Homework Soursea Construct a 95 twosided confidence interval on the mean li
(a) Construct a 95% two-sided confidence interval on the mean life.
(b) Construct a 95% one-sided lower confidence interval on the mean life.
(c) Suppose that we want to be 95% confident that the error in estimating the mean life is less than five hours. What sample size should be used?
Solution
a)
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.025
X = sample mean = 1000
z(alpha/2) = critical z for the confidence interval = 1.959963985
s = sample standard deviation = 20
n = sample size = 10
Thus,
Margin of Error E = 12.39590065
Lower bound = 987.6040994
Upper bound = 1012.395901
Thus, the confidence interval is
( 987.6040994 , 1012.395901 ) [ANSWER]
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b)
Note that
Lower Bound = X - z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.05
X = sample mean = 1000
z(alpha/2) = critical z for the confidence interval = 1.644853627
s = sample standard deviation = 20
n = sample size = 10
Thus,
Lower bound = 989.5970322
Thus, u > 989.597 [ANSWER]
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c)
Note that
n = z(alpha/2)^2 s^2 / E^2
where
alpha/2 = (1 - confidence level)/2 = 0.025
Using a table/technology,
z(alpha/2) = 1.959963985
Also,
s = sample standard deviation = 20
E = margin of error = 5
Thus,
n = 61.46334113
Rounding up,
n = 62 [ANSWER]
