A random sample of n measurements was selected from a popula

A random sample of n measurements was selected from a population with unknown mean \\mu and standard deviation \\sigma. Calculate a 90 % confidence interval for \\mu for each of the following situations

: (a) \ = 120, overline{x} = 35.1, s = 3.85

_________ < U < ________

(b) n = 85, overline{x} = 91.7, s = 3.4

_________ < U < ________

(c) n = 70, overline{x} = 11.6, s = 2.89

_________ < U < ________

(d) n = 75, overline{x} = 9.2, s = 2.81

_________ < U < ________

Solution

a)
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=35.1
Standard deviation( sd )=3.85
Sample Size(n)=120
Confidence Interval = [ 35.1 ± t a/2 ( 3.85/ Sqrt ( 120) ) ]
= [ 35.1 - 1.658 * (0.35) , 35.1 + 1.658 * (0.35) ]
= [ 34.52,35.68 ]
b)
Mean(x)=91.7
Standard deviation( sd )=3.4
Sample Size(n)=85
Confidence Interval = [ 91.7 ± t a/2 ( 3.4/ Sqrt ( 85) ) ]
= [ 91.7 - 1.663 * (0.37) , 91.7 + 1.663 * (0.37) ]
= [ 91.09,92.31 ]

c)
Mean(x)=11.6
Standard deviation( sd )=2.89
Sample Size(n)=70
Confidence Interval = [ 11.6 ± t a/2 ( 2.89/ Sqrt ( 70) ) ]
= [ 11.6 - 1.667 * (0.35) , 11.6 + 1.667 * (0.35) ]
= [ 11.02,12.18 ]
d)
Mean(x)=9.2
Standard deviation( sd )=2.81
Sample Size(n)=75
Confidence Interval = [ 9.2 ± t a/2 ( 2.81/ Sqrt ( 75) ) ]
= [ 9.2 - 1.666 * (0.32) , 9.2 + 1.666 * (0.32) ]
= [ 8.66,9.74 ]