Probability and Statistics (Coursework Sample)

Probability and Statistics
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September 2014
Part One
Confidence Interval
Confidence interval is used in Statistic to indicate the precision and accuracy of a measurement. Once applied in measurements, it shows the estimation reliability. Thus, identifying the closeness of a measurement to the estimation made (Singh & Masuku, 2012).
In this case, the teacher seeks to determine the mathematical ability of graduating high school seniors in her state. From a random sample of 15 seniors, she found a;
X= 1270 →Mean (Sample Mean)
σ = 160 →Standard Deviation
n = 15 →Sample size
The requirement here is to obtain an array of scores for the 15 seniors, where 95% will almost certainly fall, considering the Sample Mean of 1270 and the SD of 160.
Singh & Masuku, (2012) in their statistical tests used the formula;
Confidence Interval = x ± a/2 (σ/‚n)
Where x = Sample Mean
n = Sample size
σ = Standard Deviation
a = [1 – confidence level/100] ………………….1- 95/100 = 1 - 0.95 = 0.05
ʀ = Value of t as indicated in the t table …………..[0.05/2 = 2.145]
Replacing the values;
Confidence Interval = …………………………... [1270 ± 2.145(160/3.873)]
= ………………………….. [1270 ± 88.61]
= …………………………. [1181.39 to 1358.61]
=……… You get the lower and upper intervals
Result: The teacher established that 95% of the scores remained in the range of 1181.39 to 1358.61
This deduces that the teacher was 95% confident that the mathematical ability for the graduating seniors lay in this range when given the 32-item test.
Part Two
The Central Limit Theorem
This theorem states that,’ in statistics, a sampling distribution will tend to normality when an attempt is made to increase the sample size.’ Distributions tend to be closer to a normal distribution. The shape of a sample distribution approaches the normal distribution or "bell curve" shape as the sample ‘n’ increases’ as shown in figure 1.
Wooldridge, (2009) gives an illustration of one throwing a die; there is an equal chance that any of the numbers 1 to 6 would be obtained and the result could as illustrated in the left topmost square of Figure 1.
Example 1
If two dice are thrown, the range of values would be from 2 to 12 i.e., there are 11 possible outcomes instead of just 6 and the shape approaches a cone as depicted in the middle square on the left. This is because the probability of obtaining the center values is higher than getting the two extreme values 2 and 12.
With 6 dice, the range of possible values would be from 6 to 36 or 31 possible outcomes and the shape of the probability distribution now becomes that of a bell as shown in the lower right square of the diagram below. The center values would have more chances of emerging than the values at both ends of the range. In other words, there are more possible combinations of values that would give a sum of 18 from the 6 dice than a sum of 36 or sum of 6. Figure SEQ Figure \* ARABIC 1 Possible Outcomes
Example 2
Let a random variable X with the random sample say x1, x2… xn, be normally distributed with the expected value µ and variance σ2. The sample average will be given by:
17208526225500Sn=x1+x2+…+xn
n
As n approaches infinity, Sn approaches µ which therefore implies that: Lim Sn - µ=0 hence the sample mean converges to 0
n tending to infinity
The variance of the mean is given by:
170624538100081175131750077660533655000Var (Sn) = Var (x1+x2+…+xn) …...