Question

The owner of Britten's Egg Farm wants to estimate the mean number of eggs laid per chicken. A sample of 30 chickens shows they laid an average of 25 eggs per month with a standard deviation of 12 eggs per month.  

a-2. What is the best estimate of this value?

c. For a 80% confidence interval, what is the value of t? (Round the final answer to 3 decimal places.)

The value of t is  .

  

d. Develop the 80% confidence interval for the population mean. (Round the final answers to 2 decimal places.)

The 80% confidence interval for the population mean is  to  .

Answer 1

Solution :

Given that,

a-2)

Point estimate = sample mean = = 25

sample standard deviation = s = 12

sample size = n = 30

Degrees of freedom = df = n - 1 = 29

c)

At 80% confidence level the t is ,

= 1 - 80% = 1 - 0.80 = 0.20

/ 2 = 0.20 / 2 = 0.10

t _/2,df = t_0.10,29 = 1.311

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

= 1.311 * (12 / 30)

= 2.87

d)

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

- E < < + E

25 - 2.87 < < 25 + 2.87

22.13 < < 27.87

(22.13 , 22.87)