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CONFIDENCE INTERVAL FOR THE STANDARD DEVIATION σ We have a point estimate for σ: S (or S2 for σ2). Want a confidence interval for σ. Let X1, X 2 , , X n sample from N(μ, σ) population. First introduce a new distribution: chi-square distribution χ2: Statistic ( X i X )2 (n 1) S 2 2 2 has a chi-square distribution with (n-1) degrees of freedom (df), χ2(n-1). NOTE: The χ2(n-1) distribution has only positive values. It is tabulated (Table D). The χ2(n-1) is NOT symmetric. The chi-square distribution As df increase, the distribution “flattens”. 0.5 Percentiles of the chi-square distribution 0.4 df = 2 df= 4 df = 10 0.3 0.2 0.1 0.0 0 5 10 15 20 25 30 1-α α/2 α/2 Χ2 1-α/2 Χ2α/2 CONFIDENCE INTERVAL FOR THE STANDARD DEVIATION σ Using the definition of the chi-square distribution, we can construct a confidence interval for the standard deviation of a normal distribution. A C=(1-α) confidence interval for the standard deviation σ based on a sample from a N(μ, σ) population, is given by (n 1) S 2 (n 1) S 2 , , /2 2 12 /2 where χ21-α/2 and χ2α/2 are values (percentiles) from a χ2(n-1) distribution with areas α/2 below χ21-α/2 and above χ2α/2 , respectively. Example (tiger brains contd.) Sample of 16 tiger brains’ measurements yielded mean weight 10 and standard deviation of 3.2 ounces. Assuming that these weights follow a Normal distribution, find a 90% confidence interval for the true standard deviation of weights. Solution. Since C = 0.90, then α=0.1, so α/2=0.05. Since n=16, then n-1=15. We will use χ(15) distribution. The 90% CI for σ is: (n 1) S 2 (n 1) S 2 , . 0.05 2 120.05 Substituting S=3.2, n=16, and χ(15)0.05=25, and χ(15)0.95=7.26, we get (15)3.22 (15)3.22 , (2.48, 4.60). 25 2 7.26 2 We have 90% confidence that the true standard deviation of the tiger brain weights is between 2.48 and 4.6 oz. CONFIDENCE INTERVAL FOR PROPORTION Binomial experiment, X = number of successes, n=number of trials, p=probability of success is unknown. Goal: Estimate the true/population proportion of successes p in a Binomial experiment using an interval estimate. X number of successes Point estimate: p ˆ sample proportion of successes. n number of trials NOTE: For large sample size n, ˆ p is normally distributed (Normal Approx. to Binomial) with p (1 p ) p p and p ˆ ˆ n p p ˆ Thus, Z has a standard normal distribution. Now p (1 p ) proceed as for the CI for the normal mean. n LARGE SAMPLE CONFIDENCE INTERVAL FOR THE PROPORTION Binomial experiment, X = number of successes, n=number of trials, p=probability of success. X number of successes p ˆ sample proportion of successes. n number of trials A C=(1-α) confidence interval for the population proportion p is given by p (1 p ) ˆ ˆ p z / 2 ˆ , n if the sample size is large. EXAMPLE 400 people were asked if they are in favor of death penalty. 250 favored death penalty. Find a point estimate and a 95% confidence interval for the true proportion of people who are in favor of death penalty. Solution. X=number of people in the sample who favor death penalty, X=250, n=400, p= true proportion of people in the population who favor death penalty. Point estimate of p: X 250 p ˆ 0.625. n 400 95% CI: C=0.95, so zα/2=1.96, and 95% CI for p is: p (1 p ) ˆ ˆ 0.625(1 0.625) p z /2 ˆ 0.625 1.96 (0.578, 0.672). n 400 SAMPLE SIZE AND MARGIN OF ERROR Question: What is the sample size required to keep the margin of error of a given size m for a specified confidence level C=(1-α)? p (1 p ) ˆ ˆ Margin of error = half length of the confidence interval, m = z /2 . n Solving for n, we get: z /2 p (1 p ) 2 n 2 . m Since p is unknown, we have to substitute our best guess for p, say p0. We can use info from previous surveys for p0. If no previous info available, use conservative approach and plug 0.5 for p, to get: 2 2 z /2 (0.5) 2 z /2 n 2 2 . m 4m EXAMPLE. Death penalty contd. 1. Find the sample size required for the margin of error to be 0.02 for a 90% confidence interval for the proportion of people in favor of death penalty. Solution. We have prior info from the survey p0=0.625. Use it! z /2 p0 (1 p0 ) 1.6452 (0.625)(1 0.625) 2 n 2 2 1585.6. m 0.02 n=1586. 2. Find the sample size required for the margin of error to be 0.02 for a 90% CI for p assuming no prior information about p. Solution. We have no prior info about p. 2 z / 2 1.6452 n 2 2 1691.3. 4m 4(0.02) n=1692. NOTE: Without prior information, we need larger sample size to have the same margin of error.

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posted: | 8/28/2012 |

language: | English |

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