You are measuring the amount of contaminant in a lake. You take 9 samples from d

Question

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?

Explanation / Answer

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   )

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