Estimation Y 3. Confidence intervals I, Feb 11,

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1 Estimation Example: Cholesterol levels of heart-attack patients Data: Observational study at a Pennsylvania medical center blood cholesterol levels patients treated for heart attacks measurements 2, 4, and 14 days after the attack Id Y 1 Y 2 Y 3 Id Y 1 Y 2 Y Aim: Make inference on distribution of cholesterol level 14 days after the attack: Y 3 decrease in cholesterol level: D = Y 1 Y 3 relative decrease in cholesterol level: R = Y 1 Y 3 Y 3 Confidence intervals I, Feb 11,

2 Estimation Data: d 1,..., d 28 observed decrease in cholesterol level In this example, parameters of interest might be µ D = E(D) the mean decrease in cholesterol level, σ 2 D = var(d) the variation of the cholesterol level, p D = P(D 0) probability of no decrease in cholesterol level These parameters are naturally estimated by the following sample statistics: ˆµ D = 1 n d i (sample mean) n ˆσ 2 D = 1 n i=1 n (d i d) 2, (sample mean) i=1 ˆp D = #{d i d i 0} (sample proportion) n Such statistics are point estimators since they estimate the corresponding parameter by a single numerical value. Point estimates provide no information about their chance variation. Estimates without an indication of their variability are of limited value. Confidence intervals I, Feb 11,

3 Confidence Intervals for the Mean Recall: CLT for the sample mean: For large n we have X N ( ) µ, σ2 n rule: With 95% probability the sample differs from its mean µ by less that two standard deviations. More precisely, we have P ( µ 1.96 σ n X µ σ n ) = 0.95, or equivalently, after rearranging the terms, P ( X 1.96 σ n µ X σ n ) = Interpretation: There is 95% probability that the random interval X 1.96 σ n, X σ n will cover the mean µ. Example: Cholesterol levels d = 36.89, σ = 51.00, n = 28. Therefore, the 95% confidence interval for µ is 18.00, Confidence intervals I, Feb 11,

4 Confidence Intervals for the Mean Assumption: The population standard deviation σ is known. In the next lecture, we will drop this unrealistic assumption. Assumption is approximately satisfied for large sample sizes, since then ˆσ σ by the law of large numbers. Definition: Confidence interval for µ (σ known) The interval X zα/2 σ n, X + z α/2 σ n is called a 1 α confidence interval for the population mean µ. (1 α) is the confidence level. For large sample sizes n, an approximate (1 α) confidence interval for µ is given by X z α/2 ˆσ n, X + z α/2 ˆσ n. Here, z α is the α-critical value of the standard normal distribution: z α has area α to its right Φ(z α ) = 1 α f(x) α 0 1 z 2 z α 3 Confidence intervals I, Feb 11,

5 Confidence Interval for the Mean Example: Community banks Community banks are banks with less than a billion dollars of assets. Approximately 7500 such banks in the United States. Annual survey of the Community Bankers Council of the American Bankers Association (ABA) Population: Community banks in the United States. Variable of interest: Total assets of community banks. Sample size: n = 110 Sample mean: X = 220 millions of dollars Sample standard deviation: SD = 161 millions of dollars Histogram of sampled values: 20 Assets of Community Banks in the U.S. (sample of 110 community banks) 15 Frequency Assets (in millions of dollars) Suppose we want to give a 95% confidence interval for the mean total assets of all community banks in the United States. α = 0.05, z α/2 = 1.96 A 95% confidence interval for the mean assets (in millions of dollars) is , , Confidence intervals I, Feb 11,

6 Example: Cholesterol levels Sample Size Suppose we want a 99% confidence interval for the decrease in cholesterol level: α = 0.01, z = 2.58 The 99% confidence interval for µ D is , , Note: If we raise the confidence level, the confidence interval becomes wider. Suppose we want to obtain increase the confidence level without increasing the error of estimation (indicated by the half-width of the confidence interval). For this we have to increase the sample size n. Question: What sample size n is needed to estimate the mean decrease in cholesterol with error e = 20 and confidence level 99%? The error (half-width of the confidence interval) is e = z α/2 σ n Therefore the sample size n e needed is given by ( zα/2 ) σ 2 ( ) n e = = 43.16, e 20 that is, a sample of 44 patients is needed to estimate µ D with error e = 20 and 99% confidence. Confidence intervals I, Feb 11,

7 Estimation of the Mean Example: Banks loan-to-deposit ratio The ABA survey of community banks also asked about the loan-to-deposit ratio (LTDR), a bank s total loans as a percent of its total deposits. Sample statistics: n = 110 ˆµ LTDR = 76.7 ˆσ LTDR = 12.3 Frequency Loan To Deposit Ratio of Community Banks (sample of 110 community banks) LTDR (in %) Construction of 95% confidence interval: α = 0.05, z α/2 = 1.96 Standard error σ X = σ LT DR n = % confidence interval for µ LTDR : σ LT DR X zα/2, X σ LT DR + z α/2 = 74.4, 79.0 n n To get an estimation with error e = 3.0 (half-width of confidence interval) it suffices to sample n e banks, ( ) zα/2 σ 2 ( ) 2 LT DR n e = = e 3.0 Thus a sample of n e = 65 banks it sufficient. Confidence intervals I, Feb 11,