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, σ^{2}), 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, σ_{2}) 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, ∑d^{2} =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 σ^{2} is

From chi-square table with 14 degrees of freedom,

X^{2}_{0.025} = 26.12, X^{2}_{0.975} =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**

- A random variable X has a distribution with density function :

0< x <2, otherwise 0. Find the MLE of the parameter *a *(> 0).

- Consider a random sample x₁, x₂,…,x
_{n}from a normal population having mean zero. Obtain the MLE of the variance and show that it is unbiased. - Find the estimates of µ and σ in the normal populations N (µ, σ
^{2}) by the method of moments. - Find a 95% C.I. for the mean of a normal population with σ = 3, given the sample 2.3, – 0.2, 0.4 and – 0.9.
- A sample of size I 0 from a normal population produces the data 2.03, 2.02, 2.01, 2.00, 1.99, 1.98, 1.97, 1.99, 1.96 and 1.95. From the sample find the 95% C.I. for the population mean.
- The following random sample was obtained from a normal population : 12, 9, I 0, 14, µ , 8. Find the 95% C.I. for the population S.D. when the population mean is
*(i)*known to be 13,*(ii)* - 228 out of 400 voters picked at random from a large electorate said that they were going to vote for a particular candidate. Find 95% C.I. for the proportion of voters of the electorate who would in favour of the candidate.
- A study shows that 102 of 190 persons who saw an advertisement on a product on T. V. during a sports program and 75 of 190 other persons who saw it advertised on a variety show purchased the product. Construct a 99% confidence interval for the difference of sample proportions.
- A random variable X has a distribution with density function: f
*(x)*= (α +1 )x^{α }0<x<1α > -1= 0, otherwise and a random sample of size 8 produces the data: 0.2, 0.4, 0.8, 0.5, 0.7, 0.9, 0.8 and 0.9. Find the MLE of the unknown parameter a. - Consider a random sample of size
*n*from a population following Poisson distribution. Obtain the MLE of the parameter of this distribution. - Consider a random sample x₁, x₂,…,x
_{n}from a population following binomial distribution having parameters*n*and Find the MLE of*p*and show that it is unbiased. - Show that the estimates of the parameter of the Poisson distribution obtained by the method of maximum likelihood and the method of moments are identical.
- In a sample of size 10, the sample mean is 3.22 and the sample variance 1.21. Find the 95% C.I. for the population mean.
- A random sample of size 10 from N (µ, σ
^{2}) yields sample mean 4.8 and sample variance 8.64. Find 95% and 99% confidence intervals for the population mean. - The marks obtained by 15 students in an examination have a mean 60 and variance 30. Find 99% confidence interval for the mean of the population of marks, assuming it to be normal.
- In a random sample of 300 road accidents, it was found that 114 were due to bad weather. Construct a 99% confidence interval for the corresponding true proportions.

**Solutions:**

- σ
^{2}= ∑ xi^{2}/*n* - µ =
*x,*σ^{2}=*S*^{2} - [- 2.54, 3.34)
- [ 1.972, 2.008]
- (i) (1.97, 6.72],
*(ii)*(1.35, 5.30] - [0.52, 0.62]
- (0.02, 0.28)
- α = o.89oo9
- λ
*=x* *p*=*x / n.*- [2.39, 4.05]
- 95% C.I. [2.233, 7.367], 99% C.I. (1.616, 7.984)
- (55.64, 64.36)

(0.31, 0.45)

**Links of Previous Main Topic:-**

- Introduction to statistics
- Knowledge of central tendency or location
- Definition of dispersion
- Moments
- Bivariate distribution
- Theorem of total probability addition theorem
- Random variable
- Binomial distribution
- What is sampling
- Estimation

**Links of Next Statistics Topics:-**

- Statistical hypothesis and related terms
- Analysis of variance introduction
- Definition of stochastic process
- Introduction operations research
- Introduction and mathematical formulation in transportation problems
- Introduction and mathematical formulation
- Queuing theory introduction
- Inventory control introduction
- Simulation introduction
- Time calculations in network
- Introduction of game theory