Quantitative Methods: Confidence Interval and Estimator

QUANTITATIVE METHODS
Student Name
[Pick the date]

Part A
Question 1
(a) 95% confidence interval
Mean (average)
Total observation = 12
Sum of observation = 8965
Average = 8965/12 = 747.083
Standard deviation
1

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 ) ) }=749.42 90 ¿ 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
2