Statistics Assignment: Probability and Sample Size Analysis, 2019

This assignment focuses on analyzing the impact of sample size on probability calculations, using a real-world scenario involving undergraduate student salaries. The assignment begins by applying the central limit theorem to calculate the probability of a sample mean being less than a certain value, first with a sample size of 50 students, and then with a sample size of 5 students. The solution provides a step-by-step calculation of the z-score and the left-tailed p-value for each scenario, highlighting how the change in sample size affects the probability. The assignment concludes with a comparison of the probabilities obtained for the two different sample sizes, demonstrating the inverse relationship between sample size and the standard error of the mean. The student demonstrates a clear understanding of statistical concepts and their application to problem-solving.

Statistics
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12th July 2019

Question 1
A population of first year undergraduate students working part time has a mean salary of £15,572
with a standard deviation of £3,150. If a sample of 50 students is taken, what is the probability
that the mean of their salaries will be less than £15,000? How would this probability change, if
the sample had only 5 students?
Answer
According to the central limit theorem (CLM), any sampling distribution of a given sample
means tends to approach a normal distribution as the sample size increases or rather gets larger.
This is regardless of what the shape of the population distribution. This particular fact holds true
especially for sample sizes that above 30.
To answer the above given problem, we apply the concept of central limit theorem (CLM) to
analyze the given problem.
We are suppose to get the probability that the mean of their salaries will be less than £15,000 for
a population of first year undergraduate students working part time has a mean salary of £15,572
with a standard deviation of £3,150 having taken a sample of 50 students from it. So we have;
μ=15572, σ =3150 , n=50
We seek to find;
P ( meanof their salaries will be less than £ 15,000 ) =P( X <15000)
Clearly, the above problem is a left-tailed problem and we will be seeking to find a left-tailed
probability.
We obtain the z score first;
Z= X−μ
SE
SE= σ
√ n
We have;
X =15000 , μ=15572 , σ=3150 , n=50
SE= σ
√n = 3150
√50 = 3150
7.07106 =445.4773
Z= X−μ
SE = X −μ
σ
√ n
=15000−15572
445.4773 = −572
445.4773 =−1.28402
We need to find a Left-tailed p-value:

P ( X <15000 ) =P ( Z<−1.28402 ) =0.0996
Therefore, the required probability in this case where we have a sample of 50 students is 0.0996.
In the second instance, we obtain the probability from the same population but in this case we
now just consider a sample of 5 students.
Again, we try and apply the concept of central limit theorem (CLM) to analyze the given
problem.
We are suppose to get the probability that the mean of their salaries will be less than £15,000 for
a population of first year undergraduate students working part time has a mean salary of £15,572
with a standard deviation of £3,150 having taken a sample of 50 students from it. So we have;
μ=15572, σ =3150 , n=5
We seek to find;
P ( meanof their salaries will be less than£ 15,000 )=P( X <15000)
Clearly, the above problem is a left-tailed problem and we will be seeking to find a left-tailed
probability.
We obtain the z score first;
Z= X−μ
SE
SE= σ
√ n
We have;
X =15000 , μ=15572 , σ=3150 , n=5
SE= σ
√ n = 3150
√ 5 = 3150
2.2361 =1,408.7228
Z= X−μ
SE = X −μ
σ
√n
=15000−15572
1,408.7228 = −572
1,408.7228 =−0.4060
We need to find a Left-tailed p-value:
P ( X <15000 ) =P ( Z<−0.4060 )=0.342 4
Therefore, the required probability in this case where we have a sample of 5 students only is
0.3424.
The table below summarizes the probabilities based on the two different sample sizes;
Sample size Probability

50 0.0996
5 0.3424
As can be seen, the probability is highest when a small sample size is used and smallest when a
large sample size is used. This is because a large sample reduces the standard error of the mean.