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Interval Estimation

 

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, σ2) 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

The interval

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, σ2), the statistic

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

Then for 95% confidence interval for mean µ we have

Thus

is called 95% C.l. for µ

Similarly,

is called 99% C. I . for µ

 

 

(c) C.I. for variance σ2 with known mean. We know that (xi - µ )2/ σ2 follows chi-square distribution with n degrees of freedom.

 

For probability 95% we have

which is 95% confidence interval for σ2

Similarly,

∑ ( xi - µ)2 / X  ≤ σ2≤ ∑ ( xi - µ)2 / X20.995

is the 99% confidence interval for cr.

 

(d) C.I. for variance cr2 with unknown mean. In this case ns2 / σ2 = ∑ ( xi – x)2 / σ2

 

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

For probability 95% we have

which is 95% C.l. for σ2

 

Similarly, ns2 / X20.005  ≤  σ2  ≤ ns2 / X20.995   the 99% C.l. for σ2

 

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 µi 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, ∑d2 =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 σ2 is

From chi-square table with 14 degrees of freedom,

X20.025 = 26.12,   X20.975 =5.63

Therefore the C.I. is

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,…,xn 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,…,xn 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, σ2) 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, σ2 ) 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. σ2 = ∑ xi2 /  n

 

5. p = x / n.                              6. µ = x, σ2 = S2

 

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]