Listed below are measured amounts of lead in micrograms per
Solution
a)
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 = 0.025
X = sample mean = 1.588333333
t(alpha/2) = critical t for the confidence interval = 2.570581836
s = sample standard deviation = 1.888665314
n = sample size = 6
df = n - 1 = 5
Thus,
Lower bound = -0.393699359
Upper bound = 3.570366025
Thus, the confidence interval is
( -0.393699359 , 3.570366025 ) [ANSWER]
********************************
b)
D. Yes, the value 5.40 appears to be an outlier. [ANSWER]
This is a problem because we are assuming that the population is approximately normally distributed to use the t distrbution here.
