Chapter 8: Estimation
8.1 Point and Interval Estimates
Definition 8.1: A Point Estimate
The value of the sample statistics that is used to estimate a population parameter
is called a point estimate.
A sample statistics or briefly statistics is said to be the best linear estimator of
population parameters or briefly parameters, the estimator must be;
(i) Unbiased
(iii) Efficient
(ii) Consistent
(iv) Sufficient
Definition 8.2: Properties of Estimators
Unbiased Estimator
A statistics ˆ is said to be an unbiased estimator of parameter  if,
E (ˆ)  
Consistent Estimator
Let ˆ is unbiased estimator of  . If V( ˆ )  0 when sample size, n   hence
ˆ is said to be consistent estimator of  .
Efficient Estimator
An estimator is said to be more efficient from the others estimator if its variance
is smaller than variance of the others estimator.
Sufficient Estimator
 is a sufficient estimator if it makes use of all of the information in the sample.
The sample variance, for example, is more sufficient than the sample range,
since the former uses all the scores in the sample, the latter only two.
Definition 8.3: An Interval Estimate
In interval estimation, an interval is constructed around the point estimate and it
is stated that this interval is likely to contain the corresponding population
parameter.
Definition 8.4: Confidence Level and Confidence Interval
Each interval is constructed with regard to a given confidence level and is called
a confidence interval. The confidence level associated with a confidence interval
states how much confidence we have that this interval contains the true
population parameter. The confidence level is denoted by 1   100% .
8.1.1 Confidence Interval Estimates of Population Parameters
The (1-  )100% Confidence Interval for  for Large Samples (n  30)
(i)
(ii)
x  z
x  z

n
2
2
s
n
if  is known and normally distributed population
if  is not known, n large
x  z
or
2
s
s
   x  z
2
n
n
with (1-  )% CI
The (1-  )100% Confidence Interval for  for Small Samples (n<30)
(i)
(ii)
x  z
2
x  t n 1,
s
n
2
if  is known
s
if  is not known
n
The (1-  )100% Confidence Interval for p for Large Samples (n  30)
pˆ  z
2
pˆ 1  pˆ 
n
Example 8.1 :
A publishing company has just published a new textbook. Before the company
decides the price at which to sell this textbook, it wants to know the average
price of all such textbooks in the market. The research department at the
company took a sample of 36 comparable textbooks and collected the
information on their prices. This information produced a mean price RM 70.50
for this sample. It is known that the standard deviation of the prices of all such
textbooks is RM4.50.
(a) What is the point estimate of the mean price of all such college
textbooks?
(b) Construct a 90% confidence interval for the mean price of all such college
textbooks.
Solutions
(a) The point estimate of the mean price of all such college textbooks is
RM 70.50, that is
Point estimate of   x  RM 70.50
(b) It is known that, n = 36,   x  RM 70.50 and   RM 4.50 .
For 90% CI,
90% = 100(1–  )%
1–  = 0.90
 = 0.1

2
 0.05
z  z0.05  1.65
2
Hence, 90% CI
From normal distribution
table,
 
= x  z 

n
2
 4.50 
= 70.50  1.65

 36 
= 70.50  1.24
= [ RM 69.26, RM 71.74]
Thus, we are 90% confident that the mean price of all such college textbooks is
between RM 69.26 and RM 71.74
Example 8.2
Consider a survey on male students height in a certain IPTA, a random sample
of 100 male students are taken. The height of the male students is normally
distributed with mean 178.2 cm dan variance 17.75 cm2.
a) Construct a 95% CI for mean of the male students height
b) If mean of the female students height is 170.2 cm, at 98% CI, verify whether
if this can proof that male students are taller than female students.
Solutions
a) It is known that, n = 100,   x  178.2 and 2  17.75 .
For 95% CI,
95% = 100(1–  )%
1–  = 0.95
 = 0.05

 0.025
2
From normal distribution
table,
z   z0.025  1.96
2
Hence, 95% CI
 
= x  z 

n
2
 17.75 

= 178 .2  1.96
 100 
= 178.2  0.83
= [177.37, 179.03]
b) It is known that,   x  178.2 and 2  17.75 .
For 98% CI,
98% = 100(1–  )%
1–  = 0.98
 = 0.02

 0.01
2
From normal distribution
table,
z   z0.01  2.33
2
Hence, 98% CI
 
= x  z 

n
2
 17.75 

= 178 .2  2.33
100


= 178.2  0.98
= [177.22, 179.18]
We can see that mean of female students does not lies in the interval
[177.22, 179.18] hence, this indicate that male students are taller than
female students.
Example 8.3
According to the analysis of Women Magazine in June 2005, “Stress has
become a common part of everyday life among working women in Malaysia.
The demands of work, family and home place an increasing burden on average
Malaysian women”. According to this poll, 40% of working women included in
the survey indicated that they had a little amount of time to relax. The poll was
based on a randomly selected of 1502 working women aged 30 and above.
Construct a 95% confidence interval for the corresponding population
proportion.
Solution
Let p be the proportion of all working women age 30 and above, who have a
limited amount of time to relax, and let p̂ be the corresponding sample
proportion. From the given information,
n = 1502 , p̂ =0.40, qˆ  1  pˆ = 1 – 0.40 = 0.60
Hence, 95% CI
 pˆ qˆ 

= pˆ  z  
n
2

 0.40(0.60) 

= 0.40  1.96
1502


= 0.40  0.01264069
= [0.375, 0.425] or 37.5% to 42.5%
Thus, we can state with 95% confidence that the proportion of all working
women aged 30 and above who have a limited amount of time to relax is
between 37.5% and 42.5%.