Question

1) (note: Please also provide the excel formulas used to solve it )  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 18 containers. Estimate the population standard deviation at a 90% level of confidence.

12.09	12.1	11.94	12.14	12.12	11.98
11.97	11.98	11.93	11.96	11.86	11.94
12.03	12.08	11.97	12.05	12.07	12.08

(Data checksum: 216.29)

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 χ2 critical values at 90% confidence:
Lower:    Upper:

c) Report your confidence interval for σ (  ,  )

2) (note: Please also provide the excel formulas used to solve it )

If n=300 and p^=0.3, construct a 90% confidence interval about the population proportion. Round your answers to three decimal places.

Preliminary:

1. Is it safe to assume that n≤0.05n≤0.05 of all subjects in the population?
   • No
   • Yes
   
   
2. Verify nˆp(1−ˆp)≥10. Round your answer to one decimal place.
   
   nˆp(1−ˆp)=

Confidence Interval: What is the 90% confidence interval to estimate the population proportion? Round your answer to three decimal places.

< p <

Answer 1

Solution-1:

sample std dev isne xcel,s=0.078376083

alpha=1-0.90=0.10

alpha/2=0.10/2=0.05

1-alph/2=1-0.05=0.95

df=n-1=12-1=11

chi sq critical values

lower limit

=CHISQ.INV.RT(0.05,11)

19.67513757

upper limit

=CHISQ.INV.RT(0.95,11)

=4.574813079

lower limit=19.67513757

upper limit=4.574813079

90% confidence interval for stddev

sqrt(n-1)*s^2/chi sq alpha/2<sigma<sqrt(n-1)*s^2/1-chi sq alpha/2

sqrt((12-1)*0.00614281/19.67513757)<sigma<sqrt((12-1)*0.00614281/4.574813079)

0.05860315<sigma< 0.1215327

0.0586<sigma<0.1215

confidence interval(0.05860315,0.1215327)

sample std dev isne xcel,s=0.078376083

lower limit=19.67513757

upper limit=4.574813079

confidence interval(0.05860315,0.1215327)

Solution-2:

n<0.05

yes

np^*(1-p^)=300*0.3*(1-0.3)=63>10

np^*(1-p^)=63