In: Statistics and Probability
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.
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 = t0.01,39 = 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)