Quantitative Methods: Confidence Interval and Hypothesis Testing

Quantitative Methods
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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 ) =749.4290 ¿ 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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