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2.2 Interval Estimation& Estimation of Proportion
INTERVAL ESTIMATION By using point estimation ,we may not get desired  degree of accuracy in estimating a  parameter.  Therefore ,it is better to replace point estimation  by interval estimation.
INTERVAL ESTIMATION Interval estimate:- An interval estimate of an unknown parameter is an interval of the form   L1 ≤ θ≤ L2, where the end points L1 and L2 depend on the numerical value of   the statistic θ* for particular sample  on the sampling distributon of θ* . 100(1-α)% Confidence Interval:- A 100(1-α)% confidence interval for a parameter   θ is an interval of the fprm  [L1 , L2]  such that P(L1≤θ ≤L2) =1- α, 0< α <1regardless of the actual value of θ.
INTERVAL ESTIMATION Confidence limits:- The quantities L1 and L2 are called upper and lower    confidence limits Degreeof confidence (confidence  coefficient)  1-α
 Interval Estimation Suppose we have a large (n 30) random  sample from a population with the  unknown  mean  and known variance 2. We know  inequality  will satisfy with probability 1 - .
Interval Estimation   This inequality we can rewrite When the observed value       become available, we obtain Large sample  confidence interval for  -  known Thus when sample has been obtained and the value of       has been calculated, we can claim with probability (1 -  )100% confidence That the interval from
Interval Estimation   Since  is unknown in most applications, we may have  to make the further approximation of substituting for   the sample standard deviation s. Large sample  confidence interval for 
Interval Estimation   For small samples (n < 30), we assume that we are sampling from normal population and proceed similarly as before we get the (1 - )100% confidence interval formula  Small sample confi- dence interval for 
ESTIMATION OF PROPORTION There are many problems in which we must Estimate  proportion     Proportion of Defectives     Proportion of objects or things having required attributes     The mortality rate of a disease. Remark : In many of these problems it is reasonable to  assume that we are sampling a binomial population .hence that our problem is to estimate the binomial parameter p .  The probability of success in a single trial of a binomial  experiment is p. This probability is a population proportion
ESTIMATION OF PROPORTION Estimation of Proportion   Suppose that random sample of size n has been taken  from a population and that X(   n)is the number of times  that an appropriate event occurs in n trials (observations).  THEN   Point estimator of the population proportion (p) is given by
Sample proportion is an Unbiased Estimator of population proportion   If the n trials satisfy the assumption underlying the binomial  distribution ,then  mean of number of successes is   np  Variance of number of successes is  np(1-p) Expectation and variance of sample proportion  X denotes the number of successes in n trials
Estimation of Proportion   When n is large, we can construct approximate  confidence intervals for the binomial parameter p by  using the normal approximation to the binomial  distribution. Accordingly, we can assert with probability  1 -  that the inequality
Estimation of Proportion    will be satisfied. Solving this quadratic inequality for p  we can obtain a corresponding set of approximate  confidence limits for p in terms of the observed value of x but since the necessary calculations are complex, we  shall make the further approximation of substituting x/n  for p in
Estimation of Proportion   Large sample confidence interval for p     where the degree of confidence is (1 - )100%. Maximum error of estimate With the observed value x/n substituted for p we obtain an estimate of E.
Confidence Interval for p Point Estimate = X / n Confidence Interval
Estimation of Proportion   Sample size determination But this formula cannot be used as it stands unless we have some information about the possible size of p. If no much information is available, we can make use of the fact that p(1 - p) is at most 1/4, corresponding to p = 1/2 ,as can be shown by the method of elementary calculus. If a range for p is known, the value closest to 1/2 should be used. Sample size (p unknown)

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Interval Estimation & Estimation Of Proportion

  • 1. 2.2 Interval Estimation& Estimation of Proportion
  • 2. INTERVAL ESTIMATION By using point estimation ,we may not get desired degree of accuracy in estimating a parameter. Therefore ,it is better to replace point estimation by interval estimation.
  • 3. INTERVAL ESTIMATION Interval estimate:- An interval estimate of an unknown parameter is an interval of the form L1 ≤ θ≤ L2, where the end points L1 and L2 depend on the numerical value of the statistic θ* for particular sample on the sampling distributon of θ* . 100(1-α)% Confidence Interval:- A 100(1-α)% confidence interval for a parameter θ is an interval of the fprm [L1 , L2] such that P(L1≤θ ≤L2) =1- α, 0< α <1regardless of the actual value of θ.
  • 4. INTERVAL ESTIMATION Confidence limits:- The quantities L1 and L2 are called upper and lower confidence limits Degreeof confidence (confidence coefficient) 1-α
  • 5. Interval Estimation Suppose we have a large (n 30) random sample from a population with the unknown mean  and known variance 2. We know inequality will satisfy with probability 1 - .
  • 6. Interval Estimation This inequality we can rewrite When the observed value become available, we obtain Large sample confidence interval for  -  known Thus when sample has been obtained and the value of has been calculated, we can claim with probability (1 -  )100% confidence That the interval from
  • 7. Interval Estimation Since  is unknown in most applications, we may have to make the further approximation of substituting for  the sample standard deviation s. Large sample confidence interval for 
  • 8. Interval Estimation For small samples (n < 30), we assume that we are sampling from normal population and proceed similarly as before we get the (1 - )100% confidence interval formula Small sample confi- dence interval for 
  • 9. ESTIMATION OF PROPORTION There are many problems in which we must Estimate proportion Proportion of Defectives Proportion of objects or things having required attributes The mortality rate of a disease. Remark : In many of these problems it is reasonable to assume that we are sampling a binomial population .hence that our problem is to estimate the binomial parameter p . The probability of success in a single trial of a binomial experiment is p. This probability is a population proportion
  • 10. ESTIMATION OF PROPORTION Estimation of Proportion Suppose that random sample of size n has been taken from a population and that X( n)is the number of times that an appropriate event occurs in n trials (observations). THEN Point estimator of the population proportion (p) is given by
  • 11. Sample proportion is an Unbiased Estimator of population proportion If the n trials satisfy the assumption underlying the binomial distribution ,then mean of number of successes is np Variance of number of successes is np(1-p) Expectation and variance of sample proportion X denotes the number of successes in n trials
  • 12. Estimation of Proportion When n is large, we can construct approximate confidence intervals for the binomial parameter p by using the normal approximation to the binomial distribution. Accordingly, we can assert with probability 1 -  that the inequality
  • 13. Estimation of Proportion will be satisfied. Solving this quadratic inequality for p we can obtain a corresponding set of approximate confidence limits for p in terms of the observed value of x but since the necessary calculations are complex, we shall make the further approximation of substituting x/n for p in
  • 14. Estimation of Proportion Large sample confidence interval for p where the degree of confidence is (1 - )100%. Maximum error of estimate With the observed value x/n substituted for p we obtain an estimate of E.
  • 15. Confidence Interval for p Point Estimate = X / n Confidence Interval
  • 16. Estimation of Proportion Sample size determination But this formula cannot be used as it stands unless we have some information about the possible size of p. If no much information is available, we can make use of the fact that p(1 - p) is at most 1/4, corresponding to p = 1/2 ,as can be shown by the method of elementary calculus. If a range for p is known, the value closest to 1/2 should be used. Sample size (p unknown)