Question

A lab tested a random sample of 155 chicken eggs for their level of cholesterol. The average of the sample was 198 milligrams. The standard deviation of all eggs of this type is known to be 14.1 milligrams. What is the 95% confidence interval for the true mean cholesterol content of this type of chicken egg?

Answer 1

Solution :

Given that,

Point estimate = sample mean =     = 198

Population standard deviation =    = 14.1

Sample size n =155

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

Margin of error = E =   Z/2    * ( /n)
= 1.96 * ( 14.1 / 155 )

= 2.22
At 95% confidence interval
is,

- E < < + E

198 - 2.22 <   < 198 + 2.22

(195.78,200.22)