Homework SourseThe 72hour blastogenesis of chicken peripheral blood lymphoc
The 72-hour blastogenesis of chicken peripheral blood lymphocytes (10^6 cells/well) is given below. Compute the 90%, 95%, and 99% confidence intervals for the population mean. 236, 314, 471, 305, 414, 616, 304, 414, 301, 225, 369, 402.
Solution
A)
90% CONFIDENCE:
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 = 0.05
X = sample mean = 364.25
t(alpha/2) = critical t for the confidence interval = 1.795884819
s = sample standard deviation = 108.9938489
n = sample size = 12
df = n - 1 = 11
Thus,
Lower bound = 307.7446141
Upper bound = 420.7553859
Thus, the confidence interval is
( 307.7446141 , 420.7553859 ) [ANSWER]
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B)
95% CONFIDENCE:
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 = 364.25
t(alpha/2) = critical t for the confidence interval = 2.20098516
s = sample standard deviation = 108.9938489
n = sample size = 12
df = n - 1 = 11
Thus,
Lower bound = 294.9986123
Upper bound = 433.5013877
Thus, the confidence interval is
( 294.9986123 , 433.5013877 ) [ANSWER]
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c)
99% CONFIDENCE:
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 = 0.005
X = sample mean = 364.25
t(alpha/2) = critical t for the confidence interval = 3.105806516
s = sample standard deviation = 108.9938489
n = sample size = 12
df = n - 1 = 11
Thus,
Lower bound = 266.5294815
Upper bound = 461.9705185
Thus, the confidence interval is
( 266.5294815 , 461.9705185 ) [ANSWER]
