Question

A dentist finds in a routine check that six prison inmates require 2, 3, 6, 0, 4 and 3 fillings.

a) Construct a 90% confidence interval for the average number of fillings required by the inmates of this very large prison.

b) If he uses the mean of this sample to estimate the true mean of the population sampled, what can he say with 99% confidence about the maximum error?

Answer 1

a) The mean is calculated as

Mean, M= (2 + 3 + 6 + 0 + 4 + 3)/6
= 18/6
Mean = 3

and sample standard deviation as:

Standard Deviation S= √(1/6 - 1) x ((2 - 3)² + ( 3 - 3)² + ( 6 - 3)² + ( 0 - 3)² + ( 4 - 3)² + ( 3 - 3)²)
= √(1/5) x ((-1)² + (0)² + (3)² + (-3)² + (1)² + (0)²)
= √(0.2) x ((1) + (0) + (9) + (9) + (1) + (0))
= √(0.2) x (20)
= √(4)

=2

Now,

The formula for Confidence Interval:

μ = M ± t(s_M)

where:

M = sample mean
t = t statistic determined by confidence level and at degree of freedom n-1=6-1=5
s_M = standard error = √(s²/n)

M = 3
t = 2.02
s_M = √(2²/6) = 0.82

μ = M ± t(s_M)
μ = 3 ± 2.02*0.82
μ = 3 ± 1.65

90% CI [1.35, 4.65].

b) If computed at 99% comfidence level the the margin of error is calculated as

t(s_M)  

t = 4.03 computed using T table shown below
s_M = √(2²/6) = 0.82

=4.03*0.82

=3.29