Question

The health of the bear population in Yellowstone National Park is monitored by periodic measurements taken from anesthetized bears. A sample of 54 bears has a mean weight of 182.9 lb. Assuming that the population standard deviation is known to be 121.8lb, construct and interpret a 99% confidence interval estimate of the mean of the population of all such bear weights. Round the limits to 1 decimal place. State the type of interval: Z or T interval

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 182.9

Population standard deviation =    = 121.8
Sample size = n =54

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 * (121.8 / 54)

= 42.7

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

- E < < + E

182.9  - 42.7< < 182.9 + 42.7

140.2< < 225.6

(140.2 , 225.6 )