Question

You measure 24 turtles' weights, and find they have a mean weight of 71 ounces. Assume the population standard deviation is 10.5 ounces. Based on this, construct a 99% confidence interval for the true population mean turtle weight.

Give your answers as decimals, to two places

±±  ounces

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 71

Population standard deviation =    = 10.5
Sample size = n =24

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576   ( Using z table )

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

= 2.576* (10.5 / 24)

= 5.52

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

+/- E

71+/-5.52