A random sample X1 X2 X3 X36 is given from a normal distrib

A random sample X1, X2, X3, ..., X36 is given from a normal distribution with unknown mean =EXi and unknown variance Var(Xi)=2. For the observed sample, the sample mean is X¯¯¯=35.8, and the sample variance is S2=12.5.

Find and compare 90%, 95%, and 99% confidence interval for .

Find and compare 90%, 95%, and 99% confidence interval for 2.

Solution

a) WHEN Confidence Interval is 90%

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)=35.8
Standard deviation( sd )=3.5355
Sample Size(n)=36
Confidence Interval = [ 35.8 ± t a/2 ( 3.5355/ Sqrt ( 36) ) ]
= [ 35.8 - 1.69 * (0.589) , 35.8 + 1.69 * (0.589) ]
= [ 34.804,36.796 ]

b) WHEN C.I is 95
Confidence Interval = [ 35.8 ± t a/2 ( 3.5355/ Sqrt ( 36) ) ]
= [ 35.8 - 2.03 * (0.589) , 35.8 + 2.03 * (0.589) ]
= [ 34.604,36.996 ]

c) WHEN C.I is 99
Confidence Interval = [ 35.8 ± t a/2 ( 3.5355/ Sqrt ( 36) ) ]
= [ 35.8 - 2.724 * (0.589) , 35.8 + 2.724 * (0.589) ]
= [ 34.195,37.405 ]

Increasing the confidence widens the result

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