Question

You work for a soft-drink company in the quality control division. You are interested in the standard deviation of one of your production lines as a measure of consistency. The product is intended to have a mean of 12 ounces, and your team would like the standard deviation to be as low as possible. You gather a random sample of 16 containers. Estimate the population standard deviation at a 80% level of confidence.

12.09	12.1	12.01	11.9	12.24	12.06
12.06	11.81	11.92	12.03	12.14	12.14
12.05	11.82	12.09	11.9

(Data checksum: 192.36)

Note: Keep as many decimals as possible while making these calculations. If possible, keep all answers exact by storing answers as variables on your calculator or computer.

a) Find the sample standard deviation:

b) Find the lower and upper χ2χ2 critical values at 80% confidence:
Lower:    Upper:

c) Report your confidence interval for σσ: (  ,  )

Answer 1

a.
given data,
12.09,12.1,12.01,11.9,12.24,12.06,12.06,11.81,11.92,12.03,12.14,12.14,12.05,11.82,12.09,11.9
sample standard deviation =0.1173

b.
CONFIDENCE INTERVAL FOR VARIANCE
ci = (n-1) s^2 / ᴪ^2 right < σ^2 < (n-1) s^2 / ᴪ^2 left
where,
s^2 = variance
ᴪ^2 right = (1 - confidence level)/2
ᴪ^2 left = 1 - ᴪ^2 right
n = sample size
since aplha =0.2
ᴪ^2 right = (1 - confidence level)/2 = (1 - 0.8)/2 = 0.2/2 = 0.1
ᴪ^2 left = 1 - ᴪ^2 right = 1 - 0.1 = 0.9
the two critical values ᴪ^2 left, ᴪ^2 right at 15 df are 22.307 , 8.547
variance( s^2 )=0.0137
sample size(n)=16
confidence interval = [ 15 * 0.0137/22.307 < σ^2 < 15 * 0.0137/8.547 ]
= [ 0.206/22.307 < σ^2 < 0.206/8.547 ]
[ 0.009 , 0.024 ]

c.
CONFIDENCE INTERVAL FOR STANDARD DEVIATION
ci = (n-1) s^2 / ᴪ^2 right < σ^2 < (n-1) s^2 / ᴪ^2 left
where,
s = standard deviation
ᴪ^2 right = (1 - confidence level)/2
ᴪ^2 left = 1 - ᴪ^2 right
n = sample size
since alpha =0.2
ᴪ^2 right = (1 - confidence level)/2 = (1 - 0.8)/2 = 0.2/2 = 0.1
ᴪ^2 left = 1 - ᴪ^2 right = 1 - 0.1 = 0.9
the two critical values ᴪ^2 left, ᴪ^2 right at 15 df are 22.307 , 8.547
s.d( s )=0.1173
sample size(n)=16
confidence interval for σ^2= [ 15 * 0.014/22.307 < σ^2 < 15 * 0.014/8.547 ]
= [ 0.206/22.307 < σ^2 < 0.206/8.547 ]
[ 0.009 < σ^2 < 0.024 ]
and confidence interval for σ = sqrt(lower) < σ < sqrt(upper)
= [ sqrt (0.009) < σ < sqrt(0.024), ]
= [ 0.096 < σ < 0.155 ]