2
Summary of the CI’s as presented so far and in the textbook.
n
z
X
/
2
/
(6-3)
n
s
t
X
/
2
/
(6-5)
n
p
p
z
p
)
ˆ
1
(
ˆ
ˆ
2
/
(6-7)
Review Exercises
Exercise 6-5
.
The variable of interest is x = value (in $), a numerical characteristic.
We are told that x is approximately normally distributed in the population.
Population
size N not given.
Population standard deviation is given as σ = $5,500.
Random
sampling was used with sample size n = 16 with
x
=$89,673.
Find 95% CI for μ.
Ans.
Using (6-
3) with α = 0.05 gives
$89,673 ± 1.960
(
16
/
$5,500)
,
which is $89,673
± $2,695.
Can be written [$86,978, $92,368] or $86,978 ≤ µ ≤ $92,368.
Exercise 6-6
.
Find 99% CI for μ.
Ans.
Using (6-
3) with α = 0.0
1 gives
$89,673 ±
2.576
(
16
/
$5,500)
,
which is $89,673 ± $3,542.
Can be written [$86,131, $93,215]
or $86,131 ≤ µ ≤ $93,215.
Exercise 6-22
.
Find 95% CI for μ.
Ans.
Using (6-5
) with α = 0.0
5, n = 100,
x
=3.2
and s = 2.1 yields
3.2 ± 2.000
(
100
/
2.1)
,
3.2 ± 0.42.
Can be written [2.78, 3.62] or
2.78 ≤ µ ≤ 3.62.
Exercise 6-46
.
Find 90% CI for p.
Ans.
Using (6-7
) with α = 0.
10, n = 100,
p
ˆ
=
0.85
yields
100
)
15
.
0
)(
85
.
0
(
645
.
1
85
.
0
, 0.85
± 0.059.
Can be written [0.791, 0.909]
or 0.791 ≤ p ≤ 0.909 or
79.1% ≤ p% ≤ 90.9%.