Answer:
Confidence Interval - 2.290 < S < 2.965
Step-by-step explanation:
Complete question
A chocolate chip cookie manufacturing company recorded the number of chocolate chips in a sample of 50 cookies. The mean is 23.33 and the standard deviation is 2.6. Construct a 80% confidence interval estimate of the standard deviation of the numbers of chocolate chips in all such cookies.
Solution
Given
n=50
x=23.33
s=2.6
Alpha = 1-0.80 = 0.20
X^2(a/2,n-1) = X^2(0.10, 49) = 63.17
sqrt(63.17) = 7.948
X^2(1 - a/2,n-1) = X^2(0.90, 49) = 37.69
sqrt(37.69) = 6.139
s*sqrt(n-1) = 18.2
[tex]s\sqrt{\frac{n-1}{X^2 _{(n-1), \frac{\alpha }{2} } }[/tex] [tex]\leq \sigma \leq s\sqrt{\frac{n-1}{X^2 _{(n-1), 1-\frac{\alpha }{2} } }[/tex]
confidence interval:
(18.2/7.948) < S < (18.2/6.139)
2.290 < S < 2.965
