Question

marine biologist is studying the weights of newborn killer whales. He measures a sample of 34 different whales, and the mean of his sample is 305 lb, and the standard deviation is 35 lb.a) Give a point estimate for the average weight of a newborn killer whale.b) Calculate the 90% error margin c) Find the 90% confidence interval d) If we wanted to increase our confidence level from 90% to 95%, will the error margin get larger or smaller? Repeat your calculations for part b) using A=95, then use one full sentence to explain why this result makes sense.

Answer 1

a)

point estimate for the average weight of a newborn killer whale = 305

b)

sample mean, xbar = 305
sample standard deviation, s = 35
sample size, n = 34
degrees of freedom, df = n - 1 = 33

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.692

ME = tc * s/sqrt(n)
ME = 1.692 * 35/sqrt(34)
Margin of error = 10.156

c)

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (305 - 1.692 * 35/sqrt(34) , 305 + 1.692 * 35/sqrt(34))
CI = (294.844 , 315.156)

d)
Larger

sample mean, xbar = 305
sample standard deviation, s = 35
sample size, n = 34
degrees of freedom, df = n - 1 = 33

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.035

ME = tc * s/sqrt(n)
ME = 2.035 * 35/sqrt(34)
ME = 12.215

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (305 - 2.035 * 35/sqrt(34) , 305 + 2.035 * 35/sqrt(34))
CI = (292.785 , 317.215)