Confidence Intervals.
1.
To determine an average weight of a bag of apples in certain supermarket 100 bags were
examined. Assume that the wieghts of different bags are independent normal random variables
with unknown mean
μ
and standard deviation
σ
= 0
.
5
.
The mean weight of the bags under
consideration was 2.8 lbs.
(a) Construct confidence intervals for
μ
with confidence levels 90%, 95% and 99%.
(b) How many bags need to be examined so that the length of the 99% confidence interval
is less than 0.1 lbs?
Solution.
(a) The confidence interval with known variance has form
[¯
x
-
σ
√
n
z
α/
2
,
¯
x
+
σ
√
n
z
α/
2
] = [2
.
8
-
0
.
05
z
α/
2
,
2
.
8 + 0
.
05
z
α/
2
]
.
So the answers are
90% CI: [2
.
8
-
0
.
05
×
1
.
65
,
2
.
8 + 0
.
05
×
1
.
65] = [2
.
7175
,
2
.
8825]
95% CI: [2
.
8
-
0
.
05
×
1
.
96
,
2
.
8 + 0
.
05
×
1
.
96] = [2
.
702
,
2
.
898]
99% CI: [2
.
8
-
0
.
05
×
2
.
33
,
2
.
8 + 0
.
05
×
2
.
33] = [2
.
6835
,
2
.
9165]
(b) The width of the confidence interval is 2
σ
*
z
α/
2
/
√
n
= 2
×
0
.
5
×
2
.
33
√
n
=
2
.
33
√
n
.
So
n
should satisfy
2
.
33
√
n
≤
0
.
1
⇔
√
n
≥
23
.
3
⇔
n
≥
23
.
3
2
= 542
.
89
.
So we need 543 bags.
2.
50 statistics students pick up 100 bags of apples each in 50 different Maryland stores
and construct 95% confidence intervals using their data. Find the probability that exactly 3
students will come up with intervals which do not contain population mean.
Solution.
The number
N
of students who gets a wrong answer has binomial distribution
with parameters (50
,
0
.
05)
.
So
N
is approximately Poissonian with parameter 50
×
0
.
05 = 2
.
5
.
Thus
P
(
N
= 3)
≈
e
-
2
.
5
2
.
5
3
3!
≈
0
.
21
.
3.
A waiting time for a bus at John’s work has uniform distribution on the interval
[0
, θ
]
.
During the first 10 days at work the maximal time John had to wait for the bus was 12 min.
Construct 95% confidence interval for the maximal waiting time
θ.
Solution.
P
X
max
θ
≤
a
=
a
10
so
P
(
X
max
θ
∈
[
a,
1]
= 1
-
a
10
.
If 1
-
a
10
= 0
.
95
,
then
a
= 0
.
05
1
/
10
≈
0
.
74
.
Thus the confidence interval is
0
.
74
≤
X
max
θ
≤
1
.
The second inequlaity gives
θ
≥
X
max
while the fist ones gives
θ
≤
X
max
0
.
74
.
So the answer is
12
,
12
0
.
74
≈
[12
,
16
.
22]
.
4.
An avergage water consumption for a certain home during 2011 was 135 gal/day with
sample standard deviation of 25 gal/day. Compute large sample confidence interval for the
mean water consumption at that home with confidence level 95%.
1