Question

A biotechnology company produces a therapeutic drug whose concentration has a standard deviation of 0.004 g/l. A new method of producing this drug has been proposed, although some additional cost is involved. Management will authorize a change in production technique only if the standard deviation of the concentration in the new process is less than 0.004 g/l. The researchers randomly chose 10 specimens and obtained the data found below. Assume the population of interest is normally distributed.

A. Test the appropriate hypothesis for this situation with α = 0.05. provide a copy of your R input and output, state your conclusion in context.

B. Find and interpret a 95% upper confidence bound for the true standard deviation. use the interval from your R output

DATA: 16.628, 16.622, 16.627, 16.623, 16.618, 16.63, 16.631, 16.624, 16.622, 16.626

Answer 1

Claim: The standard deviation of the concentration in the new process is less than 0.004

A )

H₀ : σ = 0.004

H_a : σ < 0.004

R code :

x<-c(16.628, 16.622, 16.627, 16.623, 16.618, 16.63, 16.631, 16.624, 16.622, 16.626 )
S <- sd(x)
n <- length(x)
Sigma <- 0.004
Chisqure<- (n-1)*S^2 / Sigma^2
Chisqure
df <- n-1
pvalue <- pchisq(Chisqure,df)
pvalue

Output :

Chisqure = 9.18125

pvalue = 0.5792856

Decision : As p value is greater than α = 0.05, We fail to reject H₀

Conclusion : There is no significant evidence that the standard deviation of the concentration in the new process is less than 0.004

b) 95% confidence interval

R code:

# confidence interval for SD
S <- 0.00404
n <- 10
c <- 0.95
Alpha1 <- (1 + c)/2
Alpha2 <- (1 - c)/2
df <- n-1
LB<-qchisq(Alpha1,df)
LB
UB<-qchisq(Alpha2,df)
UB
Lowerinterval <- sqrt((df*S^2)/LB)
Upperinterval <- sqrt((df*S^2)/UB)
data.frame(Lowerinterval,Upperinterval)

Output :

Lowerinterval Upperinterval
0.002778854 0.007375465

So 95% upper confidence bound is 0.007375465