Population Sample
Size
N
n
Mean

X
S.D.
s

, general,
take
are samples
some
of
population.
Suppose
In
wewant
, parameters
are
tonot
know
 the
(say),
we
and
weknown.
will
know
X and s So, what can we say about  ?
But how
Can we say X is  ?
close
is it?
Can we say X is close to  ?
To estimate  by a number X
it is too “dangerous”!
It is much “safer” to estimate  by
an interval.
we
canon
have
sample
mean
and variance;
Based
thesome
data
from
random
samples,
suppose
find
by
an
interval
further
(L,U),
calculation,
such that P(L<  < U) = 95 % (say),
that means there’s 95% chance (L,U) traps .
We say (L,U) is a 95% confidence interval for 
In general, to estimate a parameter ,
if we can find a random interval (L,U)
such that P(L <  < U) = k%,
(L,U) is called a
k% confidence interval for 
But how to find
(L,U)?
In AL, you are required to construct confidence interval
C.I. for (1) population mean and (2) population proportion.
Let’s talk
Task:
Findabout
95% C.I.
C.I.2for
for ..

By CLT, X ~ N (  , )
n
Suppose (L,U) is a 95% C.I. for ,
P(L <  < U) = 95% --- (1)
By table, P(1.96 < z < 1.96) = 95%
X 
P (1.96 
 1.96)  95%
/ n
1.96
1.96
Rearranging,
P( X 
X 
)  95%
n
n
Comparing
(1),
95%
C.I. for
 is
1.96
1.96 
,X 
X 

n
n 

 X  1.96 , X  1.96  is a 95% C.I. for .


n
n 

How about a 99% C.I. for  ? Ans:
 X  2.58 , X  2.58 


n
n 

since P(2.58 < z < 2.58) = 99%
In general, a % C.I. for  is
 X  zc , X  zc 

 where P(zc < z < zc) = %
n
n

% is called the confidence level.
 X  zc , X  zc 

 is a % C.I. for .
n
n

Note 1:   zc , hence width of C.I. 
Reasonable! To ensure more chance to “trap”
the true , we can have wider width of C.I.
But it is close to meaningless to mention C.I.
of large range, e.g. if we claim that we have
100% confident that the true  lies on (,).
Note 2: In practice, we don’t even know ,
then we should use sample s.d. s to replace .
More precisely, use s[n/(n1)] instead of s.
E.g. 26
Masses of random sample (in g) are 182,
184, 176, 178, 181, 180, 183, 178, 179, 177,
180, 183, 179, 178, 181, 181. If this sample
came from a normal population  = 10g,
obtain a 95% C.I. for mean mass of the
population.
For the sample, X  180
Hence 95% C.I. for  is
180  1.96 10 ,180  1.96 10 


16
16 

= (175.1, 184.9)
In previous question, (175.1,184.9) is a 95%
C.I. for the true mean . Am I right in saying
that there is 95% chance that  lies in
(175.1,184.9) ?
Note 1:  is NOT a random variable! While,
the interval (L,U) is a random interval.
Note 2: We can just say that we are 95%
confident that  lies on (L,U).
How to comprehend this ?
Sample 1
Sample 2
Population
Sample n
X1
(L1,U1)
X2
..
.
(L2,U2)
Xn
(Ln,Un)
..
.
If (L1,U1), (L2,U2) , …, (Ln,Un) are 95% C.I.
then there should be 95% of theses intervals
(L1,U1), (L2,U2) , …, (Ln,Un) includes the
true mean .
For 20 95%C.I.
there should be
19 C.I. trap the
true mean.
So (175.1,184.9) is
just one of the C.I.s
and it may or may
not trap .

X
An example.
Suppose {X1, X2,…, X7} = population set.
We take 2-element samples. (n = 2)
Total possible way = 7C2 = 21
Hence we can construct 21 different C.I.s
We consider the 90% C.I.
 X  1.654s X , X  1.654s X 
See the WORDS document now.
We know 21 C.I.s, 19 of them do trap .
Please notice that 2190%  19
Also, the sample size = 2, too small!
Instead of using
sX 

n
We use the adjusted sample s.d..
N n 
sX 
N 1 n
Refer to P.81 note (ii) in text book.
E.g. 27
A certain population,  = 6. How large a
sample size => width of 95% C.I. for 
= 0.5
 X  1.96 , X  1.96 

95%C.I.= 
n
n 

Half width = 0.25
1.96 

 0.25
n
1.96  6
n
0.25
n = 2209
Do you agree?
If  is known, C.I. is
 X  zc , X  zc 


n
n

If  is unknown, C.I. is
Precisely,
 X  zc s , X  zc s 


n
n

n s
n s 

, X  zc
 X  zc

n 1 n
n 1 n 

E.g. 28
A sample of 100 plugs with mean diameter
25.10 cm. If s.d. of these plugs is 0.12,
estimate the population mean diameter at
95% confident level.
Now, we don’t know , so use sample s.d. s
n s
n s 

, X  zc
 X  zc

n 1 n
n 1 n 

100 0.12
100 0.12 

,252.10  1.96
 252.10  1.96 = (25.076,25.124)

99 100
99 100 

E.g. 31
(a) A two-stage rocket to be fired to put a satellite into
orbit. Due to variation of the specified impulse in the
second stage, the velocity imparted in this stage will be
normally distributed about 4095 ms1 with s.d. 21 ms1
Find 95% confident limits for the velocity imparted in
this stage.
v  4095 s1  21
 v  1.96 s1 , v  1.96 s1 

95% C.I. = 
n
n 

 4095  1.9621,4095  1.9621

= 
1
1 

= (4054 , 4126)
(b) In the first stage, the velocity imparted will be
normally distributed about 3990 ms1 with s.d. 20 ms1
due to variation of the specific impulse and
(independently) with s.d. 8 ms1 due to variation in the
time of burning of the change. Find 90% confident
limits for the velocity imparted in this stage.
s2 = 20, s3 = 8
Combined s.d. = s2  s3  20  8 = 21.54
2
2
2
2
 v  1.645s2 , v  1.645s2 
90% C.I. =  1

1
n
n 

= (39901.64521.54, 3990+1.64521.54)
= (3955,4025)
(c) Given that the final velocity of 8000 ms1 is
required to go into orbit and that the second stage fires
immediately after the first, find the probability of
achieving orbit.
v = 4095
s2 = 212
v1 = 3990
s12 = 202+ 82
Let V = final velocity
E(V) = 3990 + 4095 = 8085
Var(V) = 202 + 82 + 212 = 905
V ~ N(8085,905)
8000  8085
)
P(V > 8000) = P ( z 
905
= 0.9977
Prerequisite on E.g. 32
Uniform distribution
1
 r
ba
b 1
ab
 E( X )  
xdx 
ab  a
2
2
2
 Var ( X )  E ( X )  E ( X )
2
b 1
a  b

2

x dx  

ab  a
 2 
2
a  b 

12
f(x)
r
a
b
x
E.g. 32
To add 104 numbers, each of which was rounded off
with accuracy 10m degree. Assuming that the errors
arising mutually independent and uniformly
distributed on (0.510m, 0.510m), find the limits
in which the total error will lie with probability 0.99.
Let X = total error. X = X1 + X2 +…+ X10000
Since Xi is uniformly distributed,
m
m
 0.5  10  0.5  10
E( X i ) 
=0
2
m
m 2
 2m

 0.5  10  0.5  10  10
Var ( X i ) 

12
12
E ( X )  10000 E ( X i ) = 0
2 ( m2 )
10
Var ( X )  10000Var ( X i ) 
 2 ( m  2 ) 12
By CLT, X ~ N (0,10
)
12
By table,
P(2.56 < z < 2.56) = 0.99
X 0
P(2.56 
 2.56)  0.99
( m  2 )
10

 12 


Hence the limits are  2.5610( m2 )
12
Hence we can construct the 99% C.I. for total error X and this estimation is far more better! Let’s use m = 3 as an example.
|X|  0.0005104 = 5, too large for estimation! But the C.I. is (0.0739,0.0739) only, more “precise”.
Now, let’s talk about C.I. for proportion
You
have interviewed
Suppose
you want to
with
100
H.K.
and
look
into
thepeople
smoker’s
discovered
60 in
smokers.
proportion
H.K.
However, we can
construct a C.I.
to estimate the true
proportion!
Can we say the smokers’
proportion of H.K. people
is 60% ?
Let n be the sample size.
Let m be the number of “success”
(i.e. “smokers” in the e.g.)
Let p be the true proportion (of “success”)
Suppose the population is very large,
then m has a binomial distribution such that
m ~ B(n , p)
Suppose further that n is reasonably large.
We can use “normal” to approximate “binomial”.
m ~ N(np , npq)
Let Ps be the proportion on “success” in sample.
m
np
m
E ( Ps )  E ( ) 
p
Ps 
n
n
n
m Var (m) npq pq
Var ( Ps )  Var ( ) 
 2 
2
n
n
n
n
pq
Hence
Ps ~ N ( p, )
n
In practice, p is unknown.
We use Ps Qs/n to estimate pq/n.
Thus
P
Q
s
s
Ps ~ N ( p,
)
n
approximately
Hence
Ps  p
P(1.96 
 1.96)  0.95
PsQs
n
Rearranging,
PsQs
PsQs
P ( Ps  1.96
 p  Ps  1.96
)  0.95
n
n
Hence 95% C.I. for population proportion p is
PsQs
PsQs 

, Ps  1.96
 Ps  1.96

n
n 

In general, % C.I. for population proportion p is
PsQs
PsQs 

, Ps  zc
 Ps  zc

n
n 

where P(zc < z < zc) = %

n > 30 is required.
E.g. 34
4000 items, 240 defective, find 95% C.I. for the
probability p that an item is defective.
240
= 0.06 Qs = 1  0.06 = 0.94
Ps =
4000
PsQs 0.06  0.94

= 0.00375
n
4000
Required 95% C.I. is
0.06  1.96
0.00375,0.06  1.96 0.00375 
= (0.0526 , 0.00674)
E.g. 35
Suppose that we know p = 0.6 for a Bernoulli population.
How large is the size is necessary to be 95% confident
that the obtained value p lies in (0.5,0.7) ?
Let n = sample size.
(0.5,0.7) = (0.60.1,0.6+0.1)
Hence, for 95% confidence,
0 .6  0 . 4
0.1 = 1.96
n
On solving,
n  92
E.g. 37
(a) Of 50 houseflies, independently subjected to the same
insecticide, 38 were killed. Obtain an estimate of p, the
probability that a housefly is killed by the insecticide. Find
also the standard error of p.
38 19

Ps =
50 25
Ps Qs
Standard error =
n
(19 / 25)(1  19 / 25)

 0.0604
50
(b) Now conduct a larger experiment with the same
insecticide so that an estimate with standard error of
about 0.03 can be quoted. On the basis of the information
in the experiment already conducted, how many houseflies
needed ?
Ps Qs
(19 / 25)(6 / 25)

 0.03
Standard error =
n
n
So n = 203
(c) To be absolutely sure of obtaining the desired
accuracy, how many houseflies should be taken ?
 n = 203 makes standard error = 0.03 only when Ps = 19/25.
Standard error depends on Ps.
So what n to ensure s.e.  0.03 irrespective of Ps ?
Ps (1  Ps )
s.e. 
n
For fixed n, s.e. is a function of Ps.
s.e.  0.03 means max. of s.e. = 0.03.
Very easy to show that Ps(1-Ps) attains max.
when p = 0.5
0.5(1  0.5)
1
Hence s.e. 

n
4n
1
Then set
 0.03 n  279
4n
i.e. Though different samples yield different Ps, it is sure that s.e.
not greater than 0.03 if we take n = 279 (or more)