Question

A chocolate chip cookie manufacturing company recorded the number of chocolate chips in a sample of 60 cookies. The mean is 22.15 and the standard deviation is 2.08. Construct a 99​% confidence interval estimate of the standard deviation of the numbers of chocolate chips in all such cookies.

____ chocolate chips < σ < ____ chocolate chips

​(Round to two decimal places as​ needed.)

Answer 1

Solution :

Given that,

= 22.15

= 2.08

n = 60

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Margin of error = E = Z_/2* ( /n)

= 2.576 * (2.08 / 60)

= 0.69

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

- E < < + E

22.15 - 0.69 < < 22.15 + 0.69

21.46 < < 22.84

(21.46, 22.84)