Question

The health of the bear population in a park is monitored by periodic measurements taken from anesthetized bears. A sample of the weights of such bears is given below. Find a​ 95% confidence interval estimate of the mean of the population of all such bear weights. The​ 95% confidence interval for the mean bear weight is the following.

data table 80 344 416 348 166 220 262 360 204 144 332 34 140 180

Answer 1

Solution:

Given a sample of size n = 14

80 ,344 ,416 ,348 ,166 ,220 ,262 ,360 ,204 ,144 ,332 ,34 ,140 ,180

Since the population SD is unknown , we use t distribution.

First we need to find the sample mean and sample SD s.

=   

= (80 + 344 + 416 + .........+ 180)/14

= 230.7143

Now ,

s=   

Using given data, find Xi- for each term.take square for each.then we can easily find s.

s= 115.5658

construct 95% confidence interval.

c = 0.95

= 1- c = 1- 0.95 = 0.05

  /2 = 0.05 2 = 0.025

Also, n = 14

d.f= n-1 = 13

     =     = 2.160

Margin of error E = * (s / n)

= 2.160 * (115.5657/14)

= 66.7143

Required interval is

( - E , + E)

(230.7143 - 66.7143 , 230.7143 - 66.7143)

(164.0000 , 297.4286)