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

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

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

1. Consider a random sample x₁, x₂,…,xn from a normal population having mean zero. Obtain the MLE of the variance and show that it is unbiased.
2. Find the estimates of µ and σ in the normal populations N (µ, σ2) by the method of moments.
3. 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.
4. 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.
5. 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)
6. 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.
7. 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.
8. A random variable X has a distribution with density function: f(x) = (α +1 )xα   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.
9. Consider a random sample of size n from a population following Poisson distribution. Obtain the MLE of the parameter of this distribution.
10. Consider a random sample x₁, x₂,…,xn from a population following binomial distribution having parameters n and Find the MLE of p and show that it is unbiased.
11. 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.
12. 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.
13. A random sample of size 10 from N (µ, σ2 ) yields sample mean 4.8 and sample variance 8.64. Find 95% and 99% confidence intervals for the population mean.
14. 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.
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.

Solutions:

1. σ2 = ∑ xi2 / n
2. µ = x, σ2 = S2
3. [- 2.54, 3.34)
4. [ 1.972, 2.008]
5. (i) (1.97, 6.72], (ii) (1.35, 5.30]
6. [0.52, 0.62]
7. (0.02, 0.28)
8. α = o.89oo9
9. λ =x
10. p = x / n.
11. [2.39, 4.05]
12. 95% C.I. [2.233, 7.367], 99% C.I. (1.616, 7.984)
13. (55.64, 64.36)

(0.31, 0.45)

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