Confidence intervals, Lecture notes of Probability and Statistics

Download Confidence intervals and more Lecture notes Probability and Statistics in PDF only on Docsity! Confidence Intervals I. Interval estimation. The particular value chosen as most likely for a population parameter is called the point estimate. Because of sampling error, we know the point estimate probably is not identical to the population parameter. The accuracy of a point estimator depends on the characteristics of the sampling distribution of that estimator. If, for example, the sampling distribution is approximately normal, then with high probability (about .95) the point estimate falls within 2 standard errors of the parameter. Because the point estimate is unlikely to be exactly correct, we usually specify a range of values in which the population parameter is likely to be. For example, when X is normally distributed, the range of values between XX σ96.1± is called the 95% confidence interval for µ. The two boundaries of the interval, XX σ96.1− and XX σ96.1+ are called the 95% confidence limits. That is, there is a 95% chance that the following statement will we true: XX XX σµσ 96.196.1 +≤≤− Similarly, when X is normally distributed, the 99% confidence interval for the mean is XX XX σµσ 58.258.2 +≤≤− The 99% confidence interval is larger than the 95% confidence interval, and thus is more likely to include the true mean. α = the probability a confidence interval will not include the population parameter, 1 - α = the probability the population parameter will be in the interval. The 100(1 - α)% confidence interval will include the true value of the population parameter with probability 1 - α, i.e., if α = .05, the probability is about .95 that the 95% confidence interval will include the true population parameter. On the other hand, 2.5% of the time the highest value in the confidence interval will be smaller than the true value, while 2.5% of the time the smallest value in the confidence interval will be greater than the true value. Be sure to note that the population parameter is not a random variable. Rather, the probability statement is made about samples. If we drew 100 samples of the same size, we would get 100 different sample means and 100 different confidence intervals. We expect that in 95 of those samples the population parameter will lie within the estimated 95% confidence interval, in the other 5 the 95% confidence interval will not include the true value of the population parameter. Confidence Intervals - Page 1 Of course, X is not always normally distributed, but this is usually not a concern so long as N $ 30. More importantly, σ is not always known. Therefore, how we construct confidence intervals will depend on the type of information and data available. II. Computing confidence intervals. To compute confidence intervals, proceed as follows: A. Calculate critical values for zα/2 or, if appropriate, tα/2,v, such that P(Z # zα/2) = 1 – α/2 = F(zα/2), or P(Tv # tα/2,v) = 1 – α/2 = F(tα/2,v) For example, as we have seen many times, if α = .05, then zα/2 = 1.96 since F(1.96) = 1 - α/2 = .975. We divide α by 2 to reflect the fact that the true value of the parameter can be either greater than or less than the range covered by the confidence interval. B. Confidence intervals for E(X) (or p) are then estimated by the following. Again, recall that the following statements will be true 100(1 - α)% of the time, that is, for all possible sample of size N, the true value of the population parameter will lie within the specified interval 100(1 - α)% of the time. 1. Case I. Population normal, σ known: )N/* z( + x )N/* z( - x i.e., ),N/* z( x /2/2 /2 σµσ σ αα α ≤≤ ± Recall that N/σ is the true standard error of the mean. Note also that, as N gets bigger and bigger, the standard error gets smaller and smaller, and the confidence interval gets smaller and smaller too. This means that, the larger your sample size, the more precise your point estimate will be. For example, different samples of size 20 could produce widely different estimates of the sample mean. Different samples of size 1000 will tend to produce fairly similar estimates of the sample mean. This is true, not just of case 1, but of all cases in general. Confidence Intervals - Page 2 III. Examples. 1. X = 24.3, σ = 6, n = 16, X is distributed normally. Find the 90% confidence interval for the population mean, E(X). Solution: Since σ is known, this falls under Case I. Note that the true standard error of the mean = 5.14/616/6/ ===Nσ . Also, α = .10, α/2 = .05, so the critical value of Z is 1.65 (since F(1.65) = 1 - α/2 = .95). Hence, the c.i. is 26.78 21.83 i.e., 6/4),* (1.65 + 24.3 6/4)* (1.65 - 24.3 i.e., ),N/* z( x /2 ≤≤ ≤≤ ± µ µ σα Note: Again, remember that this does NOT mean that µ definitely lies between 21.83 and 26.78. Rather, we are merely saying that, 90% of the time, intervals constructed in the above fashion will include the true population value. 2. N = 100, p̂ = .40. Construct a 95% c.i. Solution. We want to construct a confidence interval for p (Case II). Since n is large, a normal approximation is appropriate. α = .05 and α/2 = .025, so the critical value for Z is 1.96 (since F(1.96) = 1 - α/2 = .975). Using the formula for the approximate binomial confidence interval, we get .496 p .304 i.e., , 100 .6* .41.96 + .4 p 100 .6* .41.96 - .4 i.e. , N qp z p /2 ≤≤ ≤≤ ± ˆˆˆ α Of course, there is no reason to use approximations when it is such a simple matter to get the real thing. So, using the Wilson formula for the binomial confidence interval, we get .4980 p .3094 i.e. , 40,000 961. + 100 .6* .41.96 + 200 961. + .4 961.+100 100 p 40,000 961. + 100 .6* .41.96 - 200 961. + .4 961.+100 100 i.e. , N4 z + N qp z 2N z + p z+N N 22 2 22 2 2 2 /2 /2 2 /2 2 /2 ≤≤ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ≤≤ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ± α α α α ˆˆˆ Note that, because N is large, the answers differ only slightly in this case. (Also note that I won’t ever make you do this by hand). Confidence Intervals - Page 5 3. X is distributed normally, n = 9, X = 4, s5 = 9. Construct a 99% c.i. for the mean of the parent population. Solution. σ is not known, and the sample size is small, hence we will have to use the T distribution (Case III) with v = 8. α = .01, so the critical value for T8 is 3.355. (See Table III, Appendix E, v = 8 and 2Q = .01). 7.355 .645 i.e. 3/3),* (3.355 + 4 3/3)* (3.355 - 4 i.e. ),N* s/ t( x v/2, ≤≤ ≤≤ ± µ µ α Confidence Intervals - Page 6