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2.2 Interval Estimation& Estimation of Proportion
INTERVAL ESTIMATIONBy 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 ESTIMATIONInterval 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 ofθ.
INTERVAL ESTIMATIONConfidence limits:-The quantities L1 and L2 are called upper and lower   confidence limitsDegreeof confidence (confidence coefficient) 1-α
 Interval EstimationSuppose 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 rewriteWhen the observed value       become available, we obtainLarge sample confidence intervalfor  -  knownThus when sample has been obtained and the value of       has beencalculated, we can claim with probability (1 -  )100% confidenceThat 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 PROPORTIONThere 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 .hencethat 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 PROPORTIONEstimation 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 ofx 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 estimateWith the observed value x/n substituted for p we obtain anestimate of E.
Confidence Interval for pPoint Estimate = X / nConfidence Interval
Estimation of Proportion  Sample size determinationBut 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 ESTIMATIONBy 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 ESTIMATIONInterval 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 ofθ.
  • 4. INTERVAL ESTIMATIONConfidence limits:-The quantities L1 and L2 are called upper and lower confidence limitsDegreeof confidence (confidence coefficient) 1-α
  • 5. Interval EstimationSuppose 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 rewriteWhen the observed value become available, we obtainLarge sample confidence intervalfor  -  knownThus when sample has been obtained and the value of has beencalculated, we can claim with probability (1 -  )100% confidenceThat 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 PROPORTIONThere 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 .hencethat 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 PROPORTIONEstimation 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 ofx 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 estimateWith the observed value x/n substituted for p we obtain anestimate of E.
  • 15. Confidence Interval for pPoint Estimate = X / nConfidence Interval
  • 16. Estimation of Proportion Sample size determinationBut 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)