Question

A chocolate chip cookie manufacturing company recorded the number of chocolate chips in a sample of 60 cookies. The mean is 22.46 and the standard deviation is 2.28. Construct a 90​% confidence interval estimate of the standard deviation of the numbers of chocolate chips in all such cookies.

Answer 1

Solution :

Given that,

= 22.46

s = 2.28

n = 60

Degrees of freedom = df = n - 1 = 60 - 1 = 59

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,59 = 1.671

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

= 1.671 * (2.28 / 60)

= 0.492

The 90% confidence interval estimate of the population mean is,

- E < < + E

22.46 - 0.492 < < 22.46 + 0.492

21.968 < < 22.952

(21.968, 22.952)