Question

An Egg producer wants to estimate the mean number of eggs produced per chicken.  A sample of 16 chickens shows they produced an average of 21 eggs per month with a standard deviation of 1.7 eggs per month.

a. Develop a 99% confidence interval for the population mean

b. What is the best estimate of the population mean?

c. What distribution would you use for this analysis?

Answer 1

solution

Given that,

= 21

s =1.7

n = 16

Degrees of freedom = df = n - 1 = 16- 1 =15

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2  df = t0.005,15 = 2.947 ( using student t table)

Margin of error = E = t/2,df * (s /n)

= 2.947 * ( 1.7/ 16) =1.2525

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

- E < < + E

21 - 1.2525 < < 21+ 1.2525

19.7475 < < 22.2525

( 19.7475,22.2525 )

point estimate= = 21