Download Estimation - Portal UniMAP

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Statistical Inference
 Estimation
-Confidence interval estimation for mean and
proportion
-Determining sample size
 Hypothesis Testing
-Test for one and two means
-Test for one and two proportions
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Statistical Inference
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Statistical inference is a process of drawing an inference
about the data statistically.
It concerned in making conclusion about the
characteristics of a population based on information
contained in a sample.
Since populations are characterized by numerical
descriptive measures called parameters, therefore,
statistical inference is concerned in making inferences
about population parameters.
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ESTIMATION
In estimation, there are two terms that firstly, should be
understand. The two terms involved in estimation are
estimator and estimate.
An estimate of a population parameter may be expressed in
two ways: point estimate and interval estimate.
Point Estimate
A point estimate of a population parameter is a single value
of a statistic. For example, the sample mean x is a point
estimate of the population mean μ. Similarly, the sample
proportion p̂ is a point estimate of the population
proportion p.
Interval estimate
An interval estimate is defined by two numbers, between
which a population parameter is said to lie.
For example, a < x < b is an interval estimate of the
population mean μ. It indicates that the population mean is
greater than a but less than b.
Point estimators
Choosing the right point estimators to estimate a parameter
depends on the properties of the estimators it selves. There
are four properties of the estimators that need to be satisfied
in which it is considered as best linear unbiased estimators.
The properties are:
 Unbiased
 Consistent
 Efficient
 Sufficient
Confidence Interval
• A range of values constructed from the sample data. So that the
population parameter is likely to occur within that range at a
specified probability.
• Specified probability is called the level of confidence.
• States how much confidence we have that this interval contains
the true population parameter.The confidence level is denoted by
• Example
:- 95% level of confidence would mean that if 100
confidence intervals were constructed, each based on the different
sample from the same population, we would expect 95 of the
intervals to contain the population mean.
 To compute a confidence interval, we will consider
two situations:
i. We use sample data to estimate,  with X and the
population standard deviation  is known.
ii. We use sample data to estimate,  with X and the
population standard deviation is unknown. In this
case, we substitute the sample standard deviation
(s) for the population standard deviation 
Example 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.
Solution:
(a) The point estimate of the mean price of all such college
textbooks is
RM70.50, that is Point estimate of μ = x = RM70.50
(b) It is known that, n = 36, μ = x = RM70.50 and RM4.50
For 90% CI
90%  1   100%
1    0.90
  0.1

2
 0.05
From normal distribution table: z  z0.05  1.65
2
Hence, 90% CI:  x  Z    
n
2


 4.5 
 70.50  1.6449 

 36 
 70.50  1.2337
  RM69.26 , RM71.73
Thus, we are 90% confident that the mean price of all such college textbooks is
between RM69.26 and RM 71.73.
Example 2:
The brightness of a television picture tube can be evaluated by
measuring the amount of current required to achieve a particular
brightness level. A random sample of 10 tubes indicated a sample
mean 317.2microamps and a sample standard deviation is
15.7microamps. Find (in microamps) a 99% confidence interval
estimate for mean current required to achieve a particular
brightness level.
Solution:
s  15.7
x  317.2
s  15.7, n  10  30, x  317.2
For 99% CI: 99%  1   100%
1    0.99
  0.01

 0.005
2
From t normal distribution table:
t ,n  1  t0.005 ,9  3.250
2
Hence 99% CI
 15.7 
 317.2  t0.005 ,9 

10


 15.7 
 317.2   3.250  

10


  301.0645,333.3355  microamps
Thus, we are 99% confident that the mean current required to
achieve a particular brightness level is between 301.0645 and
333.3355
Exercise 1:
Taking a random sample of 35 individuals waiting to be
serviced by the teller, we find that the mean waiting
time was 22.0 min and the standard deviation was 8.0
min. Using a 90% confidence level, estimate the mean
waiting time for all individuals waiting in the service
line.
Answer : [19.7757, 24.2243]
Example 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 is 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 pˆ be
the corresponding sample proportion. From the given
information,
n = 1502 , pˆ = 0.40 , qˆ =1− pˆ = 1 – 0.40 = 0.60
Hence, 95% CI :
 p̂  Z 
2
ˆpqˆ
n
 0.40  Z 0.025
 0.4  0.6   0.4  0.01264069
1502
  0.375,0.425  or 37.5% to 42%
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%.
Exercise 2
In a random sample of 70 automobiles registered in a
certain state, 28 of them were found to have emission
levels that exceed a state standard. Find a 95%
confidence interval for the proportion of automobiles in
the state whose emission levels exceed the standard.
Answer : [0.2852, 0.5148]
Exercise 3:
A paint manufacturing company claims that the
mean drying time for its paint is at most 45
minutes. A random sample of 20 trials tested. It is
found that the sample mean drying time is 49.50
minutes with standard deviation 3 minutes. Assume
that the drying times follow a normal distribution.
Construct a 99% confidence interval for the mean
drying time of the paint. Explain your answer.
Answer : [47.58, 51.42]