Interval estimation is where an interval is found which is expected to include the unknown parameter with a specific probability.
P (t₁ where (t₁, t₂) are known as confidence interval, t₁ and t₂ are called confidence limits and k is known as confidence coefficient of the interval.
Case 1: Confidence interval for mean with unknown S.D σ
In this, the sampling from a normal population N (µ, σ2), the statistic
t= where S²= follows t distribution with (n – 1) degree of freedom.
For 95% C.I. for mean µ,
–
Therefore, is called 95% C.I. for µ.
Also, is known as 99% C.I. for µ.
Case 2: Confidence interval for mean with known S.D.
Let us consider a random sample of size n from a Normal Population N (µ, σ2) in which cr₂ is known.
In order to find out C.I for mean µ, z= which follows standard normal distribution and 95% of area under standard normal curve lies between, z = 1.96 and z = -1.96
Therefore, P
That is, in 95% cases, it is seen that
is called 95% C.I. for µ.
Similarly, is known as 99% C.I for µ.
Also, is known as 99.73% C.I. for µ.
Case 3: C.I for variance cr₂ with unknown mean
In this,follows chi-square distribution with (n – 1) degrees of freedom.
For 95% probability, it is
which is 95% C.I for
Also, which is 99% C.I for
Case 4: Confidence Interval for variance with a known mean.
It is known that follows chi-square distribution with n degrees of freedom.
For a probability of 95%, it is observed that
that is 95% C.I for .
In the same manner, which is 99% C.I for cr.
Some C.I. are as follows:
(With Normal Population N (µ, cr2))
Difference of Means (µ ₁ – µ₂): (S.D is known).
95% Confidence limits = ( x₁- X₂ ) ± 1.96
99% confidence limits= ( x₁- X₂ ) ± 2.58
95% confidence limits= (
99% confidence limits= (
For Proportion P:
95% Confidence limits = p ± 1.96 (S.E. of p)
99% Confidence limits = p ± 2.58 (S.E. of p)
S.E of p=
For Difference of Proportions P₁- P₂:
95% Confidence limits = [(p₁ – p₂) ± 1.96 [S.E. of (p₁ – p ₂)]
99% Confidence limits = [(p₁ – p₂) ± 2.58 [S.E. of (p₁ – p₂)]
S.E. of (p₁ – p₂)=
Example 1:
A random sample of size 10 was drawn from a normal population with an unknown mean and a variance of 35.4 (emF, if the observations are (in ems): 55, 75, 71, 66, 73, 77. 63, 67, 60 and 76, obtain 99% confidence interval for the population mean.
Solution:
n=10,,= 68.3
Since the population S.D. σ is known, then 99% C.I. for µ is
= = 63.45, 73.15
Example 2:
A random sample of size 10 was drawn from a normal population which are given by 48, 56, 50, 55, 49, 45, 55, 54, 47, and 43. Find 95% confidence interval for mean µof the population.
Solution:
, x= 50.2, n=10, given
Let d= x-50, then the sample becomes →-2,6, 0, 5, -1, -5, 5, 4, -3, -7.
∑d =2, ∑d2 =190
S²== =18.96
S=4.35
Since, the population S.D. cr is unknown, the 95% C.I. for mean is
= = 47.09, 53.31
Example 3:
The standard deviation of a random sample of size 15 drawn from a normal population is 3.2. Calculate the 95% confidence interval for the standard deviation (a) in the population.
Solution:
n = 15, sample S.D (s) = 3.2
95% Confidence interval for σ2 is
From chi-square table with 14 degrees of freedom,
X20.025 = 26.12, X20.975 =5.63
Thus, C.I is
Example 4:
A sample of500 springs produced in a factory is taken from a large consignment and 65 are found to be defective. Estimate the assign limits in which the percentage of defectives lies.
Solution:
There’s 65 defective springs in a sample of size n = 500.
The sample proportion of defective is P= 65/500 = 0.13
The limits to the percentage of defectives refer to the C.I., which can be taken as
[p – 3 (S.E. of p), p + 3 (S.E. of p)]
S.E of p=== 0.02
Therefore, the limits are [0.13- 3 (0.02), 0.13 + 3 (0.02)] [0.07, 0.19].
Problems
0< x <2, otherwise 0. Find the MLE of the parameter a (> 0).
Solutions:
(0.31, 0.45)
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