Chapter Outline2.1 Estimation•Confidence Interval Estimates for Population Mean •Confidence Interval Estimates for the Difference Between Two Population Mean•Confidence Interval Estimates for Population Proportion•Confidence Interval Estimates for the Difference Between Two Population Proportion2.2 Error of Estimation & Determining the sample size

• Statistical inference - 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.

2.1Estimation

• In estimation, there are two terms that firstly, should be understand. The two terms involved in estimation are: i)Estimator : sample statistics used to estimate a population parameter.ii)Estimate: value that obtained from a sample to estimate a population parameter.• An estimate of a population parameter may be expressed in two ways:i) Point Estimate ii) Interval Estimate.

i) Point Estimate• A point estimate of a population parameter is

a single value of a statistic. • For example, the sample mean is a point

estimate of the population mean μ. • Similarly, the sample proportion is a point

estimate of the population proportion P.

ii) Interval Estimate• An interval estimate is defined by two

numbers, between which a population parameter is said to lie.

• For example, a < μ < b is an interval estimate of the population mean μ. It indicates that the population mean is greater than a but less than b.

x

p̂

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

X

2

1

2

1

1

n

iI XX

nS

P̂

Point estimators for mean, variance, and proportion

2.1.1 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

2

2 2

2

The (1 )100% Confidence Interval of Population Mean,

(i) if   is known and normally distributed population

or

(ii) if    is unknown, large (

x zn

x z x zn n

sx z n n

n

2 2

30)

or s s

x z x zn n

i) Confidence Interval Estimates for Population Mean,( )

1, 2

1, 1,2 2

(iii) if    is unknown, normally distributed population

and small  sample size 30     

or

n

n n

sx t

n

n

s sx t x t

n n

Example:

2

If a random sample of size 20 from a normal population

with the variance 225 has the mean 64.3, construct

a 95% confidence interval for the population mean, .

n

x

Solution:

0.025

2

It is known that, 20,  64.3 and 15

For 95% CI,

95% 100(1– )%

1–  0.95

  0.05   

0.0252

1.96

                    

n x

z z

2

Hence, 95% CI   

15 64.3 1.96

20

64.3 6.57

[57.73,70.87]

@

x zn

57.73 70.87

Thus, we are 95% confident that the mean of random variable

is between 57.73 and 70.87

The brightness of a television picture tube can be evaluated by measuring theamount of current required to achieve a particular brightness level. A

randomsample of 10 tubes indicated a sample mean microamps and a

samplestandard deviation is microamps. Find (in microamps) a 99%

confidenceinterval estimate for mean current required to achieve a particular brightnesslevel.

Solution:

For 99% CI:

From t normal distribution table:

317 2x .15 7s .

15 7 10 30 317 2s . , n , x .

99 1 100

1 0 99

0 01

0 0052

% %

.

.

.

0 005 9

2

1 3 250. ,t ,n t .

Example:

Hence 99% CI

Thus, we are 99% confident that the mean mean current required to achieve a

particular brightness level is between 301.0645 and 333.3355

0 005 9

15 7317 2

10

15 7317 2 3 250

10

301 0645 333 3355 microamps

. ,

.. t

.. .

. , .

Exercise:

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 estimate the mean waiting time for all individuals waiting in the service line.

i) Variance and are known:

ii) If the population variances, and are unknown, then the following tables shows the different formulas that may be used depending on the sample sizes and the assumption on the population variances.

21 2

2

2 2

1 21 2

1 22

X X Zn n

21 2

2

ii) Confidence Interval Estimates for the Differences Between Two Population Mean,

1 2

Equality of variances, when are unknown

Sample size

1 230 30n , n 2 2

1 2, 1 230 30n , n

2 21 2

2 2

1 21 2

1 22

s sX X Z

n n

2 21 2

1 21 22

22 21 2

1 22 22 2

1 2

1 2

1 2

1 1

,v

s sX X t

n n

s sn n

vs sn n

n n

2 21 2

1 21 22

2 21 1 2 22

1 2

1 1

1 1

2

p

p

X X Z Sn n

n s n sS

n n

1 21 22

2 21 1 2 22

1 2

1 2

1 1

1 1

2

2

p,v

p

X X t Sn n

n s n sS

n n

v n n

Two machines are used to fill plastic bottles with liquid laundry detergent. The standard deviations of fill volume are known to be and fluid ounce for the two machines, respectively. Two random samples of bottles from the machine 1 and bottles from machine 2 are selected,and the sample means fill volume are and fluid ounces.Construct a 90% confidence interval on the mean difference in fill volumes.Interpret the results.

Solution:

For 90% CI:

1 0 10. 2 0 15. 1 14n

2 12n 1 30 5x . 2 29 4x .

1 2

1 2

1 2

Machine 1: Machine 2:

30 5 29 4

0 10 0 15

14 12

x . x .

. .

n n

1 100 90

1 0 90

0 1

0 052

%

.

.

.

Example:

We are 90% confidence that the mean difference to fill volumes lies between 1.0163 and 1.1837 fluid ounces.

2 2 2 21 2

1 2 0 051 22

0 10 0 1530 5 29 4

14 12

1 1 1 6449 0 0509

1 0163 1 1837

.

. .X X Z . . Z

n n

. . .

. , .

A study was conducted to compare the starting salaries for universitygraduates majoring in computer science and engineering. A random sample of 50 recent university graduates in each major were selected and the following information was obtained.

Construct a 99% confidence interval for the difference in the mean starting salaries for two majors.

Solution:

Major Mean SD

Computer Science,

RM 2500 RM 100

Engineering RM 2800 RM 150

2 2 2 2

0 005

2

100 1502500 2800

50 50

300 2 5758 650 365 6703 234 3297

c ec e .

c e

s sX X Z Z

n n

. . , .

Example:

We are 99% confidence that the mean difference of starting salaries for to major lies between -365.6703 and -234.3297.

Exercise:18 male undergraduate students and 20 female undergraduate

students arerandomly selected from faculty of mechanical engineering.

Result for test 2SSM 3763 shown the following data:

Assume that both population are normally distributed and have equal

population variances. Construct a 95% confidence interval for the difference

in the two means.

Male : 82 8

Female : 76 6M M

F F

X , S

X , S

iii) Confidence Interval Estimates for Population Proportion,( )p

2

The (1 )100% Confidence Interval for for Large Samples ( 30)

ˆ ˆ1ˆ

p n

p pp z

n

Example:According to the analysis of Women Magazine in June 2005, “Stress hasbecome 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 be the corresponding sample proportion. From the given information,

n = 1502 , = 0.40 , = 1 – 0.40 = 0.60

p̂ pq ˆ1ˆ

p̂

Hence, 95% CI :

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%.

2

0 025

0 4 0 60 40 0 4 0 01264069

1502

0 375 0 425 or 37 5% to 42%

.

ˆ ˆpqp̂ Z

n

. .. Z . .

. , . .

Exercise:

The wedding ceremony for a couple, Jamie and Robbin will be held in

Menara Kuala Lumpur. A survey has been carried out to determine the proportion of people who will come to the ceremony. From 250 invitations, only 180 people agree to attend the ceremony. Find a 90% confidence interval estimate for the proportion of all people who will attend the ceremony.

iv) Confidence Interval Estimates for the Differences Between Two Population Proportion,

1 2p p

Two separate surveys were carried out to investigate whether or not theusers of Plus highway were in favour of raising the speed limit on highways.Of the 250 car drivers interviewed, 220 were in favour of raising the speedlimit while of the 200 motorists interviewed , 180 were in favour. Find a 95%confidence interval for the difference in proportion between the car driversand motorist who are in favour of raising the speed limit.

Solution:

Hence, 95% CI :

220 1800 88 0 9

250 200c mˆ ˆp . , p .

0 025

2

0 88 0 12 0 9 0 10 88 0 9

250 200

0 02 1 9600 0 03

0 0788 0 0388

c c m mc m .

c m

ˆ ˆ ˆ ˆp q p q . . . .ˆ ˆp p Z . . Z

n n

. . .

. , .

Example:

We are 95% confident that the difference between the car drivers and motorist who are in favour of raising the speed limits lies between -0.0788 and 0.0388.

2.2 Error of Estimation and Choosing the Sample Size

When we estimate a parameter, all we have is the estimate value from n measurements contained in the sample. There are two questions that usually arise:

(i)How far our estimate will lie from the true value of the parameter?

(ii) How many measurements should be considered in the sample?

The distance between an estimate and the estimated parameter is called the error of estimation.

For example if most estimates are within 1.96 standard deviations of the true value of the parameter, then we would expect the error of estimation to be less than 1.96 standard deviations of the estimator, with the probability approximately equal to 0.95.

n

szB

nzB

22

or

In the process of determining the sample size, we have to define the parameter to be estimated and standard error of its point estimator.

Firstly, choose the bound (B) on the margin of error and confidence coefficient (1-α).

Then, use the following equation to find suitable sample size, n:

n

qpzB

ˆˆ

2

Example:The college president asks the statistics teacher to estimate

the average ageof the students at their college. The statistics teacher would

like to be 99%confident that the estimate should be accurate within 1 year.

From theprevious study, the standard deviation of the ages is known to

be 3 years.How large a sample is necessary?

Solution:

1 3 confidence coefficient 99%, thus 1 0 99

0 01 0 0052

B , s , .

. , .

0 005 2 5758.Z .

n

n

nz

n

szB

338882.0

35758.21

31 005.0

2

6071.59

91507.0

3)38882.0(

:sideboth Square2

2

nn

n

Example:How large a sample required if we want to be 95% confident

that the error in using to estimate p is less than 0.05? If

, find the required sample size.

p̂ˆ 0.12p

Solution:

n

n

nz

n

qpzB

)88.0)(12.0(0255.0

)88.0)(12.0(96.105.0

)88.0)(12.0(05.0

ˆˆ

025.0

2

16239.162

1056.0)0255.0(

:sideboth Square2

2

nn

96.105.095.01

025.0

z

Exercise:

The diameter of a two years old Sentang tree is normally distributed with a Standard deviation of 8 cm. how many trees should be sampled if it is required to estimate the mean diameter within ± 1.5 cm with 95% confidence interval?