complete data for estimation by confidence interval

1. In the Name of Allah,
the most Gracious,
the most Merciful

2. M.SHAHAB YASEEN
BSSS-13-15

3. Presentation Topic
Estimation by Confidence Interval
(Confidence interval estimation of a population
Mean)

4. Contents
 Estimation
i. Point Estimation
ii. Interval Estimation
 Confidence Interval estimate of a population Mean
i. Normal population with σ known
ii. Normal population with σ unknown
iii. Non-normal population with known or unknown σ
 Confidence Interval for Difference of Means
i. Normal population with known Standard Deviation
ii. Normal population with unknown Standard Deviation
iii. Non-normal populations

5. Estimation
It is a procedure of making judgment about the true but unknown
values of the population parameters from the sample observations
It is further divided into two parts
 Point Estimation
 Interval Estimation

6. Confidence Interval
The concept of confidence interval was introduced in 1937 by the polish-
English-American statistician Jerzy Neyman
A confidence interval is an interval constructed from the sample values in
such a way that it has a known probability such as 95% or 99% etc. of
containing some parameter θ intended to be estimate.
Suppose θ is a parameter and L,U are two quantities derived from the
sample
P(L< θ<U)= 1 − 0< <1
 Where is the level of significance
 The probability 1- , associated with an interval estimate is called
confidence co-efficient or confidence level.

7.  The bounds L,U are called lower and upper confidence limits and the
interval L< θ<U is called 100(1- )% confidence interval of θ
 For example if =0.05 then the confidence interval is 95% confidence
interval
 The difference between U-L is called the precision of the confidence
interval.it can be increased by increasing the sample size.

8. Confidence Interval estimate of a
population Mean
To compute a confidence interval for the population mean µ,we have
to see:
i. Weather or not the population is normal
ii. Weather or not the population standard division is known
iii. Weather the sample size is small or large
We discuss these different cases below.

9. 1. Normal population with σ known
 Suppose that a random sample 𝑋1, 𝑋2, 𝑋3…., 𝑋 𝑛 of size n is drawn
from a normal population unknown mean µ and known standard
deviation σ and we drive to construct confidence interval for the
population mean µ as estimate of which is provided by the sample
mean X bar. From the sampling distribution of the mean we know
that variable X bar has mean µ and S.D =
σ
𝑛
so that statistic
Z=
𝑋−µ
σ/ 𝑛
 The normal dis. Tells us that the prob. That a value of Z will fall in the
interval from −𝑍
2
to 𝑍
2
is equal to 1 − .

10. That is we can make the following prob. statement
P[−𝑍
2
< Z< 𝑍
2
] = 1 −
P[−𝑍
2
<
𝑋−µ
σ/√𝑛
< 𝑍
2
] = 1 − [∴Z=
𝑋−µ
σ/√𝑛
]
Multiply all the terms inside the bracket by σ/√𝑛 and get
P[− 𝑍
2
σ
√𝑛
< 𝑋 − µ < 𝑍
2
σ
√𝑛
] = 1 −
By subtracting X bar from each term we have
P[− 𝑋 − 𝑍
2
σ
√𝑛
<−µ < − 𝑋+ 𝑍
2
σ
√𝑛
] = 1 −

11. We multiply all terms by -1
P[ 𝑋+ 𝑍
2
σ
𝑛
> µ > 𝑋 − 𝑍
2
σ
𝑛
] = 1 −
[−3< −2 if multiply −1 then 3>2]
By rearranging we get
P[ 𝑋 − 𝑍
2
σ
𝑛
< µ < 𝑋 + 𝑍
2
σ
𝑛
] = 1 −
Hence a100(1 − )% confidence interval for µ is
[ 𝑋 − 𝑍
2
σ
𝑛
, 𝑋+ 𝑍
2
σ
𝑛
]
Which may be expressed more efficiently as
𝑋 ± 𝑍
2
σ
𝑛

12.  Suppose we desire a 95% confidence interval,i.e.
0.95 × 100 % C.I
(1 −0.05 )100 % C.I
=0.05 and 2=0.025
From Area table we see
Z0.025=1.96
So 95% C.I for µ Is
𝑋 − 1.96
σ
𝑛
< µ < 𝑋+1.96
σ
𝑛
 Similarly for 99% we have
Z0,005=2.58
𝑋 − 2.58
σ
𝑛
<µ< 𝑋+ 2.58
σ
𝑛

13. Example 15.18
Solution:- σ=1.8 ml , n=8, find 90% C.I
X=481,479,482,480,477,478,481,482
𝑍
2
=𝑍0.05=1.645
𝑋=
481,479,482,480,477,478,481,482
8
=
3840
8
= 480 ml
The 90% C.I for µ is
𝑋 ± 𝑍
2
σ
𝑛
480 ± 1.645 (
1.8
√8
)
480 ± 1.645 (0.636)
480 ± 1.05
478.95< µ <481.05
∴ 90% C.I for µ is (478.95,481.05)

14. Normal population with σ unknown
If n is sufficiently large (i.e. n≥ 30) and σ is unknown, the sampling
distribution of 𝑋 will be approximately normal with mean µ and S.D=
𝑆
√𝑛
(where S is the sample S.D)
hence (1 − )100 % C.I for µ is
𝑋 ± 𝑍
2
𝑆
𝑛
Note: If σ is unknown and the sample size n<30 then the sampling
distribution of 𝑋 will not be normal but it has a students T-distribution
𝑋 ± 𝑡
2
𝑠
𝑛
where s=
𝑥− 𝑋 2
𝑛−1

15. Example 15.20
Solution: here 𝑋=5410 ,S=680 ,n=64
Because of 90 % confidence interval we have
𝑍0.05=1.645
hence (1 − )100 % C.I for µ is
𝑋 ± 𝑍
2
𝑆
𝑛
5410 ± 1.645
680
√64
5410 ± 1.645 (85)
5410 ± 139.8
5270.2 < µ < 5549.8
Hence 90% C.I for µ is (5270,5550)

16. Non-normal population with known and
unknown σ (large sample)
(1 − )100 % C.I for µ,mean of a non-normal population with σ known is
given by
𝑋 ± 𝑍
2
σ
𝑛
When σ is unknown
𝑋 ± 𝑍
2
𝑆
𝑛
When sample size n is greater than 5% of population size N then the C.I
estimate for µ is
𝑋 ± 𝑍
2
σ
𝑛
𝑁−𝑛
𝑁−1

17. Confidence Interval for Difference of Means
To construct the C.I for the diff. between two means µ1 − µ2,the
following three cases are to be considered
 Both the populations are normal with known S.Ds
 Both the populations are normal with unknown S.Ds
 Both the populations are non-normal, in which case, both sample
sizes are necessarily large

18. Normal population with known S.Ds
 Suppose we have two normal population having unknown means
µ1 and µ2 and known S.D σ1and σ2.suppose independent Random
of size n1 and n2 are drawn from the population respectively and let
𝑋1, X2 represent the sample means then the sampling distribution of
d=𝑋1 − 𝑋2 (i.e. the diff. between means) will be normal with mean
µ1 − µ2 and S.D=
σ1
2
𝑛1
+
σ2
2
𝑛2
Z=
𝑋1 − 𝑋2 −(µ1
−µ2
)
σ1
2
𝑛1
+
σ2
2
𝑛2
is exactly normal, no matter how the sample sizes are.
We can therefore make the following prob. Distribution

19. P[−𝑍
2
<
𝑋1 − 𝑋2 −(µ1
−µ2
)
σ1
2
𝑛1
+
σ2
2
𝑛2
<𝑍
2
]= 1 −
Multiply each term in the bracket by
σ1
2
𝑛1
+
σ2
2
𝑛2
P[−𝑍
2
σ1
2
𝑛1
+
σ2
2
𝑛2
< 𝑋1 − 𝑋2 − (µ1
−µ2)<𝑍
2
σ1
2
𝑛1
+
σ2
2
𝑛2
]= 1 −
Subtracting 𝑋1 − 𝑋2 to each term inside the bracket, we get
P[− 𝑋1 − 𝑋2 − 𝑍
2
σ1
2
𝑛1
+
σ2
2
𝑛2
< −(µ1
−µ2)< − 𝑋1 − 𝑋2 +𝑍
2
σ1
2
𝑛1
+
σ2
2
𝑛2
]= 1 −
Now multiplying each term by −1

20. P[ 𝑋1 − 𝑋2 + 𝑍
2
σ1
2
𝑛1
+
σ2
2
𝑛2
> (µ1
−µ2) > 𝑋1 − 𝑋2 − 𝑍
2
σ1
2
𝑛1
+
σ2
2
𝑛2
]= 1 −
Or
P[ 𝑋1 − 𝑋2 − 𝑍
2
σ1
2
𝑛1
+
σ2
2
𝑛2
< (µ1
−µ2)< 𝑋1 − 𝑋2 +𝑍
2
σ1
2
𝑛1
+
σ2
2
𝑛2
]= 1 −
Hence 100(1 − )% C.I for (µ1
−µ2) is
𝑋1 − 𝑋2 ± 𝑍
2
σ1
2
𝑛1
+
σ2
2
𝑛2

21. Normal population with unknown S.Ds
If prob. S.D is unknown, but sample size n≥ 30 𝑡ℎ𝑒𝑛 σ1 and σ2 is
replaced with S1 and S2
𝑋1 − 𝑋2 ± 𝑍 2
S1
2
𝑛1
+
S2
2
𝑛2

22. Non-normal population with known and
unknown S.Ds
 An Approximate 100(1 − )% C.I for (µ1
−µ2),
When the population S.Ds are known then
𝑋1 − 𝑋2 ± 𝑍
2
σ1
2
𝑛1
+
σ2
2
𝑛2
 An Approximate 100(1 − )% C.I for (µ1
−µ2),
When the population S.Ds are unknown then
𝑋1 − 𝑋2 ± 𝑍
2
S1
2
𝑛1
+
S2
2
𝑛2

23. Thank You
Any Question?