Question

A biologist measures the lengths of a random sample 45 mature brown trout in a large lake and finds that the sample a mean weight of 41 pounds. Assume the population standard deviation is 3.7 pounds. Based on this, construct a 99% confidence interval for the mean weight of all mature brown trout in the lake. Round your anwers to two decimal places. < μ

Answer 1

Solution :

Given that,

= 41

= 3.7

n = 45

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Margin of error = E = Z_/2* ( /n)

= 2.576 * (3.7 / 45)

= 1.42

At 99% confidence interval estimate of the population mean is,

- E < < + E

41 - 1.42 < < 41 + 1.42

39.58 < < 42.42