STATISTICS FOR MANAGEMENT-II
LEARNING OBJECTIVES:
 To learn how to estimate certain characteristics of a population from
sample.
 To learn the strengths and shortcomings of point estimates and
interval estimates.
 To calculate how accurate our estimates really are.
 To learn how to use the t- distribution to make interval estimates.
INSTRUCTOR’S NAME
1
CHAPTER-7
Statistical Inference:
The process of drawing inferences about population on the basis of
information contained in sample (taken from population) is called
statistical inference.
Statistical inference is divided in to two major areas,
1- Estimation of parameters.
2- Testing of hypothesis
In this chapter we deal only with estimation of parameters.
It is a process by which we obtained an estimate of the unknown population
parameter by using samples (taken from population).
For example, we may estimate the population mean (µ) by calculating the
mean of sample (taken from population).
Sample statistic used to estimate population parameter, is called estimator.
For example x is estimator of µ , S is estimator of 𝜎 etc.
2
1- Unbiasedness
2- Efficiency
3- Consistency
4- Sufficiency
1-Unbiasedness:
An estimator is unbiased if mean of its sampling distribution is equal to the
p population parameter.
For example sample mean x is unbiased estimator of population mean µ.
because mean of sampling distribution of sample mean is equal to population
mean i.e. E ( x ) = µ
2- Efficiency:
An estimator is efficient if it has smaller standard error than any other
estimator of that population parameter.
For example sample mean 𝑥 is efficient estimator of population median than
sample median because in large samples, the sample mean has small
standard error than sample median.
3
3-Consistency:
An estimator is consistence if as sample size is increases, the value of that
estimator becomes very close to the value of population parameter.
For example sample mean 𝑥 is consistent estimator of the population median
Because as sample size increases the value of sample mean becomes very
close to the population median.
4-Sufficiency :
An estimator is sufficient if it uses all information that is contained in
sample about parameter. Any estimator that is not computed from all values
in samples is not sufficient.
An estimate is the numerical value of an estimator .It is used to estimate
unknown parameter.
There are two types of estimate,
1-Point Estimate
2-Interval Estimate
1-Point estimate:
A single numerical value of an estimator is called point estimate.
Here are some estimators used to get point estimate.
x =
i)
ii)
𝑆2 =
S
∑𝑥
𝑛
∑( 𝑥− x
=√
)2
𝑛
∑( 𝑥− x
𝑛
)2
or
or
∑ 𝑥2
𝑛
∑ 𝑥2
√
𝑛
- (
∑𝑥 2
)
𝑛
− (
∑𝑥
𝑛
{𝑏𝑖𝑎𝑠𝑒𝑑 𝑒𝑠𝑡𝑖𝑚𝑎𝑡𝑜𝑟 }
)2
4
iii)
S2 =
S
iv)
∑( 𝑥− x
)2
𝑛−1
=√
∑ ( 𝑥− x
or
)2
𝑛−1
∑ 𝑥2
𝑛−1
-
𝑛( x )2
𝑛−1
∑ 𝑥2
or √
−
𝑛−1
{𝑢𝑛𝑏𝑖𝑎𝑠𝑒𝑑 𝑒𝑠𝑡𝑖𝑚𝑎𝑡𝑜𝑟}
𝑛( x )2
𝑛−1
𝑋
P =
𝑛
Ex:
Find point estimates of the mean, variance and standard
deviation of population from which the given sample was drawn.
45
32
37
46
39
36
41
48
36
Solution:
𝑥
𝑥− x
45
32
37
46
39
36
41
48
36
5
-8
-3
6
-1
-4
1
8
-4
360
a)
n
( 𝑥 − x )2
25
64
9
36
1
16
1
64
16
232
= 9
∑ 𝑥 = 360
x
=
∑𝑥
𝑛
5
360
=
S2 =
b)
=
c)
9
= 40
∑( 𝑥− x
)2
𝑛−1
232
9−1
232
=
∑ ( 𝑥− x
= √
S
= 29
8
)2
𝑛−1
= √
232
= √
9−1
232
8
=√29 = 5.39
Ex :
(EX Sc 7-1 pg 352)
Find point estimates of the mean and variance and standard deviation
of population from which the given sample was drawn.
Solution:
𝑥
𝑥− x
8.8
14.0
21.3
7.9
12.5
20.6
16.3
14.1
13.0
128.5
a)
-5.5
-0.3
7.0
-6.4
-1.8
6.3
2.0
-0.2
-1.3
( 𝑥 − x )2
30.25
0.09
49.00
40.96
3.24
39.69
4.00
0.04
1.69
168.96
n = 9
∑𝑥
= 128.5
6
x
=
=
b)
S2
=
=
c)
S
𝑛
128.5
= 14.3
9
∑( 𝑥− x
)2
𝑛−1
168.96
=
9−1
∑ ( 𝑥− x
= √
168.96
8
= 21.12
)2
𝑛−1
= √
H.W:
∑𝑥
168.96
9−1
= √
168.96
8
=√21.12 = 4.60
D0 EX 7-11 (pg 353)
2-Interval Estimate:
A range of values to estimate an unknown parameter is called interval
estimate.
Confidence Interval:
Confidence interval is the range of the estimate.
We express the confidence interval in terms of standard error rather than in
numerical values. i.e. x + 𝑧 σx̄
and x − 𝑧 σx̄
7
Confidence Limit:
The lower limit ( x − 𝑧 σx̄ ) and upper limit ( x + 𝑧 σx̄ ) of confidence interval
are called confidence limits.
Confidence Level:
The probability associated with interval estimate is called confidence level.
EX :
(EX Sc 7-5 pg 360)
Express the lower and upper limits of the confidence interval for the given
levels in terms of x and σx̄ .
a) 54%
b) 75%
c) 94%
d) 98%
Solution:
a)
P(z) = 54%
=
=
54
100
0.54
2
= 0.54
= 0.2700 = 0.74
Upper limit of confidence interval = x + 𝑧 σx̄
= x + 0.74 σx̄
Lower limit of confidence interval = x − 𝑧 σx̄
= x − 0.74 σx̄
Or
Confidence limits
= x ± 0.74σx̄
b) P(z)
= 75%
=
=
75
100
0.75
2
= 0.75
= 0.3750 = 1.15
8
Upper limit of confidence interval = x + 𝑧 σx̄
= x + 1.15 σx̄
Lower limit of confidence interval = x − 𝑧 σx̄
= x − 1.15 σx̄
Or
Confidence limits
= x ± 1.15σx̄
c) P(z)
= 94%
=
=
94
100
= 0.94
0.94
= 0.4700 = 1.88
2
Upper limit of confidence interval = x + 𝑧 σx̄
= x + 1.88 σx̄
Lower limit of confidence interval = x − 𝑧 σx̄
= x − 1.88 σx̄
or
=x
Confidence limits
± 1.88σx̄
d) P(z) = 98%
=
=
98
100
= 0.98
0.98
2
=
0.4900
= 2.33
Upper limit of confidence interval = x + 𝑧 σx̄
= x + 2.33 σx̄
Lower limit of confidence interval = x − 𝑧 σx̄
= x − 2.33 σx̄
Or
Confidence limits
H.W:
= x
± 2.33σx̄
D0 EX 7-25 (a, b, c) (pg 361)
9
𝜎
Known ,
n>30 ,
population infinite
EX :
(EX 7-28 PG 365)
From sample of 250 from a population with known 𝜎 of 13.7, the sample
mean is found to be 112.4,
Find a 95% confidence interval for the mean.
Solution:
a)
n = 250
𝜎 = 13.7
x = 112.4
σx̄ =
𝜎
√𝑛
13.7
=
P(z)
= 0.87
√250
= 95%
=
=
95
100
0.95
= 0.95
=
2
x
0.4750
= 1.96
± 𝑧σx̄
x − 𝑧 σx̄
x + 𝑧 σx̄
112.4 – 1.96(0.87)
112.4-1.71
110.69
So 95% confidence interval
𝜎
Known ,
112.4+1.96(0.87)
112.4+1.71
114.11
=
(110.69 , 114.11)
n> 30
,
population finite
EX:
From a sample of 200 from population of 1000 with known 𝜎 of 1.08, the
sample mean is found to be 69.2,
Find 95% confidence interval for the mean.
10
Solution:
n
N
= 200
= 1000
x = 69.2
𝜎 = 1.08
σx̄
x √
√𝑛
1.08
=
P (z)
𝜎
=
√200
x √
𝑁−𝑛
𝑁−1
1000−200
1000−1
X √
800
=
0.08
=
=
0.08 x 0.89
0.07
999
= 95%
=
95
100
=
0.95
2
= 0.95
= 0.4750 = 1.96
x
x − 𝑧 σx̄
± 𝑧σx̄
69.2 – 1.96(0.07)
69.2- 0.14
69.06
x + 𝑧 σx̄
69.2+ 1.96(0.07)
69.2 + 0.14
69.34
So 95% confidence interval = (69.06 , 69.34)
H.W:
Do EX 7-28 (b) pg 365.
11
𝜎
n > 30
Unknown ,
,
population infinite
EX :
A sample of 100 is taken from population with sample mean 182 and sample
standard deviation 17.29.Calculate 99% confidence interval for the mean.
Solution:
n
= 100
x
= 182
𝜎̂ = s
=17.29
𝜎
̂
̂ =
σx̄
√𝑛
=
P(z)
17.29
= 1.73
√100
= 99%
=
=
x
99
100
= 0.99
0.99
= 0.4950 = 2.58
2
̂
± 𝑧 σx̄
̂
x − 𝑧 σx̄
̂
x + 𝑧 σx̄
182 – 2.58(1.73)
182- 4.46
177.54
182+ 2.58(1.73)
182+4.46
186.46
So 99% confidence interval = (177.54 , 186.46)
𝜎
Unknown ,
n> 30 ,
population finite
EX :
(EX Sc 7-6 pg 365)
a) Find estimated standard error of the mean i.e.𝜎
b) Find 96% confidence interval for the mean
12
Solution:
a)
n = 60
N = 540
x
= 6.2
= s
𝜎̂
̂
σx̄
̂
𝜎
=
√𝑛
=
1.368
√60
= 1.368
x √
𝑁−𝑛
𝑁−1
x√
= 0.18 x √
540−60
540−1
480
539
= 0.18 x 0.94
= 0.17
b)
P(z) = 96%
=
=
x
95
100
0.96
2
= 0.96
= 0.4800 = 2.05
̂
± 𝑧 σx̄
̂ x̄
x − 𝑧 σx̄
6.2 – 2.05(0.17)
6.2 – 0.35
5.85
̂
x + 𝑧 σx̄
6.2+2.05(0.17)
6.2 + 0.35
6.55
So 96% confidence interval = (5.85, 6.55)
H.W:
a) Do EX 7-30 (Pg 365)
.
b) The height of random sample of 50 students has mean 174.5
and sample standard deviation is 6.9.Find 98% confidence interval
for the mean.
13
t- Distribution:
We use t-distribution when sample size i.e. n is less than equal to 30( n≤30)
and 𝜎 is unknown and we assume that population is normal or
approximately normal.
t – distribution is symmetrical-distribution. It is lower at the mean and higher
at the tails than the normal distribution so we can say that t-distribution is
flatter than the normal distribution. As the sample size increases then it loses
its flatness and becomes approximately equal to the normal distribution.
There is different t-distribution for every sample size.
Degree of Freedom:
Number of values in a sample that we can choose freely is called degree of
freedom.
EX :
(EX 7-44 PG 377)
For the following sample sizes and confidence levels, find the appropriate
t values for constructing confidence interval.
Solution:
a)
n=15
P(t) =90%
90
=
= 0.9
100
= 1 - 0.9 = 0.1
df = n-1
= 15-1 = 14
t0.1(14) = 1.761
b)
n= 6
p(t) =95%
95
=
= 0.95
100
14
= 1 - 0.95 = 0.05
df = n-1
= 6-1 = 5
t0.05(5) = 2.571
c)
n = 19
p(t) = 99%
99
=
= 0.99
100
= 1 - 0.99 = 0.01
df
= n-1
= 19-1 = 18
t0.01(18) =2.0878
d)
n= 25
p(t) = 98%
=
98
100
= 0.98
= 1 - 0.98 = 0.02
df = n-1
= 25-1 = 24
t0.02(24) = 2.492
e)
n = 10
p(t) = 99%
99
=
= 0.99
100
= 1 - 0.99 = 0.01
df = n-1
= 10-1 =9
t0.01(9) = 3.250
15
f)
n = 41
p(t) = 90%
90
=
= 0.90
100
= 1 - 0.90 = 0.1
df
= n-1
= 41- 1 = 40
t0.1 (40) = 1.684
H.W:
Do EX Sc 7-10 (pg 377)
𝜎
n ≤ 30
Unknown,
,
population infinite
EX :
(EX Sc 7-11 pg 377)
Solution:
n = 7
x
= 39.2
𝜎̂ = s = 3.2
̂
σx̄
=
=
̂
𝜎
√𝑛
3.2
√7
= 1.21
P(t) = 95%
95
=
= 0.95
100
= 1 - 0.95 = 0.05
df = 7-1 =6
t 0.05(6)
= 2.447
̂
x ± 𝑡σx̄
̂
x − tσx̄
̂
x + tσx̄
39.2 – 2.447(1.21)
39.2 – 2.96
36.24
39.2 + 2.447(1.21)
39.2 + 2.96
42.16
So 95% confidence interval = (36.24 , 42.16 )
H.W:
Do EX 7-46( Pg 377)
16
OBJECTIVE SECTION
Q-1 WRITE SHORT ANSWERS FOR THE FOLLOWING
1- Write the suitable name for the process stated below,
It is the process by which we obtained an estimate of unknown
Population parameter by using sample.
Answer:
Estimation.
2-Write criteria of good estimator.
Answer:
i)
ii)
iii)
iv)
Unbiasedness
Efficiency
Consistency
Sufficiency
6-Define estimator.
Answer:
A sample statistic used to estimate population parameter is
called estimator.
8-Define degree of freedom.
Answer:
Number of values in the sample that we can choose freely is
called degree of freedom.
Q-2 CHOOSE THE CORRECT ONE.
1- Statistical inference is divided in to ------- areas.
a) Two
b) Five
c) Ten
17
2- Formula for S is,
∑𝑥
a)
𝑛
∑𝑥
b)
𝑛−1
c) None of these
3- Degree of freedom can be calculated as,
a) n -1
b) N-9
c) 3n-1
4- When 𝜎 unknown ,n ≤ 30 and population assume to be normal or
approximately normal then we use,
a) t- distribution
b) F-distribution
c) Binomial distribution
Q-3 Write true or false for the following.
1-
x
Is the point estimate of µ.
2- There are two types of estimates.
3- When an estimator has larger standard error than any other estimator it
means it is efficient estimator of that estimator.
4-
S =
∑(𝑥−
x)
𝑛−1
18