Question

The values listed below are waiting times​ (in minutes) of customers at two different banks. At Bank​ A, customers enter a single waiting line that feeds three teller windows. At Bank​ B, customers may enter any one of three different lines that have formed at three teller windows. Answer the following questions. Bank A 6.4 6.6 6.7 6.8 7.1 7.3 7.4 7.7 7.7 7.7 Bank B 4.1 5.3 5.9 6.2 6.8 7.6 7.6 8.4 9.4 10

Construct a 99​% confidence interval for the population standard deviation σ at Bank A.

Answer 1

R-commands and outputs:
x=c(6.4, 6.6, 6.7, 6.8, 7.1, 7.3, 7.4, 7.7, 7.7, 7.7)
length(x)
[1] 10
## We check normality of x:
shapiro.test(x)
Shapiro-Wilk normality test
data: x
W = 0.90223, p-value = 0.2318
## p-value=0.2318 > alpha=0.01, we Accept H0.
## sample (x) follow normal distribution.

n=length(x)
n
[1] 10
s2=var(x) ## sample variance (or sample standard deviation's square)
s2
[1] 0.2426667
sqrt(s2) ## sample standard deviation
[1] 0.4926121

#Since 99% CI, 1-alpha=0.99
alpha=0.01
qchisq(1-alpha/2,df=n-1)
[1] 23.58935
qchisq(alpha/2,df=n-1)
[1] 1.734933

## Required values for computing 99% Confidence Interval (CI) for the population standard deviation σ at Bank A:
## n=10, s2=0.2426667, chi1sq=1.734933, chi2sq=23.58935

chi1sq=1.734933
chi2sq=23.58935
lower=((n-1)*s2)/ chi2sq
lower
[1] 0.09258415

upper=((n-1)*s2)/ chi1sq
upper
[1] 1.258838

## This, [lower=0.092584, upper=1.258838] gives 99% CI for population variance.
## For finding 99% CI for population standard deviation we simply take squareroot of above interval's limits

sqrt(lower)
[1] 0.3042764
sqrt(upper)
[1] 1.12198
## A 99​% confidence interval for the population standard deviation σ at Bank A is [0.3042764,1.12198]