ENGG2430A-Homework 6
Due on Mar 26, 2014 via the assignment box.
1. Ex 9.72 from textbook A random sample of 20 students yielded a mean of x̄ = 72 and a variance
of s2 = 16 for scores on a college placement test in mathematics. Assuming the scores to be normally
distributed, construct a 98% confidence interval for σ 2 .
Solution: From the question, the sample size is n = 20 and the confidence level is 0.98 = 1 − α
2
with α = 0.02. Since (n−1)S
∼ χ2 (n − 1) is chi-squared distributed with n − 1 degrees from freedom,
σ2
the desired confidence interval is given by
"
# (n − 1)s2 (n − 1)s2
(20 − 1)16 (n − 1)s2
, 2
≈
,
χ0.02/2
χ1−0.02/2
χ2α/2
χ1−α/2
≈ [8.4, 39.8]
The chi-squared values χ20.01 ≈ 36.191 and χ0.99 ≈ 7.633 can be found from Table A.5 in the textbook
or computed by octave using the command chi2inv(0.99,19) and chi2inv(0.01,19) respectively.
2. modified Ex 9.14 from textbook The following measurements were recorded for the drying time, in
hours, of a certain brand of latex paint:
3.4, 2.5, 4.8, 2.9, 3.6, 2.8, 3.3, 5.6, 3.7, 2.8, 4.4, 4.0, 5.2, 3.0, 4.8
Assuming that the measurements represent a random sample from a normal population, find a 95%
confidence interval for the true mean drying time.
Solution: Since the samples are drawn from the normal population, we can use the t-distribution to
obtain the desired confident interval. We may use Octave to carry out the computation as follows.
x = [ 3 . 4 , 2 . 5 , 4 . 8 , 2 . 9 , 3 . 6 , 2 . 8 , 3 . 3 , 5 . 6 , 3 . 7 , 2 . 8 , 4 . 4 , 4 . 0 , 5 . 2 , 3 . 0 , 4 . 8 ] ; % sample {x1 , x2 , . . . }
n=length ( x ) % sample size n
1 Pn
xbar=mean( x ) % sample mean x̄ = n
xi
i=1q
1 Pn
2
s=std ( x ) % sample standard deviation s =
i=1 (xi − x̄)
n−1
a l p h a =1−0.95 % 95% confidence level
t a 2=−t i n v ( a l p h a / 2 , n−1) % t-value tα/2 with n-1 degrees of freedom
c=t a 2 ∗ s / sqrt ( n ) % error tα/2 √sn
disp ( m a t 2 s t r ( xbar+[− t a 2 t a 2 ] . ∗ s / sqrt ( n ) ) ) % confidence interval x̄ ± tα/2 √sn
The result is n = 15, x̄ = 3.7867, s = 0.97091, α = 0.05, tα/2 = 2.1448, the (margin of) error is
c = 0.53767, and the desired confidence interval is [3.25, 4.32].
1