Question

ervar or the mean difference in body temperature? (3, 6) 2. An ornithologist in Hinckley, OH measured the wingspan of 16 turkey vultures randomly selected from a wake in the area. The mean wingspan was found to be 1.80 m, with a standard deviation of 10 cm. Determine the 90%, 95% and 99% confidence intervals for the mean, (4, 4) 3. A conservation biologist studies the endangered Mekong giant catfish,

Answer 1

sample mean, xbar = 1.8
sample standard deviation, s = 10
sample size, n = 16
degrees of freedom, df = n - 1 = 15

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.753

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (1.8 - 1.753 * 10/sqrt(16) , 1.8 + 1.753 * 10/sqrt(16))
CI = (-2.58 , 6.18)

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.131

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (1.8 - 2.131 * 10/sqrt(16) , 1.8 + 2.131 * 10/sqrt(16))
CI = (-3.53 , 7.13)

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, tc = t(α/2, df) = 2.947

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (1.8 - 2.947 * 10/sqrt(16) , 1.8 + 2.947 * 10/sqrt(16))
CI = (-5.57 , 9.17)