RESULTS OF YEAR EIGHT
STEM AND LEAFS
THE MEAN AND STANDARD DEVIATION
First calculate the mean of each group. This will be done by adding together all of the heights in one group and dividing it by 50 (the sample size). This will be repeated for each of the remaining groups.
This calculation is represented by the following formula.
Next calculate the standard deviation, which is found by taking one group e.g. Year 8 boys and subtracting the mean from one of the scores. The result of this is squared. This is repeated for every score and the mean of the squared differences is worked out. This is the variance. To find the standard deviation you must square root the variance. This calculation is represented by the following formula.
USEFUL BACKGROUND INFO
THE CENTRAL LIMIT THEOREM
If the sample size is large enough then the distribution of the sample means is approximately normal, irrespective of the distribution of the parent population.
The mean of the distribution of the sample means is approximately equal to the parent population.
The variance of the distribution of sample means is approximately the variance of the parent population divided by the sample size.
These approximations get closer as the sample size gets bigger.
These results are known as the Central Limit Theorem.
Symbolically if,
X ~ (unknown)(μ, σ²) then Χ n ~ (μ, σ²/n)
Provided n is sufficiently large. (n ≥ 30 is usually a good size).
Standard error
The standard deviation of the distribution of sample means is called the standard error.
1 s.e = σ²/n (variance of the distribution of x = σ²/n)
In previous calculations I have only worked out the mean x and variance s² of my sample. I cannot calculate confidence intervals for population mean μ because I do not know σ².
Unfortunately s² is not an unbiased estimator of σ² (i.e. the mean of the distribution of s² is not equal to σ²). However
Is an unbiased estimator of population variance, and I can use this as an estimate of σ² when calculating standard error in order to produce confidence intervals for μ.
So in order to calculate the standard errors for each of my groups I must first calculate an estimate for σ², using the above formulae.
CALCULATIONS FOR THE ESTIMATES σ² AND STANDARD ERRORS
I have previously calculated the mean (x) and standard deviation (s²) for each of my groups. I will now calculate an estimate for σ² in order to calculate the standard errors and formulate confidence intervals for each of my groups.
To estimate σ² I will use the previously stated formula.
And then using these estimates for σ² I will calculate the standard errors using the formula.
CONFIDENCE INTERVALS
If we have one sample mean x then P(μ – 1s.e < x < μ + 1s.e), but this can be rearranged
μ – 1s.e < x x < μ + 1s.e
⇒ μ < x – 1s.e μ < x + 1s.e
So P(x – 1s.e < μ < x + 1s.e) = 0.68
I.e. There is a 68% chance that μ lies within ± 1s.e of my x. This is known as a confidence interval for the population mean. In this investigation I am going to calculate confidence intervals for each of the individual groups.
I am going to use 95% confidence intervals so I will have to calculate a z score for 95%. The z score is the number of standard deviations units away from the mean value. Using numerical methods, accurate tables have been constructed for the area under the normal curve. The table gives the area to the left of (or below) any given z- value. The area under the Standard normal curve that is to the left of z is denoted by Ф(z).This is represented by the following diagram.
I will obtain a z score for 95% as follows.
The area to the left of the z = Ф(z) = 0.95 + 0.25
= 0.975
Using the normal distribution table z = 1.96.This will be used in calculating the confidence intervals for each group.
CALCULATIONS OF CONFIDENCE INTERVALS.
Using my previous calculations for the z score for 95%, the mean of each sample and the standard errors. I will now formulate confidence intervals for each of my groups.
YEAR EIGHT BOYS
x = 154.1 s.e = 1.19 z score 95% = 1.96 (this will be used for all the groups)
P(μ – 1.96s.e < x < μ + 1.96s.e) = 0.95
P(x – 1.96s.e < μ < x + 1.96s.e) = 0.95
P(154.1 – 1.96 x 1.19 < μ < 154.1 + 1.96 x 1.19) = 0.95
95% C.I for μ is (151.8 – 156.3)
.
YEAR EIGHT GIRLS
x = 158.3 s.e = 0.94
P(μ – 1.96s.e < x < μ + 1.96s.e) = 0.95
P(x – 1.96s.e < μ < x + 1.96s.e) = 0.95
P(153.8 – 1.96 x 0.94 < μ < 153.8 + 1.96 x 0.94) = 0.95
95% C.I for μ is (156.5 – 160.1)
SIXTH FORM BOYS
x = 176.7 s.e = 1.08
P(μ – 1.96s.e < x < μ + 1.96s.e) = 0.95
P(x – 1.96s.e < μ < x + 1.96s.e) = 0.95
P(176.7 – 1.96 x 1.08 < μ < 176.7 + 1.96 x 1.08) = 0.95
95% C.I for μ is (174.6 – 178.8)
SIXTH FORM GIRLS
x = 161.4 s.e = 0.99
P(μ – 1.96s.e < x < μ + 1.96s.e) = 0.95
P(x – 1.96s.e < μ < x + 1.96s.e) = 0.95
P(161.4 – 1.96 x 0.99 < μ < 161.4 + 1.96 x 0.99) = 0.95
95% C.I for μ is (159.5 – 163.3)
These results are represented by the following graphs.
ANALYSIS OF RESULTS
The first set of box plots shows the 95% confidence intervals for μ for the year eight boys and the girls. They show that the majority of girls and boys in year eight are very similar in height, but girls are more likely to be taller.
The second set of box plots shows the 95% confidence intervals for μ for the sixth form boys and girls. They show that there is a great difference in the heights of the girls and boys, and it is the majority of boys that are taller.
The third set of box plots shows the 95% confidence intervals for μ for the boys in year eight and sixth form. They show that the majority of boys are a lot taller in the sixth form.
The fourth set of box plots shows the 95% confidence intervals for μ for the girls in year 8 and the sixth form the majority of girls in both years are much more similar in height than the boys. The majority of girls in sixth form are more likely to be taller.
CONCLUSION AND ANALYSIS
My results have shown my hypothesis to be correct. They show that:
In year 8 girls and boys have more similarities in height, but girls are more likely to be taller than the boys.
In sixth form there are greater differences between the heights of boys and girls and the boys are more likely to be taller.
There is a much greater difference in the heights of boys between year 8 and sixth form than between the girls.
I believe these results are fairly reliable, as I would say that I obtained my samples of girls and boys in year eight and the sixth form as accurately as I possibly could have given the facilities and amount of time that was available. I also think that using these samples I fairly accurately estimated the confidence intervals for the population mean of each group.
I don’t think there is much I could have done to improve the method of obtaining the sample or the method I used to measure the heights.
The accuracy of my results would improve by using a larger sample size e.g. 100 girls and boys from each year group, according to the central limit theorem. However, this was not possible due to the amount of people available to measure and the amount of time allocated. I could have improved the sample further by taking groups of students from different schools in different areas, this may have given a more accurate representation of the population, as the ranges of heights in different areas for each group may be more varied. However this would have been very difficult to do and would have taken too long, also I don’t think it would have shown any great difference in my findings, as the heights of boys and girls in each group throughout the region are most likely to be fairly similar to those I measured. If I had had more time it would have been interesting to find out where exactly the changes in the heights of boys and girls actually occurs. This could have been done by taking a sample of fifty girls and fifty boys from each of the years in between year eight and the sixth form, and again calculate confidence intervals to see when the boys go from being the same height or shorter than the girls to being much taller than them.
