Download chapter7

CHAPTER SEVEN
ESTIMATION
7.1 A Point Estimate:
A point estimate of some population parameter  is a
single value  of a statistic ̂ (parameter space). For
example of the statistic x computed from a sample
of size n is a point estimate of the population
parameter µ .Similarly pˆ  X is a point estimate of the
n
true proportion P for a binomial experiment. x is an
estimator of the population parameter µ , but the
value of x is an estimate of µ .
The statistic that one uses to obtain a point estimate is
called an estimator or a decision function.
n
 (X
i
 X )2
Hence the decision function S  n  1
which is a function of the random sample is an
estimator of the population variance , σ² .The
single numerical value that results from
evaluating this formula is called an estimate of
the parameter, σ². An estimator is not
expected to estimate the population parameter
without error.
2
i 1
7.2 Interval Estimate:
An interval estimate of a population parameter  is
an interval of the form ˆ    ˆ where ˆL and ˆU depend
on the value of the statistic ˆ for a particular sample
and also on the sampling distribution of ̂ .Since
different samples will generally yield different values of ̂
and therefore different values of ˆL and ˆU . From the
sampling distribution of ˆ we shall be able to
determine ˆL and ˆU such that the P(ˆ    ˆ ) is equal to
any positive fractional value we care to specify. If for
instance we findˆL and ˆU such that:
L
U
L
U
P(ˆL    ˆU )  1  
for
0  1
, then we have a probability of (1   ) of selecting a
random sample that will produce an interval containing  .
The interval ˆ    ˆ computed from the selected sample, is
then called a (1   )100% confidence interval, the fraction(1   )
is called confidence coefficient or the degree of confidence
and the end points ˆL and ˆU are called the lower and upper
confidence limits.
Thus when   0.05 we have a 95% confidence interval and so
on, that we are 95% confident that  is between ˆL , ˆU
L
U
7.3 Estimating the Mean:
7.3.1 Confidence Interval of when is Known:
--------------------------------------------------------------If X is the mean of a random sample of size n from a
population with known variance σ² , a (1   )100%
confidence interval for µ is given by:
X  Z 1 / 2


   X  Z 1 / 2
n
n
(1)
where Z 1 / 2 is the -value leaving an area of
right.

ˆL  X  Z 1 / 2
n

, ˆU  X  Z 1 / 2
n

2
(2)
to the
EX (1):
The mean of the quality point averages
of a random sample of 36 college seniors
is calculated to be 2.6.
Find the 95% confidence intervals for
the mean of the entire senior class.
Assume that the population standard
deviation is 0.3.
Solution:
n  36 , X  2.6 ,   0.3
confidence interval for the mean µ :
------------------------------------------------At 1    0.95    0.05  2  0.025  Z  1.96
95%
1
X Z
2.6  (2.57)(
1

2

2

n
0.3
)  2.6  0.1285
36
2.4715    2.7285  P (2.4715    2.7285)  0.99
Thus,we have 95% confident that µ lies between 2.4715,2.7285
Theorem (1):
If X is used as an estimate of µ , we can then be (1   )100%

confident that the error will not be exceed Z 1  n .
2
For example (1): e= (1.96) (0.3/6)=0.098 or
e = (2.57) (0.3/6) = 0.1285
Theorem (2):
If X is used as an estimate of µ , we can be (1   )100%
confident that the error will not exceed a specified
amount, e when the sample size is:
 Z 
1
2
n 

e







2
(3)
The fraction of n is rounded up to next whole number.
EX (2):
How large a sample is required in Ex. (1) if we want to
be 95% confident that our estimate of µ is off by less
than 0.05 (the error is 0.05)?
Solution   0.05    0.025  Z
 1.96,
1
2

2
  0.3 , e  0.05
2
 (1.96)(0.3) 
n 
  138.2976  138
0.05


n is rounded up to whole number.
7.3.2 Confidence Interval of when
is Unknown n<30 :
If X and S are the mean and standard deviation of a
random sample from a normal population with
unknown variance σ² , a(1   )100 % confidence interval
for µ is given by:
X t
t
1

, n 1
2
S
S
   X t 
1 , n 1 n
n
2
(4)
where 1 2 is the value with n-1 degrees of freedom
leaving an area of 2 to the right.
EX (3):
The contents of 7 similar containers of
sulfuric acid are 9.8, 10.2, 10.4, 9.8, 10,
10.2, 9.6 liters. Find a 95%confidence
interval for the mean of all such
containers assuming an approximate
normal distribution.
Solution:
confidence interval for the mean :
-----------------------------------------------n  7 , X  10 , S  0.283 ,

at :1    0.95    0.05   0.025  t 
 t 0.025,6  t 0.975 , 6  2.447
1 , n 1
2
2
S
0.283
X t 
( )  10  (2.447)(
)  10  0.262
1 , n 1
n
7
2
9.738    10.262  P (9.738    10.262)  0.95
7.4 Estimating the Difference
between Two Means:
7.4.1 Confidence Interval for 1   2 when  12 and  22
Known:
----------------------------------------------------------------If X and X are the means of independent random
samples of size n1 and n2 from populations with
known variances  and  respectively, a (1   )100%
confidence interval for    is given by:
1
2
2
1
2
2
1
(X
where Z
right .

1
2
1
X
2
)Z
1

2
2
 12
n1

 22
n2
(5)
is the -value leaving an area of

2
to the
EX (4):
A standardized chemistry test was given to 50 girls and
75 boys. The girls made an average grade of 76, while
the boys made an average grade of 82. Find a 96%
confidence interval for the difference    where 
is the mean score of all boys and  is the mean score
of all girls who might take this test. Assume that the
population standard deviations are 6 and 8 for girls
and boys respectively.
1
2
2
1
Solution:
girls
Boys
n1  50
n 2  75
X 1  76
1  6
X 2  82
2  8
96% confidence interval for the mean 1  2 :
----------------------------------------------------
1    0.94    0.04 
(X 1  X 2 )  Z
1

2
(82  76)  (2.05)
 12
n1

2
 0.02  Z
1

 2.05
2
 22
n2
36 64

 6  2.571
50 75
3.429  1  2  8.571  P (3.429  1  2  8.571)  0.96
7.4.2 Confidence Interval for    when  12 and  22
Unknown but equal variances:
1
2
X 1 and X 2 are the means of independent random samples of
size n1 and n2 respectively from approximate normal(1   )100%
populations with unknown but equal variances, a
confidence interval for 1   2 is given by:
(X 1  X 2 )  t
, where
SP 
1

2
,v
SP
1
1

n1 n 2
(n1  1)S 12  (n 2  1)S 22
n1  n 2  2
(6)
(7)
is the pooled estimate of the population standard deviation
and t  is the t – value with v  n1  n2  2 degrees of
freedom leaving an area of 2 to the right.
1
2
, v
EX (5):
The independent sampling stations were chosen for this
study, one located down stream from the acid mine
discharge point and the other located upstream. For 12
monthly samples collected at the down stream station
the species diversity index had a mean value X 1  3.11
and a standard deviation S  0.771 while 10 monthly
samples had a mean index value X 2  2.04 and a
standard deviation S  0.448 . Find a 90% confidence
interval for the difference between the population
means for the two locations, assuming that the
populations are approximately normally distributed
with equal variances.
1
2
Solution:
Station 1
Station 2
n1  12
n 2  10
X 1  3.11
X 2  2.04
S 1  0.771
S 2  0.448
90% confidence interval for the mean 1  2 :
11(0.771) 2  9(0.448) 2
SP 
 0.646
12  10  2
at 1    0.90    0.1 

(3.11  2.04)  (1.725)(0.646)
2
 0.05  t
1

2
, n1  n 2  2
 t 0.95 , 20  1.725
1
1

 1.07  0.477
12 10
0.593  1  2  1.547  P (0.593  1  2  1.547)  0.90
7.5 Estimating a Proportion:
7.5.1 Large – Sample Confidence Interval for P:
If p̂ is the proportion of successes in a random sample
of size n and qˆ  1  pˆ an approximate (1   )100%
confidence interval for the binomial parameter is given
ˆˆ
by: pˆ  Z  pq
(8)
1
2
n
Where Z 1  is the Z- value leaving an area of
right. 2

2
to the
EX(7):
A new rocket – launching system is being considered for
deployment of small, short – rang rockets. The existing
system has p=0.8 as the probability of a successful
launch. A sample of 40 experimental launches is made
with the new system and 34 are successful. Construct
a 95% confidence interval for p .
Solution:
a 95% confidence interval for p .
n  40 ,
pˆ 
34
 0.85 , qˆ  0.15
40
at 1    0.95   0.05 
pˆ  Z
1

2

2
 0.025  Z
1

 1.96
2
ˆˆ
pq
(0.85)(0.15)
 0.85  (1.96)
 0.85  (0.111)
n
40
0.739  p  0.961  P (0.739  p  0.961)  0.95
Theorem 3:
If
p̂
is used as an estimate of p we can be (1   )100%
confident that the error will not exceed
e Z
1

2
ˆˆ
pq
n
.
EX(8):
In Ex. 7, find the error of p .
Solution:
The error will not exceed the following value:
e Z
1

2
ˆˆ
pq
(0.85)(0.15)
 (1.96)
 0.111
n
40
Theorem 4:
If p̂ is used as an estimate of p we can be (1   )100%
confident that the error will be less than a specified
amount e when the sample size is approximately:
n 
Z 12
2
2
ˆˆ
pq
e
Then the fraction of n is rounded up.
(9)
EX(9):
How large a sample is required in Ex. 7 if we want to be
95% confident that our estimate of p is within 0.02?
Solution:
e  0.02 , Z

1
2
 1.96 ,
pˆ  0.85 , qˆ  0.15
(1.96)2 (0.85)(0.15)
n
 1224.51  1225
2
(0.02)