1. CONFIDENCE INTERVAL FOR THE STANDARD DEVIATION σ We have a point estimate for σ: S (or S 2 for σ 2 ). Want a confidence interval for σ. Let sample from N(μ, σ) population. First introduce a new distribution: chi-square distribution χ 2 : Statistic has a chi-square distribution with (n-1) degrees of freedom (df), χ 2 (n-1). NOTE: The χ 2 (n-1) distribution has only positive values. It is tabulated (Table D). The χ 2 (n-1) is NOT symmetric.

2. The chi-square distribution As df increase, the distribution “flattens”. Χ 2 1-α/2 Χ 2 α/2 α/2 1-α Percentiles of the chi-square distribution

3. 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 where χ 2 1-α/2 and χ 2 α/2 are values (percentiles) from a χ 2 (n-1) distribution with areas α/2 below χ 2 1-α/2 and above χ 2 α/2, respectively.

4. 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: Substituting S=3.2, n=16, and χ(15) 0.05 =25, and χ(15) 0.95 =7.26, we get We have 90% confidence that the true standard deviation of the tiger brain weights is between 2.48 and 4.6 oz.

5. 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. Point estimate: NOTE: For large sample size n, is normally distributed (Normal Approx. to Binomial) with Thus, has a standard normal distribution. Now proceed as for the CI for the normal mean.

6. LARGE SAMPLE CONFIDENCE INTERVAL FOR THE PROPORTION Binomial experiment, X = number of successes, n=number of trials, p=probability of success. A C=(1-α) confidence interval for the population proportion p is given by if the sample size is large.

7. 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: 95% CI: C=0.95, so z α/2 =1.96, and 95% CI for p is:

8. 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)? Margin of error = half length of the confidence interval, m = Solving for n, we get: Since p is unknown, we have to substitute our best guess for p, say p 0. We can use info from previous surveys for p 0. If no previous info available, use conservative approach and plug 0.5 for p, to get:

9. 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 p 0 =0.625. Use it! 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. n=1692. NOTE: Without prior information, we need larger sample size to have the same margin of error.