Question

A scientist wishes to asses how a certain drug affects glucose levels in the anterior chambers of the eyes of dogs. The glucose levels in dogs are known to be approximately normally distributed with a standard deviation of 10.8 (mg/dLi)^2. How large of a sample size is needed to be able to estimate the true mean glucose levels in the eyes of dogs to within 2.5 (mg/dLi)^2 using 99% confidence?

Please circle the z* value used and your response to the above question.

Answer 1

Solution :

Given that,

standard deviation = =10.8

Margin of error = E = 2.5

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

(Z/2 = 2.58 ) ( Using z table ( see the 0.005 value in standard normal (z) table corresponding z value is 2.58 )  

sample size = n = [Z/2* / E] 2

n = ( 2.58* 10.8 / 2.5 )2

n =144.22

Sample size = n =144