Quantitative Methods Assignment Solution - Statistics and Probability

This document presents a comprehensive solution to a Quantitative Methods assignment, focusing on statistical concepts. Part A addresses confidence intervals and hypothesis testing, including calculations for a 95% confidence interval, and a t-test to determine if the average volume of wine bottles is exactly 750 ml. The solution also includes the calculation and interpretation of a p-value. Part B provides multiple-choice answers to additional questions related to the concepts in Part A. The solution also discusses the use of estimators, comparing the accuracy of different estimators and considering the implications of the Central Limit Theorem. The document references key statistical concepts and theorems, offering a thorough understanding of the subject matter.

A T TAT M TQU N I IVE E HODS
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Part A
Question 1
(a) 95% confidence interval
Mean (average)
Total observation = 12
Sum of observation = 8965
Average = 8965/12 = 747.083
Standard deviation
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Standard deviation= √ 1
N −1 ¿ ¿
Standard deviation= √ ( 1
12−1 ) (186.91)=4.12
The z value = 1.96 (for 95% confidence interval)
Now,
Upper limit =Mean+ z value∗ ( Standard deviation )
√(Total observation)
¿ 747.083+ {1.96∗
( 4.12
√ ( 12 ) )}=749.4290 ¿ rd error ¿ 95 % confidence itnerval
Lower limit ¿ Mean− ( z value∗Standard deviation )
√ (Total observation)
¿ 747.083− {1.96∗
( 4.12
√ ( 12 ) ) }=744.75
90 ¿ rd error ¿ 95 % confidence itnerval
95% confidence interval = [ 744.75 749.42]
(b) Hypotheses
t stat= ( x−μ)
standard deviation/ √(Total observation) =747.083−750
4.12/ √(12) =−2.45103
Degree of freedom=T otal observation−1=12−1=11
Two tailed test
The p value = 0.03218
Assume (significance level) = 0.05
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The p value is lesser than significance level and hence, rejection would be for null hypothesis
(Hillier 144). Hence, it can be said that “standard bottles of wine does not contain an average of
exactly 750 ml.”
(c) It is apparent that the confidence interval computed in part (a) does not include the value 750
ml. Hence, based on the confidence interval also, the null hypothesis would be rejected which is
exactly the outcome in part (b).
Question 2
(a) The relevant probability statement
x1, x2, x3…………..xn (random samples: normal distribution population)
Also,
x N (μ , σ 2)
It can be seen from the graph that confidence interval for estimator 1 would be 1- alpha.
(b) It would be advisable to use estimator 2 as it considers the higher accuracy which would be
obtained as there is increase in sample size and hence the sample statistics would be more close
to the population parameters. This is also recommended by the Central Limit Theorem and
provides a more accurate interval estimate of the mean as compared to estimator 1 which
provides a more conservative estimate (Harmon 96).
(c) One of the key issues with estimator (2) is that it assumes that X, the underlying variable
tends to have a normal distribution. In the absence of the same, it would not be correct to assume
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an inverse relation between sample size and standard deviation from the mean. The estimator (1)
on the contrary does not arise out of this assumption and provides a more conservative estimate
but would be more accurate if the underlying population is non- normal.
(d) In the absence of any normal distribution on the part of X, the estimator (2) can still be used
provided that the sample size is large enough which is a key requirement as per the Central Limit
Theorem (Harmon 96).
Part B
1) D
2) E
3) D
4) A
5) C
6) B
References
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Harmon, Mark. Hypothesis Testing in Excel - The Excel Statistical Master .Florida: Mark
Harmon, 2015.
Hillier, Freferick. Introduction to Operations Research. New York: McGraw Hill Publications,
2016.
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