Homework SourseYou are measuring the amount of contaminant in a lake You ta
You are measuring the amount of contaminant in a lake. You take 9 samples from different locations of lake. You find mean to be 90 and standard deviation to be 10. Plausibly normal , no prior info. What is the 98% confidence interval for true mean of contamination?
You are measuring the amount of contaminant in a lake. You take 9 samples from different locations of lake. You find mean to be 90 and standard deviation to be 10. Plausibly normal , no prior info. What is the 98% confidence interval for true mean of contamination?
Solution
Note that
Lower Bound = X - t(alpha/2) * s / sqrt(n)
Upper Bound = X + t(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.01
X = sample mean = 90
t(alpha/2) = critical t for the confidence interval = 2.896459448
s = sample standard deviation = 10
n = sample size = 9
df = n - 1 = 8
Thus,
Lower bound = 80.34513517
Upper bound = 99.65486483
Thus, the confidence interval is
( 80.34513517 , 99.65486483 )
