Quantitative Methods: Confidence Interval and Hypothesis Testing

This article covers the calculation of a 95% confidence interval for population mean and hypothesis testing. It also discusses the use of different estimators and their shortcomings in practice. The article includes a diagram and references for further reading.

Quantitative Methods
Student Name
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Part A
Question 1
(a) 95% confidence interval for population mean
Number of observation N= 12
Mean of data x= Sum of data points/ Number of observation ¿ 8965/12=747.083
Standard deviation of data
Standard deviation= √ 1
N −1 ∑ ¿ ¿ ¿
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Standard error = Standard deviation
√ ( Number of observation ) = 4.12
√ ( 12 ) =1.190
The z value for 95% confidence interval = 1.960
Upper limit of 95% confidence interval
¿ Mean+(z value∗Standard error)=747.083+ ( 1.96∗1.190 ) =749.42 90¿ rd error ¿ 95 %confidence itnerval
Lower limit of 95% confidence interval
¿ Mean− ( z value∗Standard error )=747.083− ( 1.96∗1.190 ) =744.75 90¿ rd error ¿ 95 % confidence itnerval
Therefore, the 95% confidece interval would be [ 744.75 749.42].
(b) Hypothesis testing
Null hypothesis H0 : μ=750
Alternative hypothesis H1 : μ ≠ 750
The valuetest statistic ( t value ) =( x−μ)
S . E . =747.083−750
1.190 =−2.45103
Degree of freedom = N-1 = 12 – 1 = 11
From standard normal table, the p value corresponding to t value, degree of freedom and for two
tailed test comes out to be 0.03218.
Assuming level of significance = 5%
It can be seen that p value is lower than level of significance and therefore, sufficient evidence
are present the reject the null hypothesis and to accept the alternative hypothesis (Flick 212).
Therefore, it can be concluded that standard bottles of wine does not contain an average of
exactly 750 ml.
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(C) Yes the outcome of the test in part (b) is supported by part(a) considering the fact that the
confidence interval computed does not include the value 750 ml.
Question 2
(a) Based on the given information, the following diagram can be made.
Probability Statement
Here,
Normal distribution and random points is X 1 , X 2 , X 3 , X 4 , X 5 … … … … .. Xn
x N (μ , σ 2)
The confidence interval will be 1−∝.
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(b) Estimator 2 should be used in practice considering the fact that it takes into consideration the
fact that as the sample size would increase, the deviation of the estimator from the population
mean would decrease owing to a larger sample being representative of the population in
accordance with the Central Limit Theorem (Hillier 148).
(c) With regards to estimator (2), there is a shortcoming as it is based on the assumption that the
underlying variable X is normally distributed. If this is not the case, then it would not be
appropriate to conclude that with the increase in sample size, the variation of the mean interval
estimator would tend to reduce. Therefore, it would be more suitable to use estimator (1) which
is not based on this assumption and thus would give better estimates under the situation owing to
the conservative estimates that it provides (Flick 165).
(d) Even if X is not normally distributed, it would be appropriate to use the estimator (2) when
the sample size is large enough (i.e. greater than 30) in accordance with the Central Limit
Theorem.
Part B
1. D
2. E
3. D
4. A
5. C
6. B
References
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Flick, Uwe. Introducing research methodology: A beginner's guide to doing a research project:
New York: Sage Publications, 2015.
Hillier, Freferick. Introduction to Operations Research. New York: McGraw Hill Publications,
2016.
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