Question

In Roosevelt National Forest the rangers took random samples of live aspen trees (Populus tremuloides) and measured the base circumference of each tree. Assume these measurements are normally distributed. (a) The first sample had 30 trees with a mean circumference of 15.71 inches and a sample standard deviation of 4.63 inches. Find a 95% confidence interval for the mean circumference of aspen trees from these data. (b) The next sample had 90 trees with a mean of 15.58 inches and a sample standard deviation of 4.61 inches. Find a 95% confidence interval for the mean circumference of aspen trees from these data. (c) The last sample had 300 trees with a mean circumference of 15.59 inches and a sample standard deviation of 4.62 inches. Find a 95% confidence interval for the mean circumference of aspen trees from these data. (d) What general conclusions can be drawn from (a), (b), and (c) in terms of sample size?

Answer 1

a)

sample mean, xbar = 15.71
sample standard deviation, s = 4.63
sample size, n = 30
degrees of freedom, df = n - 1 = 29

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.045

ME = tc * s/sqrt(n)
ME = 2.045 * 4.63/sqrt(30)
ME = 1.7

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (15.71 - 2.045 * 4.63/sqrt(30) , 15.71 + 2.045 * 4.63/sqrt(30))
CI = (13.98 , 17.44)

b)

sample mean, xbar = 15.58
sample standard deviation, s = 4.61
sample size, n = 90
degrees of freedom, df = n - 1 = 89

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 1.987

ME = tc * s/sqrt(n)
ME = 1.987 * 4.61/sqrt(90)
ME = 1

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (15.58 - 1.987 * 4.61/sqrt(90) , 15.58 + 1.987 * 4.61/sqrt(90))
CI = (14.61 , 16.55)

c)

sample mean, xbar = 15.59
sample standard deviation, s = 4.62
sample size, n = 300
degrees of freedom, df = n - 1 = 299

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 1.968

ME = tc * s/sqrt(n)
ME = 1.968 * 4.62/sqrt(300)
ME = 0.5

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (15.59 - 1.968 * 4.62/sqrt(300) , 15.59 + 1.968 * 4.62/sqrt(300))
CI = (15.07 , 16.11)

d)

as sampl esize increases the confidence interval would be narrower