Question

A machine is used to fill containers with a liquid product. Fill volume can be assumed to be normally distributed. A random sample of ten containers is selected, and the net contents (oz) are as follows: 12.03, 12.01, 12.04, 12.02, 12.05, 11.98, 11.96, 12.02, 12.05, and 11.99.

(d) Predict with 95% confidence the value of the 11th filled container.

(e) Predict with 95% confidence the interval containing 90% of the filled containers from the process.

*please calculate by hand (not in R)

Answer 1

Answer :

Since data is normally distributed We have obtained the answer by using Normal distribution using R.

d) To predict the 11th Filled Container paste below mentioned code in R

x=c(12.03, 12.01, 12.04, 12.02, 12.05, 11.98, 11.96, 12.02, 12.05, 11.99)
# To obtain the Mean
mu= mean(x)
mu
#To obtain the Standard deviation
sd= sd(x)
sd
n=10

#To Predict the Value of 11th Filled Container having 95 % confidence Interval

#95% CI means to be consider as Alpha=0.95
err= qnorm(0.95)*(sd/(sqrt(n)))

# As per Statistic of Normal distribution Sample mean is unbiased estimator of Population mean
#hence need to obtain the Sample mean i.e. predict11= err+mu
#Question.d

predict11= err+mu
predict11

Answer: Predicted content of 11 th Container: 12.03075

e) To Predict with 95% confidence the interval paste below mentioned code in R

#95% confidence the interval containing 90% of the filled containers from the process.
# here we need to consider the probality of error as Alpha/2 i.e. =0.975

err1= qnorm(0.975)*(sd/(sqrt(n)))

# As per Formula to derive C.I of Mean of Normal Distribution i.e. Lower_Bound= mu- err1
#and Upper_Bound= mu+ err1
lower_CI= mu- err1
lower_CI
upper_CI= mu+ err1
upper_CI

Answer : C.I.= (11.99623, 12.03377 )