35

0.90

16

0.89

38

0.88

0.87

0.86

7

Although not necessary, I thought it would be somewhat useful to depict my sample data onto a stem and leaf diagram. Other information about the sample includes the lowest value, which is 0.867g, the highest is 1.110g, and the range is 0.243g.

Sample Parameters. Mean.

Using the total sum of the fifty smarties and dividing it by fifty to obtain the mean. Variance.

The formula for variance states that you take the ‘Mean of the squares minus the square of the mean’. Standard Deviation.

The standard deviation is found by finding the square root of the variance. Population Parameters.

Estimate of the Mean of the population of smarties.

The mean is an unbiased estimator, that is, the mean of its distribution is equal to the mean of the parent population. For this reason it can be used as an estimator for the mean of the population of smarties. As the mean of my sample is 0.976, then an estimate of the mean of the population of smarties is therefore: Estimate of the Variance of the population of smarties.

The variance of the sample is a biased estimator. A biased estimator is one for which the mean of its distribution is not equal to the population value it is estimating. Therefore it must be converted to an unbiased estimator, by multiplying the sample variance by the number of smarties. Conclusion

Possible Extension.

A statistical analysis of entire tubes of smarties could be carried out. The actual weight of the smarties could be compared to the price on the tube to determine whether the manufacturers are lying about how much smartie there is in their packets. Also similar investigations looking at how many smarties per packet, average weights of packets, etc.

Weighing smarties of different colours could also be done to find if there are any differences between them. Or even counting how many smarties of different colours you get in different packets. But yet again an investigation like this would be harder to carry out, as you would need at least fifty packets of smarties to carry out a ‘small scale’ investigation…

Also, a larger sample size could be taken to determine the mean and variance more accurately, a lot more accurately in fact.

Lastly, I could have extended my confidence interval calculations; I could have included a 99% confidence of the mean varying only ± 0.001g, which would have shown I would have needed a massive sample, possibly over 20,000 to get that much confidence in such a small interval.

Charles Mallah         Mathematics Coursework (Statistics)        Deadline 18-03-02

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