Question

A chocolate chip cookie manufacturing company recorded the number of chocolate chips in a sample of 40 cookies. The mean is 22.29 and the standard deviation is 2.31.Construct a 98​% confidence interval estimate of the standard deviation of the numbers of chocolate chips in all such cookies.

Answer 1

Solution :

Given that,

= 22.29

s = 2.31

n = 40

Degrees of freedom = df = n - 1 = 40 - 1 = 39

At 98% confidence level the t is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

t _/2,df = t_0.01,39 = 2.426

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

= 2.426 * (2.31 / 40)

= 0.886

The 98% confidence interval estimate of the population mean is,

- E < < + E

22.29 - 0.886 < < 22.29 + 0.886

21.404 < < 23.176

(21.404, 23.176)