In interval estimation we find an interval which is expected to include the unknown parameter with a specified probability, i.e.,

P (t1 ≤ θ ≤ t2) = k

where, [t1, t2] is called confidence interval (C.I.),

tl , t2 are called confidence limits,

k is called confidence coefficient of the interval.

(a) C.I. 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. To find C.I. for mean µ.

We know that z = follows standard normal distribution and· 95% of the area under the standard normal curve lies between z = 1.96 and z = -1.96, Then,

i.e., in 95% cases we have

is known as 95% confidence interval for µ.

Similarly,

is known as 99% C.I. for µ

is known as 99.73% C.I. for µ.

**(b) C.I. for mean with unknown S.D. σ.**

In this case, the sampling from a normal population N (µ, σ^{2}), the statistic

follows t distribution with (n - 1) degree of freedom.

Then for 95% confidence interval for mean µ we have

is called 95% C.l. for µ

is called 99% C. I . for µ

** **

**(c) C.I. for variance σ ^{2} with known mean. **We know that (xi - µ )

^{2}/ σ

^{2}follows chi-square distribution with n degrees of freedom.

which is 95% confidence interval for σ^{2}

Similarly,

∑ ( xi - µ)^{2} / X ≤ σ^{2}≤ ∑ ( xi - µ)^{2} / X^{2}_{0.995}

is the 99% confidence interval for cr.

**(d) C.I. for variance cr2 with unknown mean. **In this case *ns ^{2 }/ *σ

^{2}= ∑ ( xi – x)2 / σ

^{2}

^{ }

follows chi-square distribution with *(n *- 1) degrees of freedom.

which is 95% C.l. for σ^{2}

Similarly, *ns ^{2 }*/ X

^{2}

_{0.005 }≤ σ

^{2 }≤

*ns*/ X

^{2 }^{2}

_{0.995 }the 99% C.l. for σ

^{2}

** **

**Some of the Confidence Limits are given below :**

(with Normal Population N (µ, cr2))

Difference of Means (µ 1 - µ 2) : (S.Ds known).

95% Confidence limits = ( x1* *- .X2 ) ± 1.96 99% Confidence limits = (x1 – X2) ± 2.58 Difference of Means (µ1 - µ 2) : (Common S.D. unknown)

For Proportion P :

95% Confidence limits = *p *± 1.96 (S.E. of *p)*

99% Confidence limits = *p *± 2.58 (S.E. of *p)*

For Difference of Proportions P 1 - P 2 :

95% Confidence limits =[(p1 - *p2) *± 1.96 [S.E. of (p1 - p 2)]

99% Confidence limits = [(p1 - *p2) *± 2.58 [S.E. of (p1 - p 2)]

**Example 8. ***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. **Given *n *= 10, ∑xi= 683, Then * *= 68 . 3

* *

Since the population S.D. σ is known, then 99% C.I. for µ is given by

**Example **9. *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, 43. Find 95% confidence interval for mean *µ*i of the population.*

**Solution. **From the given data, *∑xi *= 502, so *x *= 50.2, *n *= 10

Let *d *= *x- *50, then the samples are changed to

-2, 6,0,5,-1,-5,5,4,-3,-7.

∑d =2, ∑d^{2} =190

*s *= 4.35

Since, the population S.D. cr is unknown, the 95% C.I. for mean µi s

[47.09, 53.31].** **

**Example 10. ***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. **Here *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

**Example 11. ***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 are 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)]

Thus the limits are [0.13- 3 (0.02), 0.13 + 3 (0.02)] [0.07, 0.19).

** **

**PROBLEMS**

**1. **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..

2. A random variable X has a distribution with density function :

0< x <2

0, otherwise

Find the MLE of the parameter *a *(> 0).

3. Consider a random sample of size *n *from a population following Poisson distribution. Obtain the MLE of the parameter of this distribution.

** **

**4. **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.

5. Consider a random sample x1,x2,…,x_{n} from a population following binomial distribution having parameters *n *and *p. *Find the MLE of *p *and show that it is unbiased.

6. Find the estimates of µ and σ in the normal populations N (µ, σ^{2}) by the method of moments.

7. 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.

** **

**8. **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.

9. 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.

** **

**10. **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.l. for the population mean.

** **

**11. **A random sample of size I 0 from a N (µ, σ^{2} ) yields sample mean 4.8 and sample variance 8.64. Find 95% and 99% confidence intervals for the population mean.

** **

**12. **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) *unknown.

13. 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.

** **

**14. **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.!. for the proportion of voters of the electorate who would in favor of the candidate.

** **

**15. **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.

** **

**16. **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.

** **

**ANSWERS**

**1. **α = o.89oo9J

3. λ *=x ***4. **σ^{2} = ∑ xi^{2} / * n*

* *

5. *p *= *x / n. *6. µ = *x, *σ^{2} = *S ^{2}*

* *

8. [- 2.54, 3.34) 9. [2.39, 4.05]

**10. **[ 1.972, 2.008] 11. 95% C.I. [2.233, 7.367], 99% C.I. (1.616, 7.984]

**12. **(i) (1.97, 6.72], *(ii) *(1.35, 5.30] **13. **(55.64, 64.36)

**14. **[0.52, 0.62] 15. (0.31, 0.45)

**16. **(0.02, 0.28]