Say we have a population distribution Nmu 4 From this popula

Say we have a population distribution N(mu, 4). From this population, we have a random sample of size 30 with sample mean being 2.19. Construct 90%, 95%, and 99% confidence intervals of mu. Compare the length of these intervals, tell me your conclusion.

Solution

AT 90% CI
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)=2.19
Standard deviation( sd )=4
Sample Size(n)=30
Confidence Interval = [ 2.19 ± t a/2 ( 4/ Sqrt ( 30) ) ]
= [ 2.19 - 1.699 * (0.73) , 2.19 + 1.699 * (0.73) ]
= [ 0.949,3.431 ]

AT 95% CI
Confidence Interval = [ 2.19 ± t a/2 ( 4/ Sqrt ( 30) ) ]
= [ 2.19 - 2.045 * (0.73) , 2.19 + 2.045 * (0.73) ]
= [ 0.697,3.683 ]

AT 99% CI

Confidence Interval = [ 2.19 ± t a/2 ( 4/ Sqrt ( 30) ) ]
= [ 2.19 - 2.756 * (0.73) , 2.19 + 2.756 * (0.73) ]
= [ 0.177,4.203 ]